Mechanics. (UCSD Physics 110B) January 1, 2009

Transcription

1 Mechanics (UCSD Physics 110B) January 1, 2009

2 2 Contents 0 Vectors and Rotations Vector Identities Cross Products and Axial Vectors Rotations The Vector Derivative Homework Problems Noninertial Frames of Reference Rotating Coordinate System Calculating the Effective Force (simple rotation) Calculating the Effective Force (a more general derivation) Motion Near the Earth The Tides Example: Deflection of a Falling Object Example: Deflection of Cannon Balls Weather Systems Example: The Foucault Pendulum Homework Problems Dynamics of Rigid Bodies Calculating the Kinetic Energy The Inertia Tensor Angular Momentum Simple Example: Inertia Tensor for Dumbbell Transforming the Inertia Tensor Parallel Axis Theorem Example: The Inertia Tensor for a Cube Principal Axes Example: Cube Rotating about a Corner Proof: Principal Axes Orthogonal Proof: Roots I (a) are Real Euler Angles Body Frame Coordinates Euler s Equations

3 Example: Symmetric Top with no Torque Stability of Rigid Body Rotations Lagrange Equations for Top with One Fixed Point Homework Special Relativity Some History of Special Relativity The Michelson Morley Experiment: Some Analysis The Lorentz Transformation Checking Michelson Morley with Lorentz Transformation Phenomena of the Lorentz Transformation Minkowski Space Proper Time Causality and the Light Cone Symmetry Transformations in Minkowski Space Review of the Hyperbolic Functions Hyperbolic Function Identities Rotations in 4 Dimensions Imaginary Angles A Boost in an Arbitrary Direction Velocity Addition The Momentum-Energy 4-Vector Deriving the Momentum-Energy 4-Vector The Force 4-Vector Summary of 4-Vectors The 4D Gradient Operator x µ The Relativistic Doppler Effect The Twin Paradox Kinematics Problems in Electron Volts π 0 Decay Neutron Decay Compton Scattering Lagrange Equations in Special Relativity Covariant Electricity and Magnetism Equations Rationalized Heaviside-Lorentz Units The Electromagnetic Field Tensor

4 Lorentz Transformation of the Fields E&M is a Vector Theory Homework A Little General Relativity Geometries The Metric Tensor The Schwarzschild Metric Gravity s Effect on Time and the Gravitational Red Shift The Singularity in Schwarzschild Coordinates The Geodesic Equation Conserved Energy and Angular Momentum in the Schwarzschild Metric Orbits in the Schwarzschild Metric Orbits of Photons Deflection of Light Black Holes Homework Hamiltonian Mechanics Recalling Lagrangian Mechanics The Hamiltonian Formalism Conserved Momenta Simple Example: Particle on the Surface of a Cylinder Example: Particle in 3D Potential V(x,y,z) Example: A spherical pendulum Example: Motion in a Central Potential V(r) Phase Space and Liouville s Theorem Homework Continuum Mechanics Homework

5 5 0 Vectors and Rotations The laws of Physics are invariant under rotations of the coordinate system. Rotational symmetry of laws of Physics implies conservation of Angular Momentum. We will work with Passive Rotations where we rotate the coordinate axes rather than active rotations where we rotate the physical system and keep the axes fixed. These are essentially the same thing except the rotation angle changes sign. A vector is a mathematical object that transforms in a particular way under rotations. We know there are also physical quantities called scalars that are invariant under rotations. s = s An example is the mass of an object or a particle, m = m. Assume (for now) that we live in three dimensions. The position in three-space is an important example of a vector. x = (x,y,z) = xˆx+yŷ+zẑ = i x i ê i The ê i are three orthogonal unit vectors. ê i ê j = δ ij This the Kronecker delta, δ ij = 1 for i = j and zero otherwise. The dot product of two vectors is a scalar, and therefore invariant under rotations of the coordinate system. v x = v x x+v y y +v z z = v 1 x 1 +v 2 x 2 +v 3 x 3 = v i x i This makes sense physically since the length of a vector should not depend on a rotation of the coordinates.

6 6 We will often use the Einstein summation convention: repeated indices imply a sum over that index. v x = v i x i v i x i The repeated index is a dummy index. x = x i ê i v x = v i x i We will also use the totally antisymmetric tensor ǫ ijk. It changes sign if any of the three indices are interchanged. Its value is one for any cyclic permutation of the indices and minus one for an anticyclic permutation. It must be zero if any index is repeated since it is antisymmetric. ǫ 123 = 1 ǫ 132 = 1 ǫ 122 = 0 ǫ ijk = ǫ ikj = ǫ jik = ǫ kji ǫ ijk = ǫ kij = ǫ jki We use this antisymmetric tensor to write a cross product between two vectors. v x = v i x j ǫ ijk ê k The expression for the k component of the cross product is simpler. ( v x) k = v i x j ǫ ijk Here the sum over i and j are implied by the Einstein convention, so we can do the example of the z component of the cross product omitting all the terms in the sum for which an index is repeated and ǫ is zero. ( v x) 3 = v i x j ǫ ij3 = v 1 x 2 v 2 x 1 = v x y v y x

7 7 0.1 Vector Identities As an example, we will derive the simple vector identities using ǫ ijk. First, lets do the scalar triple product of vectors. Note that this product is completely symmetric among the three vectors once its written in our notation. Its simply cyclic combinations have a plus sign and anticyclic have a minus sign. ( ) A B C = A k B i C j ǫ ijk = A i B j C k ǫ ijk = B ( ) C A = C ( ) A B = A ( ) C B = B ( ) A C = C ( ) B A The triple vector product will require us to derive an identity involving the product of two ǫ ijk s. [ A ( B C )] k = A i ( B C ) j ǫ ijk = A i (B m C n ǫ mnj )ǫ ijk = A i B m C n (ǫ mnj ǫ ijk ) Since theǫs arewell defined andhave simple properties, we must beable tocompute a simpleexpressionforǫ mnj ǫ ijk. Notethattheindexj isrepeatedandthereforesummed over. Wejustneedtodothesumbutsomethoughtwillberequired. Nevertheless, you could do this thinking if stranded on a dessert island and in need of vector identities. We now do the sum ǫ mnj ǫ ijk = ǫ mnj ǫ ikj. First of all ǫ ijk will be zero if any of the indices are repeated. So me must have m n and i k. We will sum over j, but starting from the first ǫ, j must be the other index besides the two used up for m and n so there is only one non-zero term in the sum. This is also true for the second ǫ. Since the index j in the two tensors is the same we must have m and n using up the same two indices that i and k do. So the two possibilities are m = in = k or m = kn = i There is only one term in the sum. We can write this result in terms of Kronecker δs. ǫ mnj ǫ ikj = δ mi δ nk δ mk δ ni

8 8 ǫ mnj ǫ ikj = δ mk δ ni δ mi δ nk We can now easily compute the triple vector product identities. [ ( )] ( ) ( ) A B C = A i B m C n (ǫ mnj ǫ ijk ) = A i B m C n (δ mk δ ni δ mi δ nk ) = B k A C C k A B k ( ) A B C = B ( ) A C C ( ) A B For expressions involving derivatives in 3D, these tools become even more useful. 0.2 Cross Products and Axial Vectors Note that the cross product of two vectors behaves like a vector in many ways. Under a parity transformation in which the direction of all three coordinate axes are inverted, a vector will change sign, but the cross product of two vectors will not change sign. It is therefore actually something different from a vector. We call it an axial vector. It turns out this this type of cross product of vectors can only be treated as a vector in three dimensions. In reality it is an antisymmetric tensor. 0 L z L y ( x p) ij = x i p j x j p i = L z 0 L x L y L x 0 Since there are only three independent numbers in this tensor, it can be cast as a vector. Lets use the angular momentum as an example. We know that angular momentum is normally defined as L = r p. Both r and p are normal vectors and change sign in the coordinate system undergoes a parity inversion. L then obviously does not change sign and is an Axial vector. We can write the angular momentum as the axial vector L k = r i p j ǫ ijk or as the antisymmetric tensor. These two are easily related. L ij = r i p j r j p i L k = 1 2 L ijǫ ijk

9 9 and L ij = L k ǫ ijk We can check this by putting the last two equations together. L ij = L k ǫ ijk = ( 1 2 L mnǫ mnk )ǫ ijk = 1 2 L mn(ǫ mnk ǫ ijk ) = 1 2 L mn(δ mi δ nj δ mj δ ni ) = 1 2 (L ij L ji ) = L ij 0.3 Rotations For our passive rotations, we will change from one orthonormal, right-handed basis to another orthonormal, ê i ê j = δ ij right-handed basis. ê 1 iê2 jê3 k ǫ ijk = 1 Note that the unit vectors ê i are vectors in the original (not primed) frame. If we rotate the coordinate axes, we can compute a vector in the new (primed) coordinates from the original vector by multiplying by a Rotation Matrix. x = R x x i = R ijx j This is the normal way we write a matrix multiplication. x R 11 R 12 R 13 x y = R 21 R 22 R 23 y z R 31 R 32 R 33 z With a little thought, we can compute the elements of the rotation matrix. Lets compute the vector x, the position vector in the rotated coordinates. To get the i component of x, just take the vector and dot it into the new unit vector ê i, written in the original coordinates. x i = ê i x = ê i ê j x j Note that in the above equation, x is a vector that doesn t depend on the coordinates. The components of the vector x i in the primed frame do depend on the primed coordinate axes. So the rotation matrix is just given by

10 10 R ij = ê i ê j where ê i is written in the original coordinates. Thus, each element of the rotation matrix is simply the cosine of the angle between a new coordinate axis and an old coordinate axis. Itiseasytoseephysicallythattheproductoftworotationsisjustsomeotherrotation. R (1) R (2) = R (3) This means that the rotations form a group. We can also see physically that rotations (like matrices) do not commute. As an example, lets make a rotation through a small angle θ in the xy plane, leaving the z axis unchanged. The angle between the x and the x axis is θ. The angle between the y and the y axis is also θ. The angle between the x and the y axis is π + θ. The angle between the y and the 2 x axis is π θ. The angle between the z 2 and the z axis is 0. All the other angle with the z axis are π. 2 So we can write the rotation matrix. R(θ) = cos(θ) cos( π θ) 2 cos(π) 2 cos( π +θ) cos(θ) 2 cos(π) = 2 cos( π) 2 cos(π) cos(0) 2 cos(θ) sin(θ) 0 sin(θ) cos(θ) In three dimensions, a rotation in the xy plane can be said to be a rotation about the z axis. This is not true in four dimensions. One can write the dot product between two vectors just in terms of the lengths of vectors. u v = 1 ( u+ v 2 u 2 v 2) 2 Therefore, preserving the length of vectors implies that dot products are invariant. The invariance of dot products implies that both the lengths of vectors and the angle between vectors are unchanged in a rotation. Lets use the fact that dot products are invariant to derive a property of the

11 11 rotation matrices. x i v i = x iv i R ij x j R ik v k = x i v i R ij R ik x j v k = x i v i RjiR T ik x j v k = x i v i RjiR T ik = δ jk R T R = 1 R 1 = R T The inverse of a rotation matrix is its transpose. We call these matrices Orthogonal Matrices. The rotations in three dimensions are a representation of the Special Orthogonal Group SO(3). These matrices have determinant 1. If we include parity inversions with rotations we have the larger Orthogonal Group O(3). These matrices have detr = ±1. Any rotation in three dimensions can be written as a rotation in some plane or as a rotation about an axis orthogonal to that plane. We can describe a constantly rotating coordinate system as a rotation in a plane by and angle θ = ωt where ω is the angular velocity. We can also define ω to be a vector orthogonal to the plane with magnitude equal to that angular velocity ω = ω. The rotation matrix depends on time. Since R T R = 1, the time derivative of that product must be zero at any time so lets evaluate it at t = 0. d dt (RT R) t=0 = ṘT R(0)+R T (0)Ṙ = ṘT +Ṙ = 0 This implies that the time derivative of R is an antisymmetric matrix, thus representable by an axial vector. 0.4 The Vector Derivative The derivative operator = x i ê i transforms like a vector. We may take the gradient of a function f = f x i ê i, the divergence of a vector field B = x i B i, or the curl of a vector field B = x i B j ǫ ijk ê k.

12 Homework Problems 1. For the two vectors A = ˆx+3ŷ 2ẑ and B = 2ˆx+ŷ +ẑ, find A B, the component of B along A, the angle between A and B, A B, and ( A B) ( A+ B). 2. The gradient operator = x i ê i transforms like a vector. Use this to write the equations of electromagnetism with the Einstein summation convention, that is, with no vector symbols. To make it interesting, here are the EM equations in Rationalized Heaviside-Lorentz units. B = 0 E + 1 B c t = 0 E = ρ B 1 E c t = 1 c j ( F = e E + 1 ) c v B As an example, we can write the first equation. x i B i = 0 3. Use the operator = x i ê i to show that the divergence of a curl is zero. ( ) A = 0 Also calculate the curl of the curl. ( ) A =? 4. Use the totally antisummetric tensor to derive the identity ( A B) ( C D) = (( A B) D) C (( A B) C) D. 5. Calculate the rotation matrix for a passive rotation about the ẑ axis through an angle φ followed by a rotation about the new ˆx axis by an angle θ. 6. Calculate the rotation matrix for a passive rotation through an angle of 120 degrees about an axis making equal angles with the original three coordinte axes.

13 13 7. Calculate the time derivative of a rotation matrix through an angle ωt about the z axis. Remembering that the axial vector ω is related to an antisymmetric tensor, relate the time derivative of the rotation matrix (evaluated at t = 0) to ω. 8. Show that for matrices A and B, (AB) T = B T A T, and that (AB) 1 = B 1 A 1. 1 Noninertial Frames of Reference Newton s Mechanics is valid in any inertial frame of reference. This will be upheld and even extended when we study Special Relativity. Some convenient frames are non-inertial. The most important case for us is a coordinate system fixed on the earth. An earth fixed frame is potentially quite complex but it boils down to a rotating coordinate system. The motion of the earth includes: 1. rotation around an axis perpendicular to the equatorial plane. 2. orbital motion around the sun 3. orbital motion with the moon 4. the sun s motion in the galaxy 5. the motion of the galaxy in the cluster of galaxies and the rest of the universe. While all of these involve acceleration, 2-5 are objects in free fall in a gravitational field. The force of external gravity on an object on the earth is canceled by the acceleration of the earth due to that force. Just like an astronaut inside a satellite in orbit around the earth feels no force of gravity or fictitious force due to the acceleration of the satellite. They just cancel. (But they only cancel at the center of the earth, leaving tidal forces elsewhere.) In fact in general relativity, we understand that acceleration and gravity are the same thing and there are no forces on objects in free fall. This is not true for the rotation of the earth.

14 Rotating Coordinate System The arithmetic for rotating coordinate systems can be rather complicated, however, most of the physics can beunderstood in a relatively simple case. We will first study the simple case, then extend the equations a bit to include the general case. Since any rotation is just a rotation in a plane, the simple case includes the most general rotation. Our simplification is that we will put two of the coordinate axes in the plane of the rotation. Since we have the choice of coordinate systems in solving our physics problems this covers many cases. But sometimes, like motion on the surface of the earth, we want to define our coordinates in another way and we will deal with that in the next section. In all cases, we will set up our coordinates so that the origin of the inertial coordinate system and the rotating coordinate system coincide. Imagine we do experiments on a rotating table (rotation in the plane of the table). It is rotating about some point and we will call the point the origin of our coordinate systems. We know about the effective centrifugal force F = mv2 r ˆr from Freshman Physics. If the table is rotating with constant angular frequency ω, we have v = ωr. F = mω 2 rˆr There is another effective force that depends on velocity called the Coriolis Force. Lets understand this simply by assuming our rotating table is a frictionless air table. For our first experiment we will send a puck straight across the table through the origin. The puck goes from radius r = r 1 to radius r = 0 and back to radius r 1 in our experiment and we give the puck exactly the right velocity so that it makes this trip while the table undergoes exactly one half turn. v = 2r 1 π/ω = 2ωr 1 π So far, we are talking about the motion observed from a non-rotating frame with the same origin. The puck moves in a straight line at constant velocity from one side of the table to the other as shown in the figure below.

15 15 y y x ω t x motion of puck Now lets see what happens using the rotating coordinate system. The figure below gives a rough drawing of the puck s path in the rotating coordinate system. It does not move in a straight line. y y ω t x x motion of puck The puck starts out with an inward radial velocity as before but also has a large tangential component. The tangential component decreases as the puck gets close to the origin. It goes through the origin then returns to radius r 1 on a path symmetric with the one into the origin. (It comes back to the same point because of the velocity we have chosen for this experiment.) In the rotating frame, there was clearly some effective force that was not radially outward. That was the Coriolis force Calculating the Effective Force (simple rotation) To calculate the effective force in a simple way we will use the rotating table coordinate systems and calculate the force at t = 0 when the two coordinate systems

16 16 are aligned. The forces due to the rotation do not depend on time so this is still general. It will be best to work with the transformation from the rotating (primed) frame to the inertial frame. r = Rxy T (t) r T cosωt sinωt 0 cosωt sinωt 0 R T = sinωt cosωt 0 = sin ωt cos ωt r = R T r +ṘT r sinωt cosωt 0 Ṙ T = ω cosωt sinωt r = R T t=0 r +ṘT t=0 r r = r +ω( yê x +xê y ) r = r + ω r This last equation is an important one. It actually holds for the time derivative of any vector, not just r. So far it is only true when the coordinate axes coincide like at t = 0 but, since it is true for any vector, we can apply it to ω. ω = ω ω ω = ω So there is no additional term in the time derivative of ω. ω is also unaffected by the rotation and assumed to be constant so we have ω = ω = 0 and we have for the general vector V (at t = 0) V rotating = Vinertial ω V where ω is the angular velocity vector of the rotating coordinate system in the fixed system.

17 17 To get the acceleration (and forces), take the second time derivative. r = R T r +ṘT r r = R T r +ṘT r + R T r +ṘT r r = R T r +2ṘT r + R T r sinωt cosωt 0 Ṙ T = ω cosωt sinωt cosωt sinωt 0 R T = ω 2 sin ωt cos ωt r = R T t=0 r +2ṘT t=0 r + R T t=0 r r = r ω r ω r r = r +2ω( ẏ ê x +ẋ ê y ) ω 2 r r = r +2 ω r ω 2 r r = r +2 ω r + ω ω r F = F +2m ω r +m ω ω r F = F m ω ω r 2m ω v This gives us the actual force plus the effective force in the rotating system. F = F m ω ω r 2m ω v The three terms are the actual force, the centrifugal force, and the Coriolis force. The centrifugal force is in the plane of rotation and parallel to r. The Coriolis force is perpendicular to both the velocity and ω. Now lets go back to the first experiment. First we can compute the velocity in

18 18 the rotating frame at t = 0. (The coordinate axes correspond at t = 0.) r = r ω r v = v ω r v = 2ωr π êy v = 2ωr π êy ωrê x This is quite consistent with the path drawn in the diagram. Now we can compute the effective force in the rotating frame. F = F m ω ω r 2m ω v ( ) 2ωr F = 0 mω 2 rê y 2mωê z π êy ωrê x F = mω 2 rê y + 4 π mω2 rê x +2mω 2 rê y F = mω 2 rê y + 4 ( π mω2 rê x = mω 2 r ê y + 4 ) πêx This also agrees with the direction of acceleration shown in the diagram at t = 0. It would take some effort to integrate the equations to get the exact path. That is more easily done by just transforming the path. This problem is illustrative but not really the usual type of problem where we know the position and velocity in the rotating frame and need to get the effective forces. Note that a ball flying above the table would have the same behavior since it is only the radius perpendicular to ω that acts in the fictitious forces. A second simple experiment would be to place the puck at radius r 1 on the frictionless rotating table, with no velocity in the inertial frame. The puck would then just remain motionless in the inertial frame and therefore trace out a circle in the rotating frame. Lets check this against our equations.

19 19 v = v ω r v = 0 r ω ( ê y ) v = ωrê x F = F m ω ω r 2m ω v F = mω 2 rê y +2mω 2 rê y = mω 2 rê y This is exactly the force needed to make the puck go in a circle. The Coriolis force is in the opposite direction of the centrifugal force and is twice as large. 1.2 Calculating the Effective Force (a more general derivation) There are a lot of nasty derivations of the effective force in mechanics text books. We d like to make ours simple yet not require too much higher mathematics, so we need to set up the problem as simply as possible. Newton s laws apply in the inertial frame of reference. Lets work in that frame, but set up a set of unit vectors for an Accelerating frame. For simplicity of notation, lets call those unit vectors û i. Lets call the coordinates in the accelerating (rotating) frame x i and the position vector in the Inertial frame x I etc. Let the displacement between the origins of the Inertial and Accelerating frame be called X. We then simply have. x I = X +x j û j We will differentiate this equation twice to get accelerations and hence real and fictitious forces. Note that all the quantities in this equation are defined in the Inertial frame. d x I dt = d X dt + dx j dû j dt ûj +x j dt The change of X with time is a relative motion of the origins of the two coordinate systems. The change of the unit vectors with time is due to the rotation of the accelerating coordinate system. d 2 x I = d2 X x j dt 2 dt 2 +d2 dt ûj+ dx j dû j 2 dt dt +dx j dû j dt dt +x d 2 û j j = d2 X x j dt 2 dt 2 +d2 dt ûj+2 dx j dû j 2 dt dt +x d 2 û j j dt 2 These are all the derivatives we need.

20 20 We also need the effect of the rotation on the unit vectors û i. In an infinitesimal time dt the unit vectors rotate through an angle d θ = ωdt. dû i = d θ û i This is the change of the unit vectors, evaluated in the inertial frame. We also need the second derivative. dû i dt = d θ dt û i = ω û i d 2 û i dt 2 = d ω dt û i + ω dû i dt = d ω dt û i + ω ω û i We can now plug these into the formula for the second derivative of x I above. d 2 x I = d2 X dt 2 dt + d2 x j 2 dt 2 ûj +2 dx j dû j dt dt +x d 2 û j j dt 2 d 2 x I = d2 X dt 2 dt + d2 x j 2 dt 2 ûj +2 dx ( ) j d ω dt ω û j +x j dt û j + ω ω û j d 2 x I dt 2 = d2 X dt 2 + d2 x j dt 2 ûj +2 ω dx j dt ûj + d ω dt x jû j + ω ω x j û j We recognize x j as the position in the Accelerating frame (and x j û j as that position written in the inertial frame), dx j as the velocity in the Accelerating frame, and d2 x j dt dt 2 as the acceleration seen in the Accelerating frame. It is important to note that these are written as vectors in the Inertial system by dotting them into the û j, which is why we can do this calculation rather easily. For example, the x i give a position in the Accelerating frame, but when they are dotted into the basis vectors of the Accelerating frame, written in the Inertial frame, û i, the vector is now correctly written in the Inertial frame. a I = A+ a A +2 ω v A + d ω dt x A + ω ω x A In the equation above, a I is the acceleration the mass in the Inertial frame, A is the acceleration of the origin of the Accelerating frame, a A is the acceleration of the mass seen in the Accelerating frame, v A is the velocity of the mass seen in the Accelerating frame, x A is the position of the mass seen in the Accelerating frame, and ω is the

21 21 angular velocity of the rotation of the coordinate axes of the accelerating (rotating) frame. A and ω are properties of the transformation between the Inertial frame and the Accelerating frame. This equation is written in the inertial frame but at t = 0 the two frames coincide so the equation holds in the accelerating frame too. We can change to the accelerating frame just by removing the û j. We can multiply by the mass m to get an equation in the forces, real and fictitious. m a I = m A+m a A +2m ω v A +m d ω dt x A +m ω ω x A Now solve this equation for the force seen in the Accelerating frame. The general equation for the effective force seen in a frame that is accelerating and rotating is. F A = F I m A 2m ω v A m d ω dt x A m ω ω x A There is a simple fictitious force m A if the coordinate system is simply accelerating. To see the effect of a constant rotation, we set A = 0 and d ω dt = 0 F A = F I 2m ω v A m ω ω x A The forces are named in the equation below. F rotating = F inertial + F Coriolis + F centrifugal The centrifugal and Coriolis forces are fictitious forces that appear in the rotating frame. The Centrifugal Force grows with the perpendicular distance from the axis of rotation and is in the direction of r. F centrifugal = m ω ω r The Coriolis Force is perpendicular to the velocity of the mass and to ω. For ω pointing up, like in the northern hemisphere, the Coriolis force causes projectiles to deflect to the right. In the southern hemisphere, projectiles deflect to the left. F Coriolis = 2m ω v

22 22 Its interesting to note that Coriolis s original publication in 1835 was on the energy yield of machines with rotating parts, such as water-wheels. 1.3 Motion Near the Earth The earth rotates with and angular frequency of about 1 2π radians per (sidereal) day. The axis of the earth s rotation goes through the north and south poles of the earth. The direction of ω goes from south to north pole. The axis is at about = radians from the plane of the earth s orbit, causing the observed seasons. The primary effect of the centrifugal force is to change the magnitude and direction of the apparent acceleration of gravity near the earth. g = g ω ω r Since force only depends on latitude and the radius of the earth, it is a constant correction to gravity, given a latitude. The magnitude of the centrifugal force is biggest at the equator. The average radius of the earth is 6367 km. Lets compute the magnitude of ω here. ω = 2π (23(60)+56)60 = per second So, at the equator the apparent acceleration of gravity is. g = GM e r 2 e +ω 2 r e It is a small but easily measurable correction to gravity. Lets analyze the effect of the centrifugal acceleration at all latitudes, under the assumption that the earth is spherical. At non-zero latitude the first cross product ω r = ωrcosλ, so the magnitude of the centrifugal acceleration is ω 2 rcosλ and the direction is outward perpendicular to the axis of rotation. In the local coordinate system, this means that there is a radial component of the centrifugal acceleration which reduces the effective g g = g ω 2 r e cos 2 λ 1 During the time needed by the Earth to complete a rotation around its axis (a sidereal day), the Earth moves a short distance (around 1 ) along its orbit around the sun. Therefore, after a sidereal day, the Earth still needs to rotate a small extra angular distance before the sun reaches its highest point. A solar day is, therefore, around 4 minutes longer than a sidereal day.

23 23 and there is a component pointing along a line of longitude toward the equator of magnitude ω 2 r e cosλsinλ. This component has very small effect on the magnitude of g but changes its direction somewhat. Plumb bobs will hang in the direction of the effective g so this will locally be defined as the vertical direction. Of course if the earth were a fluid of uniform density, it would take on a shape such that the effective g would be normal to the surface of the fluid. Newton calculated that thiswould make radiusofthe earthattheequator onepart in230larger thanthe radius at the poles. The earth would take on the shape of an oblate spheroid. With much of the earth molten, it is easily deformable into this shape, however, the density is much larger at small radius than it is at the surface. Recent measurements give a somewhat smaller effect of one part in The oblate spheroid is the baseline for measurements of the earths shape. Gravity actually depends on the local mass density of the earth but the baseline model of g as a function of latitude is approximately. [ ] sin 2 λ g = sin 2 λ It is clear from this formula that objects will be lighter at the equator than at higher latitudes. Besides the forces due to rotation, there are some effective forces due to the acceleration of the coordinate system. The earth is essentially in free fall in the gravitational field of the moon and the sun, so that the net force of gravity is exactly canceled by the acceleration of the earth. Only the net force is canceled. Since the effective force varies with position on the earth, there are tidal forces that cause the oceans to rise and fall. 1.4 The Tides Tidal forces are due to the the variation of the effective force with position. The tides seen in the earth s oceans are primarily caused by the moon with a significant additional effect from the sun. Let us deal with the tidal forces from one body, the moon, first. With the earth in free fall, the effective force due to acceleration exactly cancels the gravitational attraction at the center of the earth, but, the force of gravity varies with position, while the acceleration of the earth does not. Thus, the remaining effective force comes from the difference between the gravitational attraction of the moon at some position r and the gravitational attraction of the moon at the center of the earth.

24 24 Lets set up a coordinate system. Put the moon at the origin with the center of the earth at a distance d along the x axis. We can put the point at which we are evaluating the force in the xy plane without loss of generality. As usual, call the angle from the x axis φ Let the radius of the earth be r e and of course, the vector from the center of the moon to the point at which we are evaluating the tidal force can be r. F tidal = GM moonm r 2 ˆr + GM moonm (êx ê d 2 x = GM moon m d ˆr ) 2 r 2 So there is a force due to the moon that varies with position. It is zero at the center of the earth but not zero otherwise. This tidal force tends to pull the earth apart along the x axis and push it together along y. Lets write it in terms of φ to get a more complete picture. x = d+r e cosφ y = r e sinφ r 2 = x 2 +y 2 = d 2 +2dr e cosφ+re 2 cos 2 φ+re 2 sin 2 φ = d 2 +re 2 +2dr e cosφ [ ] ê x a tidal = GM moon d (d+r e cosφ)ê x (r e sinφ)ê y 2 (d 2 +re 2 +2dr e cosφ) 3 2 (d 2 +re 2 +2dr e cosφ) 3 2 [( ) ] 1 a tidal = GM moon d (d+r e cosφ) r e sinφ ê 2 (d 2 +re 2 +2dr e cosφ) 3 x ê 2 (d 2 +re 2 +2dr e cosφ) 3 y 2 [( a tidal = GM moon 1+ re 1 cosφ ) ] r ed sinφ d ê d 2 x ê y (1+ r2 e +2 re d 2 d cosφ)3 2 (1+ r2 e +2 re d 2 d cosφ)3 2 ] a tidal GM moon d 2 a tidal GM moon d 2 [( 1 a tidal GM moon d 2 [( 1 1+ re cosφ d (1+3 re cosφ) d ) ê x r e d sinφê y ( 1 2 r e d cosφ )) ê x r e d sinφê y r e d [2cosφê x sinφê y ] ] After computing the tidal acceleration fully, we have approximated it to first order in re to make it easier to see what the primary effect is. The tidal force is proportion d to the inverse cube of the distance. The positive cosine term in the x direction tends to pull the earth apart, while the negative sine term tends to squeeze it together in the y direction. The tidal acceleration is shown in the figure below.

25 25 Its easy to see that there are two high tides and two low tides per day. What really happens to the ocean and to the earth is complicated. The earth is not solid and can be deformed even if not covered by water. To create a high tide, a lot of water has to move which takes time. Its not surprising that the horizontal component of the tidal acceleration plays an important role. With that component potentially pushing water up against land, the height of tides is not simple to calculate. Of course, we do it well by now, but there are a lot of effects, including the deformation of the solid parts of the earth that must be understood. The time it takes for a wave (say due to the horizontal acceleration) to propagate half way around the earth is about 30 hours, while the high tides are every twelve hours. In most places there is a delay between the phases of the Moon and the effect on the tide. Springs and neaps in the North Sea, for example, are two days behind the new/full Moon and first/third quarter. This is called the age of the tide. Nevertheless, we can understand the basic ocean tides rather simply. First of all we have to combine the effects of the moon and the sun. The lunar tide should be about twice as big as the solar. When the moon and the sun line up, at new and full moons, the tidal accelerations add, causing very high and low tides (spring tide). When they are at 90 degrees, they tend to cancel (neap tide).

26 26 The two tides of the day can differ significantly in height in La Jolla (for example), if the sun (or moon) is in the southern hemisphere, because we can get closer to having cosφ = 1 than we can to cosφ = 1. The two plots below show the La Jolla tides in two time scales. One can see several periods of oscillation of the highest and lowest tides. If the earth didn t rotate and the continents didn t move under the tidal forces, one could compute the height of the tides by finding equipotential surfaces. This at least gives an order of magnitude estimate of the height of the tides. Its done in Taylor and we find. gh = GM moon 3re 2 d 2d 2 This gives a height of 0.54 meters for the moon and 0.25 meters for the sun. Many famous physicists and mathematicians worked on the problem of the tides. Isaac Newton laid the foundations for the mathematical explanation of tides in the Philosophiae Naturalis Principia Mathematica (1687). In 1740, the Acadmie Royale des Sciences in Paris offered a prize for the best theoretical essay on tides. Daniel Bernoulli, Antoine Cavalleri, Leonhard Euler, and Colin Maclaurin shared

27 27 the prize. Maclaurin used Newton s theory to show that a smooth sphere covered by a sufficiently deep ocean under the tidal force of a single deforming body is a prolate spheroid with major axis directed toward the deforming body. Euler realized that the horizontal component of the tidal force (more than the vertical) drives the tide. In 1744 Jean le Rond d Alembert studied tidal equations for the atmosphere which did not include rotation. The first major theoretical formulation for water tides was made by Pierre-Simon Laplace, who formulated a system of partial differential equations relating the horizontal flow to the surface height of the ocean. The Laplace tidal equations are still in use today. William Thomson, 1st Baron Kelvin, rewrote Laplace s equations in terms of vorticity which allowed for solutions describing tidally driven coastally trapped waves, which are known as Kelvin waves. The tides can be quite complex but it is worth studying the basic phenomenon of tides Example: Deflection of a Falling Object If we drop an object from height h, the effect of the centrifugal force is already in the local value and local direction of g. The Coriolis force is not included and it is proportional to velocity so it will not affect the direction of plumb bobs. The Coriolis force will cause a deflection from a vertical path as the object falls. Lets call the z direction the local upward vertical. We need to pick two directions perpendicular to that to form a coordinate system. We can choose the y direction be be north and then we must choose east to be in the x direction to have a right handed coordinate system. For a downward velocity, the Coriolis acceleration is to the east, a = 2 ω v = 2ω(sinλẑ +cosλê north ) vẑ where v will be a negative number. a = 2ωvcosλê north ẑ = 2ωvcosλê east

28 28 One can integrate this equation to find the deflection when a cannonball is dropped from a height h. v z = gt a = 2ωgtcosλê east t v east = 2ωgcosλ t dt = ωgcosλt 2 x east = ωgcosλ h = 1 2 gt2 2h t = g x east = ωgcosλ 3 0 t 0 t 2 dt = ωgcosλ t3 3 8h 3 g = ωcosλ 8h g For a height of 100 meters, at 32 degrees latitude, the displacement is about 1.9 cm. Explain why the deflection is always to the east Example: Deflection of Cannon Balls People have been known to fire cannon balls in many direction and in many locations near the earth. Let us consider the effect of the Coriolis force on the ballistic trajectories of these cannon balls. Assume the cannon ball is fired at an angle θ from the vertical and at and angle φ from the easterly direction in the plane tangent to the earth. These are the usual angles of a spherical coordinate system. a = 2 ω v = 2ω(sinλẑ+cosλê north ) v(cosθẑ+sinθcosφê east +sinθsinφê north )

29 29 a = 2ωv( sinλsinθcosφê north +sinλsinθsinφê east cosλcosθê east +cosλsinθcosφẑ) This can be summarized in three effects. 1. cannon balls will curve to the right in the northern hemisphere and to the left in the southern hemisphere. (first two terms) 2. There is a force to the west on the way up and a force to the east on the way down, like the falling object in the example above. (third term) 3. cannon balls curve up when fired to the east and down when fired to the west. (fourth term) The magnitude of these effects can be large and were accommodated in the sighting of naval cannon by using tables Weather Systems Without the Coriolis force, the weather on earth would be completely different from the weather we know. When a low pressure system is formed, air would rapidly flow from high pressure regions to the low pressure region if the earth did not rotate. In fact pressure waves travel at the speed of sound so pressure differences would be damped out in hours at the most. With the Coriolis force, the air will be deflected to the right in the northern hemisphere as shown in the figure below. The air circulates around a low pressure system in a counter-clockwise direction with the Coriolis force pushing the air outward while the force from the pressure difference pushes the air inward toward the low pressure. The two forces balance each other (automatically) and semi-stable low (and high) pressure systems last for weeks or even months, moving across the earth while the air swirls around them. In stable weather systems, the wind actually blows along isobars, rather then perpendicular to them. The figure below shows a large low pressure system over Iceland. It is clearly swirling in the counter-clockwise direction.

30 30 Huricanes are an extreme example of a low pressure system. They are produced over warm water (but not near the equator). High pressure systems, like ones causing Santa Ana conditions in San Diego have air swirling clockwise in the northern hemisphere. Tornados are a different phenomonon due to turbulence and remanent angular momentum Example: The Foucault Pendulum A heavy weight on a long cable is suspended from a mount that is free to swing in any direction with very small friction. This is a Foucault pendulum. Its equilibrium position is along the line of a plumb bob, in the direction of effective gravity. With the long cable, the oscillations are small, so the velocity is horizontal and we do everything only to first order in r l. Its natural to put the origin of coordinates at the equilibrium point, the z axis up,

31 31 the y axis north, and the x axis east. The motion is in the xy plane. F eff = m g + T 2m ω v T a = g + 2 ω v ( m T T ê z r ) l T = mg a = g + g g r 2 ω v l ω = ω(cosλê y +sinλê z ) r = 2 ω v g l r r = 2ω( cosλv x ê z +sinλv x ê y sinλv y ê x ) g l r r = 2ωsinλ(v x ê y v y ê x ) g l r ẍ = g l x+2ωsinλẏ ÿ = g l y 2ωsinλẋ q = x+iy We add the x equation plus i times the y equation. q + g l q = 2iωsinλ q = 2iω z q q = q 0 e iat a 2 + g l = 2iω z(ia) a 2 +2ω z a g l = 0 a = 2ω z ± 4ω 2 z +4g/l 2 a = ω z ± ω 2 z +g/l g/l >> ω 2 z a = ω z ± g/l [ ] g/lt q(t) = q 0 e iωzt Ae i g/lt +Be i

32 32 Pick the initial conditions on the pendulum to give cosine for the sum of exponentials. ( ) g x+iy = q = q 0 e iωzt cos l t ( ) g x = x 0 cos(ω z t)cos l t ( ) g y = x 0 sin(ω z t)cos l t The pendulum precesses with an angular frequency ω z = ωsinλ. 1.5 Homework Problems 1. Calculate the potential due to the earth s gravity plus the tidal force of the moon. Use this to estimate the height of the high tide. Now assume that not only the ocean moves due to the tidal force but that the land is floating on a fluid which can deform due to the tidal force. What would be the height of the tide if the land masses also respond to the tidal potential? 2. A circular container of water rotates about its axis, which is vertical, with an angular frequency ω. After some time, due to viscosity, the water becomes stationaryintherotatingframe. Whatistheshapeofthewater inthecontainer (with normal gravity)? 3. A projectile is shot vertically (vertical determined by plumb bob) upward to a (small) height h, at a northern latitude λ. Show that it strikes the ground a distance of 4ωcosλ 8h 3 to the west of the point from which is was shot. 3 g 4. A cannonball is fired due east with a velocity v 0 and inclination angle θ at a latitudeλ. Showthatthecannonballisdeflectedadistanced = 4v3 0ωsinλsin 2 θcosθ g 2 to the south. 5. A cannonball is fired due east with a velocity v 0 and inclination angle θ at a latitude λ. If the range of the cannonball is R neglecting the rotation of the earth, 2R show that the range is increased by R = 3 ωcosλ tanθ ( cotθ 1 tanθ). g 3 6. Show that the angular deviation of a plumb bob, from the direction toward the center of theearth, is reω2 sinλcosλ g r eω. Findthe maximum deviation in milleradians. 2 cos 2 λ 7. Write the Lagrangian for a particle s motion in a potential U( r) in an inertial coordinate system. Now write the Lagrangian in a rotating coordinate system.

33 33 Finally calculate the Hamiltonian in the rotating frame and see if it is the total energy. 2 Dynamics of Rigid Bodies In this chapter we will consider the motion of solid objects under the application of forces and torques. We call these solid objects Rigid Bodies. Of course nothing is completely rigid. Objects deform elastically, but these deformation are negligible for a wide range of problems. For a rigid body, we will find in the equations that the motion can be separated into the motion of the center of mass and the rotation around the center of mass. In the rigid body limit, the state of a body can be described by six variables. These are the position of the center of mass and three angles to describe the orientation of the object. We will apply some of the results we have derived for transformation from an inertial frame to a rotating frame. For now we choose the body to be at rest inside the rotating (body) frame and the origin to be at the center of mass. For the purposes of calculation, we will assume that the body is made up of a set of discrete masses, labeled by an index α. In any inertial frame, the velocity of one of those masses is v α = V + ω r α In this equation, the v α is given in the inertial frame, V is the velocity of the center of mass of the object, and r α is the position of the mass in the body frame in which it is at rest. In introductory physics courses, we have learned about the moment of inertia and found that rotational motion is analogous to simple kinetics, with the analog of momentum p = m v being angular momentum L = I ω and the analog of the kinetic energy equation T = 1 2 mv2 being T = 1 2 Iω2. Thinking about the rotational physics of a dumbbell, we can see that rotational motion is not always so simple. Think about rotation of a dumb-ell about an axis at an angle φ to the axis. We find that the angular momentum is always perpendicular to the axis and thus L is not parallel to ω. Charles Stark Draper is shown below with a gyroscope used for the inertial guidance systems.

34 34 These were gyroscopes capable of measuring orientation on three angles plus accelerometers. The motion of a rocker could be measured and integrated to guide it to its target. By the time of the Minuteman missile, accuracies of 100m could be achieved on a km flight. Better accuracy can now be achieved with GPS. 2.1 Calculating the Kinetic Energy From Lagrangian dynamics, we know that we can extract the physics if we know the kinetic and potential energy, T and U. For now we will not be applying any potential so we only have the kinetic energy. T = T α = 1 m α vα 2 = 1 ( ) 2 m α V + ω rα 2 2 α α α T = 1 ( m α V ) V ( ω r α )+( ω r α ) 2 α ( T = V 21 m α + V 2 ω ) m α r α + 1 m α ( ω r α ) 2 2 α α α m α r α = 0 α T = 1 2 MV m α ( ω r α ) 2 2 T = T CM +T rot α

35 ( A B ) 2 = Ai B j ǫ ijk A m B n ǫ mnk 35 The two terms in the sum clearly separate the energy due to center of mass motion and the energy due to rotation. Now we need a vector identity. = A i B j A m B n ǫ ijk ǫ mnk = A i B j A m B n (δ im δ jn δ in δ jm ) ( ) 2 = A 2 B 2 A B T rot = 1 [ m α ω 2 rα 2 ( ω r α ) 2] 2 α T rot = 1 [ m α ωi ω j δ ij rα 2 2 (ω ir αi ω j r αj ) ] α T rot = 1 [ m α ω i ω j δij rα 2 2 r ] αir αj α T rot = 1 2 I ijω i ω j I ij [ ] m α δij rα 2 r αi r αj α The kinetic energy due to rotation is not in general T = 1 2 Iω2 but rather a more complicated inner product between the moment of inertia tensor I ij and two angular velocity vectors T = 1 ω I ω The Inertia Tensor The above calculation of the kinetic energy defines the inertia tensor. T rot = 1 2 I ijω i ω j I ij α m α [ δij r 2 α r αi r αj ]

36 36 Note that I ij is a symmetric tensor (under interchange of the two indices). We can also write the inertia tensor in matrix form. I = rα 2 x 2 α x α y α x α z α m α y α x α rα 2 y2 α y α z α α z α x α z α y α rα 2 zα 2 For a continuous mass distribution, we may use an integral rather than a sum over masses. I ij = ρ( r) [ ] δ ij r 2 r i r j dv V Angular Momentum We may compute the angular momentum for a rigid body rotating about an axis going through its center of mass in the same way. L = α r α p α = α m α r α v α = α m α r α ( ω r α ) Now use the vector identity we computed earlier. ( ) A B C = B ( ) A C C ( ) A B r α ( ω r α ) = rα ω 2 r α ( ω r α ) L = [ m α r 2 α ω r α ( ω r α ) ] α Now lets write this for the components of L. L i = [ m α ωi rα 2 ω ] [ jr αi r αj = ωj m α δij rα 2 r ] αir αj = Iij ω j α The angular momentum can be written in terms of the same inertia tensor. α L i = I ij ω j L = I ω Now we notice an important feature of rotations of rigid bodies. The angular moment will not be parallel to the angular velocity if the inertia tensor has off diagonal components.

37 Simple Example: Inertia Tensor for Dumbbell As a simple example of this phenomenon, consider two equal (point) masses m, connected by a massless rod of length l. Assume it rotates about a fixed axis at an angle θ to the rod. First, note that since L = α m α r α v α it must be perpendicular to the r α and therefore perpendicular to the rod. L is not parallel to ω. L changes direction as the dumbbell rotates so a torque is required to keep it rotating about the fixed axis ω. We might also note that if the dumbbell rotates about any axis perpendicular to the rod, L is parallel to ω. Now lets compute the inertia tensor in two coordinate systems. Both should have the origin at the center of mass, in the middle of the rod. First, with the rod in the xy plane at an angle φ from the x axis. Note that we are working in the rotating (body) frame in which the masses are at rest. I ij [ m α δij rα 2 r ] αir αj α r 2 = r 1 I ij = 2m l2 4 [δ ij ˆr iˆr j ] ˆr = (cosφ,sinφ,0) I 11 = ml2 2 [ 1 cos 2 φ ] I 12 = I 21 = ml2 2 [ cosφsinφ] I 22 = ml2 [ 1 sin 2 φ ] 2 I 33 = ml2 2 [1] I 31 = I 13 = ml2 2 [0] I 23 = I 32 = ml2 2 [0] 1 cos 2 φ cosφsinφ 0 I = ml2 cosφsinφ 1 sin 2 φ

38 38 The inertia tensor will be diagonal for φ = 0 or φ = π when the rod is along the x 2 or y axis. The diagonal element for the axis along the rod is zero because we have assumed point masses and all the mass is on the axis Transforming the Inertia Tensor The inertia tensor is called a rank two tensor because it has two indices. It illustrates the difference between a tensor and a matrix. Because the inertia tensor depends on the coordinates in a clear way, we can write down how it must behave under rotations. If we rotate the coordinate system, the r i and r j must be transformed with a rotation matrix. The r 2 is invariant since it is a dot product. So lets try the transformation I ij = R im R jn I mn = [ ] [ ] m α Rim R jn δ mn rα 2 R im r αm R jn r αn = m α Rim R jm r α 2 r αir αj α α I ij = [ ] [ ] m α Rim Rmj T r 2 α r αi r αj = m α δij r α 2 r αi r αj α α This is the same formula for the inertia tensor written in terms of the primed coordinates, so this transformation leaves the formula for the inertia tensor invariant. We can see that a rank two tensor transforms with two rotation matrices, one for each index. We also saw this the identity tensor δ ij can transform the same way but is actually invariant. δ ij = R im R jn δ mn = δ ij Like a vector, a tensor is defined by how it transforms under rotations and parity inversion. All rank two tensors will transform the same way. In summary, the inertia tensor transforms under rotations like any other rank 2 tensor. I ij = R im R jn I mn = R im I mn R T nj I = RIR T 2.3 Parallel Axis Theorem If the inertia tensor for a set of axes with the center of mass at the origin is calculated, the tensor for any set of parallel axes can be easily derived. The

39 39 translation of the coordinates is given by r = r + a where a is a constant vector. We now simply compute the inertia tensor for the new set of axes. I ij = α I ij = α m α [ δij r 2 α r αir αj ] m α [ δij ( r α + a) 2 (r αi +a i )(r αj +a j ) ] I ij = [ ( m α δij r 2 α +2 r α a+a 2) (r αi r αj +r αi a j +a i r αj +a i a j ) ] α I ij = I ij + [ ( m α δij 2 rα a+a 2) (r αi a j +a i r αj +a i a j ) ] α I ij = I ij + [ ( m ) α δij a 2 (a i a j ) ] α I ij = I ij +M [ ] δ ij a 2 a i a j This result is called the Parallel Axis Theorem. r = r+ a I ij = I ij +M [ δ ij a 2 a i a j ] It can save us a lot of time recalculating the inertia tensor for some object. Note that the parallel axis theorem shows how the inertia tensor depends on the origin. Angular momentum, torque, and kinetic energy all depend on the origin. This is physically relevant if the origin is a fixed point in the rotation. The origin should be chosen to satisfy the conditions of the physical problem being solved.

40 Example: The Inertia Tensor for a Cube We wish to compute the inertia tensor for a uniform density cube of mass M and side s. The density is simply ρ = M s 3. I 11 = M s 3 s 2 s 2 s 2 (r 2 x 2 )dxdydz I 11 = M s 3 s s s s 2 s 2 (y 2 +z 2 )dydz s 2 dx I 11 = M s 2 s s 2 2 s 2 s 2 (y 2 +z 2 )dydz s 2 s s 2 2 I 11 = M 3s ([ y 3]s 2 + [ z 3]s s 2 s 2 2) I 11 = M s 3 3s 2 = Ms2 6 I 12 = M s 3 I = M s2 6 s 2 s 2 s 2 s s s ( xy)dxdydz = 0 There are only two kinds of integrals. The rest are symmetric. The inertia tensor is diagonal so rotation about these axes will have the angular momentum parallel to the axis. The angular momentum then does not change with time and no torque is needed to rotate the cube. Now lets consider rotating a cube about one of its corners. We can compute the new inertia tensor by using the parallel axis theorem with a translation of a = ( s 2, s 2, s 2 ). I ij = I ij +M [ ] δ ij a 2 3s a i a j = Iij +M [δ 2 ij 4 s2 4 ] [ = I ij +Ms 2 3 δ ij 4 1 ] 4

41 41 Note that the final 1 has no indices so it is subtracted from all 9 terms in the tensor. 4 [ ( 1 I ij = Ms2 δ ij ) 1 ] = Ms This inertia tensor is not diagonal so torque will be required to keep a cube rotating aboutacorneriftheaxisofrotationinoneoftheedgesofthecube. (Notethatangular momentum and torque are measured relative to some origin so the location of the origin is important in the calculation of the inertia tensor.) For any rotating mechanical part, it is important that it rotate about a principal axis, otherwise it will exert torque and cause vibrations. 2.5 Principal Axes For any object (and origin), there is (at least one) a set of principal axes for which the inertia tensor is diagonal. It can be shown that he axes are orthogonal. Sometimes there is more than one set, particularly in cases of symmetry. We can find the principal axes, or the axes of rotation that do not require torque by solving an eigenvalue equation. Basically we want to pick a direction for ω so that the angular momentum is parallel to ω. I ω = I ω I ij ω j = Iω i (I ij Iδ ij )ω j = 0 For this to be zero, the determinant of the matrix has to be zero. Setting the determinant equal to zero we can find the eigenvalues I. I 11 I I 12 I 13 I 21 I 22 I I 23 I 31 I 32 I 33 I = 0 This yields a cubic equation giving three roots which are the eigenvalues for the three principal axes. Once we have the eigenvalues, we can solve the equation for the three axes or ωs that give the eigenvalues. The principal axes will be real and orthogonal. The principal axes are real and orthogonal.

42 42 For a sphere, which has the highest symmetry, the three eigenvalues will be equal. For a symmetric top, two will be equal. For an asymmetric top, they will probably all be different Example: Cube Rotating about a Corner We have computed the inertia tensor for a cube with the origin at a corner and the axes along the edges I ij = Ms Lets solve the equation to find the principal axes. 2 x x x = (The moment of inertia will be xms 2 with this definition of x.) To facilitate the solution of the cubic equation, lets get some zeros in this determinant by subtracting the second row from the first and the third x x x x 12 0 x = 0 If we let y = 11 x, we can write the determinant. 12 ( y 2 y 1 ) 14 4 y2 14 ( y2 = y 2 y 3 ) = 0 4 So there are three roots, y = 0, y = 0, and y = 3 ; corresponding to three values for 4 x of x = 11, x = 11, and x = Knowing that two equal moments of inertia along principal axes indicate a symmetry,

43 43 lets find the axis for the third value I = 1 6 Ms2. I ij ω j = 1 6 Ms2 ω i Ms ω 1 ω 2 ω 3 ω 1 ω 2 ω 3 ω 1 ω 2 ω 3 ω 1 ω 2 ω 3 = 1 6 Ms2 = 1 6 = 0 = 0 ω 1 ω 2 ω 3 ω 1 ω 2 ω 3 One can see that all three of these equations are solved for the three components of ω being equal. So this principal axis is the diagonal of the cube starting from the origin and going toward (1,1,1). The other roots just give the same equation three times ω 1 + ω 2 + ω 3 = 0. There are many possible solutions. The equation defines a plane perpendicular to the cube diagonal. We can choose and orthogonal pair of axes in that plane Proof: Principal Axes Orthogonal Take two principal axes ω (a) and ω (b). I ik ω (a) k = I (a) ω (a) i I ik ω (b) k = I (b) ω (b) i

44 44 Dot each of the two equations into the other ω. I ik ω (a) k ω(b) i = I (a) ω (a) i ω (b) i first eq. I ik ω (b) k ω(a) i = I (b) ω (a) i ω (b) i second eq. I ik ω (a) i ω (b) k = I (b) ω (a) i ω (b) i commute omegas I ki ω (a) k ω(b) i = I (b) ω (a) i ω (b) i rename indices lhs I ik ω (a) k ω(b) i = I (b) ω (a) i ω (b) i inerta tensor is symmetric In penultimate step we have renamed the indices i and k and in the last step we have used the fact that the tensor is symmetric. Now the left hand sides of the two equations are the same and we can set the right hand sides to be equal. I (a) ω (a) i ω (b) i = I (b) ω (a) i ω (b) i The dot product ω (a) i ω (b) i must be zero if the moments are not equal. If they are equal, we can just choose orthogonal axes Proof: Roots I (a) are Real I ik ω k = Iω i I ik ω k ωi = Iω iωi I ki ω i = Iω k Iki ω i = I ωk for omega along principle axis multiply by ωi same equation with k take CC Iki ω i ω k = I ωk ω k multiply by ω k I ki ω k ωi = I ωkω k inertia tensor is real, omegas commute I ik ω k ωi = I ωi ω i I symmetric, replace dummy rhs (I I )ω i ωi = 0 lhs same as 2nd line (I I ) = 0 So the roots must be real. 2.6 Euler Angles Three angles are needed to describe an arbitrary rotation. There are an infinite number of ways to do this but the Euler angles are most often used. This is a

45 45 slightly complicated problem, no matter how you define the angles. We will go from the Inertial system to the Body system, in three steps from r I to r to r to r B. The three steps are 1. rotate axes by an angle φ about the z I axis. 2. rotate axes by an angle θ about the x axis. 3. rotate axes by an angle ψ about the z = z B axis. The three rotations are shown in the figure below. z I y B z B θ φ ψ x B y I x I x The linealong thex axis iscalled theline of nodes. Itis commonto thexy planeof both the Inertial and Body coordinates and is key to finding the Euler angles needed for some rotation.

46 46 The resulting rotation matrix R = R ψ R θ R φ can be straightforwardly calculated. cosψ sinψ cosφ sinφ 0 sinψ cosψ 0 0 cosθ sinθ sinφ cosφ sinθ cosθ cosψcosφ cosθsinφsinψ cosψsinφ+cosθcosφsinψ sinψsinθ = sinψcosφ cosθsinφcosψ sinψsinφ+cosθcosφcosψ cosψsinθ sinθsinφ sinθcosφ cosθ x B cosψcosφ cosθsinφsinψ cosψsinφ+cosθcosφsinψ sinψsinθ x I y B = sinψcosφ cosθsinφcosψ sinψsinφ+cosθcosφcosψ cosψsinθ y I z B sinθsinφ sinθcosφ cosθ z I 2.7 Body Frame Coordinates A constant rotation about an axis written in the Body system is made up of the three rotations. ω = φẑ I + θˆx + ψẑ B We can get the first one from the rotation matrix we just calculated. ẑ Ii = R ij δ j3 = R i3 = (sinθsinψ,sinθcosψ,cosθ)

47 47 The ψ rotation is about the z B axis in the Body system so its just (0,0,1). The unit vector for the θ rotation can be calculated from the ψ rotation matrix. ˆx i = R i1(ψ) = (cosψ, sinψ,0) Putting this together, we get the ω corresponding to the derivatives of the Euler angles. ω = ( φsinθsinψ + θcosψ, φsinθcosψ θsinψ, φcosθ+ ψ) Euler angles may be useful to describe the motion of a rotating object. The description is in the Body system where it is easy to work with the Inertia tensor, but hard to deal with external torque that is given in the Inertial system. Some problems can be solved without using the Euler Angles but they are useful for many problems. The dynamics of rotating objects can be fairly complicated so it is useful to pick the body coordinate system to make the problem as simple as possible. The origin of the body coordinate system will be at the center of mass. The axes of the body coordinate system will be the principal axes of the object. These axes are rotating with the object. 2.8 Euler s Equations It is simple to derive a set of equations describing the motion of a rotating object in the rotating (Body) frame. If anexternal torque is applied, it may be difficult to work in the body system so this will most often be useful for problems with no torque. In the Inertial system, we have L inertial = Γ where Γ is the applied torque. This time derivative can be transferred to the body system using the equation ( dl ) ( d = ) L + ω L dt dt inertial rotating

48 48 which we will equate to the torque to get. dl B + ω L dt = Γ L Bk +ω i L j ǫ ijk = Γ k L Bk = I (k) ω k L Bk = I (k) ω k I (k) ω k +ω i I (j) ω j ǫ ijk = Γ k These are the Euler Equations. I (k) ω k +I (j) ω j ω i ǫ ijk = Γ k Note that we have used the notation I (k) with a superscript to make it clear that I is not a vector but rather three numbers that go with the three principal axes so a combination like I (k) ω k does not imply a sum over k Example: Symmetric Top with no Torque The primary application of this equation is motion with no torque since its hard to transform a torque into the body frame. We now consider a symmetric top, that

49 49 is an object with I (1) = I (2) I (12). The equations are. I (k) ω k +I (j) ω j ω i ǫ ijk = 0 I (3) ω 3 +I (2) ω 2 ω 1 I (1) ω 1 ω 2 = 0 I (3) ω 3 +(I (2) I (1) )ω 2 ω 1 = 0 ω 3 = 0 I (1) ω 1 +(I (3) I (2) )ω 3 ω 2 = 0 ( ) I (3) I (2) ω 1 + ω I (1) 3 ω 2 = 0 I (2) ω 2 +(I (1) I (3) )ω 3 ω 1 = 0 ( ) I (3) I (1) ω 2 ω I (2) 3 ω 1 = 0 ( ) I (3) I (1) Ω ω I (2) 3 ω 1 +Ωω 2 = 0 ω 2 Ωω 1 = 0 We have simple coupled oscillators. With the right initial conditions the solution is ω 1 (t) = Acos(Ωt) ω 2 (t) = Asin(Ωt) ω sweeps out a circle in the xy plane, or a cone in 3D, in the Body system where we have solved the equations. The z axis is the symmetry axis for the object. ω precesses about the symmetry axis. Since there is no torque, L is constant in the inertial system. The angle between ω and L is constant in the inertial system, since T = 1 2 ω L is constant and the magnitude of ω is constant. So ω precesses around L in the fixed system. We can choose ẑ I along L since it is constant. We can show that ω, L, and ẑ B are in a plane. This can be shown by computing L ( ω ẑ B ) and showing it is zero. The rate of precession is Ω = I(3) I (12) ω I (3) 3. This is small if the moments of inertia are nearly the same. Examples of this effect are Frisbees wobbling, a spiral pass in football, and maybe the oblate earth s precession. (The dominant term for the earth is gravity driven so we need to analyze it with an external torque.)

50 Stability of Rigid Body Rotations Consider a general rigid body with I (1) I (2) I (3), rotating about one of the principal axes. If small rotations about the other principal axes are introduced, will the rotation be stable? (What happens if you toss a book in the air?) Lets start with rotations about the first principal axis with the smallest moment of inertia. ω = ω 1 ê 1 +ǫê 2 +ηê 3 We assume ǫ and η are very small. Applying the Euler Equations for no applied torque, we get: I (k) ω k +I (j) ω j ω i ǫ ijk = 0 I (1) ω 1 +I (3) ω 3 ω 2 I (2) ω 2 ω 3 = 0 I (1) ω 1 + ( I (3) I (2)) ǫη = 0 I (2) ǫ+ ( I (1) I (3)) ω 1 η = 0 I (3) η + ( I (2) I (1)) ω 1 ǫ = 0 We may neglect terms of order of the perturbation squared so the three coupled equations become. I (1) ω 1 = 0 I (2) ǫ+ ( I (1) I (3)) ω 1 η = 0 I (3) η + ( I (2) I (1)) ω 1 ǫ = 0 ω 1 = const. ǫ = I(3) I (1) I (2) ω 1 η = 0 η = I(1) I (2) I (3) ω 1 ǫ = 0 Note that the constants in the equations are different so we must differentiate one equation again to get a solution. ǫ = I(3) I (1) ω 1 η ( I (2) I (3) I (1))( I (1) I (2)) ǫ = ω 2 I (2) I (3) 1 ǫ

51 51 If (I(3) I (1) )(I (1) I (2) ) is negative, the solutions are oscillatory. If it is positive, they can I (2) I (3) grow exponentially. The constant is negative for axis 1. (Note that the η equation has the same constant.) The equation can actually be applied to any of the three axes. For axis 2, the constant is (I(1) I (2) )(I (2) I (3) ) which is positive, and the oscillations are unstable. I (3) I (1) For axis 3, the constant is (I(2) I (3) )(I (3) I (1) ) which is negative and the oscillations I (1) I (2) are stable. So rotations about the axes with the lowest and highest moment are stable while oscillations about the axis with the intermediate moment are unstable. Try it with a book Lagrange Equations for Top with One Fixed Point We can analyze the motion of a spinning top using the Lagrange equations for the Euler angles. Let us assume that the top has its lowest point (tip) fixed on a surface. We will use the fixed point as the origin. The rotation about the origin will be described by the Euler angles so that all the kinetic energy is contained in the rotation. T = T rot Forasymmetrictop, wecanimmediately writethingsintermsoftherotation about principal axes of inertia. T = 1 I (i) ωi 2 2 (Remember we have three principal moments of inertia but they don t make up a vector.) We have already written ω in terms of the Euler angles. ω = ( φsinθsinψ + θcosψ, φsinθcosψ θsinψ, φcosθ + ψ) Recalling that θ is the angle between the Inertial z axis and the z axis in the rotating frame, the potential energy can be written, U = mghcosθ where h is the height of the center of mass of the top above the fixed tip. i

52 52 z I y B z B θ φ ψ x B y I x I x Note that we are assuming that the symmetry axis of the top is the z axis so that I (2) = I (1) I (12). We can now write the kinetic energy in terms of the Euler angles. T = 1 2 I(12)( ) ω ω I(3) ω3 2 ) 2 ω1 2 = ( φsinθsinψ + θcosψ ω1 2 = φ 2 sin 2 θsin 2 ψ +2 φ θsinθsinψcosψ + θ 2 cos 2 ψ ) 2 ω2 2 = ( φsinθcosψ θsinψ ω 2 2 = φ 2 sin 2 θcos 2 ψ 2 φ θsinθsinψcosψ + θ 2 sin 2 ψ ω 2 1 +ω 2 2 = φ 2 sin 2 θ+ θ 2 ω 2 3 = ( φcosθ + ψ ) 2 T = 1 2 I(12) ( φ2 sin 2 θ+ θ 2 )+ 1 2 I(3) ( φcosθ + ψ U = mghcosθ L = 1 2 I(12) ( φ2 sin 2 θ+ θ 2 )+ 1 2 I(3) ( φcosθ+ ψ) 2 mghcosθ ) 2 d dt ( ) L q i L q i = 0

53 53 Since the Lagrangian does not depend on φ or ψ (cyclic), so the momenta are conserved. p φ = L φ = ( I (12) sin 2 θ+i (3) cos 2 θ ) φ+i (3) cosθ ψ = const. This is the angular momentum about the ẑ I axis. p ψ = L ( ψ = ψ I(3) + φcosθ ) = const. This is the angular momentum about the ẑ B axis. This is reasonable since one can see that the torque is along the line of nodes. The actual values of p φ and p ψ are set by initial conditions in the problem. So p φ and p ψ are constants of the motion and we can solve the equations for φ and ψ. φ = p φ p ψ cosθ I (12) sin 2 θ ψ = p ψ I (p φ p ψ cosθ)cosθ (3) I (12) sin 2 θ There is a third Lagrange equation but it will be easier to understand the motion of the top by using the total energy equation, along with the two conserved momenta. E = 1 ( φ2 2 I(12) sin 2 θ + θ ) ( φcosθ 2 I(3) + ψ) +mghcosθ ( p ψ = I (3) ψ + φcosθ ) E = 1 ( φ2 2 I(12) sin 2 θ + θ ) p 2 ψ +mghcosθ 2I (3) E 1 p 2 ψ 2I = 1 ( φ2 (3) 2 I(12) sin 2 θ + θ ) 2 +mghcosθ E 1 p 2 ψ 2I = 1 (3) 2 I(12) θ2 + (p φ p ψ cosθ) 2 2I (12) sin 2 +mghcosθ θ p 2 ψ E 1 2I = 1 (3) 2 I(12) θ2 +V(θ) = const.

54 54 This is very much like a central force problem with the mass oscillating back and forth in the potential. θ goes to zero at the limits. The motion will be limited between the some angles θ 1 and θ 2 at which E 1 = V(θ). This oscillation of θ 2 I (3) as the angular momentum precesses is called nutation. p 2 ψ 2.11 Homework 1. Calculate the three principal moments of inertia for a cone of mass M, height h, and radius at the base R, using the cone s apex as the origin. 2. Transform the moments of inertia from the previous problem to an origin at the center of mass of the cone.

55 55 3. Find the height at which a billiard ball should be struck so that is will roll with no initial slipping. Find the optimum height for a rail on a billiard table. 4. A homogeneous cube of side l is balanced with one edge resting on a horizontal surface. The cube is allowed to fall. Show that the angular velocity of the ( ) 3g cube when one face hits the surface would be 2l 2 1 if the edge cannot slip on the surface. What would be the angular velocity if the edge can move frictionlessly on the surface? 5. A rigid body consists of three masses connected by nearly massless rods. m 1 = 3m at (b,0,b) m 2 = 4m at (b,b, b) m 3 = 2m at ( b,b,0) Find the inertia tensor for rotations about the given origin. Find the principal moments of inertia (but don t bother to find the axes). 6. What are the principal axes and principal moments of inertia of a uniform density solid hemisphere, about its center of mass. Assume the hemisphere has a mass M and radius R. 1 (A+B) 1 (A B) Consider the inertia tensor I = 1 (A B) 1 (A+B) 0. Perform a rotation about the z axis through an angle θ. Show that a rotation for θ = π renders 4 the tensor diagonal. What are the principal moments of inertia. 8. Show that the trace of a tensor is invariant under rotations. 9. A thin, uniform plate in the shape of an equilateral triangle hangs from the midpoint of one side. Find the frequency of small (pendulum) oscillations. 10. Show that for force-free and torque-free motion of a symmetric top, that the angular momentum, the angular velocity, and the axis of symmetry lie in a plane. 11. A symmetric top, with gravity acting on it, rotates about the vertical direction. Assume the center of mass is a height h above the table. Calculate (and sketch) the effective potential as a function of θ and find the condition on ω for the motion to be stable. 3 Special Relativity One of the biggest surprises in our understanding of Physics came in 1905 with Einstein s paper on Special Relativity. This paper completely changed our un-

56 56 Figure 1: Assistant patent agent, third class A. Einstein. derstanding of time and led to the conclusion that we live in 4-dimensional Minkowski space. ThemoststrikingpropertyofNewton slawsofmechanicsisthatthelawsarethesame in any coordinate system moving at a constant velocity. Einstein extended the symmetry that the laws of physics are the same in any inertial coordinate system to Electricity and Magnetism which had predicted the speed of light. The Einstein principle of relativity states that All of the laws of Physics are the same for every inertial observer. In particular, The speed of light is the SAME for all inertial observers, regardless of the motion of the source. But this required some big changes in our basic understanding of time. 3.1 Some History of Special Relativity The speed of light, c was the central question that gave rise to the theory of relativity. The speed of light is very large compared to the speeds we experience. We have no physical intuition about speeds approaching c. We can however measure the speed of light with a rotating mirror. In what frame is the speed of light c? Newton s laws are independent of which inertial frame is chosen. Is the speed of light going to break this symmetry of physics? Maxwell s equations predict the speed of light from some basic measurements of how fields are produced from charges and currents. c = 1 µ0 ǫ 0

57 57 Since currents are just moving charges, they also essentially predict how the fields transform as we transform from one inertial reference frame to another. These transformations were problematic. There was a simple way out. There could be one frame in which the medium on which EM waves propagate is at rest. The equations of EM were consistent if the speed of light is constant in one fixed frame. Physicists thought EM waves must propagate in some medium. Physicists postulated the ether (aether). They thought, space is filled with the ether in which EM waves propagate at a fixed speed. Ether gave one fixed frame for EM. But experiments, particularly Michelson- Morley disagreed. And we would loose the symmetry found in Newton s laws; any inertial frame. The ether theory was testable. We should see some velocity of the ether. We should see a seasonal variation. Michelson and Morley set up to be sensitive even to the motion of the earth. Albert Abraham Michelson ( ) was a German-born U.S. physicist (at Caltech) who established the speed of light as a fundamental constant. He received the 1907 Nobel Prize for Physics. In 1878 Michelson began work on the passion of his life, the measurement of the speed of light. His attempt to measure the effect of the earth s velocity through the supposed ether laid the basis for the theory of relativity. He was the first American scientist to win the Nobel Prize. Edward Williams Morley ( ) was an American chemist whose reputation as a skilled experimenter attracted the attention of Michelson. In 1887 the pair performed what has come to be known as the Michelson-Morley experiment to measure the motion of the earth through the ether. The figure below shows the Michelson interferometer on a block of granite. A beam of light split, and reflected from two mirrors will interfere.

58 58 The experiment is on a granite block floating in mercury to greatly reduce vibration and allow easy rotation. One can slowly rotate apparatus and measure the interference change. Michelson and Morley found no change as they rotated. The speed of light is the same even though the earth is moving. Oliver Heaviside ( ) was a telegrapher, but deafness forced him to retire and devote himself to investigations of electricity. He became an eccentric recluse, befriended by FitzGerald and (by correspondence) by Hertz. In 1892 he introduced the operational calculus (Laplace transforms) to study transient currents in networks and theoretical aspects of problems in electrical transmission. In 1902, after wireless telegraphy proved effective over long distances, Heaviside theorized that a conducting layer of the atmosphere existed that allows radio waves to follow the Earth s curvature. He invented vector analysis and wrote Maxwell s equations as we know them today. He showed how EM fields transformed to new inertial frames. Hendrik Antoon Lorentz ( ), a professor of physics at the University of Leiden, sought to explain the origin of light by the oscillations of charged particles inside atoms. Under this assumption, a strong magnetic field would effect the wavelength. The observation of this effect by his pupil, Zeeman, won a Nobel prize for 1902 for the pair. However, the Lorentz theory could not explain the results of the Michelson- Morley experiment. Influenced by the proposal of Fitzgerald, Lorentz arrived at the (approximate) formulas known as the Lorentz transformations to describe the relation of mass, length and time for a moving body. (Poincare did this more accurately but referred to this as the Lorentz transformation). These equations form the basis for Einstein s special theory of relativity. George Francis FitzGerald ( ), a professor at Trinity College, Dublin, was the first to suggest that an oscillating electric current would produce radio waves, laying the basis for wireless telegraphy. In 1892 FitzGerald suggested that the results of the Michelson-Morley experiment could be explained by the contraction of a body along its its direction of motion.

59 59 Einstein s On the Electrodynamics of Moving Bodies introduced Special Relativity. Einstein had read Lorentz s book and worked for a few years on the problem. He did not believe there should be one fixed frame. He had a breakthrough which he called The Step in 1905 when he published his paper. Albert Einstein ( ) grew up in Munich where his father and his uncle had a small electrical plant and engineering works. Einstein s special theory of relativity, first printed in 1905 with the title On the Electrodynamics of Moving Bodies had its beginnings in an essay Einstein wrote at age sixteen. The special theory is often regarded as the capstone of classical electrodynamic theory. Einstein did not get a Nobel prize for Special Relativity. He got one for contributions to theoretical physics including the photoelectric effect. The committee did not think Special Relativity had been proved correct until the 1940s. Einstein wanted the speed of light to be the same in every frame. This would work for E&M equations and the way the fields must transform. It would agree with experiment. Einstein did consider experiment but maybe not Michelson Morley. But velocity addition didn t make sense to anyone. How could an

60 60 observer in an inertial frame moving at 0.9c measure light to move at the same speed as we do in our frame at rest? In what he called The Step, Einstein realized that by discarding the concept of a universal time, the speed of light could be the same in every frame. In going from one inertial frame to another, both x and t transform. The time is different in different inertial frames of reference. He derived the previously stated Lorentz transformation from the requirement that the speed of light is the same in every inertial frame. 3.2 The Michelson Morley Experiment: Some Analysis Lets discuss the Michelson Morley Experiment from two points of view. First, assuming there is an ether which picks one frame in which the velocity of light is c and second, if the speed of light is the same in every frame. In the first case, we want to see if the experiment would be sensitive to the motion of the earth through the ether. In the second case we want to see what problems the constancy of the speed of light presents. A diagram of the MM apparatus is show below. Light from a single source was split with a half silvered mirror. In the rest frame of the apparatus, half the beam transited a length 2L perpendicular to the motion of the earth, and the other half transit-ted a length 2L parallel to the motion of the earth. If the earth is in the rest frame of the ether (which it can t be for the all the time), both beams go the same distance at the same velocity and are in phase.

61 61 We will use some of the standard symbols of special relativity in this calculation. β = v c γ = 1 1 β 2 If the earth is moving through the ether at a velocity v, and light moves at a velocity c through the medium of the ether, then a careful calculation is needed. In the direction parallel to the velocity of the earth through the ether, the time taken for the round trip is. ct 1 = L+vt 1 ct 2 = L vt 2 t = t 1 +t 2 = L c+v + L c v = L(c v)+l(c+v) = c 2 v 2 2L c(1 β 2 ) = 2γ2 L c = 2Lc c 2 v 2 trip out to mirror trip back to detector round trip time to v In the direction perpendicular to the velocity of the earth through the ether, the light must actually travel at an angle.

62 62 t = 2L c L = L 2 v 2L2 c 2 = L2 L = L 2 +v 2L2 c 2 L 1 β 2 t = 2γL c time for full trip to v distance of path round trip time to v The difference in time would be. t t = 2γ2 L c 2γL c = (γ 1) 2γL c β 2γL c β2l c The MM interferometer had a length of 11m. The earth s velocity is about meters persecond orabout10000timessmaller thanthespeed oflight. Thedistance difference seen by the interferometer will be about. d = β 2 L = 10 8 (11) = 110 nm This is a large fraction of the wavelength of light so it would be quite detectable. In the case that the laws of physics and the speed of light are the same in any frame, it is certainly easiest to compute what will happen in the rest frame of the experiment. In that frame, things are very simple. t = 2L c t = 2L c The interference pattern will not move as the apparatus is rotated no matter how the earth is moving. Lets look at it (with no relativity), for example, in the frame in which the sun is at rest, which is very much like the putative ether frame. Light moves at the speed of light in that frame and the calculation we had for the ether is correct, if we

63 63 assume that the lengths are the same in that frame. t t = 2γ2 L c t = 2γ2 L c t = 2L c L = L 2 v 2L2 c 2 = L2 L = 2γL c t = 2γL c L 2 +v 2L2 c 2 L 1 β 2 = (γ 1) 2γL c β 2γL c β2l c It is not consistent with the calculation in the rest frame. Fitzgerald postulated that the length along the direction of motion changes as we transform to a moving frame. L = L γ This would make the interference perfect again. We will show in the next section that this is true and is known as Lorentz contraction of lengths. 3.3 The Lorentz Transformation Einstein postulated that the speed of light is the same in any inertial frame of reference. It is not possible to meet this condition if the transformation from one inertial reference frame to another is done with a universal time, that is, t = t. Let us study a transformation from one inertial reference frame to another that is moving with a constant velocity v in the x direction. Such a transformation is usually referred to as a boost. Newton s motion with constant velocity x = vt transforming to x = v t requires that the transformation be linear, like a rotation. We therefore try a linear transformation in which both the position and the time transform. Since this linear transformation will mix x and t, it is reasonable to try to transform quantities that have the same units 2, so we will try a dimensionless transformation of x and 2 Taylor and Wheeler tell a story of a kingdom that measures east-west direction in feet, and the north-south direction in the sacred unit which is different from a foot. They have a lot of trouble surveying the land because trigonometry doesn t work for them.

64 64 ct. We will work in just two dimensions, x and t, like a rotation in the xt plane. (For the boost along the x direction, y and z are not changed.) Assume the origin of a second inertial frame (primed), is moving at velocity v in the x direction, and corresponds to the (unprimed) origin at t = 0. All of these choices can be made without loss of generality. The most general linear transformation of x and t can be written with four dimensionless coefficients. We will call these coefficients γ, u, Γ, and α. x = γ(x u c ct) ct = Γ(ct αx) By the very definition of the transformation we have that the origin of the primed frame is at x = vt in the unprimed frame. (This makes the boost clearly different from a rotation.) x = γ(x u c ct) x = 0 = γ(vt u c ct) u = v the given transformation definition of the boost So we can identify the constant u to be the velocity of the transformation. x = γ(x v c ct) ct = Γ(ct αx)

65 65 The inverse of the transformation must be the same as a transformation with v as the velocity of moving frame. γ and Γ are scale factors that depend on the velocity of the transformation. Since there is no difference between the +x and x directions in physics, we must use the same γ and Γ in the inverse transformation which has the same magnitude of velocity. x = γ(x v c ct) ct = Γ(ct αx) ( ) ( x γ v ct = γ )( ) c x αγ Γ ct ( γ v B(v) = γ ) c αγ Γ the transformation transformation as matrix eq. transformation matrix as function of v B( v) = B 1 (v) inverse is B( v) ( γ + v γ ) ( ) 1 c Γ αγ = α Γ Γ γγ vγαγ v γ γ except α depends on v c c ( γ v γ ) ( 1 c Γ v = γ ) α Γ Γ γγ(1 vα) c rhs is just inverse of 2X2 αγ γ c 1 γ = γγ(1 vα)γ = 1 γ(1 vα) upper left c c 1 Γ = γγ(1 vα)γ = 1 Γ(1 vα) lower right c c γ 2 1 = (1 vα) c Γ 2 1 = (1 vα) c 1 Γ = γ = 1 v α must be > 0 c ( γ v γ c α γ γ α = v c ) = 1 γ 2 (1 v c α) ( γ v γ c αγ γ ) ( γ v = γ c αγ γ ) plug Γ = γ upper left α = v ( c γ v B(v) = γ ) ( ) c γ βγ v γ γ = βγ γ c ( ) ( )( ) x γ βγ x ct = βγ γ ct lower left the transformation now

66 66 Now, consider a pulse of light moving in the x direction emitted at x = 0 and t = 0 in one inertial frame. Since the origins of the two systems coincide at t = 0, this light is emitted in the primed frame with x = 0, and t = 0. At a later time, the position of the light pulse will be at x = ct. By Einstein s postulate that the speed of light is independent of inertial frame, (and by the Michelson-Morley measurement). The pulse of light should be at x = ct in the primed frame. Our transformation must give this result, so lets try it. Transforming the later position of the light pulse, we get. x = γ(x v c ct) ct = γ(ct v c x) x = γ(ct v c ct) ct = γ(ct v c ct) x = ct plug in x = ct plug in x = ct from the 2 eq. The condition that the speed of light is the same in the two reference frames is met. We have shown that the most general transformation to a frame moving with a velocity v, that is consistent with Newton s laws and the isotropy of space, and that satisfies the condition that the inverse of the transformation is a transformation with velocity v, is given by: ( x ct ) = ( )( γ βγ x βγ γ ct) β = v c 1 γ = 1 β 2 y = y z = z This transformation also gives the result that the speed of light is independent of which inertial frame of reference we use. It took surprisingly little physics input to derive the Lorentz transformation for the space-time coordinates.

67 3.4 Checking Michelson Morley with Lorentz Transformation 67 Lets again look at MM in the frame in which the sun is at rest, now using the Lorentz Transformation. Light moves at the speed of light in every frame but the length parallel to motion is reduced 3. Starting in the parallel direction. ct 1 = L +vt 1 ct 2 = L vt 2 t = t 1 +t 2 = L c+v + L c v = L (c v)+l (c+v) c 2 v 2 = 2L c c 2 v 2 = 2L c(1 β 2 ) t = 2γ2 L c t = 2L c L = L 2 +v 2L2 c 2 L 2 v2l2 c = 2 L2 L L = 1 β 2 t = 2γL c L = L γ time out to mirror time back L not changed Fitzgerald contraction t t = 2γ2 L c t t = 2γL c 2γL c 2γL c = (γ 1) 2γL c using transformed length = 0 no shift Fitzgerald postulated that the length along the direction of motion changes as we transform to a moving frame. This is confirmed in the Lorentz transformation. With 3 In this calculation, we simply work in the sun s rest frame and lonly transform the length L from the lab to L in this frame.

68 68 this transformation, the speed of light is the same in any frame and the Michelson- Morley experiment s null result is expected Phenomena of the Lorentz Transformation We have learned that the Lorentz transformation of a space-time coordinate is simplest and most reasonable if the space coordinate and the time coordinate are in the same units. This is not true in our SI system. The unit of distance is one meter and the unit of time is one second. ct for one second is meters. This is one of the reasons it was hard for us to understand transformations in the xt plane. We would be better off with the foot and the nanosecond, but lets not do that. (Consider how messy rotations would be is we measured x in miles and y in microns.) The laws of Physics would be most easily understood if c = 1 but we will not make that simplification either. So we will just use the coordinate ct for time. We will also have to deal with this problem for many other quantities, like Energy for example. Working problems without a universal time is a complication to which we are not accustomed. Consider a muon (an unstable elementary particle) that is produced by cosmic rays (mainly protons) in the upper atmosphere, with a velocity of c. The muon has a mean lifetime of about τ = 2 microseconds (2000 feet in the units we don t use). So ignoring the Lorentz transformation, the muon could typically travel a distance of cτ = 600 meters, however, with the Lorentz transformation, we will find that it can travel much farther if its velocity is near the speed of light. Start with a frame with the muon is at rest at the origin (just as it is produced) at t 1 = 0. In this frame, its rest frame, it decays at t 2 = τ, at the origin. Now we can compute the lifetime in the frame of the earth, in which the muon is moving very fast. We can also compute how far it travels before it decays. In the muon s rest frame, the earth is coming toward it with a velocity of c, meaning β = for

69 69 the transformation. x = γ(x βct) ct = γ(ct βx) γ = x 2 ct 2 1 = 1 = 1 β = γ( cτ) = βγcτ) = γ(cτ β0) = γcτ) x 1 = γ(0 β0) = 0) ct 1 = γ(0 β0) = 0) 1 = transform to earth frame decay position decay time prodution position production time The lifetime viewed in the frame of the earth is γτ 70τ, about 140 microseconds. This phenomenon is called time dilation. The distance traveled in the earth s frame is x 2 x 1 = βγcτ 42 kilometers. Many of the high energy muons produced by cosmic rays can be detected as the reach the earth. They are the main component of cosmic radiation we see on the earth. Consider a stick of length l with one end at the origin x 1 = 0 and the other at x 2 = l meters in the unprimed coordinate system. It is at rest so the x positions of both ends is time independent in this frame. What is its length in a coordinate system moving inthe xdirection? How do we properly measure itslength inamoving coordinate system? This problem is a little harder to solve. A reasonable definition of the length is l = x 2 x 1 with the measurements made at same t, even though there is no simple way to make the measurement of both ends at the same time. The way to calculate the length is a little conterintuitive because of the restriction that we have to use the same t at each end. Lets do it for t = 0, x 1 = 0 and x 2 = l. Transform back to the frame in which the stick is at rest and has a length l. x 1 = γ(x 1 + v c ct ) = 0 x 2 = γ(x 2 + v c ct ) = γl x 2 x 1 = γl = l l = l γ Since γ is always greater than or equal to one, the stick is shorter in the moving frame. This is the phenomenon of length contraction (that Fitzgerald proposed to understand Michelson-Morley). Let us also consider the transverse length. Since we were transforming in the xt plane, we assumed so far that y and z are unchanged by the boost. Consider

70 70 a stick of length l along the y axis in the rest frame. At any time t, its two ends are at (x 1,y 1 ) = (0,0) and (x 2,y 2 ) = (0,l). We must face the possibility that in the primed frame, the stick will not be parallel to the y axis. At t = 0, one end will still be at the origin since both x = 0 and t = 0. The other end will in general be at (ct,x,y,z ) = (0,x,y,0). As before, we do the transformation back into the unprimed frame. The other end of the stick will be at. x = γ(x + v c ct ) = γx y = y ct = γ(0+βx ) = βγx Since x = 0 we have shown that x = 0 and the stick remains parallel to the y axis. The value of t doesn t matter because the stick is at rest in the unprimed frame. Therefore we have shown that the transverse length of the stick need not change, justifying our assumption that the transformation is in the xt plane and that y and z do not change under a boost in the x direction. It is clear from these examples that Special Relativity changes our understanding of many things, including length and time. While the length of vectors is invariant under rotations, it changes under boosts. Time intervals also change when the reference frame is boosted clearly showing that there will be no way to recover an absolute time. Even Euclidean geometry is affected. If 3 dimensional dot products and time intervals are not invariant under Lorentz boosts, what is? Lets take another look at our muon in the two frames and see if we can find anything that is invariant. In the rest frame, we have t 1 = 0 t 2 = τ x 1 = x 2 = 0 ct = cτ x = 0

71 71 In the moving frame (of the earth), we have x = γ(x βct) ct = γ(ct βx) x 2 = γ( cτ) = βγcτ decay ct 2 = γ(cτ β0) = γcτ x 1 = γ(0 β0) = 0 ct 1 = γ(0 β0) = 0 ct = γcτ x = βγcτ ( ct) 2 ( x) 2 = (γcτ) 2 (1 β 2 ) = (cτ) 2 transformation production This is the same asinthe rest frame. At least in this case ( ct) 2 ( x) 2 is invariant under Lorentz boosts 4. It is also invariant under rotations at fixed time since it then is directly related to the 3D dot product. 3.5 Minkowski Space In 1907, Hermann Minkowski proposed that special relativity could be best expressed in a 4-dimensional geometry, with a new and unusual dot product. The fourth dimension would be time if we set c = 1, but, we will continue with ct as the fourth dimension, since we are stuck with the SI units of length and time for most real problems. This is not Euclidean 4-dimensional space but rather what has become known as Minkowski space. Minkowski Space is a 4-dimensional real vector space, spanned by unit vectors with the property ê 2 x = ê2 y = ê2 z = ê2 t = 1. We may number the coordinates like we have in three dimensions, with the time dimension getting the integer number 0. So that the 4-vector position would be. x µ = (ct,x,t,z) Note that we use Greek letters for the indices in 4-vectors. The Einstein summation convention was actually made for this. For example the dot product of a position vector with itself is x µ x µ = (ct) 2 +x 2 +y 2 +z 2 4 Note this has a minus sign so there is a difference between space and time dimensions.

72 72 As we noticed above for the muon decay, this dot product is invariant under Lorentz boosts. It is also invariant under rotations so we call it a scalar. Wemayspeakofaspace-time coordinate like x µ as an event, sincetodescribean event we need to know the location and the time, particularly if we need to transform it into another reference frame. We investigated the interval between the muon s production and decay in two inertial frames and found them to be equal. s 2 = c 2 t 2 + x 2 + y 2 + z 2 This is an invariant quantity which can be used to classify the interval between two events to be spacelike for s 2 > 0, timelike for s 2 < 0 or lightlike for s 2 = 0. For timelike separations, an event can be in the future for t > 0 or in the past for t < 0. If the separation is spacelike, the events are not causally connected in either direction. That is light from one event cannot get to the other. Minkowski space seems to preserve causality. An event from the future cannot be transformed into the past or the present, even though the value of time can change under boosts. The norm of a vector can be positive, negative or zero, making the vector spacelike, timelike, or lightlike respectively. We do not perceive the world to be a 4 dimensional Minkowski space because we humans move much to slowly. We therefore move at very small angles from the time axis, and we make up our units so that velocities are bigger than they would be in natural units Proper Time For an object, a particle, or an observer, the invariant spacetime interval between two events for that object is never spacelike because the object cannot move faster than the speed of light. s 2 = c 2 t 2 + x 2 + y 2 + z 2 > 0 So the space time interval for an object can always be written as the time interval in the object s rest frame. s 2 = c 2 τ 2 So it is convenient to use the proper time τ to understand the object s motion. τ is a scalar quantity since it is directly related to the scalar spacetime interval. As the length of a vector is the square root of the dot product of the vector with itself in Euclidean space, the proper time is essentially the square root of the dot product

73 73 of a vector with itself in Minkowski space (up to the factor of c and the negative sign for timelike vectors that is conventional). WewillusethepropertimetohelpusdefinethevelocityandmomentuminMinkowski space. 3.6 Causality and the Light Cone An inertial observer sits at the origin of his rest frame. The points in 4D Minkowski space that have an (invariant) 4D spacetime interval of zero, s 2 = 0 from the observer form a 4D cone. s 2 = 0 is a lightlike separation, that is, the cone consists of the points in the future that can be reached by light emitted by the observer plus the points in the past from which light could be reaching the observer. (The 4D cone is basically a spatial 3D sphere for every time. The size of the sphere grows with the time difference from the present.) This is referred to as the light cone. Often this light cone is pictured in the time dimension plus two space dimensions to make it understandable. The cone divides space into the causally connected future, the causally connected past, and elsewhere. Events that are elsewhere, cannot affect the observer in the present, nor can the observer affect those events because he cannot move or signal faster than the speed of light. It is however possible that some past event from elsewhere can reach the observer in the future. 3.7 Symmetry Transformations in Minkowski Space In 4 dimensions, we extend the possible symmetry transformations that we had in 3D. The laws of physics are still invariant under translations in position and in time. We still have invariance under rotations. (And we still may have parity inversions and time reversal.) Now we add invariance under Lorentz boosts.

74 74 In one sense Newton already postulated that the laws of physics were the same in any inertial frame. Einstein extended this to include the speed of light. Now this (inertial frame) invariance which seemed separate from the invariance under rotations, really is part of the same group of 4D symmetry transformations. This set of symmetry transformations forms a group called the Lorentz Group. It includes rotations and boosts. The simplest rotations and boosts are transformations in a plane. We have just looked at the boost that is in the xt plane. If we add translation symmetries to the group, it is called the Poincare Group. We are not using group theory here but it can be a powerful tool to understand physics. 3.8 Review of the Hyperbolic Functions Minkowski space is called a hyperbolic geometry. Clearly the time coordinate is not treated the same as the space coordinates. We will find that when space and time coordinates are mixed in a transformation, that we use the hyperbolic functions sinh and cosh instead of sin and cos. They are very symmetrically defined. e iφ = cosφ+isinφ e iφ = cosφ isinφ cosφ = eiφ +e iφ 2 sinφ = eiφ e iφ 2i tanφ = eiφ e iφ e iφ +e iφ e φ = coshφ+sinhφ e φ = coshφ sinhφ coshφ = eφ +e φ 2 sinhφ = eφ e φ 2 tanhφ = eφ e φ e φ +e φ

75 Hyperbolic Function Identities Identities can be easily derived from the definitions. cosh 2 x sinh 2 x = 1 sinh( x) = sinhx cosh( x) = coshx tanh( x) = tanhx coth( x) = cothx The derivatives of the hyperbolic functions. d sinh(x) = cosh(x) dx d cosh(x) = sinh(x) dx d dx tanh(x) = 1 tanh2 (x) = sech 2 (x) = 1/cosh 2 (x) d dx coth(x) = 1 coth2 (x) = csch 2 (x) = 1/sinh 2 (x) Hyperbolic functions of sums. sinh(x+y) = sinhxcoshy +coshxsinhy cosh(x+y) = coshxcoshy +sinhxsinhy tanh(x+y) = tanhx+tanhy 1+tanhxtanhy sinh2x = 2sinhxcoshx cosh2x = cosh 2 x+sinh 2 x = 2cosh 2 x 1 = 2sinh 2 x+1 cosh 2 x 2 = coshx+1 2 sinh 2 x 2 = coshx 1 2

76 76 Inverse hyperbolic functions from logs. ) ( 1 coth 1 x = tanh 1 x ( sinh 1 x = ln x+ ) x 2 +1 ( cosh 1 x = ln x+ ) x 2 1 ;x 1 tanh 1 x = 1 ( ) 1+x 2 ln ; x < 1 1 x Hyperbolic sine and cosine are related to sine and cosine of imaginary numbers. e ix = cosx+i sinx e ix = cosx i sinx coshix = eix +e ix = cosx 2 sinhix = eix e ix = isinx 2 tanhix = itanx coshx = cosix sinhx = isinix tanhx = itanix 3.9 Rotations in 4 Dimensions Clearly the symmetry transformation in the xt (a boost) is not identical to that in the xy plane (a rotation) ( ) γ βγ B(β) = βγ γ ( ) cos(θ) sin(θ) R(θ) = sin(θ) cos(θ) because there is some difference in the geometry, but they are closely related. Lets try to put in the hyperbolic functions by setting tanhφ = β as the off diagonal

77 77 terms in the matrix would indicate. 1 γ = = 1 1 β 2 ( ) = 2 e 1 φ e φ e φ +e φ e φ +e φ = (e φ +e φ ) 2 (e φ e φ ) 2 eφ +e φ 4 = coshφ So we see that βγ = sinhφ and the matrix becomes something very similar to the rotation, tanhφ = β coshφ = γ ( sinhφ = βγ ) coshφ sinhφ B(φ) = sinhφ coshφ φ = tanh 1 β where the rapidity φ plays a role similar to an angle. Like an angle when two subsequent rotations are made in the same plane, the rapidity just adds if two boosts along the same direction are made. φ = φ 1 +φ 2 This can be easily demonstrated by multiplying the two matrices and using the identities for hyperbolic sine and cosine. ( )( ) coshφ1 sinhφ B(φ 1 )B(φ 2 ) = 1 coshφ2 sinhφ 2 sinhφ 1 coshφ 1 sinhφ 2 coshφ 2 ( ) coshφ1 coshφ = 2 +sinhφ 1 sinhφ 2 coshφ 1 sinhφ 2 coshφ 2 sinhφ 1 coshφ 1 sinhφ 2 coshφ 2 sinhφ 1 coshφ 1 coshφ 2 +sinhφ 1 sinhφ 2 ( ) cosh(φ1 +φ = 2 ) sinh(φ 1 +φ 2 ) sinh(φ 1 +φ 2 ) cosh(φ 1 +φ 2 ) = B(φ 1 +φ 2 ) This gives us our simplest calculation of the velocity addition formula. β = tanh(φ 1 +φ 2 ) = tanhφ 1 +tanhφ 2 1+tanhφ 1 tanhφ 2 = β 1 +β 2 1+β 1 β 2

78 78 This never becomes bigger than one and therefore no velocity can exceed the speed of light. The velocity addition formula can also be derived by considering the derivative of the position vector with respect to proper time, which is time in the rest frame. This derivative is a 4-vector while xµ t is not a 4-vector Imaginary Angles Often physicists use ict as the time component of the 4-vector to make the minus sign in the dot product automatic. x µ x µ = (ict) 2 +x 2 +y 2 +z 2 = (ct) 2 +x 2 +y 2 +z 2 It is interesting to note that if we consider the rapidity as an imaginary angle, then a rotation becomes ( ) ( ) cos(iφ) sin(iφ) cosh(φ) isinh(φ) R(iφ) = = sin(iφ) cos(iφ) i sinh(φ) cosh(φ) If we apply this to the vector with ict we get. ( ) ( )( ) x cosh(φ) isinh(φ) x ict = i sinh(φ) cosh(φ) ict ( ) ( )( ) x γ iβγ x ict = iβγ γ ict ( ) ( ) x γx+iβγict ict = iβγx+γict ( ) ( ) x γx βγct ict = iβγx+iγct ( ) ( ) x γx βγct ict = i(γct βγx) ( ) ( ) x γx βγct ct = γct βγx ( ) ( )( ) x γ βγ x ct = βγ γ ct This is the same as our Lorentz transformation. It is interesting to note that the transformation, written this way, is an antisymmetric matrix, like the rotation, while it is symmetric when written in terms of the real variable ct.

79 79 The use of ict is quite convenient for calculations on Special Relativity but is frowned upon because General Relativity requires further extensions of geometry for which the fully real version is preferred A Boost in an Arbitrary Direction More generally for a boost in an arbitrary direction (β x,β y,β z ), ct γ β x γ β y γ β z γ x y = β x γ 1+(γ 1) β2 x (γ 1) βxβy (γ 1) βxβz β 2 β 2 β 2 β y γ (γ 1) βyβx 1+(γ 1) β2 y (γ 1) βyβz z β 2 β 2 β 2 β z γ (γ 1) βzβx β 2 (γ 1) βzβy β 2 1+(γ 1) β2 z β 2 ct x y z 3.10 Velocity Addition We have derived the velocity addition formula using two Lorentz transformations (in the same direction) and rapidity. β = tanh(φ 1 +φ 2 ) = tanhφ 1 +tanhφ 2 1+tanhφ 1 tanhφ 2 = β 1 +β 2 1+β 1 β 2 We can also compute this in terms of the usual dx i dt going from one frame in which an object is moving to a frame boosted along the x direction. (This should give the same result as above if the particle is moving in the x direction.) Let the velocity in the original system be u = d x dt. u i = dx i dt u 1 = γ(dx 1 vdt) γ ( dt v c 2 dx 1 ) = u 2 = dx 2 γ ( dt v c 2 dx 1 ) = u 3 = u 3 γ ( ) 1 vu 1 c 2 dx 1 dt v 1 v c 2 dx 1 dt dx 2 dt γ ( 1 v c 2 dx 1 dt = u 1 v 1 vu 1 c 2 ) = u 2 γ ( 1 vu 1 c 2 ) This formula gives the same result as above but is more general since it allows us to add velocities that are in different directions.

80 80 u i = dx i dt u 1 = u 1 v 1 vu 1 c 2 u 2 = u 2 γ ( ) 1 vu 1 c 2 u 3 = u 3 γ ( ) 1 vu 1 c The Momentum-Energy 4-Vector It is obviously important it determine how Energy and Momentum transform in Special Relativity. A reasonable guess is that momentum is a 3-vector conjugate to position, so we need to find what the fourth component is to make a 4-vector. We again have the problem of the speed of light not being equal to one in our units. The answer, which we will derive below, is that the Momentum-Energy 4-vector is p µ = ( ) E c,p x,p y,p z where the choice of where to put the c could be made by dimensional analysis. The dot product with itself is p µ p µ = E2 c 2 +p2 x +p2 y +p2 z = E2 c 2 +p2 This quantity should be a Lorentz scalar, which we will call m 2 c 2, and we get the equation. p µ p µ = E2 c 2 +p2 = m 2 c 2 Multiplying by c 2 and rearranging. E 2 = p 2 c 2 +m 2 c 4 Again the problem of c 1 is vexing but we get the basic Energy equation of Special relativity. E = (mc 2 ) 2 +(pc) 2

81 81 We understand this as the rest energy mc 2 added in quadrature with pc. For a particle at rest we get the rest energy equation. E = mc 2 Of course any 4-vector transforms like a 4-vector so we have the transformation equations for momentum ( p x = γ p x β E ) c p y = p y p z = p z E = γ(e βp x c) Lets start in the rest frame and do a transformation. p µ = (mc,0,0,0) E c = γmc E = γmc 2 p x = βγmc = βe If we consider a boost in the minus x direction to have the particle moving in the plus x direction afterward, then the boost transformation gives. E = γmc 2 pc = βe These are very useful relations for many kinematic calculations Deriving the Momentum-Energy 4-Vector The problemwe have ishow to take a time derivative ifthetime isthecomponent of a 4-vector. We need some kind of scalar time to make sense of the equations we

82 82 know and love. A well defined time, that does not need to be transformed, is the time in the rest frame of the particle. We call this the proper time τ. We will make use of it here, but later just try to rewrite our equations so that they are covariant in 4 dimensions. The velocity 4-vector can be defined as. v µ = x µ τ = γ x µ t = γ x (ct, x) = γ(c, t t ) = (γc, v) So we see that the time-component of the usual velocity vector is γc and we have the velocity 4-vector v µ = γ(c, v) where γ is for the transformation from the rest frame to whatever frame we are defining v µ in. We can dot the velocity 4-vector into itself. v µ v µ = γ 2 ( c 2 +v 2 ) = c2 +v 2 1 β 2 = c 21 β2 1 β 2 = c2 Thisiscertainly a scalar. Itisanexample oftheproblemthatmany4d lengths are not very useful. That is, the magnitude of the velocity vector is c no matter what the velocity is. To be consistent with non-relativistic equations we will define the momentum. p µ = mv µ = γ(mc,m v) If we identify the time component as above, E c = γmc, we have the relation E = γmc 2 which looks similar to the rest energy equation but actually is true in any frame. A crucial test of this derived 4-vector is whether it gives the right physics in the non-relativistic limit. We did have some choice to make when inserting the energy into the momentum 4-vector. Start with the energy equation from above. E = (mc 2 ) 2 +(pc) 2 = mc 2 1+ ( pc mc 2 ) 2 mc 2 ( 1+ 1 ( pc ) 2 2 mc 2 ) = ( mc p 2 c 2 mc 2 ) = mc 2 + p2 2m

83 83 This is the correct non-relativistic limit. The total energy is, in the non-relativistic limit, the rest energy mc 2 plus the kinetic energy 1 2 mv2. Normally, we ignore the rest energy as being unchangeable. It does change in nuclear interactions for example The Force 4-Vector Since this is a course in mechanics, its hard to avoid Newton s definition of the force F = p. We have the same problem with time differentiation and try the same t solution, the proper time τ. F µ = p ( µ τ = γ p µ 1 t = γ E c t, p ) t Since γ gets large as the velocity approaches c, it takes an infinite amount of force to accelerate a particle to the speed of light Summary of 4-Vectors We started with the position vector in Minkowski space. x µ = (ct,x,y,z) x µ x µ = c 2 t+x 2 +y 2 +z 2 = s 2 An important dot product is that of the difference between two spacetime points. The dot product above gives the distance s in Minkowski space from the origin. The difference between spacetime points for a single particle is an important case. We use the dot product of this difference with itself. ( x µ ) 2 (x (2) µ x (1) µ )(x (2) µ x (1) µ ) = s 2 = c 2 τ 2 The time difference in the particles rest frame τ is called the proper time and is demonstrated to be a scalar quantity in the above equation. We define the velocity 4-vector with the equation.

84 84 v µ = dx µ dτ v µ v µ = c 2 We define the momentum 4-vector with. p µ = mv µ p µ p µ = m 2 c 2 We have shown that in the non-relativistic limit, the 4-momentum is consistent with. p µ = ( E c, p) We accept this as being the components of the momentum 4-vector giving us the dot product of the momentum 4-vector with itself. p µ p µ = E2 c 2 +p2 E 2 = (pc) 2 +(mc 2 ) 2 The dot product of the momentum 4-vector and the position 4-vector p µ x µ = Et+ p x is related to the phase of waves. For example in quantum mechanics, a free particle with a definite momentum is represented by the plane wave. ψ = e i( p x Et)/ We define the Force 4-vector.

85 85 F µ = p µ τ 3.13 The 4D Gradient Operator x µ This also transforms like a vector for us. For example the derivative of a scalar f function f, x µ is a 4-vector. We do need to be careful about the sign on the time derivative. In GR, we keep track of the signs using the metric tensor g µν but we do not wish to introduce this complication here. We choose to use the ict crutch to keep the calculations simple for now. If x 0 = ict, one can see that x 0 introduces a sign change due to the i in the denominator. (In GR... vectors can have upper or lower indices and we use the metric tensor to raise and lower them.) 3.14 The Relativistic Doppler Effect The sound of a train s horn shifts in frequency as the train passes by due to the relative motion of the train and the one who hears it. Similarly, the frequency of light shifts due to relative motion of the source and observer, even without relativity. Relativity modifies this Doppler Effect due to time dilation. Consider a source at rest at the origin with an observer moving in the x direction. We will consider the possibility that the observer is at some distance in y. The beginning of one wavelength is at t 1 = 0 and x 1 = y 1 = 0. The end of the wave is emitted at t 2 = τ and still at x 2 = y 2 = 0. This transforms to the observers frame to be at ct 1 = γ(ct 1 βx) = 0 x 1 = γ(x βct) = 0 y 1 = y 1 = 0 ct 2 x 2 = γ(cτ βx) = γcτ = γ(x βcτ) = βγcτ y 2 = y 2 = 0 τ = γτ

86 86 The time to emit the wave in the observers frame is dilated which decreases the frequency. If the wave travels to the observer in the y direction, the travel time is essentially the same for the beginning and the end of the wave so the frequency is not affected. That is the transverse Doppler effect gives a red-shift which is entirely a relativistic effect. ν = ν γ If the observer is moving directly away from the source we have the additional effect of the distance to the observer increasing with time which gives rise to the parallel Doppler effect. The time at which the beginning and end of the wave arrive at the observer is t 1o = t 1 x 1 c = 0 t 2o = t 2 x 2 c τ o = γ(1+β)τ = ν = 1 β 1+β ν = γτ +βγτ = γ(1+β)τ (1+β) (1+β)(1 β) τ = 1+β 1 β τ β ispositive fortheobserver moving away fromthesource andnegative iftheobserver is moving toward the source The Twin Paradox In Special Relativity, it is not possible to transform something from the future into the past or vice versa but we do have the possibility of slowing down aging by using time dilation. To illustrate this we consider a pair of twins, one of whom makes a trip to a nearby star system and returns. This twin will be much younger than the other twin who remained on the earth. This is known as the twin paradox since there is relative motion but it is not obvious which twin should be younger if the laws of physics are independent of inertial frame. Assume the astronaut twin goes a distance of 8 light years to a star at a speed of 0.8c then turns around and returns at the same speed. In the frame of the earth, the trip takes 10 years, while for the astronaut, it takes only 6 years. l = l γ = 0.6l

87 87 The twin on earth ages 20 years while the astronaut ages only 12 years for the round trip. But the relative velocities are symmetric. What does it matter that one makes the trip and one stays on earth? The difference, if there is one, must be in the acceleration at the beginning, middle, or end of the trip. We can analyze the problem using the Doppler shift. Assume the twins synchronize clocks at the beginning of the trip and that they watch each other s clocks (using a telescope) as the trip progresses. The clock frequencies can be computed using the parallel Doppler shift. For simplicity, we assume the clocks tick once per year in their rest frame. Onthe outbound tripthetwins each observe a clock frequency (of theother s clock) ν 1 β = 1+β ν = = 1 3 that is, one tick every three years. First consider what happens from the earthbound frame. The astronaut twin s clock emits 6 ticks on the trip out. The earthbound twin observes that the trip out takes 18 years receiving one tick every three years. Remember it takes 8 years for the lighttogetbacktoearthoncethespaceship hasreachedthestarsoitsreasonablethat the observed time is 10+8 = 18 years. When the ship turns around, the frequency becomes ν = 3ν. The astronaut s clock emits 6 ticks and the earth twin sees these over 2 years. Thus he sees 12 ticks over 20 years. From the space ship, The astronaut sees 2 earth clock ticks on the (6 year) trip out and sees 18 ticks on the trip back. So the Doppler shift is completely consistent with the astronaut aging 12 years while the earthbound twin ages 20 years. The difference must be in the acceleration, which we have (over) simplified in this problem Kinematics Problems in Electron Volts The electron Volt (ev) is a common unit of energy for particles, since it is directly related to how we accelerate particles. The electron volt is simply the kinetic energy a particle with the(fundamental) charge of the electron gets if it moves through a (electric) potential difference of 1 Volt (the SI unit). We take advantage of the new energy unit to define units for mass and momentum that effectively eliminate the need to multiply by c. The unit for momentum is ev c and the unit for mass is ev c 2. Often we use Million electron Volts (MeV) or Giga electron Volts (GeV) since the masses of particles are often in that range. Atoms have energy differences between

88 88 states on the order of ev. Nuclei have energy differences on the order of MeV. The proton has a mass a little lest than 1 GeV c π 0 Decay The pions are the lightest(free) particles that interact strongly. They are substantially lighter than the proton and play an important role in binding nuclei together. There are three pions, the π +, the π, and the π 0. The π 0 decays rapidly into two photons while the charged pions live for about 20 nanoseconds in its rest frame. We will analyze the decay of the π 0 into two photons. The mass of the π 0 is about 135 MeV and the photon mass is zero. c 2 In all kinematic problems, the conservation of momentum and energy is best understood as conservation of the 4-momentum. In the rest frame of the π 0, the 4-momentum is p µ = (m,0,0,0) dropping the cs from the equation. The final 4-momentum for photon-1 is (E (1), p (1) ) and the final 4-momentum for photon-2 is (E (2), p (2) ). 4-momentum conservation gives us the 4-vector equation. p (initial) µ = p µ (final) p (initial) µ = p (1) µ +p(2) µ p µ = (m, 0) = (E (1), p (1) )+(E (2), p (2) ) p (2) = p (1) m π = E (1) +E (2) E 2 = p 2 +m 2 = p 2 E γ = p γ m π = 2p γ The decay goes into two back to back photons each with energy mπ. If we want to 2 see the result in another frame, we just boost the 4-vector. This is a simple example of using the conservation of 4-momentum to solve a kinematics problem. We can analyze our solution a little to learn to count the number of variables in a problem. The initial state is given and 4-momentum is conserved giving us 4 equations. The final state has two particles. If we know the masses, we get 3 unknown variables per final state particle. That is, an unknown 4-momentum with the requirement that its norm is m 2.

89 89 In π 0 decay, that means we have 4 equations in 6 unknowns and two variables will be left undetermined. In our solution, these two variables are the polar and azimuthal angle of one final state photon. The other photon will go in exactly the opposite direction according to our solution Neutron Decay Nuclei are made up of protons and neutrons. While neutrons are stable inside many nuclei, free neutrons decay with a lifetime of about 15 minutes. This makes them a radiation problem around nuclear reactors, since they can leak out of the reactor and decay. The neutron decays into a proton, an electron, and an antineutrino of the electron type. n p + e ν e The mass of the neutron is MeV. The proton mass is MeV. The mass of the electron is MeV. The mass of the electron neutrino is nearly zero. (We think its not zero but we only measure it to be small, m < 1 ev.) This is a more complicated problem. Lets count variables. The three final state particles give us 9 variables with only 4 equations to solve. It is usual not to worry about the orientation of the whole decay (unless the initial state is polarized). The three final state particles will lie in a plane (for neutron decay in its rest frame). It takes two variables to define the orientation of the normal to the plane and the event can still be rotated inside the plane so like a rigid body, three angles are needed to describe the overall orientation of the event. Since the laws of physics are symmetric under these rotations, these three variables do not matter. So its fine to have 3 unknown variables at the end but here we have 5. That means we can have some kind of interesting physics in this decay (that we won t understand here). That is, the final state is not completely given by kinematics. There are two variables undetermined that depend on the physics of the decay. To make a problem we can solve, we ask the question what is the maximum electron energy possible? in this decay. The maximum possible energy will occur when both the proton and neutrino recoil directly opposite the electron s direction. If we pick the direction of the electron (2 variables), we have a one dimensional problem with three unknown momenta and just two equations, energy conservation and momentum conservation. We will still need to maximize the electron momentum to get the answer.

90 90 Lets work with the three independent variables p e, p p, and p ν. The three energies are. E e = m 2 e +p2 e E p = m 2 p +p 2 p E ν = p ν Defining all the momenta to be positive numbers, the two conservation equations are. m n = m 2 e +p2 e + m 2 p +p2 p +p ν 0 = p e p P p ν p ν = p e p p m n = m 2 e +p2 e + m 2 p +p2 p +p ν m 2 e +p 2 e = m n m 2 p +p2 p p e +p p m 2 e +p 2 e +p e = m n m 2 p +p2 p +p p To maximize p e, we must maximize p p. This will happen for p ν = 0. Then we have p p = p e. m 2 e +p 2 e +p e = m n m 2 p +p2 e +p e m 2 e +p 2 e = m n m 2 p +p2 e The proton is highly non-relativistic (p << m), so we may approximate its energy

91 91 by. m 2 p +p 2 e m p + p2 e 2m p m 2 e +p 2 e = m n m p p2 e 2m p m 2 e +p 2 e = m p2 e 2m p m 2 e +p2 e = ( m)2 m p2 e + p4 e m p ( 1+ m ) p 2 e p 2 e m p ( 1+ m ) 4m 2 p = ( m) 2 m 2 e + p4 e 4m 2 p ( m) 2 m 2 e m p p e ( m)2 m 2 e (1.29) ( ) = 2 (0.51) ( ) 2 = m m p E e = m 2 e +p 2 e = When the neutrino momentum is zero, the electron takes nearly all the energy available. The proton, recoiling against it, takes very little energy since it is highly nonrelativistic even though it has the same momentum as the electron Compton Scattering Compton scattered high energy photons from (essentially) free electrons in He measured the wavelength of the scattered photons as a function of the scattering angle. The figure below shows both the initial state (a) and the final state, with the photon scattered by an angle θ and the electron recoiling at an angle φ. The photons were from nuclear decay and so they were of high enough energy that it didn t matter that the electrons were actually bound in atoms. We wish to derive the formula for the wavelength of the scattered photon as a function of angle.

92 92 With two final state particles, we have 6 unknowns and 4 conservation equations, leaving two variables undetermined. One of these is the uninteresting azimuthal angle (rotation about the beam direction). The other is the scattering angle for the photon θ which is not determined but interesting. The probability distribution in θ may tell us something about the interaction responsible for the scattering but our analysis is only of the kinematics. We will simply calculate what the energy is for the scattered photon as a function of the scattering angle θ. We solve the problem using only conservation of energy and momentum. Lets work in units in which c = 1 for now. We ll put the c back in at the end. Assume the photon is initially moving in the z direction with energy E and that it scatters in the yz plane so that p x = 0. Conservation of momentum gives E = E cosθ +p e cosφ and Conservation of energy gives E sinθ = p e sinφ. E +m = E + p 2 e +m2 Our goal is to solve for E in terms of cosθ so lets make sure we eliminate the φ. Continuing from the energy equation E E +m = p 2 e +m2 squaring and calculating p 2 e from the components we eliminate φ E 2 +E 2 +m 2 2EE +2mE 2mE = (E E cosθ) 2 +(E sinθ) 2 +m 2

93 93 and writing out the squares on the right side E 2 +E 2 +m 2 2EE +2mE 2mE = E 2 +E 2 2EE cosθ +m 2 and removing things that appear on both sides and grouping 2EE +2mE 2mE = 2EE cosθ m(e E ) = EE (1 cosθ) (E E ) = (1 cosθ) EE m 1 E 1 E = (1 cosθ) m Since λ = h/p = h/e in our fine units, λ λ = h m (1 cosθ). We now apply the speed of light to make the units come out to be a length. λ λ = hc mc 2 (1 cosθ) These calculations can be fairly frustrating if you don t decide which variables you want to keep and which you need to eliminate from your equations. In this case we eliminated φ by using the energy equation and computing p 2 e Lagrange Equations in Special Relativity The basis for Special Relativity is that the laws of physics are independent of which inertial coordinate system we write them in. We wish to write equations in terms of scalars, 4-vectors, and tensors, with both sides of the equation transforming the same way under rotations and boosts. Such equations are said to be covariant, because both sides of the equation transform in the same standard way. The Lagrangian is based on the principle of least Action. The Action as we know it so far is clearly not a vector or a tensor so our obvious choice is to make it a scalar. Our definition of the Action so far is. S = t f t i Ldt = t f t i Lγdτ

94 94 We should then have γl be a scalar quantity. But what scalar do we have? For a particle, the mass is the only scalar we can use. Clearly this is not related to external potential applied to the particle. γl = mc 2 L = mc2 γ p i = L q i = mc 2 1 β 2 v i = v i c 2γmc2 = γmv i This agrees with our previous result p = γm v indicating that we have picked the right kinetic term for the Lagrangian. The (velocity independent) potential can be subtracted. L = mc2 γ V This is an important relativistic basis for our use of L = T V in nonrelativistic problems. To have fully relativistic problems we need to have potentials, for example, that are relativistically correct. We will take a look at the relativistic equations of electromagnetism, but this is really beyond the scope of this course. In field theory, there will be scalar interaction terms in the Lagrangian, like the interaction between the electric current due to particles with the vector potential, a dot product between 4-vectors. The Hamiltonian can be computed from the Lagrangian. H = v i p i L = p ip i γm L = p2 c 2 γmc 2 + mc2 γ +V = 1 γmc 2(p2 c 2 +m 2 c 4 )+V = E2 γmc 2 +V It is the total energy. H = E +V = T +mc 2 +V It is quite interesting to look at this Action which is to be minimized in Minkowski space and to try to formulate the principle of least Action in a covariant way. S = τ f τ i mc 2 dτ

95 95 What Action should we use for particles in Special Relativity? The proper time is directly related to the space-time interval. ds 2 = c 2 dτ 2 So we can write our action to be minimized in terms of the spacetime interval. S = τ f τ i mc 2 dτ = mc τ f τ i ds 2 That is the particle follows a path through 4-space that minimizes the spacetime interval. In curved space-time, this path is called a geodesic. We may take this back to an integral over proper time. S = mc τ f τ i ds2 = mc τ f τ i ds 2 dτ = mc dτ2 τ f τ i vµ v µ dτ While the factor of mc may seem useless, it becomes useful if an external potential (other than gravity) is applied Covariant Electricity and Magnetism Equations Rationalized Heaviside-Lorentz Units The SI units are based on a unit of length of the order of human size originally related to the size of the earth, a unit of time approximately equal to the time between heartbeats, and a unit of mass related to the length unit and the mass of water. None of these depend on any even nearly fundamental physical quantities. Therefore many important physical equations end up with extra (needless) constants in them like c. Even with the three basic units defined, we could have chosen the unit of charge correctly to make ǫ 0 and µ 0 unnecessary but instead a very arbitrary choice was made µ 0 = 4π 10 7 and the Ampere is defined by the current in parallel wires at one meter distance from each other that gives a force of Newtons per meter. The Coulomb is set so that the Ampere is one Coulomb per second. With these choices SI units make Maxwell s equations and our field theory look very messy. Physicists have moreoftenused CGS unitsinwhich theunit of chargeanddefinition of the field units are set so that ǫ 0 = 1 and µ 0 = 1 so they need not show up in the

96 96 equations. The CGS units are not perfect, however, and we will want to change them slightlytomakeourtheoryofthemaxwell Fieldsimple. Themistakemadeindefining CGS units was in removing the 4π that shows up in Coulombs law. Coulombs law is not fundamental and the 4π belonged there. We will correct this little mistake and move to Rationalized Heaviside-Lorentz Units by making a minor modification to the unit of charge and the units of fields. With this modification, our field theory will have few constants to carry around. As the name of the system of units suggests, the problem with CGS has been with π. We don t need to change the centimeter, gram or second to fix the problem. In Rationalized Heaviside-Lorentz units we decrease the field strength by a factor of 4π and increase the charges by the same factor, leaving the force unchanged. E E 4π B B 4π A A 4π e e 4π α = e2 c e2 4π c Its not a very big change but it would have been nice if Maxwell had started with this set of units. Of course the value of α cannot change, but, the formula for it does because we have redefined the charge e. Maxwell s Equations in CGS units are The Lorentz Force is B = 0 E + 1 B c t = 0 E = 4πρ B 1 c E t = 4π c j. F = e( E + 1 c v B). When we change to Rationalized Heaviside-Lorentz units, the equations become

97 97 B = 0 E + 1 B c t = 0 E = ρ B 1 E c t = 1 c j F = e( E + 1 c v B) That is, the equations remain the same except the factors of 4π in front of the source termsdisappear. Ofcourse, itwouldstillbeconvenient tosetc = 1sincethishasbeen confusing us about 4D geometry and c is the last unnecessary constant in Maxwell s equations. For our calculations, we can set c = 1 any time we want unless we need answers in centimeters The Electromagnetic Field Tensor The transformation of electric and magnetic fields under a Lorentz boost we established even before Einstein developed the theory of relativity. We know that E-fields can transform into B-fields and vice versa. For example, a point charge at rest gives an Electric field. If we boost to a frame in which the charge is moving, there is an Electric and a Magnetic field. This means that the E-field cannot be a Lorentz vector. We need to put the Electric and Magnetic fields together into one (tensor) object to properly handle Lorentz transformations and to write our equations in a covariant way. The simplest way and the correct way to do this is to make the Electric and Magnetic fields components of a rank 2 (antisymmetric) tensor. 0 ie x ie y ie z F µν = ie x 0 B z B y ie y B z 0 B x ie z B y B x 0 Thefieldscansimplybewrittenintermsofthevector potential, (whichisalorentz vector) A µ = (iφ, A).

98 98 F µν = A ν x µ A µ x ν Note that this is automatically antisymmetric under the interchange of the indices. As before, the first two (sourceless) Maxwell equations are automatically satisfied for fields derived from a vector potential. We may write the other two Maxwell equations in terms of the 4-vector j µ = (icρ, j). F µν = j µ x ν c Which is why the T-shirt given to every MIT freshman when they take Electricity and Magnetism should say... and God said ( A ν x ν x µ Aµ x ν ) = jµ c and there was light. Of course he or she hadn t yet quantized the theory in that statement. For some peace of mind, lets verify a few terms in the equations. Clearly all the diagonal terms in the field tensor are zero by antisymmetry. Lets take some example off-diagonal terms in the field tensor, checking the (old) definition of the fields in terms of the potential. B = A E = φ 1 c A t F 12 = A 2 x 1 A 1 x 2 = ( A) z = B z F 13 = A 3 x 1 A 1 x 3 = ( A) y = B y F 0i = A i x 0 A 0 x i = 1 ic A i t (iφ) x i = i ( 1 A i c t + φ ) ( φ = i + 1 x i x i c ) A i = ie i t

99 99 Lets also check what the Maxwell equation says for the last row in the tensor. F 0ν = j 0 x ν c F 0i = icρ x i c (ie i ) = iρ x i E i = ρ x i E = ρ Lorentz Transformation of the Fields Let us consider the Lorentz transformation of the fields. Clearly A µ just transforms like a vector. We could derive the transformed E and B fields using the derivatives of A µ but it is interesting to see how the electric and magnetic fields transform. We know that Maxwell s equations indicate that if we transform a static electric field to a moving frame, a magnetic field will be generated, because there is a current in that frame. Think of the case of a point charge in its rest frame and in a frame in which it is moving. It was clear from E&M that E and B were not simply parts of 4-vectors. The Electric and Magnetic fields are part of a rank 2 tensor and so they transform

100 100 accordingly. F µν = B µρf ρσ Bσν T γ iβγ ie x ie y ie z γ iβγ 0 0 F µν = iβγ γ 0 0 ie x 0 B z B y iβγ γ ie y B z 0 B x ie z B y B x γ iβγ 0 0 βγe x iγe x ie y ie z F µν = iβγ γ 0 0 iγe x βγe x B z B y iγe y iβγb z βγe y γb z 0 B x iγe z +iβγb y βγe z +γb y B x 0 βγ 2 E x +βγ 2 E x iγ 2 E x iβ 2 γ 2 E x iγe y +iβγb z iγe z iβγb y F µν = iβ 2 γ 2 E x iγ 2 E x βγ 2 E x βγ 2 E x βγe y +γb z βγe z γb y iγe y iβγb z βγe y γb z 0 B x iγe z +iβγb y βγe z +γb y B x 0 0 ie x iγ(e y +βb z ) iγ(e z βb y ) F µν = ie x 0 γ(b z +βe y ) γ(b y βe z ) iγ(e y βb z ) γ(b z +βe y ) 0 B x iγ(e z βb y ) γ(b y βe z ) B x 0 E = E B = B E = γ(e β B) B = γ(b + β E) Since we could choose any direction for the x axis that we boosted along, these results for the field transformation are correct for all boosts E&M is a Vector Theory A more detailed study of the E&M field shows that the free fields in the Lagrangian are the four components of the vector potential A µ. They couple to the current (charge) density vector j µ. Its a vector theory Homework For the purposes of brevity and clarity, let us define the inertial frame K to be at rest in the lab with coordinates x µ and define another inertial frame K to coincide

101 101 with K at t = t = 0 and to be moving relative to K with a velocity v = βc in the x direction. 1. Assume anobject of length l is atrest inthe K frame. Showthat the expression for the Lorentz contraction can be obtained if an observer in the K frame measures the time it takes for an object to pass and multiplies by the velocity v. 2. Two events take place at the same time but at positions differing by x in the K frame. Calculate the difference in position x and the difference in time t in the K frame. 3. Clocks located at the origins of the K and K frames are synchronized at t = 0. At a time t, an observer at the origin of the K system uses a telescope to read the the clock moving with the K system. What time does he read. 4. A relativistic rocket emits exhaust gasses with a constant velocity v e in the rest frame of the rocket and at a rate dm. Show that the equation of motion for dt the rocket is m(t) dv +v dt e dm dt (1 β2 ) 3/2 = 0, where m(t) is the mass of the rocket. 5. A muon is an unstable particle with a mean lifetime of 2 microseconds in its rest frame and a mass mc 2 = 106 MeV. What is its mean lifetime in the frame of the earth if its energy is E = 106 GeV. 6. Anearbystarisfoundtobemovingawayfromtheearthwithavelocityof m/s by measuring the Balmer α (H-alpha) wavelength normally λ = nm. How much and in what directions is the wavelength shifted? 7. An astronaut travels to a nearby star system, a distance of 11 light years away, and returns. Both legs of the trip are made at a velocity of 0.6c. How much have the astronaut and his twin on earth aged? 8. The total solar radiation at the position of the earth ( m from the sun) is measured by satellite to be 1,366 W/m 2. How much mass per second is converted to energy in the sun? 9. What is the kinetic energy of a particle of mass m and momentum p? 10. LEP collided electrons with positrons with a center of mass energy of 208 GeV. The two particles have the same energy (104 GeV) in the lab. What is β for these particles? 11. A neutral pion (mc 2 = 135 MeV) has a energy of 40 GeV. It decays into two photons. If the two photons have the same energy in the lab frame, what is the angle between them?

102 A neutron (mc 2 = MeV) decays into a proton (mc 2 = MeV) and an electron (mc 2 = MeV) and an antineutrino (mc 2 0 MeV). The decay is symmetric so that the three final state particles lie in a plane separated by angles of 120 degrees. What is the momentum of each particle? What is the kinetic energy of each particle? What trend do you notice? 13. Use the Lorentz transformation to derive the velocity addition formula. 14. A spacecraft passes Neptune with a speed of 0.9c relative to the planet. A second spacecraft, more urgently exploring the outer solar system, is observed to pass the first one (in exactly the same direction) at a speed of 0.4c. What is the speed of the second spacecraft relative to Neptune? 15. Show that the relativisticly corrected form of Newton s Second Law is. F = m dv dt (1 v2 c 2 ) Calculate the kinetic energy of a particle scattered through an angle θ by a target particle of equal mass. Assume the target particle is at rest and that the incident particle has kinetic energy T Calculate the dot product of the velocity 4-vectors for two particles, a and b. Show how it depends on the relative velocity of the two particles. 18. In modern High Energy physics, most particle accelerators collide beams of equal energy particles rather than collide a beam with a target that is at rest. Assume we want to collide protons with protons. Calculate a formula for the Center of Mass energy of the collisions for the case of a fixed target. The Large Hadron Collider will have a center of mass energy of 14,000 GeV. What energy beam would be needed to achieve this center of mass energy by colliding with a fixed target of protons (Liquid Hydrogen)? 19. Assume a neutrino beam is made by generating a beam of π + with kinetic energy T = 500 GeV. The π + decay to a positive muon and a neutrino of the muon type, π + µ + ν µ. What is the energy range of neutrinos in the beam if we define the beam to be in the direction of the original pion beam within and small angle of θ = 10 3 radians? 20. Show that Maxwell s equations written in 3D and in Rationalized Heavyside- Lorentz Units, can all be derived from the 4D Maxwell s equation in conjunction with the definition of F µν = Aν x µ Aµ x ν.

103 Assume the relativistic motion of a particle in 4D is simply given by δs 2 = c 2 τ f 2 δ ds 2 = 0. Show that this leads to the standard non-relativistic Lagrange c 2 τ 2 i equation for a particle with no forces acting on it. 4 A Little General Relativity Special Relativity extends the symmetry of the laws of physics in inertial frames of reference. Newton s laws had this symmetry but were not correct at large velocities and electromagnetism with the speed of light showed us how to extend this principle, by going to 4 dimensions. Newton s gravity also had the somewhat surprising symmetry, that the force of gravity is proportional to the same mass used in the calculation of acceleration in F = ma. We refer to this as the equality of gravitational mass and inertial mass. Why is the force of gravity proportional to the inertial mass, when other forces don t work this way. General Relativity explains this by saying that there is no difference between gravity and acceleration. This is called the general principle of equivalence. If we are inside an elevator and feel a force pushing us down against the floor, we can t tell if it is due to gravity, or the elevator accelerating upward. This explains why the gravitational mass is equal to the inertial mass. Essentially gravity is an acceleration so there is no force of gravity. Einstein extended the change of geometry started by Minkowski, to change from the flat geometry of Minkowski space to a 4D geometry which can have curvature. Curvature in spacetime geometry has the effect of producing accelerations. The Einstein Equation relates the curvature in the geometry to the mass and energy density. It is a tensor equation G µν = 8πT µν, but the math is too complicated for one chapter in this course in mechanics. The study of General Relativity is usually done in a year-long course on this topic alone. Nevertheless, it is useful to understand the implications of General Relativity and curved spacetime without trying to solve the Einstein equation.

104 Geometries 4.2 The Metric Tensor In the Riemannian geometry of General Relativity, lengths (dot products) are computed using a metric tensor which depends on the stress-energy tensor T µν