Relativistic Transformations

Transcription

1 Relativistic Transformations Lecture 7 1 The Lorentz transformation In the last lecture we obtained the relativistic transformations for space/time between inertial frames. These transformations follow mainly from the postulate that there is a limiting velocity which no information propagation can exceed. These transformations are; x = x y = y z = γ(z V 0 t) t = γ(t (v 0 /c 2 )z) with γ = 1 1 β 2 and β = V 0 /c They have an inverse because we require a one-to-one map between the inertial systems. The inverse can be obtained by solving the equation set for the unprimed variables, or just noting that a symmetry operation would change the sign of the velocity. x = x y = y z = γ(z + V 0 t ) t = γ(t + (v 0 /c 2 )z ) We need both the equation sets in the next section. 2 Velocity Transformation The velocity transformation can be obtained in several ways. The most straightforward is to differentiate the spatial positions. We expect that the velocity transformations will preserve 1

2 the maximum velocity, c, as required when the Lorentz transformations were derived. Thus; U x = dx = dx dt dt The form of the transform for U y should be similar as the equations are symmetric perpendicular to the boost direction, z. Here we need dt dt, which are evaluated using the Lorentz transformations; dt = γ(1 + (V 0 /c)u z Then as x = x the velocity transformation is; U x = U x γ(1 + (V 0 /c 2 )U z ) For the velocity in the boost direction; U z = dz = dz dt dt As we have evaluated dt we need to evaluate; dz dt = γ(u z V 0 ) Combine this with the above expression and solve for U z. The transformation is U x = U z V 0 1 (V 0 /c 2 )U z Collecting the velocity transformation equations; U x U x = γ(1 (V 0 /c 2 )U z ) U y U = y γ(1 (V 0 /c 2 )U z ) U x = U z V 0 1 (V 0 /c 2 )U z 3 Acceleration Transformation To find the transformation of an acceleration, take the derivative of the velocities. a x = du x = du x dt dt 2

3 and; a z = du z = du z dt dt As previously take the appropriate derivatives of the velocity transformation equations. This results in; a x a x = (1/γ 2 ) [ (1 (V 0 U z )/c 2 ) 2 + (U xv 0 /c 2 )a z (1 (V 0 U z )/c 2 ) 3 ] a y a y = (1/γ2 ) [ (1 (V 0 U z )c 2 ) 2 + (U yv 0 /c 2 )a z (1 (V 0 U z )/c 2 ) 3 ] a z a z = (1/γ 3 ) [ (1 (V 0 U z )/c 2 ) 3 ] In a frame instantaneously at rest we place U x = U y = U z = 0. Then; a x = a x /γ2 a y = a y /γ2 a z = a z/γ 3 4 Images of a Lorentz Contraction Consider the figure of a rod traveling perpredincular to a distant observer as shown in Fig. 1. We want to take a picture of the rod as it passes, so that light from all points on the rod must reach the observer at the same time. Because some points are further away light from these points must have been emitted earlier than points closer to the observer. We know that the length of the rod is contracted by γ. Draw in the dottted figure of the rod as in figure 1. If the length of a side of the dotted figure is L the angle between the rod and the dotted line is cos(α) = 1/γ. Light from the far side corner of the rod travels a distance d in a time d/c. During this time the rod moves a distance (d/c)v. This makes the perpendicular side have a length βd/sin(α) = d. Therefore the rod appears as if it has rotated. Note we are able to see this side of the rod perpendicular to the direction of motion. Suppose we now look at a hoop moving perpendicular to its axis as shown in fig. 2. As with the above analysis we expect to see the back side of the hoop. In a way similar to the above analysis, we require simultaneous light rays to reach the observer. The hoop then appears rotated in the same way as the rod. 3

4 L d t = d/c βd α L γ V Figure 1: Rod traveling by a distant observer. Assume parallel ligh rays reach the observer at the same time. r γ tan( α) r a r γ α V Figure 2: A hoop traveling by a distant observer. Light from all points on the surface must reach an observer at the same time. 4

5 θ k k k V Figure 3: The wave vectors involved in the doppler shift. 5 Doppler Shift Consider a plane wave propagating with wave vector k. The phase of the wave has the form; φ = k x ωt The phase of the wave should have the same form in any inertial frame. Therefore we transform to a primed frame so that; φ = φ = k x ω t Substitute the Lorentz transformations to get the equation; k x ωt = γ(v/c) z (z V t) + k x x + k y y γ(ω t (V/c) 2 z) Collect terms in x, y, z, t which because of linear independence each must be equal to zero. γ( k V + ω ) = ω k = k γ(k + (ω /c 2 )V ) = k Because of the constantcy of the light velocity k /ω = k/ω = c As k = k cos(θ) in the fig. 3, we have the Doppler shift; ω = γ ω (1 + β cos(θ)) There is then a relation between the angles in the inertial frames between the wave vector and the propagation direction; 5

6 tan(θ) = k /k = tan(θ ) = sin(θ ) γ(cos(θ ) β) sin(θ ) γ(cos(θ ) + β) Note there is a transverse Doppler shift when the angle is not 0 or π. 6 Boosted Light Cone Now assume a light source in its rest frame that emitts light isotropically with total power, P 0. Thus the intensity defined as the power per unit area of surface; I = P 0 /(r 2 dr dω ) The radiant flux is therefore ; F = I dω The radiant flux is the same in a boosted frame, so that; F = I dω = F = I dω I(θ) = (P 0 /r 2 dr) dω dω = dsin(θ )dθ sin(θ)dθ Use the angle transformation as obtained above to show that at high values of β, this is very forward peaked around the boost direction. This is called the searchlight effect which produces a narrow beam of light in the moving frame even for isotropic emission in the rest frame. dcos(θ ) dcos(θ) = 1 β 2 (1 β cos(θ)) 2 The maximum value is obtained at θ = 0 and use β = 1 1 γ 2 which one can expand in a power series when 1/γ 2 is small. The maximum value is 4γ 2 I(θ ). Take (1/2) of this and find the angle where this occurs. 2γ 2 = 1 β 2 (1 β cos(θ)) 2 Solve for θ using power expansions to obtain 6

7 θ 2/γ 2 7 Twin Paradox Suppose we synchronize 2 clocks and place one on a rocket while the other remains at rest. The rocket travels a distance L at a velocity V and returns back to the starting point with velocity V where it brought back to rest. The clock at rest, A, sends signals at a frequency f to the clock on the rocket, B. The clock on the rocket sends signals to the clock A with the same rest frequency. We now determine the number of signals sent and received by each clock. To do this we will need the doppler shifted frequencies, f = f 1 ± β. We fill out 1 β Table 1 to find that both clocks agree as to the number of signals sent, which is the same as the number received by the other clock. However, the number of signals sent by the clock in motion is less than the one that remains at rest. Table 1: Times and counts related to the twin paradox Name A (earth rest) B (onboard rocket) Time τ = 2L/V τ = 2L/(γV ) No. Signals sent f τ f τ Time τ = 2L/V τ = 2L/(γV ) Time to B s Turn t = L/V + L/c t = L/(γV ) No Signals Reveived (f 1 β L/V )(1 + β) (f 1 β L/V ) 1 β 1 β 2 1 β Remaining Time t = L/V L/c t = L/(γV ) Signals Received (f 1 + β L/V )(1 β) (f 1 β L/V ) 1 + β 1 β 2 1 β Total signals received f τ f τ The reason that the problem is not symmetric, ie one cannot swap A and B and arrive at a symmetric answer, is that B must be accelerated to move away and return while A remains at rest. 8 Energy and Momentum To develop the relativistic momentum and energy we wish to preserve momentum conservation, which is a general principle of mechanics. Thus we will define a momentum as a mass µ(m 0, U) times a vector velocity. The mass µ is allowed to be a function of a constant, m 0 and the total speed of the object, U. We set up a collision in a reference frame between 2 7

8 1 before y 1 after x 2 after 2 before Figure 4: The collision of 2 particles of equal mass and equal velocities. Symmetry allows the velocities after the collision to be directly related to those before the collision. particles having equal incoming velocities and masses. This means that the total momentum of this system is 0 and we can look at symmetry to determine the velocities after the collision. Consider Fig. 4. Velocities before the collision; U I1x = U I2x = a U I1y = U I2y = b U I1z = U I2z = 0 Velocities after the collision; U F1x = U F2x = a U F1y = U F2y = b U F1z = U F2z = 0 We now work in 2-D dropping the terms in z. Now make a transformation to a frame moving with a velocity a in the x direction. We need to apply the velocity addition equations. Before the collision; U I1x = 0 U I2x = U I1y = 2a 1 + (a/c) 2 b 1 (a/c) U 1 (a/c) 2 I2y = (a/c) 2 b 8

9 U I1 = The values of U Ij b U 1 (a/c) 2 I2z = 4a 2 + b 2 (1 (a/c) 2 ) 1 + (a/c) 2 U F1x = 0 U F2x = U F1y = U F1 = are the total velocities. After the collision; 2a 1 + (a/c) 2 b 1 (a/c) 2 U F2y = b 1(a/c) 2 U F2 = 1 (a/c) (a/c) 2 b 4a2 + b 2 (1 (a/c) 2 ) 1 + (a/c) 2 Now we apply the expression for the momentum P = µ(m 0, U) U and require that the mass, µ, reduce to m 0 as U goes to zero. Momentum conservation requires P xitotal = P xftotal. This means; µ(m 0, U I1 ) = 1 (a/c)2 1 + (a/c) 2 µ(m 0, U F2 ) With U I1 and U F2 given above. The let b = 0 so that; µ(m 0, 0) = 1 (a/c)2 1 + (a/c) 2 µ(m 0, 2a 1 + (a/c) 2 ) Now µ(m 0, 0) = m 0 and we let u = ( 2a 1 + (a/c) 2 ). This results in; µ(m 0, u) = m 0 1 (u/c) 2 The relativistic momentum is then; Pc = γm 0 c 2 β Then power is obtained from the force by the equation; de dt = u P dt In the above U is the mechanical velocity and P is the momentum. This equation can be rewritten as; de dt = d dt mc 2 1 β 2 In the above β = U/c. This integrated to obtain; 9

10 E = γ m 0 c 2 + E 0 Now we want this to reduce to the non-relativistic value defined by T = E m 0 c 2 where T is the kinetic energy. For β 0 use a power expansion, γ = 1 + β 2 /2 + ; T = (1/2)m 0 V 2 Thus we must take the constant E 0 = 0 9 Transformation of a Force Force is no longer an invarient quantity. In fact it does not, in general, remain in the same direction upon tranformation. From the above, the force is; F = d dt (γm 0 u) We apply a lorentz transformation to obtain; F z = d dt (γm 0u z ) = F x = d dt (γm 0u x ) = m 0 u z 1 (uz /c) + m 0u z (u z /c) u z 2 (1 (u z /c) 2 ) 3/2 = m 0 u z (1 (u z /c) 2 ) 3/2 m 0 u x 1 (uz /c) + m 0u x (u z /c) u z 2 (1 (u z /c) 2 ) 3/2 If we let u y = u z = 0 and F x = m 0 a x then; a z = a x /γ3 and in a similar way; a x = a y/γ 2 Thus; F x = F z F x = F x /γ 10