Examples of relativistic transformations

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1 Examples of relativistic transformations Lecture 9 1 Field transformations In the last lecture we obtained the field transformation equations. For a boost in the 1 direction E 1 = E 1 ; B 1 = B 1 E 2 = γ(e 2 βb 3 ) ; B 2 = γ(b 2 + βe 3 ) E 3 = γ(e 3 + βb 2 ) ; B 3 = γ(b 3 βe 2 ) For a boost in an arbitrary direction given by β, the transformation equations take the form; E = γ( E + β B) B = γ( B β E) γ2 γ + 1 β(β E) γ2 γ + 1 β(β B) 2 Field of a point charge Apply these to transform the field of a point charge at rest ( E = q r 2 ˆr) to a frame moving with constant velocity (Gaussian units). Restrict the E to a plane as we apply a boost in the 1 direction about which the E field is symmetric. This is shown in Figure 1. Thus obtain (assume that we take all times such that t = 0); E 1 = E 1 = qx 1 [x x 2 1 ]3/2 = qγx 1 [x γ2 x 2 1 ]3/2 E 3 = γe 3 = qγx 3 [x x 1 2 = qγx 3 ]3/2 [x γ2 x 2 1 ]3/2 Since there is no B field in the rest frame, the magnetic field is obtained from B = β E B 2 = qγβx 3 [x x 1 2 = β E ]3/2 3 B 3 = βe 2 = 0 1

2 V 1 q 3 Moving 1 q 3 Rest Figure 1: The field of a point charge in the (1,3) plane at rest and in a frame boosted in the 1 direction Put this back into spherical coordinates in the unprimed frame. E = q(1 β 2 ) r 2 [1 β 2 sin 2 3/2 ˆr (θ)] Now note that the field is radial and compressed perpendicular to the direction of motion. The paradox here is that at all points in space the E field points to the apparent position of the charge. Given causality, the question is how does an observer far away from the charge know where the charge will be? The answer is that the observed direction of the field only points to the position of where the charge is expected to be. If the charge is accelerated in some way, and the information of this acceleration does not reach the observer during the time of observation because of the maximum propagation velocity, c, the observer would believe that the charge contiuned moving in the same direction at a constant velocity. If in fact, the field lines were curved in some way this would represent radiation. A radiating charge loses energy, and a charge at rest does not. All intertial frames of reference are equivalent. Also note that a line integral; E d s 0. Of course we know that E = (1/c) B t, so that E = 0 only for static fields. 2

3 3 Wire carring a current We suppose a long straight wire carring a current, I. The curent produces a magnetic field, but there is no electric field as there is no net charge density on the wire. Suppose a charge, q is placed at rest near the wire. There is no force on the charge as its velocity is zero and there is no E field. However suppose we transform into a moving frame so that the charge has a constant velocity. In this frame there is a magnetic V B force on the charge, so that it moves either toward or away from the wire depending on the directions of the field and the velocity. As in the section above, this represents a paradox, and its resolution is instructive. Look at the current on the wire. As a model we assume that the electrons (negative charges) are moving to produce the current and the atoms of positive charge do not move, but they do cancel the charge of the negative electrons when the positive charge is at rest. Now transform to a frame so that the positive charges move. In this frame there will be a change to the positive charge density, λ + = δq/δl, because the elemental length δl is contracted to δl/γ. The negative charge density also changes, but because of the non-linear addition of velocities in the velocity transformations, it does not change the same amount. The negative charge density in the moving frame is λ = γ[1 β 0 β]. In this expression, β 0 is the initial velocity of the electrons and β the velocity of the moving frame with respect to the rest frame. Therefore the net charge density is; λ net = λ + λ = γβ 0 β Since there is now a net charge density on the wire there will be an electric field. The force due to the electric field will cancel the force due to the magentic field, so the charge does not move toward (or away from) the wire, as is the case for th erest system. 4 Proper time Suppose there are 2 space time points located at (ct 1, x 1 ) and (ct 2, x 2 ). The time for a light ray to travel between these points is; dτ 2 = (t 2 t 1 ) 2 (1/c) 2 ( x 2 x 1 ) 2 If the two events have the same space point, dτ would be the time recorded on a clock at rest in that frame of reference. Whenever two events can just be connected by a light ray that leaves one point and arrives at the other, then such events are simultaneous and dτ = 0. If the value of dτ is real then the points are time-like and events at these points can be observed. If dτ is imaginary the events would be space-like and could not be observed. Suppose we write the differential; dτ 2 = dt 2 (1/c) 2 ( d x dt )2 dt 2 = dt 2 (1 β 2 ) = (dt/γ) 2 3

4 Note here that; U α = dxα dτ = γ dxα dt = γv α which is the relativistic velocity. 5 Lorentz force As previously written, the Lorentz force is; F = d p dt = q( E + β B) (Gaussian units) Now consider the following contraction; F αδ g δβ U β = 0 E x E y E z E x 0 B z B y E y B z 0 B x E z B y B x 0 The matrix is obtained by matrix multiplication of the Field tensor F αδ by the metric g δβ. The above equation results in γ(e x V x + E y V y + E z V z F αδ g δβ U β γ(e x c + B z V y B y V z γ(e y c B z V x B x V z γ(e z c B y V x B x V y Note that dτ = dt/γ with dτ the proper time. Make the connection to the non-covarient equation by remembering that the power equals F V = q E V. The covarient form of the Lorentz force is then written; γc γv x γv y γv z (m 0 c) duα dτ = qf αβ U β The covarient force equation is then; d p dτ = γd p dt = (q/c)[ E(γc) + (γ V ) B] 6 Spin and angular momentum Consider a set of mass points which move in uniform motion with respect to each other. This system is shown in Figure 2. Locate a fixed point in the system by the vector, R. 4

5 X i X i R Figure 2: A set of mass points roatting in rigid motion with respect to each other Other points, i, in the system are located with respect to this point by the vectors x i. The energy and momentum of the points with respect to the origin are, e i and p i. Then the total energy is W = e i and total momentum P = p i. Define the angular momentum by ; L = x i p i This is an axial vector which has a dual representation by a tensor of rank 2 (matrix); M αβ = (x α i pβ i xβ i pα i ) The matrix (tensor of rank 2) is formed by the direct product of x i and p i. We wish to extend this to 4-D relativistic space, and in order to do this we need to find the time components of the angular momentum tensor. First note that M 00 = 0 because the tensor is anti-symmetric. Now consider in the rest frame; M 0k = ct p k i In the rest frame, p k i x k i e i = 0 so that; M 0k = x k i e i Take the time derivative of the above which gives, c β k i γ i m 0 i c 2 This result assumes that the total velocity of each particle β i remains constant, ie uniform rotation. The result is just the sum of the momentum which vanishes. We then choose M 0 k = 0 which satisfies this choice, and is also consistent with the CM definition in non-relativistic mechanics. Thus choose ; 5

6 R (ct, xi e i ee ) xi m i Non-relativistically γm i m i and the above spatial component approaches M. Then T xi e the 4 vector x i = x i = R which would be (0, x i ). The total angular momentum then becomes; M αβ = (x α i pβ i xβ i pα i ) + (Rp β i Rβ p α ) ei Identify the first term on the right side of the equation as the internal angular momentum of the system and the second term as the orbital angular momentum. Internal angular momentum is identified with spin, although spin in QM is not considered an internal rotation of a structureless particle, but due to the symmetry found in the equations which represent the particle. This will become more evident later. In any event the above description is classical (ignores QM) but we will call the internal angular momentum the system spin. Thus the spin can be represented by a pseudo-vector (antisymmetric tensor) whose time component is zero. (0, S 1, S 2, S 3 ). Then consider the contraction, S k p k, which is a scalar. If we evaluate this contraction in the rest frame, it equals zero. Because the contraction is a scalar it will be zero in all intetial frames. 7 Motion of Spin One way to connect spin to electromagnetism is through the magnetic moment, µ. µ = (1/2) [ r J]d 3 x Then J = ρ V and replace the charge density by the mass density multiplied by a spatially dependent function which represents the ratio of the charge density to the mass density at each spatial point. Carry out the integration to write; µ = (Qg/2M) S where Q is the system charge, M the system mass, and g is called the gyromagnetic ratio which measures the charge-to-mass ratio averaged over the system volume. Thus the spin S is related to the magnetic moment. In the rest frame, µ can be acted upon by a magnetic field. d S dt = µ B 6

7 β + δ β δ β β Figure 3: The application of 2 boosts which are not applied in the same direction Put this equation into covarient form by looking at the transformation of the time component. S 0 = 0 = γ(s 0 β S) = (1/c)U α S α In this equation U α is the relativistic 4-velocity, and S 0 = β S. Use the general transformation equation of a 4-vector and the above results to write; S = S γ γ + 1 β( β S) The primed frame is at rest. The text also shows that a covarient expression for the motion of the spin results in the BMT equation, which represents the spin precession when a particle with a magnetic moment is acted upon by an electromagnetic field; ds α dτ = (e/m)[(g/2)f αβ S β + (1/c 2 )(g/2 1)U α (S λ F λµ U µ )] Note the factor (g/2 1) which comes directly from relativity and indicates a special case. In QM a point particle satisfying a relativistic wave equation has a gyromagnetic ratio of 2, and this is no accident. Now suppose two boosts as illustrated in the Figure 3, are applied. Classically this would not be expected to change the orientation of the reference frame axes which should remain parallel. The transformation between the frames is obtained by writing the transformation matrix we previously obtained, A First boost along the 1 direction and then apply a boost which has both a component parallel to the 1 direction and a component perpendicular to the 1 direction. Then attempt to find the relation between the coordinates in the rest frames at t and t+δt. The text shows how to develop this transformation matrix. The result is; 7

8 1 γ 2 δβ 1 γδβ 2 0 A γ 2 γ + 1 δβ 1 1 = γ δβ 2 0 γδβ 2 γ + γ 1 δβ In the above, δ β i represents boosts in the i = 1 and i = 2 directions. The infinitesmal unitary operator has the form; A = I γ 1 β 2 β 1 δβ 2 S γ 2 δβ 1 κ 1 + γδβ 2 κ 2 The last 2 terms represent boosts which one might have expected. However there is also a rotation of the axes given by the 2 nd term. Thus 2 boosts which are not in the same direction, produce a rotation of the coordinates. This is called the Thomas precession, and is the relativistic effect which produces the spin precession seen in the BMT equation, for example. 8

Covarient Formulation Lecture 8

Covarient Formulation Lecture 8 1 Covarient Notation We use a 4-D space represented by the Cartesian coordinates, x 0 (orx 4 ), x 1, x 2, x 3. The components describe a vector (tensor of rank 1) in this