1 Radiation from Accelerated Charge

Physics 139 Relativity Relativity Notes 2001 G. F. SMOOT Oce 398 Le Conte DepartmentofPhysics, University of California, Berkeley, USA 94720 Notes to b e found at http://aether.lbl.gov/www/classes/p139/homework/homework.html 1 Radiation From Accelerated Charge 1.1 Intro duction You have learned ab out radiation from an accelerated charge in your classical electromagnetism course. We review this and treat it according to the prescriptions of Sp ecial Relativity to nd the relativistically correct treatment. Radiation from a relativistic accelerated charge is imp ortant in: 1 particle and accelerator physics { at very high energies 1 radiation losses, e.g. synchrotron radiation, are a dominant factor in accelerator design and op eration and radiative pro cesses are a signi cant factor in particle interactions. 2 astrophysics { the brightest sources from the greatest distances are usually relativistically b eamed. 3 Condensed matter physics and biophysics use relativistically b eamed radiation as a signi cant to ol. An example we will consider is the Advanced Light Source ALS at the Lawrence Berkeley Lab oratory.Now free electron lasers are now a regular to ol. We will need to use relativistic transformations to determine the radiation and power emitted by a particle moving at relativistic sp eeds. Lets lo ok at the concept of relativistic b eaming to get an idea b efore wego into the details which require a fair amount of mathematics. 1 1 - a "! 1 - b 1 - ` c ` ` ` ` ` ` ` ` ` ` Radiation from an accelerated relativistic particle can b e greatly enhanced. Part of this e ect is due to the ab erration of angles and part due to the Doppler e ect. 1.2 Doppler E ect From time dilation we are used to the notion that a moving clo ck or system op erating 0 at frequency in its rest frame will app ear to b e slower to a reference system. 0 t = t 1 0 0 so that if the p erio d in the rest frame is t =1= , then 0 = = 2 The factor leads to the relativistic transverse Doppler shift. The frequency shift one would observe for a clo ck or system moving transversely to the line of sight. Thus the time b etween wave p eaks crests or pulses is 0 3 t = t = 0 2 Observer 3 3 in observers frame u - u ~v - d If the sources is moving at an angle to the observer's line of sight, then the di erence in arrival times, t , of successive pulses or crests is A v d =t1 cos t = t A c c 1 v 1 cos 4 = obs 0 c obs which when inverted or multiplied by c yields: 0 v 0 = 1 cos 5 = obs obs v 1 cos c obs c 1.3 Radiation by an Accelerated Charge Near Rest In 1897 Larmor derived the formula for the radiation by an accelerated charged particle. He found for the p ower and angular distribution: 2 2 2q dP q 2 2 P = ~a ~a = jaj sin 6 3 3 3c d 4c where is the angle to the direction of acceleration. It is our task to nd the relativistically consistent and correct version of these formulae. We can rederive the Larmor formula for your education. We consider the electric eld to b e a real physical entity that p oints radially backtoacharge at rest. If wegointo a moving frame, the electric eld lines will continue to p oint radially back to the instanteous p osition of the charge. The transformation of the electric eld works out precisely that way. The Lorentz-Fitzgerald contraction along the direction of motion causes an increase by the factor of the transverse comp onent of the eld. Gauss's law continues to hold in that an integral over a closed surface, suchasa sphere, gives the net charge within. 3 Nowifacharge is diverted from uniform motion, then by our earlier arguments ab out causality, the electric eld lines out at radius R can not b e e ected by that change from uniform motion until a time t = R=c later. In fact we exp ect that the electic eld lines will change at the sp eed of light since light is an electromagnetic phenomenon. Thus at a time t after a brief t = disturbance change from one state of uniform motion to another - also called acceleration there is a critical radius R = ct. Inside of radius R c the electric eld lines p oint radially to the new instanteous p osition of the charge and outside of radius R + c the electric eld lines p oint radially to the virtual instanteous p osition of the undisturb ed charge. The virtual instanteous p osition is where the charge would have b een had it not b een disturb ed. There is a near discontinuity in the eld lines where they must makea jaunt nearly p erp endicular to radial. Nearly means that the angle b etween the eld line and p erp endicular to radial is of order c =v t where v is the velo citychange due to the disturbance. Consider: a charge moving with velo city v c abruptly, at time t =0,is decelerated at a constant rate a until it comes to rest. v 6 v o J J J J J J J - t 0 At t =0, x= 0 and at t = , x = v =2. o Now consider elds at a time t .At a distance r>ct , the eld will b e f f that of a uniformly moving charge, emanating from the \virtual present p osition" the p oint where the particle would have b een, x = v t , if it had continued unaccelerated. 0 f At a distance r<ct , the eld will b e that of a charge at rest with x = v =2. f o There is a transition region which is nearly a spherical shell v cA o particular eld line L de nes a cone of angle, , inside, which contains a certain 0 ux. Its continuation L de nes another cone whichmust contain the same ux by 0 reason of Gauss's law relating the eld ux and the enclosed charge. Thus = and 0 L is parallel to L. Consider the p ortion connecting these two regimes. insert gure The radial comp onent of the electric eld, E must b e the same in the shell as r just outside of it on either side Gauss's law. q q q = = 7 E = r 2 2 r ct r ct f f 4 By the geometry of the situation E v t sin o f = 8 E c r q v t sin v t sin qv sin o f o f o E = E = = : 9 r 2 3 c c ct c t f f Now ct = r , and a = v =t , so that f o f q asin E = 10 2 c r 2 The signi cance of this result is that E / 1=r while E / 1=r .At a large distance r the tangential electric eld E will dominate. From our general knowledge of varying vacuum elds we know that there will ~ ~ ~ B of strength equal to E and p erp endicular b oth to E and ~r . b e a comp onentof The energy density energy p er unit volume in the transition layer is 2 2 2 2 2 Energy E B E q a sin u = = + = = 11 4 2 Volume 8 8 4 4c r 2 The volume of the shell is it area 4r times its thickness c and the average of 2 sin =2=3, Z Z Z Z 2 1 1 1 1 1 1 2 2 2 2 < sin > = sin dcos d = sin dcos = 1 x dx 4 2 2 0 1 1 1 1 2 1 3 1 x x j = 12 = 1 2 3 3 so that the energy in the transition layer is 2 2 2 q a : E = 3 3 c The radiated p ower is then the energy p er unit time: 2 2 2 q a P = : 13 3 3 c which is precisely the formula derived by Larmor in 1897. 1.4 Radiation from Circular Orbit ~ Supp ose that ~a ? B , giving motion in a circle and that 1: 0 insert gure: Reference frame S Lab oratory and reference frame S which is the electron instantaneous rest frame two column 5 ~ The lab oratory reference frame S has B p erp endicular to the plane of the ~ circular orbit B = B = 0, and B = B and E =0 x y z 2 m v o ~ F =ej~vBj= 14 r 0 2;3 By transformation law F = F column 2 Transform elds: 0 E = E B 15 or more precisely 0 ~ ~ ~ ~ E = E B = 0 + B e^ x z y = Be^ 16 y Thus 0 ~ ~ F =eE =+ qBe^ = m ~a 17 y o Thus qB a = 18 m o The p ower emitted is 2 2 2 2 4 2 2 2 q a q B = 19 P = 3 2 3 3 c 3 m c o This is the correct relativistic form! b ecause P is energy p er unit time and eachis the 0-comp onentofa4-vector!! To make a relativistic generalization we employ the concept of covariance. The Poynting vector is the 0-0 comp onent of the electromagnetic stress tensor and the radiated electromagnetic energy is the 0-comp onent of a Lorentz 4-vector. Time is the 0-comp onent of a Lorentz 4-vector.

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