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UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

LECTURE NOTES 17

Proper and Proper   As you progress along your {moving with “ordinary” velocity u in lab frame IRF(S)} on the ct vs. x Minkowski/-time diagram, your runs slow {in your rest frame IRF(S')} in comparison to on the wall in the lab frame IRF(S).

 The clocks on the wall in the lab frame IRF(S) tick off a time interval dt, whereas in your 2 rest frame IRF( S ) the time interval is: dt  dtuu1 dt

 n.b. this is the exact same formula that we obtained earlier, with:

2 2  uu11uc  11 and: u  uc

  We use uurelative speed of an object as observed in an inertial reference frame {here, u = speed of you, as observed in the lab IRF(S)}.

  We will henceforth use vvrelative speed between two inertial systems – e.g. IRF( S ) relative to IRF(S):

 Because the time interval dt occurs in your rest frame IRF( S ), we give it a special name: ddt  = interval (in your rest frame), and:   t = proper time (in your rest frame).

 The name “proper” is due to a mis-translation of the French word “propre”, meaning “own”.

 Proper time   is different than “ordinary” time, t.

Proper time   is a Lorentz-invariant quantity, whereas “ordinary” time t depends on the choice of IRF - i.e. “ordinary” time is not a Lorentz-invariant quantity.

 222222 The Lorentz-invariant interval: dI dx dx dx  dx  ds  c dt  dx  dy  dz 

Proper time interval: d  dI c2222222  ds c  dt  dx  dy  dz cdtdt22 = 0 in rest frame IRF(S)

 22t Proper time: ddtttt       21 t 21 11

 Because d and   are Lorentz-invariant quantities: dd    and:    {i.e. drop primes}.

 In terms of 4-D space-time, proper time is analogous to S in 3-D .

 Special designation is given to being in the rest frame of an object.

 The rest frame of an object = the proper frame.

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 1 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

Consider a situation where you are on an airplane flight from NYC to LA. The pilot comes on the loudspeaker and announces in mid-flight that the jet stream is flowing backwards today, and that the plane’s velocity is uc 0.8 u  0.8!! , due west.  What the pilot means by “velocity” is the spatial d  per unit time interval dt .

The pilot is referring to the plane’s velocity relative to the ground (n.b. here, we make the simplifying assumption that the is non-rotating/non-moving, so that we can use IRF’s…)  Thus, d  and dt are quantities as measured by an observer on the ground (e.g. an airplane flight controller, using RADAR) in the ground-based (lab) IRF(S).

   d  Thus: u  = “ordinary” velocity in the lab IRF(S) d  and dt are measured in the dt ground-based (lab) IRF(S)

You, on the other hand are in your own rest frame IRF(S') in the airplane, sitting in your seat.

You know that the distance from NYC to LA is L  2763 miles (as measured on the ground, referring to your trusty Rand-McNally Road Atlas {back pages} that you brought along with you).

So you, from your perspective, might be more interested in the quantity known as your   , defined as:   d  Spatial displacement, as measured on the ground Proper 3-Velocity:   = hybrid measurement = (in lab IRF(S)) per unit time interval, as measured d in your (or an object’s) rest frame (in IRF(S')).

1 2 1 Since: ddtdt 112 dt  ucdt and:   ,   uc  u u 2 u u 1 u     dd d   d   11 n.b. Then:   , but: u   uuu  u ddt 1 u dt 220  dt 1 u 1 uc u  u

225   5 5 4 4 If uc 0.8 u  0.8 , then: uu11 110.8 3 , hence: u uc3 0.8 3 5 cc 3 !!!   Of course, for non-relativistic speeds uc , then:   u to a high degree.  From a theoretical perspective, an appealing aspect of proper 3-velocity  is that it Lorentz- transforms simply from one IRF to another IRF.   = 3-D spatial component(s) of a relativistic 4-vector,  

dx The {contravariant} proper 4-velocity is:    whose zeroth/temporal/ component is: d 1 0  u  0 dx cdt dt dt cc 2 cc   c  with: 1 u dd 1 u dt u 22 dt 1 u 1 uc u  uc  u

2 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

The proper 4-velocity vector is: dx0  d 0    c 1 u c dx   1  u u   dx 0    ux x d  ,,  c  or:    u 22u  d   uyu uy dx  3  u u d  uz z  dx3  d

The numerator of the proper 4-velocity dx is the displacement 4-vector (as measured in the ground-based (lab) IRF(S). The denominator of the proper 4-velocity d = proper time interval (as measured in your (or an object’s) rest frame IRF(S').

The of a Proper 4-Velocity   :

Suppose we want to Lorentz transform your proper 4-velocity from the lab IRF(S) to another (different) IRF(S") along a common xˆ -axis, in which IRF(S") is moving with  vvx ˆ with respect to lab IRF(S):

  Most generally, in tensor notation:   v with  v = Lorentz boost tensor. Thus:

01  00 00   0   1   11  1 10 2     00    1         v  22  2  with:  0010   2 v    33  3    0001  3 c    dx  dx Where:    and:    d d

Compare this result to the same Lorentz transformation of “ordinary” 3-, along a common xˆ -axis. We use the Einstein velocity addition rule:

 dx uv uuxuyuzˆˆˆ u  x x yz x  2 dt 1 uvx c  dy u 1 v uuxuyuz ˆˆ   ˆ u  y with:   and:   x yz y dt 2 2 c  1 uvx c 1  dz u u  z z dt 2  1 uvx c

{See Griffiths Example 12.6 (p. 497-98) and Griffiths Problem 12.14 (p. 498)}

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 3 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

Now we can see why Lorentz transformation of “ordinary” velocities is more cumbersome than Lorentz transformation of proper 4-velocities:    d   numerator,dd  For “ordinary” 3-velocities u  , we must Lorentz transform both  dt denominator,dt dt dx   For proper 4-velocities    we only need to transform the numerator, dd  . d

Relativistic Energy and - Relativistic 4-Momentum:  In , the 3-D vector linear momentum p= velocity v , i.e. p  mv . How do we extend this to ?   d   Should we use the “ordinary” velocity u  for v , dt   d   or should we use the proper velocity   for v ?? d   In classical mechanics,  and u are identical.   In relativistic mechanics,  and u are not identical.   We must use the proper velocity  in relativistic mechanics, because otherwise, the law of conservation of momentum would be inconsistent with the {the laws of  physics are the same in all IRF’s} if we were to define relativistic 3-momentum as: p  mu . No!!

Thus, we define the relativistically-correct 3-momentum as:

  mu mu 1 u pmu mu   with:    and: u   22 2 c 1 u 1 uc 1 u

 Relativistic 3-momentum: p mmu  u is the spatial part of a  relativistic 4-momentum vector: p  m  , i.e. p  pp0 , .

The temporal/zeroth/scalar component of the relativistic 4-momentum vector is: p0  Ec

00 mc mc 1 But: pm mc   with:   and:   uc u 22 2 u 1 u 1 uc 1 u

cc Thus: p00Ec m  mc where: 0 c  u u 22 1 u 1 uc   Since: p mmu  u , then: p pmumumcmcEcuuuuuuu     .

4 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

mc22 mc 1 Relativistic Energy: Emc 2  with:   and:   uc u 22 2 u 1 u 1 uc 1 u

p0 Ec  1 p  p x Therefore, the components of the relativistic 4-momentum are: p  2  p py  3  p pz

 The 4-vector dot/scalar product p p is a Lorentz-invariant quantity (same in all IRF’s):

 222222 2 p p Ecppp xyz    Ecp    mc

2 2 2 2 This can be rewritten in the more familiar form as: Epcmc22 or: E pc mc2 .

2 22222 2 2 222 1 Since: Emc  u then:  mc pc mc or: pcmc 1 . But:   u   u  2 1 u 22 2222 2 221 211uu 2 2 hence: pcmcmcmcmc u 11 222     111uuu

2 2  or: pcmc uu . However: Emc  u Thus, we {again} also see that: p pEcu .

Note that the relativistic energy E of a massive object is non-zero even when that object is 2 stationary - i.e. in its own rest frame – when: p = 0, u  0 and: uu11 1. 2 2 Then: Emcrest  = rest energy = rest mass * c .  Einstein’s famous formula!

If u  0 , then the remainder of the relativistic energy E is attributable to the motion of the particle

– i.e. it is relativistic , Ekin .

2 2 Total Relativistic Energy: E EEEtot kin rest  u mc but: Erest  mc

22 2  EEEkin tot rest  u mcmc   u 1 mc

 2211 2 Relativistic Kinetic Energy: Emcmcmckin u 11     1 1  22 u 1 uc

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 5 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

13mu4 1 In the non-relativistic regime uc , then: Emu22 mu (classical formula). kin 28c2 2 p2 However, for uc then: p  mu and thus: E  (classical formula). kin 2m

 Note that total relativistic energy, Etot and total relativistic 3-momentum, ptot p tot are separately conserved in a closed system.  If the system is not closed, (e.g. external forces are present) then Eptot and tot will not {necessarily} be conserved.  Simply expand/enlarge the definition of the “system” until it is closed {e.g.  include what’s producing the external forces}, then the (new) Eptot and tot will be conserved.

Note the distinction between a Lorentz-invariant quantity and a conserved quantity.

Same in all inertial reference frames Same before vs. after a process/an “”

Rest mass m is a Lorentz-invariant quantity, but it is not {necessarily} a conserved quantity.

Example: The {unstable} charged pi-meson decays (via weak charged-current interaction, with mean/proper lifetime   26.0 ns ) to a muon and muon neutrino:  v . The charged    pion mass m  is not conserved in the decay {mmm ( ) }, however the relativistic  v

22 24 energy of the charged pion Epcmc  is a conserved quantity: EEE,   v but E is not a Lorentz-invariant quantity.  

Since the scalar product of any relativistic 4-vector a with itself is a Lorentz-invariant quantity  (i.e. = same numerical value in any IRF): then here, for  v decay:

2  2 22 p pppp02 ppEcp Ec p2 mc   But:  

2 2 pp  p2 mc p2  mc Thus:     

6 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

Griffiths Example 12.7: Relativistic

Two relativistic lumps of clay {each of rest mass m} collide head-on with each other. 3 Each lump of clay is traveling at uc 5 as shown in the figure below:

 3  3 ucx1 5 ˆ ucx2 5 ˆ xˆ m m The two relativistic lumps of clay stick together (i.e. this is an inelastic collision).

What is the total mass M of the composite lump of clay after the collision?

Conservation of momentum - before vs. after:

Since the two lumps of clay have identical rest and equal, but opposite velocities:

   1  pbefore pp but: p pmu  where:    pbefore  0 tot 12 12u 1 u 2 tot 1 u Conservation of energy - before vs. after: mc22 mc Before: Each lump of clay has total energy: Emc22   mc uu22 1 u 1 uc mc222 mc mc 5  Emc2 2 9164 3 1 1  5 25 25

55 Thus: Ebefore E E22 mc222  mc mc tot tot12 tot u 42 5 However, E is {always} conserved in a closed system.  EEafter before mc2 tot tot tot 2

 after before And ptot is also {always} separately conserved in a closed system.  pptot tot 0

after after after 11  u  0 since: pMutot u 0 . n.b.   u  1 after after 1  22 uafter 1 ucafter

after225 2 before 5 Then: EMcMcmcEtot u ToT  M mm2 !!! Does this sound crazy?? after 2 2

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 7 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

This is what happens in the “everyday” world of particle physics! It’s perfectly OK !!! e.g. The production of a neutral rho meson in electron-positron collisions: ee 0 . M  770 MeV c2 mMeVc 0.511 2 The rest mass of the neutral rho meson is: 0 Electron rest mass: e  p  0   0  p1 pxˆ p2 pxˆ xˆ m m e M e  0 Run the collision process backwards in time, e.g. the decay of a neutral rho meson:  0 ee  p  0   0  p1 pxˆ p2 pxˆ xˆ

me me M 0 The production of a neutral rho meson ee 0 manifestly involves the EM interaction. Similarly, the time-reversed situation: the decay of a neutral rho meson  0 ee manifestly also involves the EM interaction.

The EM interaction is invariant under time-reversal, i.e. tt , thus {in the rest frame of the neutral rho meson} the transition rate ee 0 (#/sec) vs. the decay rate  0 ee (#/sec) are identical {for the same/identical electron / positron momenta in neutral rho meson 0181  production vs. decay}. Experimentally:  ee 7.02 KeV  1.70  10 sec .

For our above macroscopic inelastic collision problem, microscopically what would the new 5 1 matter of the macroscopic mass M be made up of, since M Mmmmm2222 ???

In a classical analysis of the inelastic collision of two relativistic macroscopic lumps of clay 5 {each of mass m} the composite / stuck-together single lump of clay of mass M 2 mm2 would be very hot – it would have a great deal of thermal energy in fact !!!

225 2 2 2 Mc mc2 mc  Ethermal  Ethermal = 0.5mc !!! E = mc = Einstein’s energy-mass formula 2 classical mass of composite lump

8 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

Conserved Quantities vs. Lorentz-Invariant Quantities in Collisions/Scattering Processes:

    Before: pEcpbefore  before, before After: pEcpafter  after, after . Neither is a Lorentz invariant  quantity. However, total relativistic energy E and total relativistic momentum p are separately 2  conserved quantities: EafterEMc before and: ppafter before 0 . The scalar 4-vector dot- product is a Lorentz invariant quantity, which is also a conserved quantity – i.e. its value is the same before vs. after the collision/scattering process:

 02  2 2 22 22 ppppp   ppEcp    Mc 0  Mc

 Griffiths Example 12.8: Relativistic Kinematics Associated with  v Decay.

mMeVc139.57 2  26.033 nsec 26.033 109 sec Pion rest mass:   Pion mean lifetime:   2 Muon rest mass: mMeVc 105.66 Muon neutrino rest mass: m  0 (assumed).  vu In the rest frame of the   meson:  p  0     p  pxˆ p pxˆ  v xˆ

m  m  0  v M   Energy Conservation: Momentum Conservation:  Emcbefore  2 pbefore  0 Before: tot   tot

after 2 after  After: E EEmc ppp  0  p  ppx  ˆ tot v tot v  v    But: Epcpc since: m  0 .  p  p = p  p vv v  v   vv Epcmc22224 p22cE 2 mc 24 p cEmc224 And:    or: u  

  224  p  p = p  p = Emcc   vv 

224 Emc after  Then: EEEEpc but: pp tot v v v  c after 224before 2  EEEEpcEEmcEmc      tot v v  tot

2242  EEmcmc  Solve for E :    2 224 2 24 2 2 22424 E mc mc   E  mc 2 mcE   E  or: 2mcE mc  mc        

24 24 222 224 mc mc mmc Emc    Thus: E   and: pp with: p  p  22mc2 m v  c  v  as viewed from the rest frame of the   meson.

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 9 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede   In classical collisions, total 3-momentum ptot and total mass, mtot are always conserved: before after before after ptot p tot , mmtot tot .

tot  In classical collisions, if total kinetic energy Ekin is not conserved  inelastic collision.

 An inelastic (i.e. a “sticky”) collision generates heat at the expense of kinetic energy.

 An inelastic collision of an electron (e) with an atom {initially in its ground state} may leave the atom in an excited state, or even ionized, kicking out a once-bound atomic electron!  Internal {quantum} degrees of freedom can be excited in inelastic e - atom collisions.

 An “explosive” collision generates kinetic energy at the expense of chemical (i.e. EM) energy, or nuclear (i.e. strong-force) energy, or weak-force energy. . . .

 If kinetic energy is conserved (classically),  elastic (i.e. billiard-ball) collision.

 In relativistic collisions, total 3-momentum and total energy are always conserved (in a closed system) but total mass and total kinetic energy are not in general conserved.

* Once again, in relativistic collisions, a process is called elastic if the total kinetic energy is conserved  total mass is also conserved in relativistic elastic collisions.

* A relativistic collision is called inelastic if the total kinetic energy is not conserved.  Total mass is not conserved in a relativistic inelastic collision.

Griffiths Example 12.9:

Compton Scattering = Relativistic Elastic Scattering of with Electrons.

00 An incident of energy E  pc elastically scatters (i.e. “bounces” off of/recoils) from an electron, which is initially at rest in the lab frame. Determine the final energy E of the outgoing scattered photon as a function of the scattering angle  of the photon:

Consider conservation of relativistic momentum in the transverse (  ) (i.e. yˆ -axis) direction:

pbefore0 p after tot tot

before before before ppp000 tot  e yyˆˆ direction direction after after after after after ppp0  pp tot  e  e

10 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

yyˆˆ direction direction after after Since: pp  e

after after Or: pp  e ppsin  sin Or:  e

But: p Ec

E  sin  p  sin c e E Solve for sin : sin   sin pc e

Conservation of relativistic momentum in the longitudinal (i.e. xˆ ) direction gives:

E0 before  before  before p  (n.b. p   0 , since e initially at rest, hence p   0 ) tot c e e

after after after pppp cos  p cos tot  e e

before after 0 Since: p  p then: Ec pcos p cos TOT TOT  e

2 E  But: sin   cos thus: cos  1 sin22  1 sin pc  c e e

2 0  EE2  ppcos  1 sin  cpc e  e

2 2 p22cEE 0cos  E 2 sin 2  E 0 2 EE 0 cos   E 2 Or: e  

Ebefore after tot  Etot  EEbefore after EmcEE0 2 E pcmc 22 24 Conservation of Energy: tot tot  eeee

02 002 224  EmcEee E2cos EE  Emc

1 Solve for E (after some algebra): E    20 1cos mce E

0 E = energy of recoil photon in terms of initial photon energy E , scattering angle of photon θ 2 and rest energy/mass of electron, mce .

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 11 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

We can alternatively express this relation in terms of photon wavelengths:

00 0 Before: Ehfhc  Useful constants:

After: Ehf hc   hc1239.841 eV - nm 1240 eV - nm

hc 26 Get:  0 2 1cos  mce  0.511 MeV 0.511  10 eV mce

hc 12 Define the so-called Compton wavelength of the electron: e 2 2.426 10 m mce

Then:  0 e 1cos  

The Compton Differential Scattering Cross Section:

As we learned in P436 Lecture Notes 14.5 (p. 9-22) non-relativistic photon-free electron 02 scattering E  mce is adequately described by the classical EM physics-derived {unpolarized} differential Thomson scattering cross section:

unpol d, 2 Classical T  1 e e 22 15 electron  re 1cos  where: rme  2.82 10  2 radius d 2 4 oemc

02 However, when Emc  e from the above discussion of the relativistic kinematics of photon- free electron scattering, it is obvious that the classical theory is not valid in this regime. The fully-relativistic quantum mechanical theory – that of quantum electrodynamics (QED) – is required to get it right... Without going into the gory details, the results of the QED calculation associated with the two Feynman graphs {the so-called s- and u-channel diagrams} shown on p. 5 of P436 Lect. Notes 14.5 for the Compton differential scattering cross section – known as the Klein-Nishina formula for relativistic {unpolarized} photon-free electron scattering is:

unpol 2 2 dC , x 1cos  e 1122   re 1cos 2  1  d 2 1cos2  1x  1cos 11cosx  

x Emchfmc0202 where: ee  . In the non-relativistic limit x  0 , the relativistic Compton scattering cross section agrees with the classical Thomson scattering cross section, as shown in the 1 unpol figure below of the normalized differential scattering cross section 2 dd C cos vs.  . re e

Note that as x  the relativistic Compton differential scattering cross section becomes increasingly sharply peaked in the forward direction,   0 .

12 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

The Relativistic Doppler Shift – for Photons/Light:

A rapidly moving atom isotropically emits monochromatic light (photons of frequency f ) in its own rest frame IRF. What is the frequency f of the emitted photons as observed in the lab frame  IRF as a function of the lab angle  between the atom’s velocity vz  ˆ and the direction of  observation { = photon’s momentum vector p } in the lab frame?

Rest frame of atom, IRF: Lab frame, IRF:  yˆ p

  vz ˆ

 Without any loss of generality, we can choose the lab velocity v of the atom to be along the zˆ  axis in the lab frame IRF {note that: zzvˆˆ }.

 hf  The energy of the photon in the atom’s rest frame IRF is: Epchf  where: ppc is the magnitude of the photon’s momentum in the atom’s rest frame IRF. We can also assume  without loss of generality that the emitted photon’s momentum vector p lies in the y -z plane of the atom’s rest frame IRF. In the atom’s rest frame IRF, the emitted photon makes an angle  with respect to the zˆ axis.

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 13 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

Hence: ppz cos  Ec  cos  and: ppy sin  Ec  sin  and: px  0 .

The 4-momentum vector of the emitted photon in the atom’s rest frame IRF is thus:

hf hf hf  hf  pEcppp,xyz , ,cc ,0, sin  , c cos   c 1,0,sin  ,cos

We then carry out a 1-D Lorentz transformation from the atom’s rest frame IRF to the lab frame  IRF, boosted along the zzvˆˆ axis (see e.g. Physics 436 Lect. Notes 16, p. 11), where:   vc and:  11 2 :

Ec00  E c  00 1    pp0100 0100 0 xxv   hf  ppv   ppyy0010  c 0010 sin    ppzz00    00 cos

 cos  1cos   hf0 hf 0  ccsin sin   cos   cos 

Thus, in the lab IRF, the emitted photon’s 4-momentum vector is:

hf  pEcppp ,xyz , ,  1 cos ,0,sin  , cos  c

The emitted photon’s energy as observed in the lab IRF is: Ehfhf1cos   . The frequency of the emitted photon observed in the lab IRF is: ff  1cos .

Experimentally, the atom’s rest frame photon emission angle  is {often} not measureable; the lab frame photon emission angle  is what is measured experimentally. Hence, in order for this formula to be useful, we must re-write this expression in terms of the lab frame photon emission angle  . The relationship between the atom’s rest frame photon emission angle  and the lab frame photon emission angle  can be obtained by analyzing the 3-momentum components of the   photon in the atom’s rest frame p vs. the lab frame p , as shown in the figures below:

Rest frame of atom, IRF: Lab frame, IRF:   yˆ p yˆ p   p p   ppy  sin ppy  sin    zˆ zvˆ 

ppz  cos   ppz  cos

14 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

In the lab frame IRF, the 4-vector momentum components of the emitted photon are:

Ehfhf   1 cos E 1  cos pp 0  xx ppyysin p sin  p

ppz cos p  cos 

We see that: ppz cos p  cos . But for photons: p  Ec and: p Ec

Thus: pEcz cos  Ec cos . But: EE 1cos

Hence: pEz  1 cos c cos   Ec    cos    Or:  E 1coscos E   cos  1 cos cos cos .

cos cos  Thus: cos . 1cos1cos 

However, we need an expression for cos in terms of cos . Solve for cos :

1 cos cos cos  cos  cos cos  cos    cos cos cos  cos   cos 1 cos  cos

 cos cos  cos   cos 1 1 cos

cos 1 v Thus: ff  1cos  f 1 where:   and:   1cos   1  2 c

1 n.b. RHS expressed 1cos11cos   entirely in lab frame  IRF variables – i.e. Or: f 11cosfff 2   1cos 1 experimentally measured quantities Similarly, we can also obtain a relation for sin using:

pyyppsin  sin  p. But for photons: p  Ec and: p Ec

Thus: pEcy sin  Ec sin But: EE 1cos

Hence: pEcy    1cossin Ecsin   1cossinsin  

sin Thus: sin   1cos  cos   And: cos  1cos   sin sin Hence: tan  cos  cos  

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 15 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

cos   Since: cos  1cos  

Then: sin sin sin  sin   1cos  cos 1cos  cos   2 1   1cos   1cos 

sin  2 1cossin    1cossin 1  2    1cos 

sin Thus: sin   1cos  cos   And: cos  1cos   sin sin Hence: tan  cos  cos 

Comments:

cos   cos From: cos  and: ff  1cos  f 1 1cos   1cos  

1.) The photon will be emitted in the forward direction in the lab frame IRF { 090  } when the numerator: cos  0 , i.e. when:  cos {n.b. cos  0 for 90  180 }.

2.) The photon will be emitted in the backward direction in the lab frame IRF {90  180 } when the numerator: cos  0 , i.e. when:  cos .

3.) When  1 vc, all photons are emitted in the forward direction in the lab frame IRF.

4.) When:   0 , then:   0  , and: ff  1 . n.b. so-called 2 transverse 5.) When:   90 , then: f   f  . When:   90 , then: f   ff 1 .  Doppler shift(s) 6.) When:  180 , then:  180 , and: ff  1 .

16 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

Special Relativity and Stellar Luminosity:

In its own rest frame IRF, a star isotropically radiates {thermal/black-body} photons. For simplicity’s sake {here}, we will assume that all radiated photons in the star’s rest frame IRF have the same energy E . The total rate of emission of such photons into 4 steradians in the rest frame IRF of the star {assumed to be constant} is: R dN dt # sec . Note that in the star’s rest frame IRF, the temporal interval dt d the proper time interval. The total luminosity of the star in its rest frame IRF is: LE R E dN dt   Joules sec  Watts .

The differential rate of emission of such photons into solid angle element ddcos d in the star’s rest frame IRF is:

dR d dN ddNdt   # sec sr dddtd  

The differential luminosity of the star as measured in its rest frame IRF is:

dLd E R dR d dN d dN dt  E E  E   Joules sec sr  Watts sr dd d   ddtd    

Suppose that you are an astronomer on earth, observing this star through a telescope. Your inertial reference frame is the lab frame IRF. {For simplicity’s sake here, we neglect/ignore the  motion of the earth}. If the star is moving with velocity vzz ˆˆ , then what is the differential luminosity dL d in the earth’s lab frame IRF – i.e. how is dL d related to dL d ?

There are three inertial reference frame effects that must be taken into account here:

 Time dilation: dt dt Rate of emission in rest frame IRF  rate of emission in lab frame IRF  Angle transformation: dd  Lorentz transformation from rest frame IRF of star to lab frame IRF.  Doppler effect: EE  dt 1 The time dilation effect is: dt  dt or:  , where:  11 2 and:   vc. dt 

The 1-D Lorentz transformation from the star’s rest frame IRF to the lab frame IRF on earth, for  vzˆˆ z is the same as that for the above relativistic Doppler shift example:

The 4-momentum vector associated with a photon emitted from the surface of the star in the rest EE E  E  frame IRF of the star is: pEcppp,xyz , ,cc ,0, sin  , c cos   c 1,0,sin  ,cos

The Lorentz transformation is:

Ec00  E c   1cos      pp0100  E 0 ppxxv     v  ppyy0010  c sin       ppzz00     cos 

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 17 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

Thus, the photon’s 4-momentum vector as seen by an astronomer in the lab frame IRF is:

E pEcppp ,xyz , ,  1 cos ,0,sin  , cos  c

Here {again}, we need to express this result in terms of lab frame IRF measured variables {only}:

cos   sin cos  , sin  and: EE  1cos 1cos    1cos 

Now, the lab frame IRF vs. the star’s rest frame IRF solid angle elements are, respectively:

dd cos d and: ddcos d .

The infinitesimal solid angle element dd cos d and the infinitesimal area element 2 da R d Rdcos  Rd {where the infinitesimal  and  arc lengths SRd  cos and

SRd   , respectively} associated with the lab frame IRF are shown in the figure below.  As per the above discussion of the relativistic Doppler shift, for vzˆˆ z, without any loss of  generality we can choose the observation point Pr Rrˆ to lie in the y-z plane (  90 ):

zˆ da R2 d Rdcos  Rd

rzyˆˆcos ˆ sin

dd cos d  v R ˆ cosyˆ sin zˆ

 ˆ  xˆ

Star yˆ  90

xˆ

 For the choice of observation point Pr Rrˆ lying in the y-z plane (  90 ), note that ˆ  xˆ is  to rzyˆˆcos ˆ sin . Thus, since transverse components of a 4-vector are unaffected by a Lorentz transformation e.g. from the rest frame IRF to the lab frame IRF, then ˆˆ  , hence    and dd   . Thus, in order to determine the relationship between solid angle element dd cos d and dd cos d , we only need to determine how d cos is related to d cos. From above, we already have the relation:

cos   cos  1cos  

18 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

Thus, using the chain rule of differentiation:

ddcos  cos 1 cos  2 ddcos cos  1  cos 1  cos 1cos 

1cos  cos   2 111  2   1cos 222 1cos 2  1cos Hence: ddcos dd cos 1 1 2 2 dd cos dd cos 1cos    1

We also already have the relationship between E and E from the Lorentz transformation result:

E 1 E  1cosE or:  E  1cos 

The lab frame vs. rest frame differential luminosity of the star are related to each other by:

dL dR d  dN dt d E d  dN EE  E  d d d dt dt d  E d   dt

dt d E dR   dt d E dL   Watts  E    dt d E d  dt d  E  d  sr  Thus: dL dt d E dL 11 1 1 dL   Watts   2 2     ddtdEd1cos  1cos    d  sr 

dL11 dL  Watts n.b. peaks sharply in  = 0 (forward) Or:  4 3   direction, and  0 in  =  (backward) ddsr 1cos    direction as   1 (  ).

The differential luminosity of the star in its own rest frame IRF is thus:

n.b. RHS expressed entirely in lab dL 4 3 dL Watts 1cos   frame IRF variables – i.e. ddsr  experimentally measured quantities dL dL The total luminosity of the star in its own rest frame IRF is: L d4  Watts  dd since the emission of photons in the star’s own rest frame is isotropic.

Hence the total luminosity of the star in its own rest frame IRF is:

n.b. RHS expressed entirely in lab dL 4 3 dL L 441cos  Watts  frame IRF variables – i.e. dd experimentally measured quantities

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 19 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

Relativistic Dynamics

Newton’s 1st Law of Motion: “An object at rest remains at rest, an object moving with speed v continues to move at speed v, unless acted upon by a net/non-zero/unbalanced force” – the Law of Inertia – is built/incorporated into in the Principle of Relativity.

Newton’s 2nd law of motion {classical mechanics} retains its validity in relativistic mechanics, provided that relativistic momentum is used:   dp r, t Frt,, mart dt

Griffiths Example 12.10: 1-D Relativistic Motion Under a Constant Force.

 A particle of (rest) mass m is subject to a constant force: F r, t F Fxˆ  constant vector.

If the particle starts from rest at the origin at time t = 0, find its position xt as a function of t. dp t dp t Since the relativistic motion is 1-D, then: F  = constant, or:  F = constant. dt dt  pt Ft constant of integration . The particle starts from rest at t = 0.  pt00  constant of integration = 0.  ptFt {here}

mu t 1 Relativistically: pttmut    Ft where:  t  u 2 u 2 1 ut c 1 ut c

Solve for ut: mu2 2 Ft 221  u 2 c 2  Ft 22 Ft 22 c 2 u 2  mFtcuFt2222222

22 2 2 Ft Ft m Ft m Or: u   ut = velocity for constant mFtc2222 1 Ft mc 2 2   1 Ft mc applied force F n.b. when: Ft mc 1 i.e. Ft m c then: ut Ftm  Classical dynamics answer.

Note also that as t → ∞: ut  c !!! (Relativistic denominator ensures this!)

Ft m dx t ttt Since: ut Then: xtutdtFm    dt  2 002 1 Ft mc dt 1 Fmct2

2 t 2 Fmc22 mc The motion is hyperbolic: xt111 Fmct22 Fmct     mF0  F n.b. Had we done this in classical dynamics, the result would have been parabolic motion:

F 2 xtt  2m

20 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede  Thus, in relativistic dynamics – e.g. a charged particle placed in a uniform electric field E ,  the resulting motion under a constant force FqE is hyperbolic motion (not parabolic motion, as in classical dynamics) – see/compare two cases, as shown in figure below:

Relativistic Work:   Relativistic work is defined the same as classical work: WFd  

The Work-Energy Theorem (the net work done on a particle = increase in particle’s kinetic energy) also holds relativistically:

    dp dp d dp   d  WFd  d   dtudtE   since: u  dt  dt dt  dt kin dt

  dp d mu mu mu But: uu since: pmu u  dt dt 2 22 1 uc 1 u 1 uc

Thus:

 mu 2  2 u muu dp m  du c   du  mu duc du uu 33   u  dt22 dt2222 dt dt dt 11uc 11uc   uc  uc  

2 2 2   1 uc  du 1 uc  uc  du mu du  3 mu  33mu 2 2 2 dt 2222dt dt 1 uc 1 uc 11uc  uc     mu du d mc2 3   2 2 dt dt 2 1 uc 1 uc  

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 21 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede  1 dp d 2 dEtot But:  u   umc u 1 uc2 dt dt dt    dp dp d dp  W F d  d   dt   udt  E    kin Thus: dt dt dt dt dE tot dt  Efinal  E initial   E  dt tot tot tot

But: EEE  Emc2 n.b. Emcmcmc2221 , Emc 1 2 tot kin rest kin tot u u kin u Ekin

final initial final 2 initial 2 final initial (final-initial) difference in  EEtot tot E kin  mcEmckin EEkin kin       total energy = (final-initial) Etot Ekin difference in kinetic energy final initial final initial = work done on particle. i.e. WEEtot  tot  E tot  EE kin  kin  E kin

As we have already encountered elsewhere in E&M, Newton’s 3rd Law of Motion (“For every (force) there is an equal and opposite reaction”) does NOT (in general) extend to the relativistic domain, because e.g. if two objects are separated in 3-D space, the 3rd Law is incompatible with the relativity of simultaneity.

 As Suppose the 3-D force of A acting on B at some instant t is: Frt ,, Frt AB BB  observed e.g. in lab and the 3-D force of B acting on A at the same instant t is: Frt ,, Frt BA AA  IRF(S)

Then Newton’s 3rd Law does apply in this reference frame.

However, a moving observer {moving relative to the above IRF(S)} will report that these equal-but-opposite 3-D forces occurred at different as seen from his/her IRF(S'), thus in     rd   his/her IRF(S'), Newton’s 3 Law is violated (the two 3-D forces FrtAB B , and FrtBA A , at the same time t in IRF(S') are quite unlikely to be equal and opposite, e.g. if they are changing in time in IRF(S)). Only in the case of contact interactions (i.e. 2 point particles at same point in space-time =     A A   A (x , t )) where the two 3-D forces FrtAB B , and FrtBA A , are applied at the same point (x ) at the same time, and in the {trivial} case where forces are constant, does Newton’s 3rd Law hold!

   dp r, t Frt,  dt

  The observant student may have noticed that because Frt, is the derivative of the  (relativistic) momentum prt, with respect to the ordinary (and not the proper) time t, it “suffers” from the same “ugly” behavior that “ordinary” velocity does, in Lorentz-transforming  dp r, t it from one IRF to another: both numerator and denominator of must be transformed. dt

22 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

Thus, if we carry out a Lorentz transformation from IRF(S) to IRF(S′), along the xˆ -axis   where vvx ˆ is velocity vector of IRF(S′) as observed in IRF(S), and u is the velocity vector of a particle of mass m as observed in IRF(S):

1 v  Then:   where:   with: vvx ˆ 1  2 c

The  and  factors are needed for the Lorentz transformation of kinematic quantities from IRF(S) → IRF(S′).

  First, let us work out the yˆand zˆ (i.e. the transverse) components of the 3-D force Frt, as seen in IRF(S′) {they are simpler / easier to obtain. . . }:    dp  dp  dx Noting that: F  , F  and that: dt  dt dx and: u  dt dt c x dt

dpy dp dp F In IRF(S′): F yy dt  y y  dt  dx 1 ucx  dt dx  1 c cdt

dpz dp dp F Similarly: F zz dt  z z  dt  dx 1 ucx  dt dx  1 c cdt

  Now calculate the xˆ -component of the force Frt, in IRF(S′): 0 dpx dp dEtot 0 Fc dp  dp   dp x  E In IRF(S′): F x x dt dt dt where: p0  tot x dt    dx 1 uc c  dt  dx 1 x c cdt   dE dE dp   dp But we have calculated tot above / earlier: tot uFuuF since: F  dt dt dt dt

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 23 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

  Relativistic “ordinary” x, y, z force components FuFcx     Fx  observed in IRF(S′) acting on particle of mass m, for a 1 ucx Lorentz transformation from lab IRF(S) to IRF(S′).  F IRF(S′) moving with velocity vvx ˆ relative to F  y y 2 1 ucx IRF(S) (as seen in IRF(S)),  11 ,   vc.

Fz Particle of mass m is moving with “ordinary” velocity Fz   1 ucx u as seen in IRF(S).

We see that only when the particle of mass m is instantaneously at rest in lab IRF(S)   (i.e. ut 0 ) will we then have a “simple” Lorentz transformation of the “ordinary” force FF :

FFx  x  FF  n.b. || force components are same/identical !!!  u  0 : FFyy   Where the subscripts  (  ) refer to the parallel FF   FF   (perpendicular) components of the force with respect to zz   the motion of IRF(S′) relative to IRF(S), respectively.

  Note that for u  0 , the component of F  to the Lorentz boost direction is unchanged.   For u  0 , the component of F  to the Lorentz boost direction is reduced by the factor 1  .

Proper Force – The Minkowski Force:

In analogy to the definition of the proper time interval d and the proper velocity       dd  versus the “ordinary” time interval dt and the “ordinary” velocity uddt  ,  we define a proper force K (also known as the Minkowski force), which is the derivative of the  relativistic momentum p with respect to proper time d :

  dp dt dp dt dt 11 K  but:  u  dddt ddt  22 1 u 1 uc   dp dt dp  dp  KF  u where: F  dddt dt 11   11 u Thus: KF F  F where:   and:   u 22u 22u c 1 u 1 uc 1 u 1 uc

24 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

We can “4-vectorize” the Minkowski Force, because it’s plainly / clearly a 4-vector:

dp0 1 dE E Proper rate at which energy of particle ct: K 0 tot since: p0  tot  K 0 = increases (or decreases) dcd  c = (Proper power delivered to the particle)/c ! dp1 11 x: KF11  F 1  F 1 d u 22 1 u 1 uc dp2 11 11 y: KFF222   F 2 with:   d u 22 u 22 1 u 1 uc 1 u 1 uc dp3 11 z: KFF333   F 3 d u 22 1 u 1 uc

dp Thus: K   ← Minkowski 4-vector force = proper 4-vector force. d

Relativistic dynamics can be formulated in terms of either “ordinary” quantities or “proper” (particle rest frame) quantities. The latter is much neater / elegant, but it is (by its nature) restricted to the particle’s rest frame IRF(S′) {n.b. We can always Lorentz boost this “proper” result to any other inertial reference frame. . . }

There is a very simple reason for this! Since we humans live in the lab frame IRF(S) – we want to know everything about particle’s trajectory, the forces acting on it, etc. in the lab because this is the only IRF that we can (easily) carry out physical measurements in – often, it is not possible to make physical measurements e.g. in a particle’s rest frame / proper frame, especially if the particles are in relativistic motion (e.g. at Fermilab/LHC/… hadron colliders).

In the long run, we will (usually) be interested in the particle’s trajectory as a function of “ordinary” time, so in fact the “ordinary” 4-force Fdpdt is often more useful, even if it is more painful / cumbersome to calculate / compute…

We want to obtain the relativistic generalization of the classical law    FqEquB {u = particle’s “ordinary” velocity in IRF(S)}. Does the classical formula F C   C correspond to the “ordinary” relativistic force F , or to the proper / Minkowski force K ?

    Thus, for the relativistic Lorentz force, should we write: F qEquBqEuB ???     Or rather, should the relativistic Lorentz force relation be: K qEquBqEuB ???

Since proper time and “ordinary” time are identical in classical physics / Euclidean / Galilean 3-space, classical physics can’t tell us the answer.

  It turns out that the Lorentz force law is an “ordinary” relativistic force law: FqEuB We’ll see why shortly… We’ll also construct the proper / Minkowski EM force law, as well . . .

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 25 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

But first, some examples:

Griffiths Example 12.11: Relativistic Charged Particle Moving in a Uniform Magnetic Field

We’ve discussed this before, from a classical dynamics point of view:

The typical trajectory of a charged particle (charge Q, mass m) moving in a uniform magnetic field is cyclotron motion.   If the velocity of particle (u ) lies in the x-y plane and B  Bzo ˆ ,  then F Qu B QuBoo  rˆˆ QuB r as shown on the right:

The magnetic force points radially inward – it provides the centripetal needed to sustain the circular motion. However, in the centripetal force is not mu2 R dp d R d1  Rd  u (as it is in classical mechanics). Rather, it is: Fppp    p. dt dt R dt R dt R Top View: Vector Diagram:

 u   u2 Fp rˆ n.b. Classically: p  mu thus, classically: Fm rˆ R R

u u Thus, relativistically: QuB rˆˆ p  r or: QuB p or: p  QBo R o R o R

The relativistic cyclotron formula is identical to the classical / non-relativistic formula!

  However here, p is understood to be the relativistic 3-momentum: p mmu  u .

26 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

Griffiths Example 12.12: Hidden Momentum  Consider a magnetic dipole m modeled as a rectangular loop of wire (dimensions  w) carrying a steady current I. Imagine the current as a uniform stream of non-interacting positive charges flowing freely through the wire at constant speed u. (i.e. a fictitious kind of  superconductor.) A uniform electric field E is applied as shown in the figure below:

 The application of the external uniform electric field E  Eyo ˆ changes the physics – the electric charges are accelerated in the left segment of the loop and decelerated in the right segment of the loop. [n.b. admittedly this is not a very realistic model, but other more realistic models do lead to the same result – see e.g. V. Hnizdo, Am. J. Phys. 65, 92 (1997)].

Find the total momentum of all of the charges in the loop.

The momenta associated with the electric charges in the left and right segments of the loop   cancel each other (i.e. p (in left segment) =  p (in right segment), so we only need to consider the momenta associated with the electric charges flowing in the top and bottom segments of the loop.

Suppose there are N charges flowing in the top segment of the loop, moving in xˆ direction  with speed uuE 0 {because they underwent acceleration traveling on the LHS segment} and N charges flowing in the bottom segment of the loop, moving in the xˆ direction with  speed uuE 0 {because they underwent deceleration traveling on the RHS segment}. Note that the current I  u must be the same in all four segments of the loop, otherwise charges would be piling up somewhere.

In particular: I  I (top segment of loop) = I (bottom segment of loop), i.e. I II.

Qtot NQ Q Q Since:   then: I uN u  = I uN u    

QQ  I  NuNuI   Nu Nu    Q  Classically, the linear momentum of each electric charge is pclassical mu Q where mQ = mass of the charged particle.

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 27 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

The total classical linear momentum of the charged particles flowing to the right in the top N segment of the loop is: ptop segment mu  Nmu  xˆ classical  QQ i1

The total classical linear momentum of the charged particles flowing to the left in the bottom N segment of the loop is: pbottom segment mu Nmu  xˆ classical  QQ i1

The net (or total) classical linear momentum of the charged particles flowing in the loop is:

   pptot  left segment pptop segment right segment pppbottom segment top segment bottom segment classical classical classical classical classical classical classical NmuxQQˆˆ  Nmux   Nu   Nu mx Q ˆ  I Q  I Qmx Q ˆ 0!!! 

 tot Thus, pclassical  0 as we expected, since we know the loop is not moving.

However, now let us consider the relativistic momentum:   11 p   mu (even if uuc  ) where:   rel u Q u 22 1 u 1 uc

The total relativistic momentum of the charged particles flowing to the right in the top

 top segment 11 segment of the loop is: p  NmuQ   xˆ where:    . rel u u 1  22  1 uc

The total relativistic momentum of the charged particles flowing to the left in the bottom

 bottom segment 11 segment of the loop is: p  NmuQ  xˆ where:    . rel u u 1  22  1 uc The net / total relativistic momentum is:

tot top segment  bottom segment p p p  NmuNmuxNuNumx ˆˆ    relrel rel uu  Q Q uu Q

  I I tot ˆ But I II gave us: Nu Nu   pmxrel Q  0 because    !!! Q uu Q uu

Charged particles flowing in the top segment of the loop are moving faster than those flowing in the bottom segment of the loop.

2 The gain in energy  u mc of the charged particles going up the left segment of the loop

= the work done on the charges by the electric force (WQEw o ) (w = height of the rectangle).

Thus, for a charged particle going up the left segment of the loop, the energy gain is:

QE w Emcmc22  mcWQEw 2 o uuQQ uu  Q o  uu 2 mcQ

Where Eo = the magnitude of the {uniform/constant} electric field.

28 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

 I Q Ewo I EI w tot ˆ ˆˆo  pmxrel Q  2 m Q x  2 x uu Q  c  m Qc Q 

 EIA   mE But: wA = area of the loop.  pTOT  o xˆ but: mmIA  pTOT  o xˆ rel c2 rel c2

    But: mm zˆ (see picture above) and: EEy o ˆ i.e. m  {here}.  xˆ tot 1  Thus, vectorially we {actually} have: prel mE where: mE mEzy  ˆ ˆ  c2  o   Thus a magnetic dipole moment m in the presence of an electric field E carries relativistic  linear momentum p , even though it is not moving !!!   n.b. it also (therefore) carries relativistic angular momentum Lrelrp rel . How big is this effect? Explicit numerical example - use “everyday” values:

Eo = 1000 V/m I = 1 Amp A = (10 cm)2 = 0.01 m2 m = IA = 0.01 A-m2

23 tot mEo 10 10 16 2 prel 2 2 10kg - m / s Tiny !!! The 1/c factor kills this effect !!! c 310 8

This so-called macroscopic hidden linear momentum is strictly relativistic, purely mechanical  But note that it precisely cancels the electromagnetic linear momentum stored in the EB and fields!!! (Microscopically, the momentum imbalance arises from the imbalance of virtual photon emission on top segment of the loop vs. the bottom segment of the loop.)

Likewise, the corresponding hidden angular momentum precisely cancels the  electromagnetic angular momentum stored in the EB and fields.

→ Now go back and take another look at Griffiths Example 8.3, pages 356-57. (The coax cable carrying uniform charge / unit length λ and steady current I flowing down / back cable.)

Let’s pursue this problem a little further…

Suppose there is a change in the current, e.g. suppose the current drops / decreases to zero. dI For simplicity’s sake, assume K (i.e. the current decreases linearly with time) dt

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 29 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

Classically: I tItIt   (as before)

Q QQ  ItNt  u  Nt u Nt  u   

Q We assume that u, u+ and u are ItNt  u   unaffected by the change in the current with time.

dI dI t dI t QQdN t  dN t Then:   uuK  dt dt dt dt  dt dN t dN t K  uu = constant (no time dependence on RHS of equation) dt dt Q

Then:  dp  t dN t KmQQ Km classical  mu xˆˆˆ x x dt dtQ  Q Q  Constant dp  t dN t KmQQ Km classical  mu xˆˆˆ x x dt dtQ  Q Q

 The net / total classical time-rate of change of linear momentum is:

 tot   dp t dp t dp t KmQQ Km Fttot classical classical classical xˆˆ x0 classical dt dt dt Q Q

  dptot  t Thus: Fttot classical 0 as we expected, since the loop is not moving. classical dt

Now, let’s investigate this situation relativistically:

For individual charges Since: p   mu  p    mu and: p    mu rel u Q rel u Q rel u Q with mass mQ

 top segment 11 Then: ptNtmux  Q   ˆ  where:    . rel u u 1  22  1 uc

 bottom segment 11 And: ptNtmux  Q ˆ  where:    . rel u u 1  22  1 uc

 top segment dp  t dN t Km Q rel mu xˆˆ  constant   x And: uu Q     dt dt Q  bottom segment dp  t dN t Km Q rel mu xˆˆ constant x And: uu Q     dt dt Q

30 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

The net / total time rate of change of relativistic linear momentum is:

 tot top segment bottom segment dp t dp t dp t KmQQ Km rel rel rel xˆˆx uu dt dt dt Q Q

Km  Q xˆ 0     uu  u  u  Q

QE w o From above (p. 20): uu 2 where: Eo = electric field amplitude mcQ

 tot dp t Q Ewo Km Q EK w EKA rel xˆˆˆooxx  Aw = cross- dt 2 cc22sectional area of the loop m Qc Q

dI dm dI Now: K and: mIA →  A (Since A = constant). dt dt dt

dm dI  KA  A = time rate of change of the magnetic dipole moment of the loop. dt dt

 dptot  t1 dm t   rel ˆ   2 Exo but: mmzˆ EEyo ˆ dt c dt 

 tot dprel  t1  dm t   Ftrel 2   E 0 (assuming external E -field is constant in time) dt c dt

Thus,  a net “hidden” force acting on the magnetic dipole, when dI dt  0 .

One might think that this net “hidden” force would be exactly cancelled / compensated for by a countering force due to the electromagnetic fields, as we saw in the static case ( dI dt  0 ), with a steady current I. But it isn’t!! Why??

As we saw for M(1) magnetic dipole radiation, a time-varying current in a loop produces EM radiation. Essentially there is a radiation reaction / back-force that acts on the “antenna” – a radiation pressure – much like the recoil / impulse from firing a bullet out of a gun – the short explosive “pulse” launches the bullet, but the gun is also kicked backwards, too.

The same thing happens here when dI dt  0 - the far zone EM radiation fields are produced (i.e. real photons) while dI dt  0 and carry away linear momentum, and since dI dt  0 ,  a net force imbalance on the radiating object! (n.b. – e.g. by linear momentum conservation, a laser pen has a recoil force acting on it from emitting the laser radiation – a radiation back reaction)

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 31 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

Likewise, the net “hidden” time rate of change of relativistic angular momentum is:

  dLtot t dp tot  t1 dm t relrErx rel  ˆ dt dt c2 dt o

Which will also not be exactly cancelled either, for the same reason – the EM radiation field can / will carry away angular momentum… dLtot  t In reality, in order to calculate rel , we need to go back dt and integrate infinitesimal contributions along the (short) segments of upper and lower / top and bottom segments of the   loop because rp rpsin ,    between rp and .   dp dp Same for rr sin . dt dt

Will get result that has geometrical factor of order  1. → Conclusions won’t be changed by this, just actual #.  dL  As we know, the time rate of change of angular momentum:  = torque. dt  Thus, the time rate of change of the net / total “hidden” relativistic angular momentum dLtot  t  rel = net “hidden” relativistic torque,  tot t . dt rel

  tot tot  totdLrel t dp rel  t tot 1  dm t Thus:  reltrrFtrE rel  2 0 dt dt c dt

Which is not completely / exactly cancelled when dI t dt  0 !!! Linear momentum, angular momentum, energy, etc. are all conserved for this whole system, it’s just that the EM radiation emitted from the antenna is free-streaming, carrying away all these quantities with it!

In the static situation I = constant, the “hidden” relativistic linear momentum and angular momentum is exactly cancelled by the linear momentum and angular momentum (respectively)  carried by the (macroscopic) static electromagnetic fields EB and . Microscopically, the field linear and angular momentum is carried by the static, virtual photons associated with the  macroscopic E and B fields, cancelling the (macroscopic) “hidden” linear and angular  relativistic momentum of the magnetic dipole in a uniform E -field.

In the non-static situation dI t dt  0 , virtual photons undergo space-time rotation, becoming real photons, which carry away {real} linear and angular momentum. “Hidden” relativistic linear and angular momentum is no longer exactly cancelled by the (now) real field linear and angular momentum associated with the EM radiation fields. It is only partially cancelled by remaining / extant virtual / near-zone / inductive zone EM fields.

32 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

Griffiths Problem 12.36: Relativistic “Ordinary” Force

  In classical mechanics Newton’s 2nd Law is: F  ma .   dp The relativistic “ordinary” force relation; F  cannot be so simply expressed. rel dt

   dp d d 1 1 Fmumu  where:   reldt dt u dt 2 u 2 1 uc 1 uc  du1  du 2u    1 2  du dt c dt a  Fmrel  u 3 where: = “ordinary” acceleration. 2 2 2 2 dt 1 uc 1 uc     mmuua uua  Farel   a Q.E.D. 22cuc2 1 222cu 11uc  uc 

Griffiths Problem 12.38:

We define the proper four-vector acceleration in the obvious way, as:

2   ddx 0   dx 0    2 , where: ,, uucu = proper four-velocity dd d

  a.) Find  0 and  in terms of ua and ( = “ordinary” velocity, “ordinary” acceleration):

 dddt00 1 d c 1 dt 1  0    since: ddt    ddtd 22 dt  d u 2 11uc uc u 1 uc 1 1    2 2ua   0 c 2 c 1 ua  du   32 where: a  2 2 2 c 2 dt 1 uc 1 uc 1 uc

Similarly:

   dddt 1 d u   1     since:    u and:   ddtd 22 dt u u 2 11uc uc 1 uc 1 1     2 2ua  1 a  2 c  u  223 11uc uc 1 uc2 2 

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 33 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

   1  uua Frel a u  ← see Problem 12.36 above. 222 m 1 uc cu 

  b.) Express  in terms of ua and : 2 2 2  11ua 2 1   0    aucuua1     2 44 2  cc11uc22 uc 

11222 2 222  1   ua  a2211  uc   uc ua  u ua 4 224    1 uc2 ccc

2 22 2 1 2 2 ua uu  auc1122        2 4 2    1 uc ccc   2  1 2 ua Or:  a ← n.b. Lorentz-invariant quantity – same in all IRFs. 2 4 22 1 uc cu

 c.) Show   0 .

Recall that the “dot-product” of any two relativistic four-vectors is a Lorentz-invariant quantity.

 0 2  Thus, if we deliberately/consciously choose to evaluate   in the rest  frame of an object, where   0 ,   0 ,  1 and  0  c , then:

22   002   c    = constant. 0

d Note that    is also the “dot-product” of two relativistic four vectors {  and  }.  d 

 dd d Note also that:   2 ddd

dd  c2  2   0 But:  = constant (from above). Thus:  c 0   . dd

dp d.) Write the Minkowski / proper force version of Newton’s 2nd law, K   in terms of the d proper acceleration  .

dp d d Kmmm     dd d 

34 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved. UIUC Physics 436 EM Fields & Sources II Fall Semester, 2015 Lect. Notes 17 Prof. Steven Errede

 e.) Evaluate the Lorentz-invariant 4-vector “dot product” K  :

  Km   but:   0 from part c.) above.

  K   0

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 35 2005-2015. All Rights Reserved.