Special Relativity - Santa Cruz Institute for Particle …scipp.ucsc.edu/~dine/ph214/214_relativity_lecture.pdfSpecial Relativity Physics 214 2011, Electricity and Magnetism Michael

Special Relativity

Physics 214 2011, Electricity and Magnetism

Michael DineDepartment of Physics

University of California, Santa Cruz

January 2011

Physics 214 2011, Electricity and Magnetism Special Relativity

Consider, first, boosts along the x axis. These should preserve:

c2t2 − x2. (1)

Take units with c = 1, and call xo = ct ; this is similar to theconservation of vector lengths by ordinary rotations, except forthe funny minus sign. So we write the transformation in theform:

x = cosh(ω)x ′ + sinh(ω)xo′ (2)

xo = sinh(ω)x ′ + cosh(ω)xo′.

We can determine ω by considering the motion of the origin ofthe primed system, viewed in the unprimed coordinates.


Exercise (a.): Show that tanh(ω) = v (units with c = 1), andderive the standard formulas for the Lorentz transformation.


This analogy with rotations suggests generalizing the notion ofa vector (something which transforms in a definite way underrotations) to a four-vector (an object which transforms in adefinite way under Lorentz transformations). Calling

xµ = (t , ~x) = (xo, ~x) (3)

we have the rule:xo = γ(vx ′ + xo′). (4)

x = γ(x ′ + vxo′).


It is not hard to generalize these rules to boosts by a velocity ~vin an arbitrary direction:

xo = γ(~v · x ′ + xo′). (5)

~x = vγ(v · x ′ + vxo′) + (~x ′ − v~x ′ · v).

Here v = ~v/v .Exercise (b): Check this.


The vectors xµ, and any vectors which transform in the sameway, are called contravariant vectors. It is convenient to definecovariant vectors,

xµ = (xo,−~x). (6)

Note that xµxµ = t2 − x~x2. This is the quantity preserved byLorentz transformations. We can write the transformations in amatrix form:

xµ = Λµνxν′. (7)


Exercise (c): Work out the components of Λ for thisparticular transformation.


In special relativity, xµyµ is invariant. A particularly importantfour vector is the differential, dxµ. This transforms like any otherfour vector; it is, after all, the difference of two four vectors.From the differential, we can construct an invariant

dxµdxµ = dτ2 (8)

= +dt2 − d~x2

In a particle’s rest frame, dτ is just the time which elapses. Sodτ is called the “proper time." While it refers to a special frame,it is a Lorentz-invariant notion.A useful way to write the proper time in a general frame is:

dτ2 = dt2(1−(

d~xdt

)2

) (9)

or dτ = γ−1dt . We see again the time dilation effect.


Lagrangian and Hamiltonian in Special RelativityWe can try to write a lagrangian for a free particle. In order thatthe equations of motion for the particle take the same form inany frame, we can try to find a lagrangian which is LorentzInvariant. We have seen that dτ is Lorentz invariant. So we try,as action, the integral over dτ , multiplied by a constant, whichwe call −mc2.

S = −m∫

dτ = −m∫

dt√

1− v2 (10)

=

∫dtL

L = −m

√1− (

d~xdt

)2 (11)

Note that for small velocities,

L = −m +12

mx2i (12)


Using the Hamiltonian construction, the momenta are:

pi =∂L∂x i =

mx i√

1− v2(13)

The Hamiltonian is:H = pi x i − L (14)

=mv2 + m(1− v2)√

1− v2

=m√

1− v2

So we have derived E = γmc2 just following our rules fromordinary mechanics.


We are really supposed to express the Hamiltonian in terms ofthe momenta. This is easy to do here, and gives the importantrelation:

H2 = E2 = m2 + p2. (15)


Four-Velocity and Four-MomentumWe can construct another four-vector by dividing dxµ by dτ .The result is the four-velocity:

uµ =dxµ

dτ(16)

with components:uo = γ; ui = γv i . (17)

The components of uµ are not all independent;

uµuµ = (uo)2 − ~u2 = γ2(1− v2) = 1. (18)

This is a consequence of the definition of uµ.


The four velocity is closely related to another four-vector, theenergy-momentum four-vector,

pµ = muµ. (19)

The components are:

po = mγ; pi = mv iγ. (20)

At low velocities, pi is the ordinary three-momentum. po is E ,the relativistic energy.


The constraint on uµ is closely related to theenergy-momentum relation:

(po)2 − ~p2 = m2. (21)

We see that this is a relativistically invariant relation.One very valuable fact to note: γ is the energy divided by themass of the particle. This is almost always the easiest way tofind the γ factor for the motion of a particle. If you absolutelyneed to know the velocity for some reason, you can then solvefor v in terms of γ. (E.g. SLAC; high energy gamma rays).


One can compute γ quickly by remembering that γ = Em .

Interesting examples are provided by the LHC (E = 7 TeV, soγ ≈ 7000; SLAC: E = 50 GeV, so γ = 105; also cosmic rays(think about muon lifetime):


More Four-VectorsAnother very useful four-vector is the gradient. We write this as:

∂µ =∂

∂xµ. (22)

This transforms under Lorentz transformations as a covariantvector (indices downstairs). One way to see this is to note that

∂µxµ = 4 (23)

i.e. it is a number, which is Lorentz invariant. But otherwise,you can use the chain rule to check.


With this observation,

2 = ∂µ∂µ =

∂2

∂x0 2 − ~∇2 (24)

is Lorentz invariant!Note, also, that the energy-momentum in quantum mechanicsis a four vector:

pµ = i~∂µ. (25)

(Check that this form, with the i instead of −i , is correct).


Still another useful four vector is the current, jµ. It’scomponents are:

jo = cρ;~j = ~J. (26)

I won’t prove it now, but let’s think about it. If we have a chargeat rest, then under a boost, we have a charge density and acurrent. The charge density is increased by a γ factor as aresult of the Lorentz contraction. The current is the chargedensity times the velocity. As one further piece of evidence, thismakes the equation of current conservation,

∂µjµ = 0 (27)

a Lorentz-invariant equation, i.e. true in any frame. [In anotherpart of your homework, you will prove that the current of apoint particle is a four vector]


The Vector Potential as a Four-VectorIf we define a four-vector Aµ by

Ao = V ; ~A = ~A (28)

then the Lorentz gauge condition is:

∂µAµ =∂V∂t

+ ~∇ · ~A = 0. (29)

In other words, it is a Lorentz-invariant condition. Moreover, inthis gauge, the equations for the potentials can be written in amanifestly Lorentz-covariant form:

2Aµ = jµ. (30)


Mechanics againSo far, we wrote the lagrangian for a free particle. We can tryand guess a lagrangian for a particle in an electromagnetic fieldfrom. We might expect that the lagrangian should involve V . Itshould also be Lorentz covariant. So we try:

S = −m∫

dτ −∫

dxµAµ. (31)

At low velocities, the last term is −∫

dtφ. In general, we canwrite this as:

S =

∫dt(−mγ − qφ+ q~A · ~v). (32)


Our goal is to show that this lagrangian gives the Lorentz forcelaw. We need to work carefully with the Euler-Lagrangeequations. First, we work out:

∂L∂xi

= γmvi + qAi . (33)

To write the equations of motion, we need to take the totalderivative with respect to time. Ai has, in general, both explicittime dependence and time dependence coming from itsdependence on xj(t):

ddt

(∂L∂xi

)=

dτdt

mddτ

ui + qAi + qvj∂jAi . (34)


In the Euler-Lagrange equations, we also need:

∂L∂xi

= −q∂iφ+ q(∂iAj)vj . (35)

Putting this all together:

mui = −q(Ai + ∂iφ)− qvj(∂jAi − ∂iAj) (36)

= q~E + q~v × ~B.

This is the relativistic generalization of the Lorentz force law:

mui = γq[~E + ~v × ~B]. (37)


More on the Matrices ΛNow that we are a bit more used to four vector notation, let’sreturn to Λ. Write:

xµ = Λµνxν′; yµ = y ′ρΛρµ. (38)

So in order that yµxµ be invariant, we need:

ΛρµΛµν = δρν . (39)

It also follows from their definitions that

Λµν = gµρgνσΛσρ (40)

SogµρgνσΛσρΛνρ = δµρ. (41)


This is the case for our simple Lorentz transformation along thez axis:

Λ =

γ γv 0 0γv γ 0 00 0 1 00 0 0 1

(42)

Λ =

γ −γv 0 0−γv γ 0 0

0 0 1 00 0 0 1

(43)

(note here that to make the matrices easy to write, I have takenthe four vector xµ to be:

xo

x1

x2

x3

. (44)

It is true for the more general transformation we have writtenabove.


In terms of Λ, it is easy to understand the covariance ofequations. For example,

∂2Aµ = jµ (45)

is covariant, since both sides transform in the same fashionunder Lorentz transformations.


Infinitesimal Lorentz TransformationsWriting

Λµν = δµν + ωµν (46)

equation 39 becomes

ωµνδρµ + δµρω

ρµ = 0. (47)

Soωρν + ω ρ

ν = 0 (48)

orωρν + ωνρ = 0 (49)

i.e. ωρν is antisymmetric.


Exercise (d): Determine the components of ω for aninfinitesimal Lorentz boost along the z axis.


Covariant Equations, Covariant Tensors

Aµ = Bµ (50)

Since both sides of this equation transform in the same wayunder Lorentz transformations, this equation is true in anyframe. Similarly,

Tµνρ = J µνρ. (51)


Following Einstein, we rewrite Maxwell’s equations in acovariant form. The scalar and vector potentials can begrouped as a four vector,

Aµ = (φ, ~A) (52)

Then introduce the field strength (sometimes called Faraday)tensor:

Fµν = ∂µAν − ∂νAµ. (53)


Foi = Ei ; Fij = −εijkBk . (54)

These transform as:

Fµν = ΛµρΛνσF ρσ′. (55)

There is not much to do but multiply these out. You find thetransformation laws for the ~E and ~B fields in your text.What about the Lorentz force law? This should involve uµ,which is a four-vector which would generalize the acceleration.It should be linear in the fields ~E and ~B, i.e. linear in Fµν . Aguess is:

muµ = quνFµν . (56)


Exercise e: Check that the components of this equation yieldthe Lorentz force law.


Other Interactions: Strong, Weak Interactions Look SimilarStrong interactions: F a

µν , a = 1, . . .8. Equations for the fields:

∂νFµνa = Jµa. (57)

Currents: color of the quarks.Weak interactions: same thing! (a = 1 . . . 3). Currents: weakcharge. But includes an extra piece, due to the Higgs field,which leads to a mass:

m2Ai2µ .


More Exercises:Exercise (f): Show that Foi = Ei , Fij = εijkBk .Exercise (g): Fµν , being a tensor, transforms like the productxµxν under Lorentz transformations. Work out thetransformation of ~E and ~B under Lorentz transformations alongthe z axis using this fact.


Exercise (h): Verify that Maxwell’s equations take the form:

∂µFµν = jν .

Explain why this result shows that Maxwell’s equations areLorentz covariant.Exercise (i): The charge and current density, together form afour vector, jµ = (ρ,~j). Write the equation for currentconservation in a covariant form.Exercise (j):For a particle of mass m, what is p2 = pµpµ?


Covariant derivation of the equations of motion

We have written the lagrangian for a particle in anelectromagnetic field in a covariant form. We now do the samething for the electromagnetic fields, and derive the fullequations of electromagnetism covariantly.First, let’s write the matter lagrangian in a slightly different way.We write the electromagnetic current in a manner whichexhibits its four vector character.

jµ(x) = q∫

dτuµ(τ)δ(x − x0(τ)) (58)

where the δ function is a four dimensional δ function. Evaluatingthe components:

j0 = q∫

dτdx0

dτδ(x − x0(τ)) (59)

= q∫

dtδ(x − x0(τ)) = qδ(~x − ~x0(t))

which is the usual expression for the charge density.Physics 214 2011, Electricity and Magnetism Special Relativity

Exercise(k): Check the other components.This immediately generalizes to many charges, and allows usto write:

S = −∑

a

ma

∫dτa −

∫d4xjµ(x)Aµ(x). (60)


Let’s derive the Lorentz force equation more covariantly.Starting with the action in the form

S = −m∫

dτ − q∫

dxµAµ(x) (61)

we make the variation:

xµ → xµ + δxµ. (62)

Then

dτ =√

dxµdxµ →√

(dxµ + dδxµ)(dxµ + dδxµ) (63)

= dτ +1

dτdxµdδxµ.

(We drop terms of order δx2.)


So the variation of the action is:

δS = −m∫

dτdxµ

dτdδxµ−

∫dτ

dxµ

dτ∂Aµ

∂xνδxν−

∫dδxµAµ. (64)

In the last term we integrate by parts (as in the usual derivationof the Euler Lagrange equations), giving

δS = −m∫

dτdxµ

dτdδxµ −

∫dτuµ

∂Aµ

∂xνδxν +

∫dτuν∂νAµδxµ.

(65)Integrating the first term by parts, from the requirement that thevariation vanish, we read off:

dpµ

dτ= quνF νµ. (66)


So how do we write an action for the fields? Any such action, tobe sensible, and to be invariant under Lorentz transformations,should be an integral over all space time over a Lorentz scalar.The simplest scalar we can construct out of Fµν is

S = − 116π

∫d4xF 2

µν −∫

d4xjµAµ. (67)

Let’s derive the Maxwell equations. To perform the variation, wework in terms of Aµ. Just like we vary ~x or xµ, we vary Aµ:

Aµ → Aµ + δAµ. (68)


We proceed in two ways. First we choose the Lorentz gauge,∂µAµ = 0. Then we write∫

d4xF 2µν =

∫d4x(∂µAµ − ∂νAµ)(∂µAν − ∂νAµ) (69)

which we can rewrite, using the antisymmetry of F :

2∫

d4x∂µAν(∂µAν − ∂νAµ) (70)

Now we integrate by parts, and use the gauge condition torewrite as:

−2Aµ∂2Aµ. (71)


Performing the variation, then

δS =1

4π

∫d4x∂2AµδAµ −

∫d4xjµδAµ (72)

yielding∂2Aµ = 4πjµ (73)

the Maxwell equations in Lorentz gauge.


We can work in a more gauge independent setting, startingwith:

δ

∫d4xFµν2 = 2

∫d4x(∂µAν − ∂νAµ)(∂µδAν − ∂νδAµ) (74)

Integrating by parts, and taking advantage of the antisymmetryyields

= −4∫

d4x4∂µFµνδAν .

So we obtain the covariant form of the equations:

∂µFµν = 4πjµ. (75)

Exercise (`): Work through the derivation of the Maxwell’sequations and the Lorentz force equations in the covariantformulation. This is all done above, but make sure youunderstand where the indices go, etc.


Green’s function Calculation in a Manifestly LorentzCovariant setup

We seek a solution of the equation:

∂µ∂µG(x − x ′) = 4πδ(x − x ′) (76)

We first Fourier transform both sides of the equation:

δ(x) =

∫d4q

(2π)4 e−iq·x (77)

G(x) =

∫d4q

(2π)4 g(q)e−iq·x (78)

So, plugging in equation 76,

g(q) = −4πq2 (79)


So it’s easy to get the Fourier transform of the Green’s function.Inverting the Fourier transform is trickier. First we note:

G(x) = −4π∫

d4q(2π)4

e−iq·x

q0 2 − ~q2 . (80)

We will do the q0 integral first. At this point, the expressionlooks completely Lorentz-invariant (like d4x , d4q is Lorentzinvariant if we interpret q = (q0, ~q) as a four-vector). Singlingout the q0 integral looks like it might violate this, but we will seethat our final answer is completely invariant. The integral overq0 is slightly singular. We will deal with this by viewing theintegral as an integral in the complex plane. As we will seeshortly, how we define the contour corresponds to a choice ofboundary conditions in our solution of the wave equation. Wewill take the integration contour to run slightly above the realaxis, and thus slightly above the poles at q0 = ±|~q|.


Suppose, first, that x0 < 0. In that case, we can close thecontour in the upper half plane, since Re[−iq0x0] < 0.. But inthis way, we avoid both poles, so the contour integral gives zero.In the case x0 > 0, we must close below. In this case, we pickup the residues associated with each of the poles. These are:

q0 = −|~q| : R = − 12|~q|

ei|~q|x0q0 = |~q| : R =

12|~q|

e−i|~q|x0(81)

The remaining integral over ~q then looks like (note that becausewe close below, we pick up an extra minus sign):

G(x) = 4πΘ(x0)

∫d3q

(2π)3i

2q

[−eiqx0+i~q·~x + e−iqx0+i~q·~x

]. (82)


Introducing spherical coordinates, ~q · ~x = qr cos θ. The θintegral is elementary, and the φ integral trivial, yielding (notethe q2 in dqq2 cancels the q2 factor in the denominator):

G(x) =4πr

Θ(x0)

∫ ∞0

dq(2π)2 (83)

12

[−eiqx0+iqr + eiqx0−iqr + e−iqx0+iqr − e−iqx0−iqr

].

The integration limits can be changed to −∞ to∞ bycombining the first and fourth terms in the braces, and thesecond and third:

G(x) =4πr

Θ(x0)

∫ ∞−∞

dq(2π2)

12

[−eiqx0+iqr + eiqx0−iqr

]. (84)


Now we recognize these integrals as δ-functions, yielding:

G(x) = Θ(x0)1r

[δ(x0 − r)− δ(x0 + r)

](85)

The second term in brackets can be set to zero, due to the θfunction, and we have

G(x − x ′) = Θ(x0 − x ′0)1

|~x − ~x ′|δ(x0 − x ′0 − |~x − ~x ′|). (86)


In terms of G, we can write the solution for the four-vectorpotential in Lorentz gauge:

Aµ(x) =

∫d4x ′G(x − x ′)jµ(x ′) (87)

=

∫d3x ′

1|~x − ~x ′|

jµ(t ′ret , ~x′)

where t ′ret = t − |~x − ~x ′|


Documents

Special Relativity - Santa Cruz Institute for Particle …scipp.ucsc.edu/~dine/ph214/214_relativity_lecture.pdfSpecial Relativity Physics 214 2011, Electricity and Magnetism Michael