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Physics 214 2011, and

Michael Dine Department of 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 the conservation of vector lengths by ordinary rotations, except for the funny minus sign. So we write the transformation in the form: x = cosh(ω)x0 + sinh(ω)xo0 (2) xo = sinh(ω)x0 + cosh(ω)xo0. We can determine ω by considering the of the origin of the primed system, viewed in the unprimed coordinates.

Physics 214 2011, Electricity and Magnetism Special Relativity Exercise (a.): Show that tanh(ω) = v (units with c = 1), and derive the standard formulas for the .

Physics 214 2011, Electricity and Magnetism Special Relativity This analogy with rotations suggests generalizing the notion of a vector (something which transforms in a definite way under rotations) to a four-vector (an object which transforms in a definite way under Lorentz transformations). Calling

xµ = (t, ~x) = (xo, ~x) (3) we have the rule: xo = γ(vx0 + xo0). (4) x = γ(x0 + vxo0).

Physics 214 2011, Electricity and Magnetism Special Relativity It is not hard to generalize these rules to boosts by a ~v in an arbitrary direction:

xo = γ(~v · x0 + xo0). (5)

~x = vˆγ(vˆ · x0 + vxo0) + (~x0 − vˆ~x0 · vˆ). Here vˆ = ~v/v. Exercise (b): Check this.

Physics 214 2011, Electricity and Magnetism Special Relativity The vectors xµ, and any vectors which transform in the same way, are called contravariant vectors. It is convenient to define covariant vectors, o xµ = (x , −~x). (6) µ 2 2 Note that x xµ = t − x~x . This is the quantity preserved by Lorentz transformations. We can write the transformations in a form: µ µ ν0 x = Λ νx . (7)

Physics 214 2011, Electricity and Magnetism Special Relativity Exercise (c): out the components of Λ for this particular transformation.

Physics 214 2011, Electricity and Magnetism Special Relativity µ In special relativity, xµy is . A particularly important four vector is the differential, dxµ. This transforms like any other four vector; it is, after all, the difference of two four vectors. From the differential, we can construct an invariant

µ 2 dx dxµ = dτ (8)

= +dt2 − d~x2 In a particle’s rest frame, dτ is just the which elapses. So dτ is called the “." 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:

d~x 2 dτ 2 = dt2(1 − ) (9) dt or dτ = γ−1dt. We see again the effect.

Physics 214 2011, Electricity and Magnetism Special Relativity Lagrangian and Hamiltonian in Special Relativity We can try to write a lagrangian for a free particle. In order that the for the particle take the same form in any frame, we can try to find a lagrangian which is Lorentz Invariant. We have seen that dτ is Lorentz invariant. So we try, as action, the over dτ, multiplied by a constant, which we call −mc2.

Z Z p S = −m dτ = −m dt 1 − v 2 (10) Z = dtL r d~x L = −m 1 − ( )2 (11) dt Note that for small , 1 L = −m + mx˙ 2 (12) 2 i

Physics 214 2011, Electricity and Magnetism Special Relativity Using the Hamiltonian construction, the momenta are:

∂L mx˙ i pi = = √ (13) ∂x˙ i 1 − v 2 The Hamiltonian is: H = pi x˙ i − L (14) mv 2 + m(1 − v 2) = √ 1 − v 2 m = √ 1 − v 2 So we have derived E = γmc2 just following our rules from ordinary .

Physics 214 2011, Electricity and Magnetism Special Relativity We are really supposed to express the Hamiltonian in terms of the momenta. This is easy to do here, and gives the important relation: H2 = E2 = m2 + p2. (15)

Physics 214 2011, Electricity and Magnetism Special Relativity Four-Velocity and Four- We can construct another four-vector by dividing dxµ by dτ. The result is the four-velocity:

dxµ uµ = (16) dτ with components: uo = γ; ui = γv i . (17) The components of uµ are not all independent;

µ o 2 2 2 2 u uµ = (u ) − ~u = γ (1 − v ) = 1. (18)

This is a consequence of the definition of uµ.

Physics 214 2011, Electricity and Magnetism Special Relativity The four velocity is closely related to another four-vector, the -momentum four-vector,

pµ = muµ. (19)

The components are:

po = mγ; pi = mv i γ. (20)

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

Physics 214 2011, Electricity and Magnetism Special Relativity The constraint on uµ is closely related to the energy-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 the mass of the particle. This is almost always the easiest way to find the γ factor for the motion of a particle. If you absolutely need to know the velocity for some reason, you can then solve for v in terms of γ. (E.g. SLAC; high energy rays).

Physics 214 2011, Electricity and Magnetism Special Relativity E One can compute γ quickly by remembering that γ = m . Interesting examples are provided by the LHC (E = 7 TeV, so γ ≈ 7000; SLAC: E = 50 GeV, so γ = 105; also cosmic rays (think about muon lifetime):

Physics 214 2011, Electricity and Magnetism Special Relativity

More Four-Vectors Another very useful four-vector is the gradient. We write this as: ∂ ∂ = . (22) µ ∂xµ This transforms under Lorentz transformations as a covariant vector (indices downstairs). One way to see this is to note that

µ ∂µx = 4 (23) i.e. it is a number, which is Lorentz invariant. But otherwise, you can use the chain rule to check.

Physics 214 2011, Electricity and Magnetism Special Relativity With this observation,

∂2 2 = ∂ ∂µ = − ∇~ 2 (24) µ ∂x0 2 is Lorentz invariant! Note, also, that the energy-momentum in is a four vector: µ µ p = i~∂ . (25) (Check that this form, with the i instead of −i, is correct).

Physics 214 2011, Electricity and Magnetism Special Relativity Still another useful four vector is the current, jµ. It’s components are: jo = cρ;~j = ~J. (26) I won’t prove it now, but let’s think about it. If we have a charge at rest, then under a boost, we have a and a current. The charge density is increased by a γ factor as a result of the Lorentz contraction. The current is the charge density the velocity. As one further piece of evidence, this makes the equation of current conservation,

µ ∂µj = 0 (27) a Lorentz-invariant equation, i.e. true in any frame. [In another part of your homework, you will prove that the current of a is a four vector]

Physics 214 2011, Electricity and Magnetism Special Relativity The Vector Potential as a Four-Vector If we define a four-vector Aµ by

Ao = V ; A~ = A~ (28) then the Lorentz gauge condition is:

∂V ∂ Aµ = + ∇~ · A~ = 0. (29) µ ∂t In other words, it is a Lorentz-invariant condition. Moreover, in this gauge, the equations for the potentials can be written in a manifestly Lorentz-covariant form:

2Aµ = jµ. (30)

Physics 214 2011, Electricity and Magnetism Special Relativity Mechanics again So far, we wrote the lagrangian for a free particle. We can try and guess a lagrangian for a particle in an electromagnetic field from. We might expect that the lagrangian should involve V . It should also be Lorentz covariant. So we try: Z Z µ S = −m dτ − dx Aµ. (31)

At low velocities, the last term is − R dtφ. In general, we can write this as: Z S = dt(−mγ − qφ + qA~ · ~v). (32)

Physics 214 2011, Electricity and Magnetism Special Relativity Our goal is to show that this lagrangian gives the Lorentz law. We need to work carefully with the Euler-Lagrange equations. First, we work out:

∂L i = γmvi + qA . (33) ∂x˙i To write the equations of motion, we need to take the with respect to time. Ai has, in general, both explicit time dependence and time dependence coming from its dependence on xj (t):   d ∂L dτ d i ˙ i i = m u + qA + qvj ∂j A . (34) dt ∂x˙i dt dτ

Physics 214 2011, Electricity and Magnetism Special Relativity In the Euler-Lagrange equations, we also need:

∂L = −q∂i φ + q(∂i Aj )vj . (35) ∂xi Putting this all together:

i ˙ i mu˙ = −q(A + ∂i φ) − qvj (∂j Ai − ∂i Aj ) (36)

= qE~ + q~v × B~ . This is the relativistic generalization of the law:

mu˙ i = γq[E~ + ~v × B~ ]. (37)

Physics 214 2011, Electricity and Magnetism Special Relativity More on the Matrices Λ Now that we are a bit more used to four vector notation, let’s return to Λ. Write:

µ µ ν0 0 ˜ρ x = Λ νx ; yµ = yρΛ µ. (38)

µ So in order that yµx be invariant, we need:

˜ρ µ ρ Λ µΛ ν = δ ν. (39)

It also follows from their definitions that

˜µ µρ σ Λ ν = g gνσΛ ρ (40)

So µρ σ ν µ g gνσΛ ρΛ ρ = δ ρ. (41)

Physics 214 2011, Electricity and Magnetism Special Relativity This is the case for our simple Lorentz transformation along the z axis:  γ γv 0 0 γv γ 0 0 Λ =   (42)  0 0 1 0 0 0 0 1  γ −γv 0 0 −γv γ 0 0 Λ˜ =   (43)  0 0 1 0 0 0 0 1 (note here that to make the matrices easy to write, I have taken the four vector xµ to be: xo x1   . (44) x2 x3 It is true for the more general transformation we have written above. Physics 214 2011, Electricity and Magnetism Special Relativity In terms of Λ, it is easy to understand the covariance of equations. For example,

∂2Aµ = jµ (45) is covariant, since both sides transform in the same fashion under Lorentz transformations.

Physics 214 2011, Electricity and Magnetism Special Relativity Infinitesimal Lorentz Transformations Writing µ µ µ Λ ν = δ ν + ω ν (46) equation 39 becomes

µ ρ µ ρ ω νδµ + δ ρωµ = 0. (47)

So ρ ρ ω ν + ων = 0 (48) or ωρν + ωνρ = 0 (49) i.e. ωρν is antisymmetric.

Physics 214 2011, Electricity and Magnetism Special Relativity Exercise (d): Determine the components of ω for an infinitesimal Lorentz boost along the z axis.

Physics 214 2011, Electricity and Magnetism Special Relativity Covariant Equations, Covariant

Aµ = Bµ (50) Since both sides of this equation transform in the same way under Lorentz transformations, this equation is true in any frame. Similarly, T µνρ = J µνρ. (51)

Physics 214 2011, Electricity and Magnetism Special Relativity Following Einstein, we rewrite Maxwell’s equations in a covariant form. The and vector potentials can be grouped as a four vector,

Aµ = (φ, A~ ) (52)

Then introduce the field strength (sometimes called Faraday) : Fµν = ∂µAν − ∂νAµ. (53)

Physics 214 2011, Electricity and Magnetism Special Relativity Foi = Ei ; Fij = −ijk Bk . (54) These transform as:

µν µ ν ρσ0 F = Λ ρΛ σF . (55)

There is not much to do but multiply these out. You find the transformation laws for the E~ and B~ fields in your text. What about the Lorentz force law? This should involve u˙ µ, which is a four-vector which would generalize the acceleration. It should be linear in the fields E~ and B~ , i.e. linear in F µν.A guess is: µ µν mu˙ = quνF . (56)

Physics 214 2011, Electricity and Magnetism Special Relativity Exercise e: Check that the components of this equation yield the Lorentz force law.

Physics 214 2011, Electricity and Magnetism Special Relativity Other Interactions: Strong, Weak Interactions Look Similar a Strong interactions: Fµν, a = 1,... 8. Equations for the fields:

µνa µa ∂νF = J . (57)

Currents: color of the quarks. Weak interactions: same thing! (a = 1 ... 3). Currents: weak charge. But includes an extra piece, due to the Higgs field, which leads to a mass: 2 i2 m Aµ .

Physics 214 2011, Electricity and Magnetism Special Relativity More Exercises: Exercise (f): Show that Foi = Ei , Fij = ijk Bk . Exercise (g): Fµν, being a tensor, transforms like the product xµxν under Lorentz transformations. Work out the transformation of E~ and B~ under Lorentz transformations along the z axis using this fact.

Physics 214 2011, Electricity and Magnetism Special Relativity Exercise (h): Verify that Maxwell’s equations take the form:

µν ν ∂µF = j .

Explain why this result shows that Maxwell’s equations are Lorentz covariant. Exercise (i): The charge and , together form a four vector, jµ = (ρ,~j). Write the equation for current conservation in a covariant form. 2 µ Exercise (j):For a particle of mass m, what is p = pµp ?

Physics 214 2011, Electricity and Magnetism Special Relativity Covariant derivation of the equations of motion

We have written the lagrangian for a particle in an electromagnetic field in a covariant form. We now do the same thing for the electromagnetic fields, and derive the full equations of covariantly. First, let’s write the matter lagrangian in a slightly different way. We write the electromagnetic current in a manner which exhibits its four vector character. Z µ µ j (x) = q dτu (τ)δ(x − x0(τ)) (58)

where the δ function is a four dimensional δ function. Evaluating the components: Z dx0 j0 = q dτ δ(x − x (τ)) (59) dτ 0 Z = 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 us to write: Z Z X 4 µ S = − ma dτa − d xj (x)Aµ(x). (60) a

Physics 214 2011, Electricity and Magnetism Special Relativity Let’s derive the Lorentz force equation more covariantly. Starting with the action in the form Z Z µ S = −m dτ − q dx Aµ(x) (61) we make the variation:

xµ → xµ + δxµ. (62)

Then q p µ µ µ dτ = dx dxµ → (dx + dδx )(dxµ + dδxµ) (63)

1 = dτ + dxµdδxµ. dτ (We drop terms of order δx2.)

Physics 214 2011, Electricity and Magnetism Special Relativity So the variation of the action is: Z dxµ Z dxµ ∂Aµ Z δS = −m dτ dδxµ − dτ δxν − dδxµA . (64) dτ dτ ∂xν µ

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

Z dxµ Z ∂Aµ Z δS = −m dτ dδxµ − dτuµ δxν + dτuν∂ A δxµ. dτ ∂xν ν µ (65) Integrating the first term by parts, from the requirement that the variation vanish, we read off: dpµ = qu F νµ. (66) dτ ν

Physics 214 2011, Electricity and Magnetism Special Relativity So how do we write an action for the fields? Any such action, to be sensible, and to be invariant under Lorentz transformations, should be an integral over all space time over a . The simplest scalar we can construct out of Fµν is

1 Z Z S = − d 4xF 2 − d 4xjµA . (67) 16π µν µ

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

Aµ → Aµ + δAµ. (68)

Physics 214 2011, Electricity and Magnetism Special Relativity We proceed in two ways. First we choose the Lorentz gauge, µ ∂µA = 0. Then we write Z Z 4 2 4 µ µ ν µ d xFµν = d x(∂ A − ∂ A )(∂µAν − ∂νAµ) (69) which we can rewrite, using the antisymmetry of F: Z 4 µ ν 2 d x∂ A (∂µAν − ∂νAµ) (70)

Now we integrate by parts, and use the gauge condition to rewrite as: µ 2 −2A ∂ Aµ. (71)

Physics 214 2011, Electricity and Magnetism Special Relativity Performing the variation, then

1 Z Z δS = d 4x∂2AµδA − d 4xjµδA (72) 4π µ µ yielding ∂2Aµ = 4πjµ (73) the Maxwell equations in Lorentz gauge.

Physics 214 2011, Electricity and Magnetism Special Relativity We can work in a more gauge independent setting, starting with: Z Z 4 µν2 4 µ ν ν µ δ d xF = 2 d x(∂ A − ∂ A )(∂µδAν − ∂νδAµ) (74)

Integrating by parts, and taking advantage of the antisymmetry yields Z 4 µν = −4 d x4∂µF δAν.

So we obtain the covariant form of the equations:

µν µ ∂µF = 4πj . (75)

Exercise (`): Work through the derivation of the Maxwell’s equations and the Lorentz force equations in the covariant formulation. This is all done above, but make sure you understand where the indices go, etc.

Physics 214 2011, Electricity and Magnetism Special Relativity Green’s function Calculation in a Manifestly Lorentz Covariant setup

We seek a solution of the equation:

µ 0 0 ∂µ∂ G(x − x ) = 4πδ(x − x ) (76)

We first Fourier transform both sides of the equation:

Z d 4q δ(x) = e−iq·x (77) (2π)4

Z d 4q G(x) = g(q)e−iq·x (78) (2π)4 So, plugging in equation 76,

4π g(q) = − (79) q2

Physics 214 2011, Electricity and Magnetism Special Relativity So it’s easy to get the Fourier transform of the Green’s function. Inverting the Fourier transform is trickier. First we note:

Z d 4q e−iq·x G(x) = −4π . (80) (2π)4 q0 2 − ~q2

We will do the q0 integral first. At this point, the expression looks completely Lorentz-invariant (like d 4x, d 4q is Lorentz invariant if we interpret q = (q0, ~q) as a four-vector). Singling out the q0 integral looks like it might violate this, but we will see that our final answer is completely invariant. The integral over q0 is slightly singular. We will deal with this by viewing the integral as an integral in the complex plane. As we will see shortly, how we define the contour corresponds to a choice of boundary conditions in our solution of the wave equation. We will take the integration contour to run slightly above the real axis, and thus slightly above the poles at q0 = ±|~q|.

Physics 214 2011, Electricity and Magnetism Special Relativity Suppose, first, that x0 < 0. In that case, we can close the contour in the upper half plane, since Re[−iq0x0] < 0.. But in this way, we avoid both poles, so the contour integral gives zero. In the case x0 > 0, we must close below. In this case, we pick up the residues associated with each of the poles. These are:

1 0 1 0 q0 = −|~q| : R = − ei|~q|x q0 = |~q| : R = e−i|~q|x (81) 2|~q| 2|~q|

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

3 Z d q i h 0 0 i G(x) = 4πΘ(x0) −eiqx +i~q·~x + e−iqx +i~q·~x . (82) (2π)3 2q

Physics 214 2011, Electricity and Magnetism Special Relativity Introducing spherical coordinates, ~q · ~x = qr cos θ. The θ integral is elementary, and the φ integral trivial, yielding (note the q2 in dqq2 cancels the q2 factor in the denominator):

4π Z ∞ dq ( ) = Θ( 0) G x x 2 (83) r 0 (2π)

1 h 0 0 0 0 i −eiqx +iqr + eiqx −iqr + e−iqx +iqr − e−iqx −iqr . 2 The integration limits can be changed to −∞ to ∞ by combining the first and fourth terms in the braces, and the second and third: ∞ 4π Z dq 1 h 0 0 i ( ) = Θ( 0) − iqx +iqr + iqx −iqr . G x x 2 e e (84) r −∞ (2π ) 2

Physics 214 2011, Electricity and Magnetism Special Relativity Now we recognize these as δ-functions, yielding:

1 h i G(x) = Θ(x0) δ(x0 − r) − δ(x0 + r) (85) r The second term in brackets can be to zero, due to the θ function, and we have 1 G(x − x0) = Θ(x0 − x00) δ(x0 − x00 − |~x − ~x0|). (86) |~x − ~x0|

Physics 214 2011, Electricity and Magnetism Special Relativity In terms of G, we can write the solution for the four-vector potential in Lorentz gauge: Z Aµ(x) = d 4x0G(x − x0)jµ(x0) (87)

Z 1 = d 3x0 jµ(t0 , ~x0) |~x − ~x0| ret 0 0 where tret = t − |~x − ~x |

Physics 214 2011, Electricity and Magnetism Special Relativity