darvin l

Last Latexed: April 6, 2010 at 10:40 1

Physics 504, Lecture 18April 5, 2010

Copyright c2010 by Joel A. ShapiroDarwin and Proca Lagrangians

1 Darwin Lagrangian

Newtonian classical mechanics is based on the idea that motion is describedby the positions ~xj of particles at a given time t, and these evolve according toforces given by the positions of all other particles ~xk(t) at that instant. Whenthese forces are conservative we may specify the forces by potential energieswhich are again functions of the several positions ~xj(t) at that instant. Thisnotion violates the constraint of special relativity that information cannottravel faster than light, so there is no way the particle at ~xj(t) can feel a forcethat depends on where particle k is at this instant, unless ~xj(t) = ~xk(t). Sothe only potential consistent with relativity is a delta function!

Nonetheless we know that Lagrangians with potential energies are a veryeective way of describing physics if the relevant velocities remain smallcompared to c.

Let us see what we can do for charged particles interacting electromag-netically. We learned as freshmen how to do the lowest order (c ! 1):V (~xj ; ~xk) = qjqk=j~xj ~xkj and T = 12

Pmj~v

2j . This encapsulates the eect

of the electric eld produced by one charge on the motion of the other. In ourrelativistic treatment the interaction lagrangian (of charges with the elds)

Lint =P

j qj(~xj) + 1c~uj ~A(~xj)

, and (~xj) =

Pk qk=j~xj ~xkj is the

c ! 1 limit for the scalar potential. Magnetic forces are only produced tonext order in v=c, and these produce forces only proportional to the velocityof the second particle, so these only enter to order v2=c2. At that order, the

expressions for and ~A will depend on the choice of gauge, and it is usefulhere to use not the Lorenz gauge but the Coulomb gauge ~r ~A = 0, becausein that gauge1 r2 = 4, and (~r; t) = R d3r0(~r 0; t)=j~r ~r 0j really isinstantaneous. From @F

j = 4J j=c we have

1

c2@2

@t2~Ar2 ~A+ ~r

1

c

@

@t + ~r ~A

!= 4 ~J=c:

1Gaussian units.

504: Lecture 18 Last Latexed: April 6, 2010 at 10:40 2

The ~r ~A is zero. Working accurate to order (v=c)2 we may drop the 1c2

@2

@t2~A

term, as ~A is already order (v=c)1. Thus we may take

r2 ~A = 4c~J +

1

c~r @@t

:

The contribution of particle j to ~J(~x 0) is qj~vj3(~x 0 ~xj). Its contributionto (~x 0) is qj=j~x 0~x 0jj, so it contributes qj~vj (~x 0 ~xj)=j~x 0 ~xj j3 to @[email protected] the Greens function for Laplaces equation is 1=j~x~x 0j, we have

~A(~x) =Z d3x0j~x~x 0j

1

c~J(~x 0) 1

4c~r 0 @@t

(~x 0)

!

=Z

d3x0

j~x~x 0j qjvjc3(~x 0 ~xj) qj

4c~r 0 ~vj (~x 0 ~xj)j~x 0 ~xj j3

!!

=qj~vj

cj~x ~xj j +qj

4c

Zd3x0

~vj (~x 0 ~xj)j~x 0 ~xj j3

!~r 0 1j~x~x 0j

where we have integrated by parts and thrown away the surface at innity.The gradient ~r 0 ~r action on a function of ~x~x 0, so we can pull ~r outof the integral. Let ~r = ~x ~xj and ~y = ~x 0 ~xj . Then

~A(~x) =qj~vjcj~rj

qj4c

~rZd3y

~vj ~yj~yj3

1

j~y ~rj=

qj~vjcj~rj

qj4c

~rZ 10

y2dyZ 0d sin

Z 20

dy(cos vjz + sin cosvjx)

y31p

y2 + r2 2yr cos where we have chosen z in the ~r direction and ~vj in the xz plane. The integral kills the vjx term and, writing vjz = ~vj ~r=r, we have

~A(~r) =qjc

"~vjj~rj

1

2~r ~vj ~rr

!C

#;

where the integral giving C is

C =Z 10

dyZ 11du

upy2 + r2 2yru = 1;


though this integral is not as straightforward as Jackson claims. Then

~Aj(~xk) =qj

2cj~xj ~xkj"~vj +

(~xk ~xj)~vj (~xk ~xj)j~xk ~xj j

#:

Multiplying by qk~vk=c to get the appropriate contribution to Lint, and cor-recting the free-particle Lagrangian, mc21 +mc2 1

2mv2 + 1

8mv4=c2, we

get the Darwin Lagrangian

LDarwin =1

2

Xj

mjv2j +

1

8c2Xj

mjv4j

+Xj 6=k

qjqkrjk

1

2+

1

4c2[~vj ~vk + (~vj r^jk)(~vk r^jk)]

;

where of course ~rjk := ~xj ~xk, rjk := j~rjkj, and r^jk = ~rjk=rjk.We mostly experience slightly relativistic particles in atomic physics,

though the electrons are best described by the Dirac formalism, so the ve-locities are replaced by ~. It is also of use in plasma physics.

2 Proca Lagrangian

As we saw, our lagrangian density for electromagetic elds,

LEM = 116

F F 1cJA

;

gives equations of motion which do not completely determine the evolutionof the elds A. Let us consider adding a term proportional to A2:

LProca = 1

16F F +

2

8AA

1cJA

;

known as the Proca Lagrangian, which as we shall see describes a eld whichhas quanta of mass rather than the massless photons whose classical limitis Maxwell theory. We still mean F to be shorthand for @A@A ratherthan an independent eld, so the homogeneous Maxwells equations still hold,as they are consequences of F = dA. The extra term does not contributeto P , as it depends only on A and not on derivatives thereof, so the extracontribution to the equations of motion is just from @L=@A = (2=4)A,and

@F + 2A =

4

cJ:


One consequence comes from taking the 4-divergence of this equation:

@@F + 2@A =

4

c@J:

The rst term is identically zero by symmetry, and if the current density stillrepresents a conserved charge, the right hand side is also zero, so the Lorenzcondition @A = 0 is now a consequence of the equations of motion and notan arbitrary choice. As a consequence, we now have @F = A, so

+ 2

A =

4

cJ:

In the absence of sources, this has solutions as before,

X~k

A~k +e

i~k~xi!~kt + A~kei~k~x+i!~kt

;

but with !2 = c2(~k 2 +2). Quantum mechanically we know ~p = ih~r h~kand E = ih@=@t = h!, so this eld represents particles for which E2 =P 2c2 + 2h2c2. Of course quantum eld theoriests take h = 1 and c = 1, sothis represents a massive photon with mass .

If we consider a point charge at rest and look for the static eld it wouldgenerate, we need to solve

r2 + 2 = 4q3(~r)or

@

@r

r2@

@r

!+ r22 = q(r):

Away from r = 0 this clearly requires r(r) = Cer and Gausss law tellsus

4q = 4R2 d=drjR + 2Z

r


2.1 Superconductors

In the BCS theory of superconductivity, electrons form pairs, and each pairacts like a boson. So the quantum mechanical state that each pair is incan be multiply occupied, and superconductivity occurs when states developmacroscopic occupation numbers, 1. The wave function (~x) describingthese particles is a complex function, with the density of particles n(~x) = , so = n(~x)ei(~x). We may approximate n(~x) as being roughly constant.

The velocity of these particles is related to the canonical momentum by

~v =1

m

~P q

c~A

which can be viewed as an operator acting between and . It is thecanonical momentum ~P which acts like ih~r. Thus the current density is

~J = q ~v =nq

m

hr q

c~A:

If we take the curl of both sides, we get

~r ~J = nq2

mc~r ~A = nq

2

mc~B; (1)

as ~r ~r = 0. This equation doesnt quite say

~J = nq2

mc~A; (2)

but it does say, in a simply connected region, that the dierence is the gra-dient of something, and as such a gradient could be added to ~A by a gaugetransformation, we might as well assume (2), which is known as the Londonequation. This gauge is still compatible with Lorenz (which can be viewedas determining A0), so we have

r2 ~A 1c2@2 ~A

@t2= 4

c~J =

4nq2

mc2~A;

which is the Proca equation with 2 = 4nq2/mc2.At the boundary of the superconductor, if no current is crossing the

boundary, we must have ~n ~A = 0. If we look for a static solution for a


planar boundary ? z, uniform along the boundary, we have A / ez. TheLondon penetration depth is

L :=1

=

smc2

4nq2:

With q = 2e and m = 2me for the electron pair, and taking n as thedensity of valence electrons, the penetration depth is of the order of tensof nanometers. As the A eld is not penetrating further than that into themedium, any external magnetic eld has been excluded.

But magnetic eld lines can enter the medium if our assumption of beingable to do away with ~r ~A by a gauge transformation is not correct. Thatcould happen if the region of the superconductor is not simply connected |that is, a flux line could enter and destroy the superconducting region aroundwhich is incremented by a multiple of 2. This is called a vortex line, andcorresponds to a quantized amount of flux, as

I~A d = 2Nhc=q =

ZS

~r ~A = B;

with N 2 Z. With q = 2e, the quantum of flux is hc=2e.

Documents

darvin l