I 12 1 f d3rdV w ix

Note that the second term in ¿int, which we will refer to as ¿>nt, is zero once the Coulomb gauge is taken into account, and therefore this term will make no contribution to the particle equations or to the total energy.

Equations of Motion

Adding the particle coordinates to the Lagrangian does not change the derivation of the equations for A' given in Sec. 2.2. We have, as before,

DAl + ViV-A = j\ , where A0 is shorthand for the solution of the Poisson equation and the gauge condition V • A — 0 is imposed. The only new feature is that the current is now specified in terms of the particle coordinates.

The equations for the motion of the two particles become

maqi + J d3rjt [pa(r) A'{r,t)] + j d3r (Vr), {qa pa(r)} -A(r,t)

1 r rl3r rl3r'

~J ^r-r>\ Pair) p(r') + Pair') P(r)} = 0 , (2.29)

where use was made of

TTlPa (\qa ~ /"I) = ~{yr)jPa i\9a ~ »"I)

and the fact that the second interaction term, L^, integrates to zero.

Now simplify the equation (2.29) and extract the Lorentz force law. First, the second and third terms in (2.29) are reduced by integrating by parts and using current conservation:

/ ||(p„(r)^(r, t)) + V, (<& Pa(r))A*(r, t) j

However, the second of these terms is recognized as related to v xB (where v = q)\ vxB = vx(VxA)-> ejimvitmtkVtAk = vi(VjAi -ViAj) . The fourth term in (2.29) can be simplified by integrating by parts and using

X (pair) [pi(r') + p2(r')\ - pa(r') ^(r) + ^(r)]) .

Note that regardless of the value of a, the only terms in the square brackets which survive are those for particle b ^ a, giving

Combining all terms gives the Lorentz force law:

maqa = J d3r {pa(r) Ea(r, t) + pa{r) qa x B(r, t)} , where no summation over the repeated index a is implied and F(rt\- dA{r't] V f dV Pb*a{r>)

Note that the Ea field which enters the force law only includes the longitudinal Coulomb part due to the other charges, so there is no self-force.

Note that the first of these tells us that each component of A satisfies the wave equation, and the second places a restriction on the three components. Therefore the solutions we found in Chapter 1 can be immediately applied to the EM field if they are generalized to:

• three-dimensional space and

• two independent vector degrees of freedom (only two because V • A — 0 constrains the third).

In addition, the wave velocity is now that of light, so that v = c = 1.

Referring back to Eq. (1.15), the vector field must therefore have the form

where the sum is over three integers n = (nx,ny,nz), corresponding to the requirement that the solutions of the wave equation satisfy periodic boundary conditions in each of the three space dimensions (referred to as box normalization), and the integer a — 1 or 2, corresponding to the two independent vector degrees of freedom of the vector potential which are not constrained by the Coulomb gauge. Specifically, the momenta of the plane wave solutions are given by the three-dimensional generalization of Eq. (1.6),


Was this article helpful?

0 0

Post a comment