130_notes.dvi

Note that~p 6 =m~v. The momentum conjugate to~rincludes momentum in the field. We now time
differentiate this equation and write it in terms of the components ofa vector.

dpi

dt

=m

dvi

dt

−

e

c

dAi

dt.

Similarly for the other Hamilton equation (in each vector component)p ̇i=−∂H∂xi, we have

dpi

dt

= ̇pi=−

e

mc

(

~p+

e

c

A~

)

·

∂A~

∂xi

+e

∂φ~

∂xi

.

We now have two equations fordpdtiderived from the two Hamilton equations. We equate the two
right hand sides yielding

mai=m

dvi

dt

=−

e

mc

(

~p+

e

c

A~

)

·

∂A~

∂xi

+e

∂φ

∂xi

+

e

c

dAi

dt

.

mai=−

e

mc

(m~v)·

∂A~

∂xi

+e

∂φ

∂xi

+

e

c

dAi

dt

.

Thetotal time derivativeofAhas one part fromAchanging with time and another from the
particle moving andAchanging in space.

dA~

dt

=

∂A~

∂t

+

(

~v·∇~

)

A~

so that

Fi=mai=−

e

c

~v·

∂A~

∂xi

+e

∂φ

∂xi

+

e

c

∂Ai

∂t

+

e

c

(

~v·∇~

)

Ai.

We notice the electric field term in this equation.

e

∂φ

∂xi

+

e

c

∂Ai

∂t

=−eEi

Fi=mai=−eEi+

e

c

[

−~v·

∂A~

∂xi

+

(

~v·∇~

)

Ai

]

.

Let’s work with the other two terms to see if they give us the rest ofthe Lorentz Force.

e

c

[

(

~v·∇~

)

Ai−~v·

∂A~

∂xi

]

=

e

c

[

vj

∂

∂xj

Ai−vj

∂Aj

∂xi

]

=

e

c

vj

[

∂Ai

∂xj

−

∂Aj

∂xi

]

We need only prove that
(
~v×B~

)

i

=vj

(

∂Aj

∂xi

−

∂Ai

∂xj

)

.

To prove this, we will expand the expression using the totally antisymmetric tensor.

(

~v×B~

)

i

=

(

~v×

(

∇×A~

))

i

=vj

(

∂An

∂xm

εmnk

)

εjki=vj

∂An

∂xm

(εmnkεjki)

=−vj

∂An

∂xm

(εmnkεjik) =−vj

∂An

∂xm

(δmjδni−δmiδnj) = +vj

(

∂Aj

∂xi

−

∂Ai

∂xj

)

.