1
$\begingroup$

consider a coulomb gauge and the following volume integration: $$\int d^3x{\dot{A}.\nabla A}$$ How can we show that this is zero in coulomb gauge? (A is a vector potential) this is my attempt at solution. $$\dot{A}.\nabla A=\dot{A_l}\nabla_lA_m=\nabla_l (A_lA_m)-(\nabla_l\dot{A_l})A_m$$ but in the second term since we use coulomb gauge $$\nabla_l\dot{A_l}=\frac{d}{dt}(\nabla.A)=0$$ then I'm left with $$\int d^3x\nabla_l(A_lA_m)$$ but I don't know how to show this one is equal to zero? any help is appreciated.

$\endgroup$
  • 2
    $\begingroup$ It's a surface term. $\endgroup$ – Brian Moths Apr 6 '16 at 16:20
0
$\begingroup$

Assuming your calculations are accurate, you could apply Gauss Theorem to get:

$\int d^{3}x \vec{\nabla}\cdot \vec{A} = \int_{S} \vec{A}\cdot \hat{n} \thinspace dS$

Since your integration occurs at the entire space, the surface S will be "at infinity" where your fields would enevntually go to zero, so the Right Hand Side would be null, proving your statement.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.