# Coulomb gauge and vector identites

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.

• It's a surface term. – Brian Moths Apr 6 '16 at 16:20

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