Question

Suppose that in a certain region of space there is an electrostatic field and also a magnetostatic field. Show that, although the Poynting vector can be nonzero, the surface integral of is annulled on a closed surface within this region, which does not contain charges.

Answer 1

The Poynting vector can be written as

The Pointing vector can be written as vector S = vector E times vector B where I have neglected the 1/mu_0 term for brevity as that is a constant term and I will not need it for the proof The closed surface integral can be written as follows using the divergence theorem contourintegral_a (vector S middot n^) dA = integral_v (nabla middot vector S) dv where the surface integral is over closed surface A and the volume integral is over the volume enclosed in the surface A. The divergence of S can be be written as nabla middot vector S = nabla middot (vector E times vector B) = vector B (nabla times vector E) - vector E (nabla times vector B) Now note that for static field configuration and the space without free charges, using the Maxwell's equations nabla times vector E = -partial differentials vector B/partial differential t = 0 because E and B are static fields(means they are independent of time)

where I have neglected the $1/\mu_o$ term for brevity as that is a constant term and I will not need it for the proof

The closed surface integral can be written as follows using the divergence theorem

$\oint_A (\vec S\cdot \hat n) dA=\int_V (\nabla \cdot \vec S)dV$

where the surface integral is over closed surface A and the volume integral is over the volume enclosed in the surface A. The divergence of S can be bewritten as

$\nabla \cdot \vec S=\nabla \cdot(\vec E\times \vec B)=\vec B\cdor (\nabla \times \vec E)-\vec E(\nabla \times \vec B)$

Now note that for static field configuration and the space without free charges, using the Maxwell's eqations

$\nabla \times \vec E=-\frac{\partial \vec B}{\partial t}=0$

because E and B are static fields(means they are independent of time)

And

$\nabla \times \vec B=\frac{1}{c^2\epsilon_o}\vec J+\frac{1}{c^2}\frac{\partial \vec E}{\partial t}=0$

Because there is no free current in the enclosed space over which the integral is performed and E is static

Therefore

$\nabla \cdot \vec S=0$

and hence the closed surface integral of the Poynting vector within a region, which does not contain charges or currents is

$\oint_A (\vec S\cdot \hat n) dA=\int_V (\nabla \cdot \vec S)dV=0$