Tensors for Physics

8.3 Volume Integrals, Gauss 137

Fig. 8.16The potato of
Prof. Muschik as integration
volume,nis the outer
normal vector, perpendicular
to its peel

The isotropic part of the tensor equation involves the trace

Vμμ≡

∫

V

∇μvμd^3 r=

∮

∂V

vμnμd^2 s. (8.72)

This equation can also be written as

∫

V

∇·vd^3 r=

∮

∂V

v·nd^2 s, (8.73)

which is the standardGauss theorem.
Again a remark on parity is in order. The nabla-operator∇and the outer normaln
both are polar vectors. Thus parity is “conserved” in the generalized Gauss theorem
(8.70) and its special cases, e.g. in (8.73). Notice that the surface elementnd^2 s
occurring in connection with the Gauss theorem is a polar vector, whereas the surface
element dsoccurring in the Stokes law (8.35) is a pseudo vector.
A proof for the Gauss theorem is not given here, however, its validity shall be
verified next with a simple example. A sphere with radiusRis chosen as integration
volume. The origin of the coordination system coincides with the center of the sphere.
The radial vector fieldv=ris inserted in (8.71). Due to∇μrν=δμν, the left hand
side of this equation yields the isotropic unit tensor times the volume of the sphere:

Vμν≡

∫

V

∇μrνd^3 r=δμν

∫

V

d^3 r=δμν

4 π

3

R^3.

Sincen=̂rfor the sphere, the surface integral standing on the right hand side of the
Gauss theorem is equal to

∮

∂V

rν̂rμR^2 d^2 ̂r=R^3

∫

̂rν̂rμd^2 ̂r.

The sphere does not possess any preferential direction. Thus the integral

∫

̂rν̂rμd^2 ̂r
over the unit sphere must be of the formcδμν, due to symmetry arguments. The
proportionality coefficientcis obtained from the trace relation

∫

̂rμ̂rμd^2 ̂r= 4 π=
3 c, notice thatδμμ=3. Thus