Tensors for Physics

400 Appendix: Exercises...

(i)Homogeneous Field, wherev=e=const., witheparallel to thex-axis.
The expectation is

∮

v·dr=I=0 since the vector field is the gradient of a scalar
potential. The explicit calculation of the line integralsI 1 andI 2 along the curves
C 1 andC 2 yields

I 1 =

∫

C 1

v·dr=

∫ρ

−ρ

dx= 2 ρ,

I 2 =

∫

C 2

v·dr=−ρ

∫π

0

sinφdφ=ρcosφ|π 0 =− 2 ρ.

ThusI=I 1 +I 2 =0, is found, as expected.
(ii)Radial Field, wherev=r.
Again

∮

v·dr=I=0 is expected since the vector field possesses a scalar potential
function, viz.:Φ=( 1 / 2 )r^2. Here the integration alongC 1 yields

I 1 =

∫

C 1

r·dr=

∫ρ

−ρ

xdx=( 1 / 2 )x^2 |ρ−ρ= 0.

The integralI 2 , performed alongC 2 , also gives zero since one hasr·dr= 0
on the semi-circle. Thus again, the explicit calculation confirms the expectation
I=

∮

v·dr=0.
(iii)Solid-like Rotation or Vorticity Field, wherev=w×r, with the constant axial
vectorwparallel to thez-axis.
In this case, the curl∇×vis not zero and no scalar potential exists. So

∮

v·dr=
I=0 is expected.
For the explicit calculation of the integrals, the scalar productv·dris needed.
Withwz≡w, the vector fieldvhas the components{−wy,wx, 0 }and consequently
v·dr=w(−ydx+xdy). For the integralI 1 this impliesI 1 =−w

∫ρ
−ρydx=0,
sincey=0 along the lineC 1. The integration along the semi-circle yields

I 2 =

∫

C 2

v·dr=wρ^2

∫π

0

(sin^2 φ+cos^2 φ)dφ=wρ^2

∫π

0

dφ=πwρ^2.

Thus the non-zero resultI=I 1 +I 2 =πwρ^2 is obtained here for

∮

v·dr.

8.2 Surface Integrals Over a Hemisphere(p.124)
Consider a hemisphere with radius R and its center at the origin. The unit vector
pointing from the center to the North pole isu.Surface integralsSμν=

∫

vνdsμ
are to be computed over the hemisphere.
(i)Homogeneous Vector Field vν =vv̂ν=const. By symmetry, the integral is
proportional touμv̂ν. The ansatz

Sμν=c 1 uμv̂ν