Tensors for Physics

Appendix: Exercises... 401

is made, with a scalar coefficientc 1. Multiplication of this equation byuμv̂νleads to

c 1 =uμ̂vνSμν=v

∫

uμdsμ.

For the hemisphere, located on thex–y-plane, one hasuμdsμ=R^2 cosθsinθdθdφ.
Withζ =cosθ, whereθis the angle betweenuand the vector̂r, the integral for
c 1 is

c 1 =v 2 πR^2

∫ 1

0

ζdζ=vπR^2.

The fluxS=Sμμis found to be

S=v·uπR^2 ,

which, as expected, is equal to the flux through the circular base of the hemisphere.
(ii)Radial Field vν=rν. Here the ansatz

Sμν=c 2 uμuν

is made. Multiplication of this equation byuμuνyields an expression for the scalar
coefficientc 2 ,viz.

c 2 =

∫

uνrνuμdsμ= 2 πR^3

∫ 1

0

ζ^2 dζ=

2

3

πR^3.

The resulting flux is just the area of the hemisphere.

8.3 Verify the Stokes Law for a Vorticity Field(p.127)
The vector field is given byv=w×rwithw=const.The curl of the field is
∇×v= 2 w,cf. Exercise7.1.
Since the surface elementdsis parallel tow, one has

S≡

∫

(∇×v)·ds= 2 w

∫

̂w·ds,

wherewis the magnitude ofw. Using the planar polar coordinatesρandφyields

S= 2 w

∫R

0

dρ

∫ 2 π

0

dφρ=^2 πR^2 w.

The line integral to be compared with is

∮

v·dr.Nowvanddrare parallel to each
other andv·dr=RwRdφ. This leads to