Tensors for Physics

392 Appendix: Exercises...

The result is

Bμν(α)=

⎛

⎝

cos 2α−sin 2α 0

sin 2α cos 2α 0

000

⎞

⎠. (A.8)

Consequently. one has

Bμν(α)=

1

3

cos 2α

⎛

⎝

10 0

01 0

00 − 2

⎞

⎠, (A.9)

andBμμ=2 cos 2α. Thus one hasBμμ=0forα=π/4, or 45◦. For this angle,
the two tensors (A.7) are ‘orthogonal’ in the sense that the trace of their product
vanishes.

3.3 Angular Momentum in Terms of Spherical Components(p.43)
Compute the z-component of the angular momentum in terms of the spherical
components.
For a particle with massm,thez-component of the angular momentum isLz=
m(xy ̇−yx ̇), in cartesian coordinates. In polar coordinates, cf. Sect.2.1.4, one has
x=rsinθcosφ,y=rsinθsinφ,z=rcosθ. The time change ofxandyis

x ̇= ̇rr−^1 x+θ ̇rcosθcosφ− ̇φrsinθsinφ,

y ̇= ̇rr−^1 y+θ ̇rcosθsinφ+ ̇φrsinθcosφ.

In the calculation ofLz, the terms involvingr ̇andθ ̇cancel, the remaining terms add
up to

Lz=mr^2 φ. ̇

3.4 Torque Acting on an Anisotropic Harmonic Oscillator(p.44)
Determine the torque for the force

F=−kr·ee−(r−r·ee),

where the parameter k and unit vectoreare constant. Which component of the
angular momentum is constant, even for k= 1?

The torque isT=r×F=−(k− 1 )(r·e)r×e. Clearly, the torque vanishes fork=1.
Fork=1, the torque still is zero, whenr×e=0or(r·e)=0 hold true. The first
case corresponds the one-dimensional motion along a line parallel toewhich passes
through the originr=0. This is a one-dimensional harmonic oscillator. The second
case is a motion in the plane perpendicular toe. This corresponds to an isotropic
two-dimensional harmonic oscillator.