Day: October 26, 2020

Posted on

202010260718 Exercises 2.1 (Q1)

(Sketch of a proof)

\displaystyle{\frac{\mathrm{d}^2u^{i}}{\mathrm{d}t^2}} + \Gamma_{jk}^{i}\displaystyle{\frac{\mathrm{d}u^{j}}{\mathrm{d}t}}\displaystyle{\frac{\mathrm{d}u^{k}}{\mathrm{d}t}} = h(s) \displaystyle{\frac{\mathrm{d}u^{i}}{\mathrm{d}t}}

where h(s)=-\displaystyle{\frac{\mathrm{d}^2t}{\mathrm{d}s^2}\bigg( \frac{\mathrm{d}t}{\mathrm{d}s}\bigg)^{-2}}

will reduce to Eq. (2.11):

\displaystyle{\frac{\mathrm{d}^2u^{i}}{\mathrm{d}t^2}} + \Gamma_{jk}^{i}\displaystyle{\frac{\mathrm{d}u^{j}}{\mathrm{d}t}}\displaystyle{\frac{\mathrm{d}u^{k}}{\mathrm{d}t}}=0

if and only if t=As+B.

That is to say,

h(s) \displaystyle{\frac{\mathrm{d}u^{i}}{\mathrm{d}t}}=0 if and only if t=As+B.

(if-part) Assume t=As+B, then \displaystyle{\frac{\mathrm{d}^2t}{\mathrm{d}s^2}}=0. From h(s)= - \big( 0\big) \bigg( \displaystyle{\frac{\mathrm{d}t}{\mathrm{d}s}} \bigg)^{-2}=0, we have h(s)\displaystyle{\frac{\mathrm{d}u^{i}}{\mathrm{d}t}}=0.

(only-if part). Assume h(s)\displaystyle{\frac{\mathrm{d}u^i}{\mathrm{d}t}}=0, then some one of the following should be true:

i. \displaystyle{\frac{\mathrm{d}^2t}{\mathrm{d}s^2}}=0;

ii. \bigg(\displaystyle{\frac{\mathrm{d}t}{\mathrm{d}s}}\bigg)^{-2}=0;

iii. \displaystyle{\frac{\mathrm{d}u^i}{\mathrm{d}t}}=0.

Situation ii. implies that \displaystyle{\frac{\mathrm{d}s}{\mathrm{d}t}}=0, which is impossible for \displaystyle{\frac{\mathrm{d}t}{\mathrm{d}s}}\neq \infty and t=f(s) cannot have a point at infinity.

Situation iii. is impossible because it is only for some, but not any, i‘s in spherical coordinates (i.e., r, \theta, \phi), that u^i=0. It is also for some i‘s that \displaystyle{\frac{\mathrm{d}u^i}{\mathrm{d}t}}=0. As i‘s are to be chosen arbitrarily, the equality cannot hold.

As the second and the third were ruled out, the first situation is what that could be left possible.

The proof is as yet incomplete. It remains to be shown that

\displaystyle{\iint\bigg(\frac{\mathrm{d}^2t}{\mathrm{d}s^2}\bigg)\,\mathrm{d}s\,\mathrm{d}s}=0 \qquad \Longrightarrow \qquad t=As+B.