- #1

physlosopher

- 30

- 4

$$\frac {d \vec t} {ds} = \vec 0,$$ where ##\vec t = \frac {d\vec r}{ds}## and ##\vec r(s)## is the position vector parameterized by arc length.

Then they just write out the derivative ##\frac {d\vec t}{ds} = t^{i}{}_{;k} \frac {du^{k}}{ds} \vec e^{}{}_{i} = \vec 0##. I'm totally new to all of this so I'm not sure how standard their notation is, but ##t^{i}{}_{;k}## is the covariant derivative. This all comes from differentiating ##\vec t = t^{i}\vec e^{}{}_{i}## with respect to arc length. Some chain rule gets you to the final expression.

Writing out the covariant derivative, the equation is $$(\frac{dt^{i}}{ds} + \Gamma^{i}{}_{jk}t^{j} \frac{du^{k}}{ds})\vec e^{}{}_{i} = 0.$$

Up to that point I follow, but then to simplify they assert that ##t^{j} = \frac{du^{j}}{ds}##, which substituting for ##t^{i}## and ##t^{j}## gives their final equation for a geodesic: $$\frac{d^{2}u^{i}}{ds^{2}} + \Gamma^{i}{}_{jk}\frac{du^{j}}{ds}\frac{du^{k}}{ds}=0.$$

My problem is specifically on the use of ##t^{j} = \frac{du^{j}}{ds}##. Is this always true, even in general curvilinear coordinates? Why doesn't the fact that basis vectors aren't constant throughout the space complicate that derivative so that you can't always just take the derivative of the components and ignore the basis vectors? Why don't we have to take an absolute derivative with respect to arc length of ##\vec r(s)## to get the components ##t^{i}## of ##\vec t##?

When I treat the position vector ##\vec r(s)## as I would any other vector, I get $$\vec t = \frac{d\vec r}{ds} =\frac{d}{ds}(u^{i}\vec e^{}{}_{i})= \frac{du^{i}}{ds}\vec e^{}{}_{i} + u^{i}\frac{d\vec e^{}{}_{i}}{ds} = \frac{du^{i}}{ds} + u^{m}\Gamma^{i}{}_{mk}\frac{du^{k}}{ds}.$$

Why does that last term in the last expression go to zero? I know I must be overlooking something (probably obvious) to do with the fact that I'm differentiating a position vector with respect to arc length. Any help would be much appreciated, and thanks a ton in advance. Please let me know if I can clarify or expand on anything!