Homework Help: Geodesic problem

1. Feb 19, 2012

quasar_4

1. The problem statement, all variables and given/known data

I'm working my way through Wald's GR book and doing this geodesic problem:

Show that any curve whose tangent satisfies $u^\alpha \nabla_\alpha u^\beta = k u^\beta$, where k is a constant, can be reparameterized so that $\tilde{u}^\alpha \nabla_\alpha \tilde{u}^\beta = 0$.

2. Relevant equations

Geodesic equation

3. The attempt at a solution

If we assume that the curve was originally parameterized so that $u = d/d\lambda$ and we reparameterize so that $\tilde{u} = d/dt$ with $t = t(\lambda)$ then it follows immediately that $u^\alpha = \frac{dt}{d\lambda}\tilde{u}^\alpha$.

So then $u^\alpha \nabla_\alpha u^\beta = u^\alpha \nabla_\alpha \left(\frac{dt}{d\lambda}\tilde{u}^\beta \right)$. Now I know I need to expand this out into two terms, and I was thinking I could use the product rule to do that, but I'm a bit confused about how to do that. It seems I'd get
$u^\alpha \nabla_\alpha u^\beta = u^\alpha \nabla_\alpha \left(\frac{dt}{d\lambda}\tilde{u}^\beta \right) = \frac{dt}{d\lambda}u^\alpha \nabla_\alpha \tilde{u}^\beta + u^\alpha \tilde{u^\beta} \nabla_\alpha \frac{dt}{d\lambda}$
The first term I can relate to my definition $u^\alpha = \frac{dt}{d\lambda}\tilde{u}^\alpha$ to get $\left(\frac{dt}{d\lambda}\right)^2 \tilde{u}^\alpha \nabla_\alpha \tilde{u}^\beta$, and I know I'm supposed to relate this and the other term to the other side of the geodesic equation (with the constant k) to get a differential equation to solve for t(lambda). But I'm confused about how to get there from here, mostly from term $\nabla_\alpha \frac{dt}{d\lambda}$. I don't understand what it means mathematically (isn't dt/dlambda a scalar?), let alone how to do anything with it. Can anyone explain how to take the next step? Or was my "product rule" guess total nonsense?

2. Feb 21, 2012

sgd37

transform all vectors and remember that the covariant derivative becomes a partial derivative for scalars