# 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