Solve Geodesic Problem: Show ku^β = u^α∇αu^β

[quasar_4]

Homework Statement

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 $\tilde{u}^\alpha \nabla_\alpha \tilde{u}^\beta = 0$.

Homework Equations

Geodesic equation

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?

 

[sgd37]

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