[Math] One of Wald’s General Relativity problems

general-relativitymathematical physics

I've been working through Wald's book "General Relativity", and I'm fairly stuck on question 4. a) of chapter 6:

"Let $(M,g_{ab})$ be a stationary space-time with time-like Killing field $\xi^a$. Let $V^2 = -\xi^a \xi_a$. Show that the acceleration $a^b = u^a \nabla_a u^b$ of a stationary observer is given by $a^b = \nabla^b \log (V)$."

Firstly I used that a stationary observer is one whose 4-velocity is the time-like Killing vector $\xi^a$, and I introduced coordinates $t,x,y,z$ so that $\xi^a = (\partial / \partial t)^a$ and the remaining three are arbitrarily defined. Then $\xi^a$ is the vector $(1,0,0,0)$ so
\begin{align}
a^b &= \xi^a \nabla_b \xi^a \\
&= \xi^a \partial_a \xi^b + \xi^a \Gamma^b \text{}_{ac} \xi^c \\
&= \xi^0 \partial_0 \xi^b + \xi^0 \Gamma^b \text{}_{00} \xi^0 \\
&= 0 + \Gamma^b \text{}_{00} \\
&= (1/2) g^{ab} ( 2 \partial_t g_{0a} – \partial_a g_{00}) \\
&= -(1/2) \nabla^b g_{00},
\end{align}
where to reach the last line the independence of $g_{ab}$'s coordinates from the Killing coordinate $t$ and the property $\nabla_a f = \partial_a f$ for any scalar $f$ were both used. Furthermore
\begin{align} V^2 = – \xi^a \xi_a = – g_{ab} \xi^a \xi^b = -g_{00},
\end{align}
so in fact $a^b = (1/2) \nabla^b (V^2) = V \nabla^b V$. However, $\nabla^b \log (V) = (1/V) \nabla^b V$ by the chain rule, so it appears necessary and sufficient, for $a^b$ to equal $\nabla^b \log(V)$, that $V = \pm 1$; indeed it could only be $+1$ since otherwise $\log(V)$ is undefined. But this would imply the highly unphysical situation that $a^b = 0$.

Can anybody help me resolve my quandary? Many thanks in advance!

Best Answer

  • We require $u^a u_a = -1$ for 4-velocities, but $\xi^a \xi_a = -V^2$ is not generally like this.
  • $g_{00}$ is not a scalar, it's a component of a tensor.

Let's not mess about with specific frames.

We have: $V^2 = \xi^a\xi_a$, $a^b = u^a\nabla_a u^b$, $\nabla_{(a}\xi_{b)}=0$, $u^a = f \xi^a$.

Firstly, $u^a u_a = f^2 \xi^a \xi_a = -V^2 f^2$ so $f=\pm1/V$.

Let us write $$a_b = 1/V\,\xi^a\nabla_a(\xi_b/V) = \frac{1}{V^2}\xi^a\nabla_a\xi_b-\frac{1}{V}\xi^a\xi_b\nabla_a \frac{1}{V}$$ But then using the antisymmetry property $$\xi^a\nabla_a(\xi_b) = -\xi^a\nabla_b(\xi_a) = -\frac{1}{2}\nabla_b(\xi^a\xi_a) = \frac{1}{2}\nabla_b V^2$$ and again $$\xi^a\xi_b\nabla_a V^2 = -\xi^a\xi_b\nabla_a (\xi_c \xi^c) = -2\xi_b\xi^a\xi^c\nabla_a \xi_c = 0$$

Hence

$$a_b = \frac{1}{2V^2}\nabla_b V^2 = \nabla_b \log V$$

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