# Wald Lemma 8.1.1-help with proof

1. Jun 22, 2013

### WannabeNewton

Consider the following Lemma from Wald (Lemma 8.1.1): Let $(M,g_{ab})$ be a time-orientable space-time. Then there exists a (highly non-unique) smooth nonvanishing timelike vector field $t^{a}$ on $M$.

Here is his proof: Since $M$ is paracompact, we can choose a smooth Riemannian metric $k_{ab}$ on $M$. At each $p \in M$ there will be a unique future directed timelike vector $t^{a}$ which minimizes the value of $g_{ab}v^{a}v^{b}$ for vectors $v^{a}$ subject to the condition that $k_{ab}v^{a}v^{b} = 1$. This $t^{a}$ will vary smoothly over $M$ and thus prove the desired vector field.

Now he doesn't justify any of his claims so I'm left here trying to see why any of what he says is true (everything other than the existence of the Riemannian metric on $M$ which is a trivial consequence of paracompactness and partitions of unity).

He never rigorously defined what it means for $(M,g_{ab})$ to be time-orientable in the discussion preceding this Lemma; all he says in fact is that in a time-orientable space-time there exists a continuous designation of the future and past half of a lightcone in $T_p M$ as one varies $p$. I'm assuming this means that there exists a continuous time-like vector field $X^{a}$ on $M$ in analogy with the usual definitions of orientability.

I can figure out the first "half" of his proof. Let $S\subseteq T_p M = \{v^{a}\in T_p M: k_{ab}v^{a}v^{b} = 1\}$. $S$ is compact and $g_p:S\rightarrow \mathbb{R},v^{a} \mapsto g_{ab}v^{a}v^{b}$ is continuous because any bilinear map on a finite dimensional normed space is continuous. Hence there exists some $t_p^{a}\in S$ such that $g_{ab}t_p^{a}t_p^{b}$ is a a minimum; note that if $v^{a}\in T_p M$ is any time-like vector (meaning $g_{ab}v^{a}v^{b} < 0$) then $u^{a} = \frac{v^{a}}{(k_{ab}v^{a}v^{b})^{1/2}}$ is also time-like by bilinearity and $u^a\in S$ by construction so we know there necessarily exists a time-like vector in $S$ hence the minimizing vector $t_p^{a}$ must be time-like. Hopefully these are the justifcations Wald had in mind when he wrote that "proof".

What I don't get is why this minimizing vector must necessarily be future directed, where in the world the time-orientability of $(M,g_{ab})$ is even used, and why the vector field $t^{a}$ defined as $t^{a}(p) = t_p^{a}$ at each $p \in M$ is necessarily smooth. Could anyone kindly explain those parts. Thanks in advance.

Last edited: Jun 22, 2013
2. Jun 22, 2013

### micromass

Staff Emeritus
We can choose it to be future directed. Let v be a minimizing vector, then -v is also a minimizing vector. And these are the only of two choices. So we pick the vector that is future directed.

We use it because we have to make a choice somewhere (we choose the direction to be future directed). Locally, we can always make such a choice to be smooth. Then we want to paste together the smooth bits. This can only be done if the smooth bits are all compatible. To show that things are compatible, we will need to use time-orientable.

Basically, you need to prove that if $N$ is compact and if $f:M\times N\rightarrow \mathbb{R}$ is smooth, then $p\rightarrow \min_{x\in N} f(p,x)$ is smooth. Let me think if I can prove this.

3. Jun 23, 2013

### WannabeNewton

Thanks! Only thing left now is a rigorous proof of smoothness and where time-orientability rigorously comes up in the proof of smoothness (i.e. how does giving a continuous designation of future-directed from point to point explicitly show up when proving smoothness of the minimal vector field so obtained).

Last edited: Jun 23, 2013
4. Jun 23, 2013

### micromass

Staff Emeritus
Sketch:

Smoothness is local. So we can let $M$ and $N$ be Euclidean spaces.

Let $E_p\subseteq T_p M$ be such that $df_p(E_p) = 0$. Then we can apply the Frobenius theorem on $E_p^\bot$. Thus there exists a map $T:\mathbb{R}^k\rightarrow M\times N$ such that $T_*(\frac{\partial}{\partial x^i}) \in E_p^\bot$. Then consider $J = pr_M(T(\mathbb{R}^k))$. Then $f\circ pr^{-1}_M: J\rightarrow \mathbb{R}$ coincides with $S$. And thus it is smooth.

5. Jun 23, 2013

### WannabeNewton

Thanks, I understand the sketch more or less but I'm having trouble connecting it back to the minimal time-like field $t^{a}$. Could we prove continuity of $t^{a}$ first as that might get the ball rolling for me. Intuitively it seems like the choice of having $t^{a}(p) = t^{a}_p ,\forall p\in M$, where $g_{ab}t^a_p t^b_p$ is a minimum for $v^{a}\in S$, is for the purpose of having the length of this time-like vector field vary as slowly as possible as one goes from point to point in space-time. So it would seem that since $t^{a}$'s notion of future directed is continuously designated by $X^{a}$ meaning more or less that $t^{a}$'s direction varies continuously from point to point, if $t^{a}$ automatically having minimum length under $g_{ab}$ at each $p \in M$ does imply its length varies as slowly as possible from point to point, then intuitively it should be continuous at the least. Is this intuition correct or incorrect?

By the way, no book I have consulted thus far proves this theorem. It either assumes the orientation is smooth to begin with, starts with a continuous orientation and refers the reader to the above proof in Wald which is not a proof even by the loosest of standards, doesn't mention a proof at all, or doesn't mention the smoothness existence at all (e.g. Hawking and Ellis).

Last edited: Jun 23, 2013