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Suppose at some point$$R_{ab}v^av^b = 0$$ for arbitrary timelike vectors $v^a$, where $R_{ab}$ is the Ricci tensor. I am wondering how to show that this implies $R_{ab} = 0$ at this point.

Ray d'Inverno (Introducing Einstein's Relativity, p. 144) suggests this can be shown as follows: let $v^a = t^a + \lambda s^a$, where $t^at_a = 1$, $s^as_a = -1$, $t^as_a = 0$, $\lambda\le0<1$, and with $\lambda$ arbitrary. Then one can test special coordinate systems such as $t^a = \delta^a_0$, $s^a = \delta^a_1$, followed by similar ones.

I have two questions with this approach:

  1. Note $v^av_a = 1 - \lambda^2$. And yet, $v_a$ is supposed to be a unit tangent vector, correct?
  2. What good does it do to evaluate $R_{ab}v^av^b$ in various special coordinate systems (which at best implies that only some components vanish), when we need to show that all components of $R_{ab}$ vanish in a single coordinate system.
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I copy-paste here the exercise for completeness:

If at some point $P$, the symmetric tensor $R_{ab}$ satisfies $$R_{ab}v^{a}v^{b} = 0$$ for an arbitrary timelike vector $v^{a}$, then deduce that $R_{ab}$ must vanish at $P$.

[Hint: let $v^a = t^a + \lambda s^a$, where $t^at_a = 1$, $s^as_a = -1$, $t_a s^a = 0$, $0\leq\lambda<1$, $\lambda$ arbitrary, and consider a special coordinate system in which $t^a=\delta^{a}_0$ and $s^a=\delta^a_{1,2,3}$ in turn.]

So let's go carefully through the assignment. This exercise is a about a (generic) symmetric two-index tensor $R_{ab}$ (that he then applied to the Ricci), which vanishes contracted with timelike vectors (i.e. with vectors $v^a$ such that $v^av_a > 0 $); it is a general identity about tensors rather than a peculiarity of the Ricci.

Let's now follow the hint. As you correctly state, you get that $v_a v^a = 1- \lambda^2$ (this vector does not have unit norm, and there is no need why it should have); in particular it remains timelike for all values of $\lambda$.

Then we just compute what the contraction of $v$ with $R$ is: $$ R_{ab} v^a v^b = R_{ab} (t^a + \lambda s^a ) ( t^b + \lambda s^b) = R_{ab} t^at^b + 2 \lambda R_{ab}t^a s^b + \lambda^2 R_{ab} s^a s^b,$$ where, in the last step, we used the symmetry of $R_{ab}$ and we renamed the two indices. Now we compute the previous expression for $t^a=\delta^{a}_0$ and $s^a = \delta^{a}_i$, being $i = 1,2,3 $; we get $$ R_{ab} v^a v^b = R_{00} + 2 \lambda R_{0i} + \lambda^2 R_{ii} . $$ Since we observed that $v^a$ is timelike for any $\lambda$, we know by hypothesis that this quantity vanishes, i.e. $$ R_{00} + 2 \lambda R_{0i} + \lambda^2 R_{ii} = 0.$$

Now we recall that $\lambda$ can assume any value between $0$ (comprised) and $1$; this is enough to say that all the components $R_{00}$, $R_{0i}$ (and by symmetry $R_{i0}$ as well) and $R_{ii}$ are equal to zero. Starting with $\lambda = 0$, we get $R_{00} = 0$, which holds for any choice of $\lambda$; then the equation (restoring $\lambda \neq 0$) reduces to $$ 2 \lambda R_{0i} + \lambda^2 R_{ii} = 0,$$ which can clearly be satisfied for any $0<\lambda <1$ only if the two $R$'s vanish.

We have to show now that the mixed-spatial component $R_{ij}$ vanish; this is achieved by considering $s^a = (\delta^a_i + \delta^a_j)/\sqrt{2}$. Going through the same steps, and making use of the already-know-to-be-vanishing components we get $ 0 = R_{ab} v^a v^b = \lambda^2 R_{ij}/2.$

When a tensor vanishes in one frame of reference, it vanishes in all; this is a general property that worth being discussed more. The change of coordinates is reflected on tensors in a linear nonsingular transformation: we contract indices with the jacobian of the transformation; being the change of coordinates a diffeomorphism, the jacobian is injective, and therefore a tensor vanishes in one system of coordinates if and only if vanishes in all.


For completeness, I underline why the hypothesis of the symmetry of $R_{ab}$ is necessary. If it were not the case, we could not simplify the contraction as we did; we would just remain with $$ R_{ab} v^a v^b = R_{ab} t^at^b + \lambda ( R_{ab} + R_{ba}) s^a t^b + \lambda^2 R_{ab} s^a s^b, $$ and here we have nontrivial cancellations between antisymmetric components of $R$: indeed, following the same steps, we would end up with just the symmetric part of $R_{0i}$ (i.e.\ $R_{(0i)} = (R_{0i}+R_{0i})/2$) vanishing. (But I am not 100% sure of this very last comment since it is quite late in the night and I had better to go to sleep)

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  • $\begingroup$ Note that all frame-depending statements are not covariant, hence there are no sum implied! $\endgroup$ – user139175 Dec 14 '16 at 1:33
  • $\begingroup$ Thanks, very helpful! I see that we only need one (not many) special coordinate systems. But in the derivation of the geodesic deviation equation, and in the whole surrounding context, $v^a$ is necessarily a unit vector. Why is this suddenly no longer the case? $\endgroup$ – Doubt Dec 14 '16 at 2:54
  • $\begingroup$ I have not gone through the book by d'Inverno, so I am not really sure of his notation and his conventions. However, it looks to me that the fact that $v^a$ usually represent velocities is not important for the solution of this exercise. Note moreover that you can always rescale the vectors $v^a$ in this exercise to get norm 1, because $1-\lambda^2$ is always non vanishing. $\endgroup$ – user139175 Dec 14 '16 at 8:50
  • $\begingroup$ To complete my previous comment, note that $R_{ab}v^a v^b$ vanishes if and only if $R_{ab} (\alpha v)^a (\alpha v)^b$ ($\alpha \neq 0$) vanishes as well, so what really matters is the fact that $v$ is timelike (and the multiplying factor is non zero). $\endgroup$ – user139175 Dec 14 '16 at 8:55
  • $\begingroup$ We can further investigate the physical meaning of a rescaling of $v$: it corresponds to an affine transformation of the proper time. It is easy to see that this transformation leaves the geodesic equations invariant. $\endgroup$ – user139175 Dec 14 '16 at 8:58

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