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My question concerns the following definition

Definition: The timelike (resp. null) generic condition in GR is fulfilled if $$u_{[\alpha} R_{\rho]\mu \nu [\sigma}u_{\beta]}u^\mu u^\nu \ne 0$$ at some point of each timelike (resp. null) geodesic with tangent vector $\vec u$. ($R_{\rho \mu \nu \sigma}$ is the Riemann curvature tensor.)

It is written in many places that this is the right condition to impose if one wants to assume that every freely falling (or light) particle encounters some form of matter or radiation in its history (or something to that effect).

But I don't understand, why the particular tensor $u_{[\alpha} R_{\rho]\mu \nu [\sigma}u_{\beta]}u^\mu u^\nu \ne 0$ is the right thing to look at in this context. For example, why don't we assume that $R_{\rho\mu\nu\sigma}u^\mu u^\nu \ne 0$ at some point? Or maybe that $R_{\mu \nu}u^\mu u^\nu \ne 0$, or perhaps that $G_{\mu \nu} u^\mu u^\nu \ne 0$?

I'm guessing that the last two condition could be too weak to derive the singularity theorems we want, so that something stronger must be assumed. But the expression $u_{[\alpha} R_{\rho]\mu \nu [\sigma}u_{\beta]}u^\mu u^\nu \ne 0$ really looks a bit strange to me (i.e. I don't understand its significance). Could someone explain to me why this is the right condition to impose?

Thanks!

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  • $\begingroup$ Wald has on p. 227: "A spacetime ... is said to satisfy the timelike generic condition if each timelike geodesic possesses at least one point at which $R_{abcd}\xi^a\xi^d\ne 0$." I haven't dug into it enough to understand what's going on, but I think $\xi$ is just an arbitrary vector, not a tangent to the geodesic. Wald also introduces a separate condition $R_{ab}\xi^a\xi^b\ne 0$. $\endgroup$
    – user4552
    Commented Jul 6, 2013 at 20:34
  • $\begingroup$ @BenCrowell : $\xi^a$ is the tangent (beginning of the last paragraph, page 224) $\endgroup$
    – Trimok
    Commented Jul 6, 2013 at 20:53
  • $\begingroup$ $u_{[\alpha} R_{\rho]\mu \nu [\sigma}u_{\beta]}u^\mu u^\nu \ne 0$ and $R_{\rho\mu\nu\sigma}u^\mu u^\nu \ne 0$ are equvalent. $\endgroup$
    – MBN
    Commented Jul 7, 2013 at 10:11
  • $\begingroup$ @MBN - They are not equivalent. The first one applies to timelike geodesics and the second applies to null geodesics. $\endgroup$
    – Prahar
    Commented Jul 8, 2013 at 1:09

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Carroll has this to say about this condition

These fancy conditions simply serve to exclude very special metrics for which the curvature consistently vanishes in some directions - Carroll P. 242-243

While that answers your question, it doesn't give much insight on what the curvature vanishing in some directions has to do with singularity theorems. (The generic condition on the metric is required to prove Hawking's and Penrose's singularity theorems) I hope someone else can give more insight on that.

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  • $\begingroup$ You can do fine tunings to get around a lot of the general theorems. Stuff like sets of measure zero in the space of collapsing matter metrics violating cosmic censorship. $\endgroup$
    – Zo the Relativist
    Commented Aug 7, 2013 at 2:08

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