How can we immediately see that the Riemann tensor and Ricci tensor in Rindler space are zero?

I know that the Rindler metric is given by:


and what I just did was compute the Christoffels and then the Riemann and Ricci tensors according to the usual definition, giving me zero.

However you are supposed see immediately that they vanish. Why?

  • $\begingroup$ I must admit that when I was faced by this I did exactly the same as you. $\endgroup$ Oct 20, 2017 at 14:58
  • $\begingroup$ @JohnRennie Yes, but supposedly there is a way of seeing this without doing calculations $\endgroup$ Oct 20, 2017 at 14:59
  • 4
    $\begingroup$ Rindler coordinates are just a set of coordinates for describing Minkowski space. Since Minkowski space is flat, its curvature tensors vanish in all coordinate systems. $\endgroup$
    – user4552
    Oct 20, 2017 at 17:40
  • $\begingroup$ In fact this is the obvious reason why curvatures vanish in Rindler space: it is just part of Minkowski space and curvatures are tensors. I thought the OP wanted a different, say more direct, answer. $\endgroup$ Oct 20, 2017 at 21:06

1 Answer 1


It's more obvious if you're familiar with the tetrad formalism. From the metric provided, we can define an orthonormal basis by simply reading off, $e^{(t)} = a x \mathrm dt$ and $e^{(i)} = \mathrm dx^i$.

Now all $\mathrm d e^{(i)} = 0$, and $\mathrm de^{(t)} = -a \mathrm dt \wedge \mathrm dx = -\frac{1}{x} e^{(t)} \wedge e^{(x)}$ meaning the only non-zero connection is $\omega^t_x = -a \mathrm dt$ which is a constant and so $R = d\omega + \omega \wedge \omega = 0$.

It's easy to conclude any single-variable function replacing $a^2 x^2$ will lead to vanishing curvature.

  • $\begingroup$ Could there also be a way without using this tetrad formalism? I haven't seen it before and it is not part of the material that I am covering $\endgroup$ Oct 20, 2017 at 15:36
  • $\begingroup$ The notation $e^t$ looks like you're using the exponential function (at least it did to me, at first glance). Perhaps $e_t$ would be better, or $e^{(t)}$... $\endgroup$
    – Danu
    Oct 20, 2017 at 16:07
  • $\begingroup$ Regarding "immediate" understanding: I think it's because the metric is conical, and we all know cones are flat (expect at the point). $\endgroup$
    – JEB
    Oct 20, 2017 at 16:16
  • $\begingroup$ @JEB That sounds quite good, why exactly is Rindler conical? $\endgroup$ Oct 20, 2017 at 17:35
  • $\begingroup$ @Danu Agreed it can look odd; $e_t$ however doesn't quite fit as the vielbein is $e^a_\mu$ and the orthonormal index on the bottom has a different meaning. I'll go with $e^{(t)}$. $\endgroup$
    – JamalS
    Oct 21, 2017 at 23:17

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