In spherical coordinates in a $(1+3)$-dimensional space-time one has the $\mathrm{AdS}_4$ metric $$ds^{2}=-\left(1+k^{2}r^{2}\right)dt^{2}+\left(1+k^{2}r^{2}\right)^{-1}dr^{2}+r^{2}d\Omega ^{2}.$$

Does it mean that for any massless particle moving radially $d\Omega^{2}=0,$ its velocity


is greater than the speed of light?

Even when the cosmological constant $|\Lambda|\sim k^{2}$ is very small, still $v^{2}\geq1$?


1 Answer 1


The expression you have written down is the coordinate speed of light. That is, you have selected some coordinate system in which to write down your metric, then calculated the speed in those coordinates. This is a perfectly reasonable thing to do, but it means the speed you get is coordinate dependent and will be different for observers using other coordinate systems. In particular the coordinate speed of light is not necessarily equal to $c$. Indeed depending on the geometry and the coordinate system it can have any value from zero upwards without limit.

If you're interested this issue has recently been discussed in GR. Einstein's 1911 Paper: On the Influence of Gravitation on the Propagation of Light.

The invariant speed is the local speed of light i.e. the speed at the observer's location. In this case your metric is using a coordinate system in which the observer is stationary at $r=0$. If we substitute this value into your expression for the speed of light we get:

$$ \frac{dr}{dt} = 1 $$

as expected.

I won't go into detail here since the issue is discussed in great detail in the question I linked. However as a general rule the local speed of light is always $c$ while the coordinate speed of light may differ from $c$.


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