# Even light would take an infinite time

Thus wrote PAM Dirac in his little monograph, "General Theory of Relativity", p.34t. More fully, "Even a light signal would take an infinite time to cross the boundary r = 2m, as we can easily check."

Yes, we can, and I did easily by setting ds = 0 in the Schwarzchild solution for a particle described by radially symmetric coordinates with a static metric. The time for transit is governed by a log function of the radius, r, for an observer in the fixed coordinate system and is infinite for him (or her!)

Please someone, kindly explain how this result is consistent with the constancy of the speed of light. The distance to the point at which r = 2m is a determined quantity. If the speed of light does not change, how can the time to traverse it be other than finite for the fixed observer?

Staff Emeritus
The constancy of the speed of light is true locally.

Vanadium 50: Thanks you your reply, but I find it too concise for my understanding. I was under the impression that one of the two principles of relativity in the constancy of the speed of light. I have not, until now, read that the speed of light is constant locally. I do not know what locally means in this regard. In the context of my question, the fixed observer is using a coordinate system that describes events in the three space r > 2m. In this locality, the time of transit grows without limit as the light approaches r = 2m + . I am making no reference to what happens inside this region, because the coordinates are not used therein - though the solution is analytically continued. Why is this external sphere not in his locality?

DrGreg
Gold Member
If the speed of light does not change, how can the time to traverse it be other than finite for the fixed observer?
Because the Schwarzschild t coordinate does not measure local time, and the Schwarzschild r coordinate does not measure local distance. Both need to be converted (using the metric) before you can measure the local speed of light, which is always c.

To calculate local radial distance (for a local hovering observer), put $dt=d\phi=d\theta=0$ in the metric.

To calculate local time (for a local hovering observer), put $dr=d\phi=d\theta=0$ in the metric.

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Dr Greg, I think I see what you mean: we can say dx = c dt, but to integrate to finite distances we must use the metric, and that is where the time interval grows without bound. Thanks much!