Hello there,(adsbygoogle = window.adsbygoogle || []).push({});

We know that for lightlike paths, there are circular geodesics at ##r = 3GM## in Schwarzschild geometry. Suppose an observer flashes his flashlight at ##r=3GM## and after some time the light reappears from the other side of the black hole. The time he measures is ##6 \pi GM##. I accidentally obtained this correct result, but I don't know why my method worked. I doubt it's a coincidence, and I'd like to understand why this is right. Basically by assuming that the light travels round a circle of length ##2\pi r## at a speed ##c = 1##. Write ##\Delta \tau = \frac{2 \pi r}{c} = 6 \pi GM ##

Why does this work? I expected this to only work using the coordinate time, i.e. i expected ## \Delta t = 6 \pi GM##, not ## \Delta \tau= 6 \pi GM##.

Thanks.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# A GR: Why shouldn't this orbital time surprise me?

Have something to add?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**