General Relativity - Circular Orbit around Earth

[TSny]

Recall that the correct expression for the light pulse is $$ds^2 = -\left( 1 - \frac{2GM}{c^2r} \right) c^2 dt^2 + \left( 1 + \frac{2GM}{c^2r} \right) dr^2$$

$ds = 0$, but $dt \neq 0$ and $dr \neq 0$.

But you just need to find $\Delta \tau = \int d\tau$ for the total round trip of the light. So, you don't need to worry about $dt$ or $dr$.

 

[unscientific]

Recall that the correct expression for the light pulse is $$ds^2 = -\left( 1 - \frac{2GM}{c^2r} \right) c^2 dt^2 + \left( 1 + \frac{2GM}{c^2r} \right) dr^2$$

$ds = 0$, but $dt \neq 0$ and $dr \neq 0$.

But you just need to find $\Delta \tau = \int d\tau$ for the total round trip of the light. So, you don't need to worry about $dt$ or $dr$.

Yeah, but $d\tau = 0$ so $\int d\tau = 0$. So from the light's perspective, no time has passed at all. Does this even make sense?

 

[TSny]

Yeah, but $d\tau = 0$ so $\int d\tau = 0$. So from the light's perspective, no time has passed at all.

That's right. Of course, no physical observer can travel at the speed of light. But, in a thought experiment, you can imagine the observer moving at 0.99999...c (add as many 9's as you wish). The time elapsed for this observer will approach zero. It's the ultimate "twin paradox" scenario where the traveling twin would not age at all!

Does this even make sense?

It's strange, but it's a natural consequence of SR and GR.