# I GR observer at infinity

1. Nov 1, 2017

### Kara386

We were shown the answer to this question as a worked example: A photon is emitted from a radius $r_2$ and travels radially inward to $r_1$ until it's reflected by a fixed mirror and travels back to $r_2$. Calculate the time taken for the photon to travel in and back, according to a stationary watcher at infinity. Use the Schwarzschild metric.

The answer then stated $ds^2=0$ for a photon, and as it is radially infalling $d\theta$ and $d\phi$ are also zero. Subbing these into the metric and integrating gave an expression for $\Delta t$. I thought that would be the time measured by the photon. But the calculation ends there, and states that this is time as measured by the observer at infinity.

Is that because for the observer, spacetime is flat? So $ds^2 = d\tau^2 = dt^2-dx^2-dy^2-dz^2$, but since the observer is stationary you just get $d\tau = dt$. It just seems a bit strange that someone infinitely far away would measure the same time as the photon. But then I'm a week into my first GR course, so at the minute most of it seems odd!

Last edited: Nov 1, 2017
2. Nov 1, 2017

### vanhees71

No, you have to use the Schwarzschild line element rather than the Minkowskian. In the usual coordinates $\mathrm{d} t$ is the time increment of an observer very far from the mass responsible for the Schwarzschild gravitational field. You just have to calculate the radial light-like (null) geodesics in a Schwarzschild metric.

3. Nov 1, 2017

### Ibix

Photons don't measure time. That's why $ds=cd\tau=0$ in this case.

The $\Delta t$ you calculated was the difference in time coordinate between the emission and reception events. In curved spacetime, the difference in time coordinates isn't necessarily anything to do with the time experienced by anyone (just as, on the curved surface of the Earth, the same latitude but a difference of 1° in our longitudes could mean anything from millimetres to hundreds of miles between us). However, the Schwarzschild time coordinate is chosen to be the time for an observer at infinity ($d\tau=dt$ for an observer stationary at r=∞). So in this case, the coordinate time difference is the answer you were looking for. It's not the same as the time experienced by someone waiting at the emitter for the reflection to return.

I'm not sure it ever stops being odd, by the way. It does become more understandable.

4. Nov 1, 2017

### Kara386

Right ok, so for r tending to infinity, you get from the Schwarzschild metric (for a stationary observer) that $dt=d\tau$? And that's different to the time experienced by someone at $r_2$. If you had a stationary observer at $r_2$, would this be the time $d\tau$ that they experience (setting c=1)?
$ds^2 = d\tau^2 = \left(1-\frac{2GM}{r}\right)(dt)^2$

So there's a factor of $\sqrt{1 - \frac{2GM}{r}}$ compared to the coordinate time, and that takes you to proper time?

That's very reassuring, I'm quite glad it stays odd. My lecturer is treating it as somewhat mundane - but then he's a cosmologist, and he's also said things like galaxies are tiny. Which I suppose they might be, in the context of cosmology!

I didn't know coordinate time wasn't necessarily a quantity experienced by someone. Thanks for the answers, they're really helpful!

5. Nov 1, 2017

### Ibix

Yes.
That should be $r_2$ rather than r in the bracket. And strictly you should integrate $d\tau$ to get a finite length of time if you want to talk about the amount of time that's passed, but the integral is trivial in this case and you can just replace d with Δ. But basically yes.

Some aspects of it are (my analogy to coordinates on Earth is quite literally mundane, and I find it very useful). But the consequences of even the mundane stuff can be surprising.
Coordinates are just a way of "labelling" events in spacetime for easy reference. If you're going to be systematic about that labelling then it's reasonable that your time coordinate increasing should have some relationship to everyone's clock readings increasing. But it doesn't have to be the exact same thing - and that's where the $\sqrt {1-2GM/r}$ comes in, as you said.

6. Nov 1, 2017

### Kara386

Fantastic, thanks for your help, that's cleared up some misconceptions I had!