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Kara386
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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!
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!
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