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An astronaut is travelling towards a black hole in a space ship traveling at 0.8 C. The guy’s position is not fixed relative to the black hole. He is travelling towards it from a distance. How to calculate the gravitational time dilation in this case, as the "relative time" keeps getting slower as one approaches the event horizon?

Please note that I have already factored the ‘time dilation’ due to the space ship traveling at 0.8 C. I just want to understand the ‘gravitational time dilation ‘ part of it.

Can some form of integration be applied on the gravitational time dilation equation for the case of motion in gravitational field?

Vismay Harani
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1 Answers1

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A radial free fall to a Schwarzschild black hole from rest a distance $R$ in geometrized units is given by the geodesics

$$ \tau=\dfrac{R}{2}\sqrt{\dfrac{R}{2M}}\left(\arccos\left(\dfrac{2r}{R}-1\right)+\sin\left(\arccos\left(\dfrac{2r}{R}-1\right)\right)\right) $$

And

$$ t=\sqrt{\dfrac{R}{2M}-1}\cdot\left(\left(\dfrac{R}{2}+2M\right)\cdot\arccos\left(\dfrac{2r}{R}-1\right)+\dfrac{R}{2}\sin\left(\arccos\left(\dfrac{2r}{R}-1\right)\right)\right)+ $$

$$ +\, 2M\ln\left(\left|\dfrac{\sqrt{\dfrac{R}{2M}-1}+\tan\left(\dfrac{1}{2}\arccos\left(\dfrac{2r}{R}-1\right)\right)}{\sqrt{\dfrac{R}{2M}-1}-\tan\left(\dfrac{1}{2}\arccos\left(\dfrac{2r}{R}-1\right)\right)}\right|\right) $$

You can figure the time dilation by differentiating these equations by $r$

$$ \dfrac{d\tau}{dt}=\dfrac{d\tau}{dr}\dfrac{dr}{dt}=\dfrac{d\tau}{dr}\cdot \dfrac{1}{\dfrac{dt}{dr}} $$

safesphere
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