# Gravitational Time dilation for motion in a gravitational field [duplicate]

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?

## marked as duplicate by John Rennie black-holes StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 12 '18 at 15:43

• Time as measured by what observer ? – StephenG Sep 12 '18 at 2:13
• The observer is not affected by the gravity of the black hole (ideal condition: zero time dilation). He has super-vision: he looks at this astronaut moving towards the black hole till the time he reaches event horizon. If my understanding is correct, the gravitational time dilation equation applies for a fixed position in space, and not to a moving astronaut in an increasing gravitational field. – Vismay Harani Sep 12 '18 at 3:01

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 $r$

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

• That's the time dilation, but it's not the "gravitational time dilation". The time dilation is caused by two things—the astronaut's near-relativistic speed and the gravity well of the black hole—and this doesn't separate them out. But maybe this is what the OP wanted in the first place. – Peter Shor Sep 12 '18 at 11:55
• @PeterShor I didn't say "gravitational". It is a time dilation of a free falling observer in a gravitational field. Its two underlying causes are clear in the metric containing temporal and spatial components. However, they are not linearly independent in the geodesics equation, so the full time dilation is not clearly split into gravitational and speed related despite obviously being caused by two effects. In other words, general relativity doesn't easily split into "pure general" plus special. It is just general. And yes, the OP clearly stated "motion" in the title, not a static solution. – safesphere Sep 12 '18 at 14:03
• The OP's title was "Gravitational Time dilation ...". If you answer a question other than the one the OP asks, you need to point it out, rather than leaving them to infer this from your saying "time dilation" rather than "gravitational time dilation". – Peter Shor Sep 12 '18 at 17:56
• @PeterShor The title includes "motion" and so is my answer. However, you are more than welcome to suggest an edit to improve the answer. Thanks! – safesphere Sep 12 '18 at 18:22
• Actually, it isn’t a free fall into the black hole. The size of the black hole is 20,000 times the mass of sun. The guy is brought just to the edge of the black hole by a space ship traveling at 0.8 speed of light. And he stays at the edge for around 15 minutes. So I had divided my calculations is 3 parts. The ‘time dilation’ due to traveling at 0.8 C. The ‘gravitational time dilation’ because he was standing at the edge, now I want to calculate the gravitational time dilation for the span of the journey in the spaceship as he moves towards increasing gravity. – Vismay Harani Sep 13 '18 at 5:41