<p>In general relativity, then notion of space and time is <em>local</em>. As underlined in the OP, it means that at a radius $r$ of a star of mass $M$, time will run at $$\sqrt{1-\frac{2GM}{rc^2}}$$ the rate at an infinite distance of the object. For example imagine that $1-2GM/rc^2 = 1/4$, so that at a distance $r$, time runs at half the rate compare to an observer at infinity. If you are initially with an observer at infinity, you travel to $r$ (we neglect the time distortion during the travel), stay here an hour and then go back. When you'll reach the observer once again, your local clock will be one hour late, because 1 hour at $r$ will have passed in two hours at $r=\infty$.</p>

<p>So this distortion of space and time is <em>only local</em>. A massive object, like a star is only impacting in its neighborhood space and time. A distant observer is not impacted by the distortion, so that time and distances are unaffected.</p>

<p>To answer your question, an observer far away (for example on Earth) measuring by some meaning the speed of the two stars will actually see $c/2$ for both of them.</p>

CC BY-SA 3.0
2016-09-20T16:04:08.957Z
so the speed is $c/2$ because the space is distorted locally. But the projection of the distorted space onto a fixed coordinate system would then be $c/4$ because time passes slower? So in that fixed coordinate system they will have different speeds and the distance between them would increase?
asdf
CC BY-SA 3.0
2016-09-20T09:15:21.340Z
2016-09-20T09:15:21.340Z
cphyc
<p>What does a distant observer see if two masses with a given velocity are close enough to each other that time is dilated ?</p>

<p>Scenario (using 2 point masses for simplicities sake):
Mass a: 1,00E+030 kg
Mass b: 2,00E+030</p>

<p>Kinetic energy:</p>

<p>mass a: 1,39037910021726E+046</p>

<p>mass b: 2,78075820043453E+046</p>

<p>this would mean a velocity of
Velocity: 149896229 m/s [ = c/2]
for both masses</p>

<p>distance between masses: 3960 m</p>

<p>using the formula for time dilation <img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/fc63dfccb6a0a7379475600a65b36b65fb7b88e5" alt="formula"> the time would pass at a rate of ~0,5 at mass a and at a rate of ~0,8 at mass b</p>

<p>if both masses are flying parralel in the same direction does the observer see mass a flying with a speed of 
74948115 m/s [= c/4] and mass b flying with a speed of 119916983 m/s [= c/2 *0.8]
?</p>

CC BY-SA 3.0
2016-09-15T12:13:23.393Z
Given the mass you give and the distance between your two points, you are certainly speaking about two black holes, so you cannot 'see them', You're also misunderstanding the meaning of the time dilatation. This is the time dilatation *around the object*, but if you measure its speed, it'll always be c/2 in your case.
cphyc
CC BY-SA 3.0
2016-09-15T12:14:53.977Z
Also, it is worth noting that given the masses and the distance you give, you will have two black holes one very close to each other's horizon. FYI the radius of a black hole is given by $2GM/c^2$. For the Sun, you find $R \approx 2954m$ and for the other star, you'd find $1477m$.
cphyc
CC BY-SA 3.0
2016-09-15T19:53:23.170Z
Yes i know they would probably be black holes and you couldnt "see" them, just pretend the observer can see it somehow. From how i understood it object a would measure its own speed at $c/2$ and object b would also measure its speed at $c/2$ , but since time passes faster for object b the observer would see object b beeing faster than object a, but both with speeds $<c/2$
asdf
CC BY-SA 3.0
2016-09-16T08:04:27.407Z
It really depends on how you measure speed. If you for example use redshift techniques, you'd get both speeds at $< c/2$. So which way of measuring speed do you want to use? The 'physical speed' of both your objects will always be $c/2$ though, even though measuring it may be challenging.
cphyc
CC BY-SA 3.0
2016-09-16T17:56:40.210Z
speed as in rate of increase of distance to a fixed origin point the object is coming from. so if object a starts at $(0,0,0)^T$ in direction $(1,0,0)^T$  speed $v$ would be the distance from $(0,0,0)^T$ the object has after 1 second as seen by the observer that is unaffected by any time dilation effects which in this example would be $(v,0,0)^T$
asdf
CC BY-SA 3.0
2016-09-19T07:41:30.743Z
From far away from the two stars, their is no time shift and no space distortion, so that the velocity of both stars is exactly $c/2$ as measured by a remote observer. If however, the observer is getting close to the star and comoving with them and is measuring its speed relative to far away stars, you have to take into account the time distortion. The observed speed will be something smaller than $c/2$.
cphyc
CC BY-SA 3.0
2016-09-19T23:30:44.230Z
ok just to make sure i understood it correctly: in a fixed coordinate system that does not account for any distortions all objects move with the speed derived from their kinetic energy and only time is affected? so the objects will actually stay parallel to each other on the axis they move along?
asdf
CC BY-SA 3.0
2016-09-20T09:20:38.143Z
There is no such thing as a fixed coordinate system in general relativity. In general relativity, a force free trajectory is defined as a geodesic, a.k.a. a straight line. Time and space are distorted so that geodesics don't look straight and you effectively fall onto massive objects. At long range, space and time is said to be 'flat', which means it is not distorted. So if you take such a coordinate system, set an observer far away and measure the speed of the stars you'll measure its $c/2$.
cphyc
<time-dilation>
absolute speed of bodies as observed by a distant observer
asdf
# absolute speed of bodies as observed by a distant observer

(Asked by: username_0 on 2016-09-14)

What does a distant observer see if two masses with a given velocity are close enough to each other that time is dilated ?

Scenario (using 2 point masses for simplicities sake):
Mass a: 1,00E+030 kg
Mass b: 2,00E+030

Kinetic energy:

mass a: 1,39037910021726E+046

mass b: 2,78075820043453E+046

this would mean a velocity of
Velocity: 149896229 m/s [ = c/2]
for both masses

distance between masses: 3960 m

using the formula for time dilation ![formula](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc63dfccb6a0a7379475600a65b36b65fb7b88e5) the time would pass at a rate of ~0,5 at mass a and at a rate of ~0,8 at mass b

if both masses are flying parralel in the same direction does the observer see mass a flying with a speed of
74948115 m/s [= c/4] and mass b flying with a speed of 119916983 m/s [= c/2 \*0.8]
?

**Tags:** time-dilation

> **Comments:**
> **username_1**: Given the mass you give and the distance between your two points, you are certainly speaking about two black holes, so you cannot 'see them', You're also misunderstanding the meaning of the time dilatation. This is the time dilatation *around the object*, but if you measure its speed, it'll always be c/2 in your case.  
> **username_1**: Also, it is worth noting that given the masses and the distance you give, you will have two black holes one very close to each other's horizon. FYI the radius of a black hole is given by $2GM/c^2$. For the Sun, you find $R \approx 2954m$ and for the other star, you'd find $1477m$.  
> **username_0**: Yes i know they would probably be black holes and you couldnt "see" them, just pretend the observer can see it somehow. From how i understood it object a would measure its own speed at $c/2$ and object b would also measure its speed at $c/2$ , but since time passes faster for object b the observer would see object b beeing faster than object a, but both with speeds $<c/2$  
> **username_1**: It really depends on how you measure speed. If you for example use redshift techniques, you'd get both speeds at $< c/2$. So which way of measuring speed do you want to use? The 'physical speed' of both your objects will always be $c/2$ though, even though measuring it may be challenging.  
> **username_0**: speed as in rate of increase of distance to a fixed origin point the object is coming from. so if object a starts at $(0,0,0)^T$ in direction $(1,0,0)^T$  speed $v$ would be the distance from $(0,0,0)^T$ the object has after 1 second as seen by the observer that is unaffected by any time dilation effects which in this example would be $(v,0,0)^T$  
> **username_1**: From far away from the two stars, their is no time shift and no space distortion, so that the velocity of both stars is exactly $c/2$ as measured by a remote observer. If however, the observer is getting close to the star and comoving with them and is measuring its speed relative to far away stars, you have to take into account the time distortion. The observed speed will be something smaller than $c/2$.  
> **username_0**: ok just to make sure i understood it correctly: in a fixed coordinate system that does not account for any distortions all objects move with the speed derived from their kinetic energy and only time is affected? so the objects will actually stay parallel to each other on the axis they move along?  
> **username_1**: There is no such thing as a fixed coordinate system in general relativity. In general relativity, a force free trajectory is defined as a geodesic, a.k.a. a straight line. Time and space are distorted so that geodesics don't look straight and you effectively fall onto massive objects. At long range, space and time is said to be 'flat', which means it is not distorted. So if you take such a coordinate system, set an observer far away and measure the speed of the stars you'll measure its $c/2$.  

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## 1 answers

### Answer by username_1 on 2016-09-20. Score: 1 | Accepted answer

In general relativity, then notion of space and time is *local*. As underlined in the OP, it means that at a radius $r$ of a star of mass $M$, time will run at $$\sqrt{1-\frac{2GM}{rc^2}}$$ the rate at an infinite distance of the object. For example imagine that $1-2GM/rc^2 = 1/4$, so that at a distance $r$, time runs at half the rate compare to an observer at infinity. If you are initially with an observer at infinity, you travel to $r$ (we neglect the time distortion during the travel), stay here an hour and then go back. When you'll reach the observer once again, your local clock will be one hour late, because 1 hour at $r$ will have passed in two hours at $r=\infty$.

So this distortion of space and time is *only local*. A massive object, like a star is only impacting in its neighborhood space and time. A distant observer is not impacted by the distortion, so that time and distances are unaffected.

To answer your question, an observer far away (for example on Earth) measuring by some meaning the speed of the two stars will actually see $c/2$ for both of them.

> **Comments:**
> **username_0**: so the speed is $c/2$ because the space is distorted locally. But the projection of the distorted space onto a fixed coordinate system would then be $c/4$ because time passes slower? So in that fixed coordinate system they will have different speeds and the distance between them would increase?  

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