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 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] ?

  • $\begingroup$ 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. $\endgroup$
    – cphyc
    Sep 15 '16 at 12:13
  • $\begingroup$ 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 $1477m$. $\endgroup$
    – cphyc
    Sep 15 '16 at 12:14
  • $\begingroup$ 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 $<c/2$ $\endgroup$
    – asdf
    Sep 15 '16 at 19:53
  • $\begingroup$ 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. $\endgroup$
    – cphyc
    Sep 16 '16 at 8:04
  • $\begingroup$ 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 $(v,0,0)^T$ $\endgroup$
    – asdf
    Sep 16 '16 at 17:56

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.

  • $\begingroup$ 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? $\endgroup$
    – asdf
    Sep 20 '16 at 16:04

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