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How much temporal dilation would a relativistic orbit about a frame-dragging massive object create? Would the black hole or other objects orbiting it look weird (relatively--hehe--speaking).



It's not really until you get up near 90% of the speed of light, where dilation starts getting insane. That's when the curve starts to sharply jump straight up.


As a rule of thumb, relativistic effects are very small below 5-10% of the speed of light. You may still need to calculate them if you want very high precision, like for GPS.


There is the relativistic effects of gravity to consider in this case too


Right, but for that orbit to be so slow, it must be a very distant orbit.


Using 1% of speed of light, it will slow down by 0.005% (That's much less than I expected)


If anyone is curious, the reason is that many/most special relativistic effects are proportional to gamma = 1/sqrt(1-v^2/c^2), with gamma = 1 being non-relativistic. Even at v=20% of c, gamma is only 1.02 (a mere 2% effect).

http://www.wolframalpha.com/input/?t=crmtb01&f=ob&i=Plot%5B1...


General relativity also accounts for gravitational/accelerational effects, in addition to the speed-related effects. For example, in the Schwarzschild metric[1], effects such as time dilation are on the order of 1-phi/c^2 where phi=GM/r. Or expressed in terms of "g-force": phi=gr.

[1] https://en.wikipedia.org/wiki/Schwarzschild_metric


Yes, but doesn't that ignore the gravitational acceleration? Time dilation isn't just speed, it's also acceleration. Even if you could hover very near the event horizon the time dilation would be severe.


    Yes, but doesn't that ignore the gravitational
    acceleration
Gravitational acceleration doesn't exist (in GR). All bodies are stationary in GR and observe other things moving in reference to themselves.

What we perceive as gravitational acceleration is just objects traveling in straight lines (at constant speeds) though 4D space-time. Circular orbits are just straight lines of constant velocity bent around by the local curvature of space-time (which can be caused by a local density of mass-energy).

The idea of acceleration assumes a force is acting on a body. Gravity isn't a force in General Relativity. It is the shape of the universe.


> Gravitational acceleration doesn't exist (in GR).

A better way of stating this would be that objects moving solely under gravity have zero acceleration in GR, because in GR, acceleration means proper acceleration--what is measured by an accelerometer. Objects moving solely under gravity are in free fall.

> All bodies are stationary in GR and observe other things moving in reference to themselves.

This isn't correct. There is no requirement that an observer in GR must use coordinates in which they are at rest.


> Time dilation isn't just speed, it's also acceleration.

That's not correct. Gravitational time dilation isn't a function of acceleration, it's a function of gravitational potential, i.e., how deep you are in the potential well of the source.

(Also, an object that is in a free-fall orbit has zero acceleration in GR, since in GR acceleration is proper acceleration--what is measured by an accelerometer.)


As far as I know, there's only speed in the formula of time dilation, not acceleration.

See also [1].

[1] http://www.edu-observatory.org/physics-faq/Relativity/SR/clo...


Special relativity gives you dilation from velocity. General relativity gives you dilation from acceleration.


> General relativity gives you dilation from acceleration.

Not from acceleration, from depth in a gravitational potential well. See my other posts upthread.


That's for special relativity, which is only valid if you're not accelerating. This star is, so we'd need general relativity...

I don't know what that formula looks like.


Seeing as the star is in close proximity to a blackhole it would likely be similar to the Kerr-Metric [1] (as we _should_ assume the black hole as some spin).

The white star has non-trivial mass of its own which greatly complicates the curvature of the local space-time region and render the Kerr-Metric equations lacking.

GR is really complicated FYI.

[1] https://en.wikipedia.org/wiki/Kerr_metric


> it would likely be similar to the Kerr-Metric

Unless the hole is rotating extremely rapidly (which I don't see mentioned), the difference between the Kerr metric and the Schwarzschild metric (which is what I used for the formula I posted upthread) is too small to matter unless you are really close to the hole's horizon.


You got me curious, and I wanted to put numbers around this. It's worth noting that the actual paper claims only this is a black hole candidate, and is not certain it's a black hole (despite the Science alert headline).

The paper guesses that the black hole is ~ 1 solar mass, so the Schwarzschild radius would be about 3 km.

The distance is about 2.5x the earth-moon, which is about 10^6 km.

btw, paper here: https://arxiv.org/pdf/1702.02167.pdf


The formula for an object at radius r in the field of a non-rotating gravitating mass M is:

sqrt(1 - 2M/r - v^2)

instead of

sqrt(1 - v^2)

(where I am using units where G = c = 1). For an object in a free-fall circular orbit, v^2 = M/r, so the above formula can be simplified to

sqrt(1 - 3M/r)




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