4 ms·

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

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.

valarauca1 10y ago
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 https://en.wikipedia.org/wiki/Kerr_metric
- pdonis 10y ago
  > 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.
  - jpmattia 10y ago
    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 https://arxiv.org/pdf/1702.02167.pdf
pdonis 10y ago
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)

4 ms·

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

by Filligree 10y ago

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.

valarauca1 10y ago
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 https://en.wikipedia.org/wiki/Kerr_metric
- pdonis 10y ago
  > 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.
  - jpmattia 10y ago
    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 https://arxiv.org/pdf/1702.02167.pdf
pdonis 10y ago
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)