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- Thread starter Daleri Mc Rileda
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Nugatory

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We have corrected the thread tag for you.

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As you seemed to have intuitively understood, you cannot just lay a measuring rod to measure that distance. And because of the curvature of spacetime it would not mean the same thing anyway.With length contraction being ever increased while approaching the mass of a black hole singularity, how does one measure the distance to the virtual infinite contracting frames of reference of time and space /distance of the singularity?

Instead, what is measured is the area of the sphere around the singularity. The unmeasurable radial coordinate is related to that measurable area by ##A=4\pi r^2##

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With length contraction being ever increased while approaching the mass of a black hole singularity

This is not correct.

the virtual infinite contracting frames of reference of time and space /distance

I don't know what you mean by this.

How can one measure the size of a galaxy passing through it's center

You can't, because the "center" of the galaxy (assuming that you mean the singularity at ##r = 0## of the black hole at the center of the galaxy) is not a point in space. It's a moment of time, which is to the future of anyone inside the hole's horizon.

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Nugatory

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The quick answer is that you don't - the "distance" from the central singularity to anywhere else is undefined, as is the "diameter" passing through the singularity.how does one measure the distance to the virtual infinite contracting frames of reference of time and space /distance of the singularity? Related question : How can one measure the size of a galaxy passing through it's center /singularity?

The longer answer: although the distance to the singularity is undefined, and therefore there is no such thing as a radius or diameter, there is a quantity that we can use outside the event horizon. Consider two imaginary spheres centered on the black hole. The first one has surface area ##A_1=4\pi{r_1}^2## and the second has surface area ##A_2=4\pi{r_2}^2## - we can measure these surface areas and calculate ##r_1## and ##r_2## from them. These values ##r_1## and ##r_2## are not distances from the center, and the difference between them is not the distance between the surface of the smaller sphere and the larger sphere - but we can use them to label points around the black hole.

If you have a spherical galaxy with a black hole in the center, and someone says that it is ##X## lightyears across... They are really saying that the surface area of the galaxy is ##X\pi## square lightyears. If the black hole weren't there then the galaxy would be ##X## lightyears across... But it is there, so the distance across is undefined.

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Thank youThe quick answer is that you don't - the "distance" from the central singularity to anywhere else is undefined, as is the "diameter" passing through the singularity.

The longer answer: although the distance to the singularity is undefined, and therefore there is no such thing as a radius or diameter, there is a quantity that we can use outside the event horizon. Consider two imaginary spheres centered on the black hole. The first one has surface area ##A_1=4\pi{r_1}^2## and the second has surface area ##A_2=4\pi{r_2}^2## - we can measure these surface areas and calculate ##r_1## and ##r_2## from them. These values ##r_1## and ##r_2## are not distances from the center, and the difference between them is not the distance between the surface of the smaller sphere and the larger sphere - but we can use them to label points around the black hole.

If you have a spherical galaxy with a black hole in the center, and someone says that it is ##X## lightyears across... They are really saying that the surface area of the galaxy is ##X\pi## square lightyears. If the black hole weren't there then the galaxy would be ##X## lightyears across... But it is there, so the distance across is undefined.

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ThanksAs you seemed to have intuitively understood, you cannot just lay a measuring rod to measure that distance. And because of the curvature of spacetime it would not mean the same thing anyway.

Instead, what is measured is the area of the sphere around the singularity. The unmeasurable radial coordinate is related to that measurable area by ##A=4\pi r^2##

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It should be no space and no time.This is not correct.

I don't know what you mean by this.

You can't, because the "center" of the galaxy (assuming that you mean the singularity at ##r = 0## of the black hole at the center of the galaxy) is not a point in space. It's a moment of time, which is to the future of anyone inside the hole's horizon.

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It should be no space and no time.

Do you mean the singularity? If so, your statement is incorrect. It would be correct to say that the singularity itself, strictly speaking, is not part of spacetime. But it can still be treated as a limit, a boundary of spacetime inside the horizon, and in that sense, it is a moment of time, not a place in space. (In more technical language, it is a spacelike line, not a timelike one.)

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OK, I see what you are saying, frozen time.Do you mean the singularity? If so, your statement is incorrect. It would be correct to say that the singularity itself, strictly speaking, is not part of spacetime. But it can still be treated as a limit, a boundary of spacetime inside the horizon, and in that sense, it is a moment of time, not a place in space. (In more technical language, it is a spacelike line, not a timelike one.)

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I see what you are saying, frozen time.

No, that's not what I was saying. Objects falling into the singularity are not "frozen in time". According to the idealized classical model we are discussing, they hit the singularity and are destroyed.

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I don't understand what you mean by, "destroyed".No, that's not what I was saying. Objects falling into the singularity are not "frozen in time". According to the idealized classical model we are discussing, they hit the singularity and are destroyed.

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I don't understand what you mean by, "destroyed".

It means that, according to this idealized model, objects that hit the singularity cease to exist. Bear in mind that this is an idealized model and physicists don't believe it's actually realistic. The problem is that we don't have a more realistic model to replace it with at this point.

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OKIt means that, according to this idealized model, objects that hit the singularity cease to exist. Bear in mind that this is an idealized model and physicists don't believe it's actually realistic. The problem is that we don't have a more realistic model to replace it with at this point.

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