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Here's why I'm asking.

The gravitational time dilation factor at the surface of a star or planet can be expressed as $$gr/c^2$$

In theory, the same equation should apply to the gravitational length contraction factor.

If the gravitational length contraction factor reaches 1.0 then all lengths at the surface are reduced to 0 and the mass becomes invisible.

This will occur when $$gr = c^2$$

Now I have read that the surface gravity, g, on the largest known neutron star is about $$7*10^{12} m/s^2$$

Which means it should become invisible when $$r=c^2/7*10^{12}$$

Or $$r=12.84km$$

But that's the approximate radius of the largest known neutron star. So, in theory, it is on the verge of invisibility.

If it was any bigger it would probably be seen as a black hole.

UPDATE

As my formula for Gravitational Length Contraction has been questioned, I'd like to explain it at least, without attempting to prove it:

In that formula, $g$ is the rate of length contraction from the centre to the surface of the mass in units of $m/s^2$.

But the contraction happens at the speed of light, so the $c^2$ term is a conversion from time squared to length squared.

So $g/c^2$ is the rate of length contraction from the centre to the surface in units of $m/m^2$ or $m/m$ per $m$ radius.

Multiplying this by the radius gives the length contraction factor at the surface in $m/m$.

UPDATE END

So is a black hole a neutron star with a gravitational length contraction factor of 1.0, or is it pure coincidence that the largest known neutron star has a gravitational length contraction factor very close to 1.0?

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    $\begingroup$ Are you sure about your formula? How did you derive it? $\endgroup$
    – Anders Sandberg
    Commented May 3, 2022 at 17:36
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    $\begingroup$ Your time dilation formula (for $1-t_0/t_f$) is an approximation. It's ok when $gr/c^2$ is small, but not when GR effects become significant. Instead, use $t_0 = t_f\sqrt{1 - \frac{r_s}{r}}$, where $r_s=2GM/c^2$, which can be found at en.wikipedia.org/wiki/… $\endgroup$
    – PM 2Ring
    Commented May 3, 2022 at 22:37
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    $\begingroup$ Here's a Schwarzschild radius & time dilation calculator. You can put expressions in the input fields, eg 10*au gives a distance of 10 AU. $\endgroup$
    – PM 2Ring
    Commented May 3, 2022 at 23:06
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    $\begingroup$ @PM2Ring Schwartzchild got this wrong. Mass is decelerated away from the surface by gravity and therefore has an escape velocity. Light is accelerated away from the surface by gravity, and can therefore escape regardless of the surface gravity. This not so subtle difference between mass and light appears to have been quite readily ignored by Schwartzchild and many others since. $\endgroup$
    – Alan Gee
    Commented May 5, 2022 at 7:59
  • $\begingroup$ @AlanGee - Sorry, but this does not make much sense. Light is not "accelerated away from the surface": unless pointed radially it follows a curved null geodesic and as the Schwarzschild radius is approached most such geodesics end up captured by the surface. And "rate of length contraction factor" is an entirely nonstandard term that doesn't make much sense. Are you trying to measure the difference in radial geodesic distance from coordinate distance? $\endgroup$
    – Anders Sandberg
    Commented May 7, 2022 at 23:19

1 Answer 1

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I'm not sure what you mean by gravitational length contraction, and the time dilation formula you are using is only approximately correct for weak gravitational fields.

Observed neutron stars are indeed only a small factor larger than their Schwarzschild radii. If they were just a little smaller (say $<1.4 r_s$) then they would become unstable and likely collapse to become black holes.

Is this a coincidence? Well yes and no. It is the reason why we do not see neutron stars with masses of 3 times the sun or more. Such objects have indeed collapsed to have become black holes.

I suppose you could argue that it is an interesting twist of astrophysics that the massive stars that end up producing neutron stars finish their lives with iron cores that are too big to be supported by electron degeneracy pressure, but still small enough that they can avoid becoming a black hole. i.e. That the Chandrasekhar mass is smaller than the maximum mass for a neutron star. It wouldn't take much of a weakening of the strong nuclear force between closely packed neutrons to make forming neutron stars practically impossible. This is because neutron degeneracy pressure alone can only support neutron stars up to 0.75 solar masses and these could not be produced during stellar evolution without becoming stuck as stable white dwarfs. Anything more massive would then collapse directly to a black hole.

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    $\begingroup$ Of course there is gravitational length contraction. That's why Einstein talked about space-time warping and not just time warping. It's also why a photon doesn't lose or gain the ability to do work when it passes through a gravitational field. I would need to read up on electron and neutron degeneracy forces to understand your answer better, but I certainly find it interesting. $\endgroup$
    – Alan Gee
    Commented May 4, 2022 at 16:41
  • $\begingroup$ High mass stars are supported by nuclear fusion tho, which neutron stars are not, so I'm not sure what the twist is. $\endgroup$
    – Daddy Kropotkin
    Commented May 8, 2022 at 19:13
  • $\begingroup$ @DaddyKropotkin the cores of high-mass stars at the ends of their lives are supported (briefly) by electron degeneracy pressure. It is an accident of nuclear physics that the Chandrasekhar mass is smaller than the maximum mass of a neutron star. $\endgroup$
    – ProfRob
    Commented May 8, 2022 at 20:09
  • $\begingroup$ Oh I see, it was not clear to me that you meant at the end of nuclear fusion... $\endgroup$
    – Daddy Kropotkin
    Commented May 9, 2022 at 7:21

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