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You're observing a massive object (probably a neutron star), and it is moving at a significant fraction of the speed of light relative to you. The mass of the object is just below the mass necessary to form a black hole of the corresponding size (ie, if a relatively small amount of mass was added, or the current mass was compressed, it would form a black hole). In the moving reference frame of the object, it is not observed to be a black hole, and it doesn't have sufficient density to form one.

The interesting conundrum is that from your point of view, the object undergoes length contraction. In this case, if an object of the same mass were to have the size that you're observing due to length contraction, it would be of sufficient density to form a black hole. Obviously, you don't observe the moving object becoming a black hole, because it isn't actually doing that, but you observe what appears to be an object of sufficient density to become a black hole, but is not a black hole.

What makes this possible? Is the observed mass from your reference frame different for some reason? Does what you observe not matter? Is it something else?

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The black hole solution you are referring to is the Schwarzschild solution that applies to a static centrally symmetric object. If the object is moving in your reference frame, it is not static, so this solution does not apply in your coordinates. In other words, the Schwarzschild spacetime is not Lorentz invariant just like most anything in General Relativity.

The easiest way to resolve this is to describe the object in its rest frame and then use the equivalence principle that physics does not depend on the reference frame. If the object is not a black hole in its own rest frame, then, according to the equivalence principle, this object is not a black hole in any reference frame.

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    $\begingroup$ That makes so much sense; I hadn't thought of the problem from that perspective! $\endgroup$
    – Blapor
    Oct 7, 2019 at 8:26
  • $\begingroup$ If I understand what you're saying correctly, the length contraction doesn't change anything because the Schwarzchild radius would also undergo length contraction? $\endgroup$
    – Blapor
    Oct 7, 2019 at 8:36
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    $\begingroup$ Yes, the length contraction is a projection of one frame onto another in hyperbolic geometry. It is not something that physically happens to the object, but only to our view of it. A neutron star flying to you is equivalent to you flying to it and clearly nothing would change to it because of this. The length contraction is similar to your shadow getting shorter as the Sun goes up. A real effect, a projection, but not something actually happening to you. Also, projections in a curved spacetime are tricky, so in general the Lorentz transformation can be applied only locally. $\endgroup$
    – safesphere
    Oct 7, 2019 at 14:48
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What makes this possible?

It is called relativistic mechanics, special relativity mathematics and definitions have to be used.

Is the observed mass from your reference frame different for some reason?

What one observes in an inertial frame, where you are at rest, is the so called relativistic mass. This is a function of velocity so it is not invariant. What is invariant is the length of the four vector that characterizes the star :

invmass

where $m_0$ is the invariant mass and the relativistic mass is connected to the invariant mass with

$m_{rel}/m_0=γ$

and it is $γ$ that gives the function of velocity.

Does what you observe not matter? Is it something else?

What you observe matters, the something else is that one needs the mathematics of special relativity to get the invariant mass of a system from the observed velocity of that system.

Density is defined in the rest frame of a system. General relativity equations turn to special relativity in flat spaces, and you problem is in a flat space as far as masses and velocities go.

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It is not only the object that undergoes length contraction. But also the Schwarzschild radius will be length contracted in the direction of the velocity of the observer. Basically there will be a transverse Schwarzschild radius and also a longitudinal Schwarzschild radius. But your longitudinal and transverse neutron star radius will not be lower than the corresponding Schwarzschild radius. So the observer will not see your neutron star becoming a black hole.

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I observe an object the mass of which in my reference frame is large enough to make a black hole if it were its rest mass. However, when observing it, I realise it has a huge velocity. So I compute back its rest mass, that is, its mass in its own reference frame, find it is below the threshold and conclude it is not a black hole.

Indeed, as you say yourself, what you observe does not matter, only what is happening in the frame of the object itself.

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    $\begingroup$ This is not what the OP is asking. $\endgroup$
    – safesphere
    Oct 7, 2019 at 7:28
  • $\begingroup$ @Alfred It's not about mass but density. Mass is invariant. Even if you use relativistic longitudinal or transverse mass, the circumstances become worse, and the density is measured much greater from the viewpoint of the moving observer. $\endgroup$
    – Mohammad Javanshiry
    Oct 7, 2019 at 7:30
  • $\begingroup$ @Mohammad Javanshiry Mass or mass density, it is all irrelevant. The only thing that matters is what is the case in the reference frame of the object. What the observer sees is unimportant, he just has to compute back what is the situation in the frame of the observed objet. $\endgroup$
    – Alfred
    Oct 7, 2019 at 7:57

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