# Neutron Star Becomes Black Hole: How Can 2 Realities Coexist?

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• a1call
In summary, the neutron star can form a black hole in some inertial frames of reference, but not in others. This inconsistency means that something is wrong with your reasoning, and either one of your descriptions is wrong.
a1call
• If a neutron star can be seen dense enough in some inertial frames of reference (due to relativistic length contraction) to form a Black-Hole, but not in others, how could the two realities coexist?

You've answered your own question. The "two realities" are inconsistent, so there's something wrong with your reasoning. There's only one reality, so at least one of your descriptions is wrong.

In general relativity the source of gravity is the stress-energy tensor, not merely mass or density. This is a rather complex entity, and does not lead to an expectation that a neutron star will collapse into a black hole just because it is moving.

russ_watters and Dale
The Schwarzschild radius formula gives a radius For any given mass below which the escape velocity would be greater than the speed of light. It is conceivable that the surface of a neutron star would be below this radius in some inertial frames of reference, while not in others due to relativistic length contraction or lack thereof. I know there is a conflict but don't see the oversight or resolution to the conflict. Hence posting this thread to be pointed out the oversight.

No. The Schwarzschild radius only has that interpretation when the massive object is not moving. If it's moving, the source term of the gravitational field has elements that do not depend solely on the mass, and you can't interpret the Schwarzschild radius the way you are doing.

An approximate way to think of things is this: the neutron star is length contracted, but so is its gravitational field.

Before someone rips me apart for that statement (length contraction is a phenomenon in flat spacetime only), imagine a cube around the neutron star large enough to enclose its entire gravitational field to a given accuracy; this will length contract and must continue to enclose the gravitational field.

Dale

The poster grant hutchison on that forum is correct. Length contraction of the gravity field does not correctly describe the maths; it doesn't correctly describe what happens to the neutron star either. The reason for this is that length contraction is a phenomenon in flat spacetime. The global Lorentz transforms you use to define it do not apply in the curved spacetime near a strongly gravitating mass.

That was why I was careful to mention a large cube enclosing the mass, where spacetime is flat to a good enough approximation. The cube will be measured to length contract, and will always contain the volume where there is significant gravity. Essentially, I've hidden all the difficult stuff inside the cube. Consistency requires that there be no significant spacetime curvature outside the box for any observer, and one could make rigorous claims about the angular separation of light rays grazing the neutron star and meeting at points on the cube as measured by stationary and moving observers.

At this point, I'll just summon @PeterDonis to criticise.

If you want to know what's going on deeper in the gravitational field then you'll have to solve Einstein's field equations. As far as I'm aware there's a known solution when the mass is stationary (the Schwarzschild solution) and one where it's moving at near lightspeed (the Aichelberg-Sexl ultraboost), but only numerical solutions for masses moving at less extreme speeds.

Note that the second paragraph of grant hutchison's quote from Baez's "If you go too fast, do you become a black hole?" says what I said in the first paragraph of my last post, in a bit more detail.

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Special relativity is based on the concept of filling spacetime with a regular grid of clocks at rest with respect to you. I do the same. If we're in relative motion then we'll measure length contraction, time dilation and relativity of simultaneity. But add a gravitational field and your inertial clocks tick at different rates and keep crashing into each other and the neutron star. You can't build the conceptual framework you need for special relativity, so you can't borrow results from it without great care.

That's what my giant cube with observers stationary and in motion with respect to it are about. I'm hiding the bits of spacetime where my clocks keep crashing, and I'm cutting my experiment short before I can notice that there's gravity even outside my cube. That's why SR is good enough at a distance. (Note I don't need an actual cube - I just define a volume of spacetime, write "here be dragons" and don't send any clocks there).

But I can't do this close to the neutron star. My clocks tick at different rates and rapidly crash into each other and/or the star. I can't even approximately construct SR over a patch of spacetime including the star, so I have to use the full machinery of GR.

I recommend Sean Carroll's lecture notes and Ben Crowell's book on GR.

Trying to understand as much as I can. So let's go one step at a time.
Let's define volumetric mass density as mass/volume
https://en.m.wikipedia.org/wiki/Density

Would the volumetric mass density of a body such as our sun be higher than that measured in our frame of reference, than one measured in an inertial frame of reference where the sun would be observed as contracted in length?
Another way of asking the same question is, is relativistic length contraction associated with reduced volume our not?

[
Ibix said:
If you want to know what's going on deeper in the gravitational field then you'll have to solve Einstein's field equations. As far as I'm aware there's a known solution when the mass is stationary (the Schwarzschild solution) and one where it's moving at near lightspeed (the Aichelberg-Sexl ultraboost), but only numerical solutions for masses moving at less extreme speeds.

Note that the stationary, spherically symmetric character of a neutron star solution is global, invariant feature, irrespective of whether an observer is moving very fast relative to it (which is obviously the same as the neutron star moving fast relative to some observer). Presence or absence of a horizon is also a global, invariant feature. It is perfectly feasible to come up with an exact coordinate description of a moving 'neutron star model'. By model, I mean some simple representation of the neutron star as a classical fluid. There are numerous published exact solutions of perfect fluid balls with exterior vacuum being the Schwarzschild metric greater than some r. Any of these is easily cast to a cartesian style metric instead of spherical. Then, apply some transform with the general character of a Lorentz boost, computing metric in the new coordinates. Clearly, invariant features cannot change in this procedure. However, these coordinates would not be very physically meaningful near the neutron star (but, of course, all physics computed in them would be fine).

To answer physical questions about a neutron star as measured by a flyby observer near lightspeed, a different procedure would be better and simpler. Use vanilla Schwarzschild coordinates, and just compute fly by geodesic with initial 4-velocity that is near lightlike in these coordinates. Then, in principle, you can compute any observable for such an observer. In particular, you can deduce without computation that this observer can send and receive light signals to the surface of the ball. Thus, trivially, there is no horizon and no BH.

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a1call said:
Trying to understand as much as I can. So let's go one step at a time.
Let's define volumetric mass density as mass/volume
https://en.m.wikipedia.org/wiki/Density

Would the volumetric mass density of a body such as our sun be higher than that measured in our frame of reference, than one measured in an inertial frame of reference where the sun would be observed as contracted in length?
Another way of asking the same question is, is relativistic length contraction associated with reduced volume our not?
The whole framework of this question is not valid. Consider a much more mundane question in pure special relativity. Consider that a gas at some temperature becomes a liquid if compressed. Consider a cannister of such gas moving at near c relative to you. Do you think it is liquid per you while being gas per an observer in the cannister? If that seems nonsensical, it is. To make the silliness apparent: no matter how fast you fly be the cannister, its state is not affected by your choice of motion.

If this question is silly for state change in SR, it is equally silly for BH formation in GR.

As to your specific question, length contraction is associated with volume reduction. But the statement that state change occurs at density D for fluid X, at temperature T is a statement for being at rest relative to the COM of the fluid system. The same law changes form for moving coordinates by a transform law that preserves all invariants.

In GR, the requirement that all laws be expressed in terms of covariant quantities achieves the guarantee that invariants are preserved. And physical measurements are invariants. So here, you intuition that fluid's liquid vs. gas state cannot observer dependent is correct and built into the mathematical structure of SR. Similarly for neutron star vs. BH in GR.

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I have not made any statements regarding matters' phase change from gas to liquid. But thank you for the pointer.
All I was expecting was a yes or no answer.
Thank you for supplying that as well.

Is there any mass reduction associated with relativistic speeds compared to rest mass? Please note I have no interest in the notion of relativistic mass increase. All I am asking is that if there is mass decrease. I am still trying to get an answer to my question regarding density. If there is volume decrease (as stated by PAllen) and no mass decrease, it would have to result in a density increase according to the volume mass density formula linked to earlier in the Wikipedia article.

Is the escape velocity formula linked above a frame dependent formula or not?

a1call said:
Is the escape velocity formula linked above a frame dependent formula or not?
Yes, it is a coordinate dependent formula. Since one is referring to gravity, with high speeds, one needs GR. In GR, frames are strictly local quantities. The equivalent question in GR is whether the formula is coordinate dependent, and it is. Alternatively, you could make it invariant by defining a precise procedure for measuring the variables which would, for example, involve deriving the radius from the circumference measured by an observer on the planet. However, if construed it as invariant in this way, then none of the quantities in it are affected by motion of the planet relative to some observer.

Thank you for the reply. So if I understand it correctly such an invariant formula would yield equal energy requirement for a projectile to just escape the gravity of a given planet in all inertial frames of reference. And this equality in consideration of relativistic effects such as length contraction can be the means to formulate such a formula..
Is that Correct?

a1call said:
Schwarzschild radius formula gives a radius For any given mass below which the escape velocity would be greater than the speed of light
The Schwarzschild formula also assumes that spacetime has spherical symmetry. A moving neutron star would not. The result does not apply

Dale said:
The Schwarzschild formula also assumes that spacetime has spherical symmetry. A moving neutron star would not. The result does not apply
Actually, it would, as long as it is isolated. The symmetry would not be apparent in coordinates in which it is moving, but could be recovered by analyzing the killing vector fields.

Nugatory and Dale
PAllen said:
Actually, it would, as long as it is isolated. The symmetry would not be apparent in coordinates in which it is moving, but could be recovered by analyzing the killing vector fields.
Oops, yes you are right. The killing vector fields are invariant.

Of course, if you are using coordinates that do not have the symmetry then the coordinate dependent results will not apply in those maladapted coordinates. It is only once you go to invariant quantities that you recover the symmetry.

Thank you for all the replies. I am closer than before to being able to resolve the apparent conflict in my understanding. But I am not 100% there Just yet. I am trying to understand how a constant amount of energy/fuel would be required to achieve escape velocity anywhere on a spherical planet in the rest frame of the planet. And to make an observer in a relativistic frame agree that equal energy/fuel is required for the same anywhere on a spheroidal planet as seen by the relativistic observer. I hope that is clear.

The energy would not be the same in all frames. Energy is a frame variant quantity, even in Newtonian physics. Energy is conserved, but not invariant.

Yes I thought about that but whatever the energy/fuel required is and considering that it is different in each of the frames, still it is common sense to be variable on different spots on the Spheroidal planet and constant on a spherical planet, regardless of what frame you are observing from. In other words I don't see how two observers can agree on the constancy(not amount) of fuel required to escape the planet's gravity anywhere on the planet.

ETA spheres are a subset of spheroids. So in this context I am referring to non spherical spheroids seen as a result of length contraction.

ETA II To clarify further, if it takes n Jules to escape (on any spot) from planet X in the rest frame of planet X, in which it is spherical. And it takes m != n Jules to escape (on some spot) from planet X in an inertial frame of reference in which planet X is a non spherical spheroid. Why is there not o != m Jules required on another spot on the non-spherical spheroidal planet where the gravity is different as seen In the latter frame of reference?

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a1call said:
another spot on the non-spherical spheroidal planet where the gravity is different
I am trying to think of a way to make this statement make sense in terms of GR. The problem is that GR is formulated in terms of invariant quantities, and in terms of invariant quantities the gravity is the same at all points on the spheroidal surface.

If it were Newtonian gravity on a spheroidal planet then yes, gravity would be different at different points, but GR doesn't work that way.

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Dale said:
The energy would not be the same in all frames. Energy is a frame variant quantity, even in Newtonian physics. Energy is conserved, but not invariant.
But the mass of the fuel is invariant. Could one argue that this mass is converted into proper acceleration of a rocket such that escape velocity is achieved?

timmdeeg said:
But the mass of the fuel is invariant. Could one argue that this mass is converted into proper acceleration of a rocket such that escape velocity is achieved?
Yes, that is the type of argument one would need to make. Everything in terms of invariants like invariant mass and proper acceleration

timmdeeg said:
But the mass of the fuel is invariant. Could one argue that this mass is converted into proper acceleration of a rocket such that escape velocity is achieved?
Sure, but to do the calculation properly you'll have to account for all the energies involved, including the kinetic and potential energy of the rocket exhaust which does not escape and the rocket which does escape. These energies are all frame dependent, but when you add them all up you'll find that the difference between the pre-launch energy and the post-launch energy is frame-independent and what you'd expect from burning the invariant amount of fuel consumed.

Before you follow this line of thought further, you'll want to be sure that you have a solid understanding of a much simpler classical "paradox": a gun fires a two kilogram projectile with a muzzle velocity of 1000 m/sec. We use the classical ##E=mv^2/2## to calculate the before and after energies. If we're at rest relative to the gun, we calculate that the explosive charge released ##1\times{10}^6## joules (velocity of projectile went from 0 to 1000 m/sec). If the gun is approaching us at 1000 m/sec we calculate that the explosive charge released ##3\times{10}^6## joules (velocity of projectile went from 1000 to 2000 m/sec). But it's the same amount of the same explosive, so the energy released has to be the same.

(It turns out that the first calculation is correct, and the second one is an example of how to manufacture a paradox by choosing coordinate systems that obscure the physics).

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Dale
PAllen said:
Actually, it would, as long as it is isolated. The symmetry would not be apparent in coordinates in which it is moving, but could be recovered by analyzing the killing vector fields.
Thanks for pointing this out to me. The more I think about it the more powerful a concept this becomes.

Nugatory said:
Sure, but to do the calculation properly you'll have to account for all the energies involved, including the kinetic and potential energy of the rocket exhaust which does not escape and the rocket which does escape. These energies are all frame dependent, but when you add them all up you'll find that the difference between the pre-launch energy and the post-launch energy is frame-independent and what you'd expect from burning the invariant amount of fuel consumed.
I had the idealized assumption to neglect the mass of the rocket in my mind but should have mentioned that.

Thank you for all the replies and insight.

## 1. What is a neutron star and how does it become a black hole?

A neutron star is the densest and smallest type of star, formed when a massive star explodes in a supernova. It is made up of tightly packed neutrons and has an extremely strong gravitational pull. When a neutron star reaches a certain mass, it can collapse under its own gravity and become a black hole.

## 2. How can a neutron star exist alongside a black hole?

In the scenario of a neutron star becoming a black hole, both objects exist in the same space but have different properties. The neutron star still retains its original mass and density, while the black hole has a much stronger gravitational pull due to its smaller size and higher mass.

## 3. Can a neutron star turn into a black hole without a supernova?

No, a supernova is necessary for a neutron star to form in the first place. The explosion of a massive star provides the energy and pressure needed for the remaining core to collapse into a neutron star. Without this process, a neutron star cannot form and therefore cannot become a black hole.

## 4. What happens to the matter in a neutron star when it becomes a black hole?

When a neutron star collapses into a black hole, the matter is compressed to an infinitely small point known as a singularity. This is where the gravitational pull is so strong that not even light can escape, making it impossible to observe the matter inside the black hole.

## 5. How do we know that neutron stars can become black holes?

Scientists have observed the process of a neutron star collapsing into a black hole through the detection of gravitational waves. These ripples in space-time are created by the intense gravitational pull of the black hole. Additionally, theoretical models and simulations support the idea that a neutron star can turn into a black hole under certain conditions.

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