Objects turning into a black hole near the speed of light?

JesseM
pmb_phy said:
Keep in mind that although a single moving body will never become a black hole in a moving frame if its not one it its rest frame it does not follow, and is incorrect to say, that the gravitational field of the body does not increase. The fact is that it does increase.
Not as seen by an observer at rest relative to the center of mass of the object though, which is why it is no more likely to form a black hole--that's what I thought the question was about. If eNathan was asking about the gravity experienced by an observer who sees the object's center of mass moving at velocity v relative to himself, then I assume you're correct that he will experience a greater field the larger v is.

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JesseM
JesseM said:
OK, so I assume this is true even if you have a box containing only a single particle? If so, what is the crucial difference between increasing the velocity of a particle bouncing around in a box and increasing the velocity of a particle flying in a straight line through empty space? Is it just because the particle in the box has to accelerate rapidly to change the direction of its velocity vector each time it hits a wall? Is a black hole likely to form only at the moments and locations that the particle hits the wall?
pmb_phy said:
The mass-energy will never remain within a small enough region to form a black hole and that is crucial.
Wait, are you saying that a single particle bouncing around in a box will never form a black hole no matter how great its speed, but multiple particles bouncing around in a box can form a black hole if their average speed is increased high enough? That doesn't make much sense to me--what would the threshold be? Anyway, if a single particle is moving fast enough, why wouldn't the energy of the collision with the wall be high enough to form a black hole?

pervect said:
One fine point - one might wonder why the increase in the mass of the box is E/c^2 since pressure causes gravity, and the box is under pressure. The answer to this riddle is that (assuming weak fields and small boxes) the intergal of the pressure of the gas over the volume of the box is equal to the intergal of the tension in the container over the volume of the container, so the two terms cancel out. The easiest case to analyze is a spherical container (there's still a few tricks to it, though, I wound up asking some mech E types for how to calculate the stress).
I'm going to have to disagree with ya here pervect. While this is true for a box it is not true for a star, which has no containing wall, for which there is non-zero pressure inside.

I mentioned a while back, and many times since, that "mass" is the ratio of momentum over velocity. Use that definition to find the mass of something like a rod which is under stress and moving along the length of the rod. The mass of the rod will then be a function of $\gamma m_0$ as well as the stress (i.e. pressure) in the rod. (see http://www.geocities.com/physics_world/sr/inertial_energy_vs_mass.htm). Notice that $\gamma m_0$ is not the mass in this case since the rod is not isolated from that which causes the stress.

Now apply this kind of thinking for a gas with no walls. It appears to me (a hunch/guess) that to get the result I just add one p/c term for each spatial dimension. This will reduce the equation to

$$\rho_{active} = \rho + 3p/c$$

If you take a look at Einstein's field equations in the week field limit then you'll see that this rho on the left, i.e. what I called "mass" above" will be a function of the pressure in the star.

Pete

JesseM said:
Wait, are you saying that a single particle bouncing around in a box will never form a black hole no matter how great its speed, but multiple particles bouncing around in a box can form a black hole if their average speed is increased high enough? That doesn't make much sense to me--what would the threshold be? Anyway, if a single particle is moving fast enough, why wouldn't the energy of the collision with the wall be high enough to form a black hole?
I never used the term "particle" since that would be troublesome. A particle has zero radius and is therefore a blackhole.

No matter how fast that object moves there will always be a frame in which that single object is at rest and does not fit in the given small region. For two particles not moving in the same direction this arguement does not apply though so that's why two objects can clash to form a black hole but a single one can't.

Pete

JesseM
pmb_phy said:
I never used the term "particle" since that would be troublesome. A particle has zero radius and is therefore a blackhole.

No matter how fast that object moves there will always be a frame in which that single object is at rest and does not fit in the given small region. For two particles not moving in the same direction this arguement does not apply though so that's why two objects can clash to form a black hole but a single one can't.
Well, doesn't the wall of the container count as an "object"?

JesseM said:
Well, doesn't the wall of the container count as an "object"?
Yes. And if you want to take that into account then you don't have a single particle system.

Pete

pervect
Staff Emeritus
pmb_phy said:
I'm going to have to disagree with ya here pervect. While this is true for a box it is not true for a star, which has no containing wall, for which there is non-zero pressure inside.
I agree that this formula is not true for a star, or any system that has significant _gravitationally induced_ pressures.

That's why I made a point of saying:

(assuming weak fields and small boxes)
When the box is small, though, the tension in the box walls contributes a negative term to the mass of the box which exactly cancels out the positive terms due to the pressure. In short, the pressure terms in the stress-energy tensor T11, T22, and T33 do not contribute to the mass of the box - the intergal over the entire volume is zero. This is why the mass of the box is E/c^2.

I suppose I should have added that specifically the assumption that the box is an isolated system to avoid your "stressed rod" example, but that was inherent in the original problem definition.

A rod under external stress is not a closed system. E/c^2 works only for a closed system
Right. But it does work for a closed system.

Now apply this kind of thinking for a gas with no walls. It appears to me (a hunch/guess) that to get the result I just add one p/c term for each spatial dimension. This will reduce the equation to

$$\rho_{active} = \rho + 3p/c$$

If you take a look at Einstein's field equations in the week field limit then you'll see that this rho on the left, i.e. what I called "mass" above" will be a function of the pressure in the star.

Pete
The answer that I have confidence in for finding the mass of a large isolated, static box (static star) with significant gravitational self-energy via a volume intergal I've posted earlier, and is from Wald:

$$M = 2 \int_{\Sigma} (T_{ab} - \frac{1}{2} T g_{ab}) n^a \xi^b dV$$

Here Tab is the stress-energy tensor, T is the trace of the stress-energy tensor, gab is the metric tensor, na is the unit future normal to the volume element, and $$\mbox{\xi^b}$$ is the Killing vector representing the time translation symmetry of the static system, sutiably normalized so that $$|\xi^a \xi_a|$$ equals unity at infinity.

Now if we assume that $$n^a$$ and $$\xi^b$$ both point in the time direction, which is denoted by the superscript or subscript zero, we can simplify this result a lot -- i.e. we assume that the only non-zero comoonents of n and $$\xi$$ are $$n^0$$ and $$\xi^0$$, and we know that the former has the value of 1. Then we write

$$M = 2\int_{\Sigma} (T_{00} - \frac{1}{2}T g_{00} )\xi^0 dV$$

[correction]
T is just $$T^0_0 + T^1_1+T^2_2 + T^3_3 = g^{00} T_{00}+ g^{11} T_{11} + g^{22} T_{22} + g^{33} T_{33}$$

and we wind up with

$$M = \int_{\Sigma} ((2-g_{00}g^{00}) T_{00} - (g^{11} T_{11}+g^{22} T_{22}+g^{33} T_{33})g_{00}) \xi^0 dV$$

Unfortunately, last I checked, $$\xi$$ which is a nice unit vector outside the star isn't necessarily unity inside the star :-(, so you have to solve Killings equation to find $$xi^0$$ :-(

I suppose I should add that when g_00 = 1 and g_11 = g_22 = g_33 =-1 (flat spacetime with a +--- metric), which has the timelike Killing vector $$\xi^0 = 1$$ the above formula reduces to

$$M = \int_{Sigma} (T_{00}+T_{11}+T_{22}+T_{33})dV$$

which is the same result given in Carlip's paper
http://arxiv.org/abs/gr-qc/9909014

for weak fields (nearly flat space-time).

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pervect said:
When the box is small, though, the tension in the box walls contributes a negative term to the mass of the box which exactly cancels out the positive terms due to the pressure.
I think you missed my point. You're referring to the entire gas-box system whereas I am referring only to the gas. Also it makes no difference as to the size of the box as to whether F/A is a tension or a pressure term.
In short, the pressure terms in the stress-energy tensor T11, T22, and T33 do not contribute to the mass of the box - the intergal over the entire volume is zero. This is why the mass of the box is E/c^2.
The mass of the box is not E/c^2. It is the total mass of the mass-box system that is E/c^2.
I suppose I should have added that specifically the assumption that the box is an isolated system to avoid your "stressed rod" example, but that was inherent in the original problem definition.
Recall the statement that I was responding to "one might wonder why the increase in the mass of the box is E/c^2 since pressure causes gravity, and the box is under pressure."
Right. But it does work for a closed system.
The answer that I have confidence in for finding the mass of a large isolated, static box (static star) with significant gravitational self-energy via a volume intergal I've posted earlier, and is from Wald...
The purpose of my weak field example was to help the layman since it is easier to follow. It more readily shows the relationship of the inertial mass of the gas to the active gravitational mass of the gas. If you have Schutz's new book then look up his derivation as to why the inertial mass of a gas (not box-gas, just the gas) is a function of the pressure of the gas.

(I've been through the rest in the past too manyh times to bother re-addressing it again here - plus I'm getting lazy )

Pete

pervect
Staff Emeritus
pmb_phy said:
I think you missed my point. You're referring to the entire gas-box system whereas I am referring only to the gas.
If you measure the "mass of the gas" by measuring the mass of the box empty, and the mass of the box full (which is both a logical means of measuring the mass of the gas AND the means specified by the original poster), you must take account the contribution of the stresses in the box.

The box was unstressed in the first measurement, and stressed in the second. This makes a difference to the calculation.

When you do this procedure, you get E/c^2 for the mass of the gas, at least for small boxes, which leads us to the second point.

Also it makes no difference as to the size of the box as to whether F/A is a tension or a pressure term.
I'm not even sure what you are trying to say here. (After I can figure out what you are saying, I can start to think about whether or not I agree with it and what your basis for saying it is).

My second point is this - gravity is nonlinear, and with a large enough box this is an important effect. This means that the mass of a gravitationally bound system is not equal to the sum of the masses of its parts. It's also, unfortunately, a rather advanced topic.

With a small enough box, the non-linearity is unimportant. Since this is by far the easiest case to analyze, and interesting in its own right, I think it's the one to address. This is the case that I calculated.

I'd also like to point out that even in our sun, the contribution of the pressure terms to the mass is negligible, so one can have a pretty darn big box without the non-linear effects of gravitation becoming imporant.

Clearly, the approximations I made are good when the box is small enough that the pressure at the center of the box is not significantly different than the pressure at the edge of the box because of the box's self-gravity.

In practice, the approximations are even better than that, because the pressure terms are extremely small anyway, thus a small error in a small term becomes an extremely small error in the final result. The magnitude of the term in standard units will be 3*pressure*volume/c^2, and c^2 is a very big number.