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

[pmb_phy]

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]

I'm going to have to disagree with you 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:

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

Here T_ab is the stress-energy tensor, T is the trace of the stress-energy tensor, g_ab is the metric tensor, n_a 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

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

[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

<br /> 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<br />

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 :-(

[add]
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

<br /> M = \int_{Sigma} (T_{00}+T_{11}+T_{22}+T_{33})dV<br />

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

for weak fields (nearly flat space-time).

 

[pmb_phy]

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.

your point being?

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]

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.