The Schwarzschild Radius: Mass vs. Energy

[dsaun777]

When calculating the Schwarzschild radius are we supposed to be using the rest mass of the object or its total energy?

 

[Nugatory]

They're the same thing, because we're assuming a spherically symmetrical and static distribution of mass/energy.

 

[PeterDonis]

When calculating the Schwarzschild radius are we supposed to be using the rest mass of the object or its total energy?

To the extent that the concept of "Schwarzschild radius" makes sense for an object that is not a black hole, you would use the object's rest mass.

 

[PeterDonis]

They're the same thing, because we're assuming a spherically symmetrical and static distribution of mass/energy.

This doesn't make total energy the same as rest mass without qualification, because total energy is frame dependent but rest mass is not. In the object's rest frame, they are the same, but not in other frames.

 

[PAllen]

When calculating the Schwarzschild radius are we supposed to be using the rest mass of the object or its total energy?

Roughly speaking, you use its rest mass = invariant mass. This includes all forms of energy, as measured in center of momentum frame. This is a simplification, since invariant mass is not really well defined in GR. But the gist is that moving rapidly past an object doesn’t somehow change its Schwarzschild radius.

 

[Nugatory]

In the object's rest frame, they are the same, but not in other frames.

This is true... but are we not effectively committing to using that frame when we assume a static and spherically symmetric distribution?

The event horizon is an invariant property of the spacetime, but calculating $2Gm/c^2$ doesn't tell us much about which region of spacetime it is unless we're using the object's rest frame.

 

[dsaun777]

Roughly speaking, you use its rest mass = invariant mass. This includes all forms of energy, as measured it center of momentum frame. This is a simplification, since invariant mass is not really well defined in GR. But the gist is that moving rapidly past an object doesn’t somehow change its Schwarzschild radius.

In terms of the stress tensor what is rest mass?

 

[PAllen]

In terms of the stress tensor what is rest mass?

Well, if you leave the realm in which SR concepts are a good approximation, it gets complicated. The stress energy tensor is a tensor field with values at points. You cannot talk about its value for a body of some size. If you can assume the body is stable, not currently collapsing, you would integrate the stress energy tensor using the Komar mass formula to get the Schwarzschild radius that would result if it collapsed in its entirety without radiating.

 

[PeterDonis]

are we not effectively committing to using that frame when we assume a static and spherically symmetric distribution?

No, of course not. You can describe such a distribution in any frame you like. It will look more complicated in a frame that isn't the rest frame, but "looks more complicated" isn't the same as "isn't valid".

 

[dsaun777]

Well, if you leave the realm in which SR concepts are a good approximation, it gets complicated. The stress energy tensor is a tensor field with values at points. You cannot talk about its value for a body of some size. If you can assume the body is stable, not currently collapsing, you would integrate the stress energy tensor using the Komar mass formula to get the Schwarzschild radius that would result if it collapsed in its entirety without radiating.

If you are calculating the rest mass for a stress tensor of a perfect fluid would you integrate over the volume?

 

[PAllen]

If you are calculating the rest mass for a stress tensor of a perfect fluid would you integrate over the volume?

Yes, but you would have to use the Komar integral. The stress energy tensor values reflect local measurements, but different parts of a body are at different gravitational potentials relative to each other. The Komar integral accounts for this. It also accounts for how to take pressure as well as energy density into account.

 

[pervect]

When calculating the Schwarzschild radius are we supposed to be using the rest mass of the object or its total energy?

The Schwarzschild mass parameter M in the Schwarzschild metric would pretty much be one of severral possible GR replacement for the Newtonian idea of "rest mass", for a static body.

A few posters have mentioned some other, closely related and widely used defintions of mass in GR, such as the Komar mass. But the details may not be of interest, other than the fact that there are more than one.

Typically, one calculates the mass of an astronomical body by its gravitational effects, rather than knowing it's internal structure. So people don't actually try to measure the volume and density of other planets, for instance, they make various orbital observations and infer the mass from studying the orbits.

So I'm a bit uncertan how you'd know "the rest mass" in the first place, as you're talking about a basically Newtonian quantity (rest mass) and trying to compare it to a GR quantity (the Schwarzschid mass parameter).

If you were given a basically Newtonian problem, and wanted the Schwarzschild radius from the Newtonian problem to estimate the importance or lack thereof of GR effects, you'd generally want to use the rest mass of the body in conjunction with a coordinate system in which the body was also at rest.

We could give a more detailed and better answer if we had a better idea of the context of your question.

 

[pervect]

When calculating the Schwarzschild radius are we supposed to be using the rest mass of the object or its total energy?

The Schwarzschild mass parameter M in the Schwarzschild metric would pretty much be one of serveral possible GR replacement for the Newtonian idea of "rest mass", for a static body.

Typically, one calculates the mass of an astronomical body by its gravitational effects, rather than knowing it's internal structure.

So I'm a bit uncertan how you'd know "the rest mass" in the first place, as you're talking about a basically Newtonian quantity (rest mass) and trying to compare it to a GR quantity (the Schwarzschid mass parameter).

If you were given a basically Newtonian problem, and wanted the Schwarzschild radius to estimate the importance or lack thereof of GTR effects, you'd generally want to use the rest mass of the body and a coordinate system in which the body was at rest.

We could give a more detailed and better answer if we had a better idea of the context of your quesiton.

 

[PAllen]

Just a note that there were specific reasons I brought up the Komar mass integral, and not other notions of mass in GR. The thread started in elementary terms where it made sense to answer per SR notions, as all the first several posts did. Then, the question was raised of how to relate the stress energy tensor to the Schwarzschild radius. For this, there is only one answer, so far as I know - the Komar mass integral, which applies as long as the body is stable, as I put it - technically, as long as the manifold has a timelike KVF. In the special case of spherical symmetry under these conditions, the Komar mass, the Schwarzschild mass parameter, the ADM mass, and the Bondi mass are all equal. But only the Komar mass is evaluated from the stress energy tensor. All the others are evaluated from vacuum region outside the body. Thus, I considered that only the Komar mass was pertinent to the question asked.

 

[dsaun777]

Yes, but you would have to use the Komar integral. The stress energy tensor values reflect local measurements, but different parts of a body are at different gravitational potentials relative to each other. The Komar integral accounts for this. It also accounts for how to take pressure as well as energy density into account.

The Komar integral will get the total "rest" energy of a perfect fluid in a stationary symmetric spacetime by integrating over pressure and energy density? The integral is essentially over the trace of the stress tensor?

 

[pervect]

The Komar integral will get the total "rest" energy of a perfect fluid in a stationary symmetric spacetime by integrating over pressure and energy density? The integral is essentially over the trace of the stress tensor?

It's bit more complicated than taking the trace of the stress energy tensor.

From Wald, "General relativity", pg 289, as a surface integral it would be

$$M = -\frac{1}{8 \pi} \int_S \epsilon_{abcd} \nabla^c \xi^d$$

S is a 2 sphere bounding the region whose mass is to be computed, $\xi$ is the time-like Killing vector associated with the stationary (or more generally, static), space-time for which the mass is to be computed, and $\epsilon$ is the Levi-Civita tensor.

The above expression is a two-form, the surface integral can be regarded as the "natural" integral of a two-form over a surface, if you like.

But you probably want the volume integral form, which is

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

where the 2-sphere S above is the boundary of the spacelike (3 dimensional) hypersurface $\Sigma$, so we can think of the 2-sphere S being the surface of the 3-volume $\Sigma$. We introduce $n^a$, the "unit future" orthogonal to $\Sigma$. In the Schwarzschild metric, $n^a$ would be parallel to $\xi^a$, but of unit length.

There's probably a simpler way of putting it that's also correct, but this is straight from the textbook. Though I do hope I haven't confused the Levi-civita tensor with the Levi-civita symbol, or made any typos.

Note that the trace $T^a{}_a$ is equal to the trace of $T^{ab} g_{ab}$. So if we take an orthonormal basis for physical insight, rather than computational convenience, for a perfecc fluid we'd have $T^{ab}$ = diag($\rho, P, P, P)$. With a -+++ sign convention, the trace is $\rho - 3P$, because g = diag(-1,1,1,1).

If we make $\xi^a$ have components k,0,0,0 and the unit future have components (1,0,0,0), we only need to consider the first term.

$$\int_\Sigma k(2\,T_{00} - T)$$

So we're basically taking the volume integral of k ( 2 $\rho$ - ($\rho$ - 3P) , i.e the volume integral of k($\rho$+3P). Note the sign difference of what we're integrating - it's not the trace, because of the sign of P. Also, we have the weighting factor k.

Computationally, it's ugly to use an orththonormal basis, because it's hard to determine where the boundaries are. But it's hard to get physical insight from the coordinate basis. So, imagine chopping the volume up into a bunch of small pieces (say, cubes), then using a local orthonormal basis for each cube, rather than a coordinate basis. Each cube contributes k ($\rho$ + 3P) V to the integral, where $\rho$ is the local density in the local coordinates of the cube, P is similarly the local pressure, V is the local 3-volume element, and k is the length of the Killing vector, which is often called the "redshift" factor.