Negative Schwarzschild mass parameter?

[mersecske]

I know that a Schwarzschild metric cannot have negative mass parameter.
But what about regions?
It is possible to have negative mass parameter between two spherical shells?
If not, what is the argument?

 

[zhermes]

To my knowledge there is nothing expressly forbidding negative energy density---but we don't know of anything or any situation (at least non-quantum mechanical) for which this is the case. Most likely the answer to your question will remain unknown until there is a quantum theory of gravity.

 

[bcrowell]

I know that a Schwarzschild metric cannot have negative mass parameter.
But what about regions?
It is possible to have negative mass parameter between two spherical shells?
If not, what is the argument?

I don't think you should worry about making the distinction between a negative mass parameter in the S. metric and a negative mass in some region. The S. metric doesn't have to represent a black hole. The Earth's external gravitational field is described by a S. metric. A S. metric with a negative mass parameter would be a valid description of the field external to a spherically symmetric region of negative energy.

This kind of thing goes by the name of "energy conditions." There are various different energy conditions with a variety of different strengths: http://www.lightandmatter.com/html_books/genrel/ch08/ch08.html#Section8.1 (subsection 8.1.3). The negative-m S. metric you're describing would violate all of them, even the weakest. We know that some energy conditions *are* violated in our universe. E.g., we know that we have a nonvanishing cosmological constant, and it violates the strong energy condition. Neutron stars violate some of the energy conditions. On the other hand, most energy conditions appear to work well in most situations.

If the weaker energy conditions get violated too egregiously, it has distasteful consequences. For example, you lose the ability to interpret geodesics as world-lines of test particles. (Negative mass falls up.)

 

[mersecske]

A am not talking about negative energy density!
I am talking about negative Schwarzschild mass parameter,
it is just a parameter (parameter of the metric)!
In thin shell formalism it is possible to construct
spherical shell around positive central mass (mc>0)
with positive surface energy density (sigma>0)
but the shell itself has negative gravitational mass
(mg=M-mc, where M is the total mass),
moreover the negative mass is so big,
that the outer Schwarzschild mass parameter:
mc+mg becomes negative.

 

[bcrowell]

A am not talking about negative energy density!
I am talking about negative Schwarzschild mass parameter,
it is just a parameter (parameter of the metric)!

It's a parameter that you really can't avoid interpreting as the mass-energy. Mass-energy isn't normally a well defined scalar in GR, because Gauss's theorem fails in curved spacetime when applied to a flux that is a vector. But mass-energy is a well defined scalar in certain special cases. This is one of those special cases, because the S. metric is stationary. For example, the external field of the Earth is well approximated by the S. metric, and the m appearing in the equations it the mass of the earth. The same applies to a black hole. It doesn't matter if some of the mass is at a singularity at r=0 and some of the mass is in spherical shells.

 

[mersecske]

but mass energy of what?
We have only the shell, with positive energy density, and S. vacuo!

 

[bcrowell]

but mass energy of what?
We have only the shell, with positive energy density, and S. vacuo!

In the exterior region, you have a metric of the S. form. It's parametrized by m, which equals the total mass in the interior region.

 

[mersecske]

so what?

mass parameter of the Schwarzschild is just a parameter...

and we have lots of mass definition in gen.rel.

 

[Passionflower]

so what?

mass parameter of the Schwarzschild is just a parameter...

bcrowell tries to explain to you why in the Schwarzschild solution it is not 'just a parameter', honestly I do not understand what you are arguing against.

 

[mersecske]

What is the statement? Negative mass parameter is possible between two shells?

 

[pervect]

As far as I know, you need to assume that there isn't any such thing as exotic matter to rule out negative masses.

You said that you "knew" there was no such thing, it's not clear how you "knew" this.

It's not entirely clear what you mean by "negative mass parameter". If I were to guess, by "mass parameter" you mean the integral of 4*pi*r^2 * rho(r) * dr, where rho = T_00 which is assumed to be spherically symmetrical and thus only a function of r.

So I would say that you could have negative mass parameter in a shell, and also negative masses as a whole, if and only if you assume you have the right sort of exotic matter (one with a negative energy density).

This assumes I've correctly figured out what you mean by "mass parameter", which is a bit vague.

 

[mersecske]

mass parameter = Schwarzschild mass parameter,
which is M, where f(r)=1-2M/r is the metric function.
This is just a parameter which can be negative.
The only non-zero energy-momentum is concentrated on the shell
in he above spacetime, and the the shell has positive surface energy density (sigma).
The rest mass of the shell is m=4*pi*r^2*sigma > 0.
This latter one cannot be negative with a suitable energy condition.
But energy condition doe not say anything about the mass parameter.

 

[Dickfore]

If the whole Universe is of uniform positive mass density and there is a spherical vacuum hole, wouldn't this behave as a negative mass?

 

[mersecske]

Not. In the hole there is Minkowski spacetime (M=0).
But this is not the subject.

 

[Dickfore]

What about in the uniform distribution of matter?

 

[mersecske]

I do not understand. Outside the hole the Schwarzschild mass parameter is positive, if the density is positive. If there is no distributional surface term.

 

[Dickfore]

The mass determines the curvature (Ricci) tensor, and not the metric tensor. To find the metric tensor, you still need boundary conditions.