What would GRT look like if negative masses existed?

[WannabeNewton]

I believe it is ADM energy that includes radiation and Bondi energy that excludes it, so the Bondi energy of a binary system decreases while the total ADM energy remains constant.

Yes indeed you're right, sorry!

 

[stevendaryl]

By Birkhoff, it would be impossible for spherical, non-rotating, body to produce such a metric, since the solution is unique (for cosmological constant zero). With a cosmological constant, you could get repulsive effects, but they would not appear to originate with the central body.

If you just take the Schwarzschild metric, which has a parameter $M$, and you replace that parameter by $-M$, that will still be a solution to Einstein's field equations, right? But on the other hand, the resulting metric would be equivalent to the Schwarzschild metric under the coordinate transformation $r \rightarrow -r$. So maybe the sign of $M$ doesn't make any difference?

 

[PAllen]

If you just take the Schwarzschild metric, which has a parameter $M$, and you replace that parameter by $-M$, that will still be a solution to Einstein's field equations, right? But on the other hand, the resulting metric would be equivalent to the Schwarzschild metric under the coordinate transformation $r \rightarrow -r$. So maybe the sign of $M$ doesn't make any difference?

The sign of M does make a difference. With negative M, there is no horizon, but there is still a singularity (in asymptotically flat spacetime). In asymptotically flat spacetime, a negative M SC geometry can only result from collapse of exotic matter (or be eternal).

 

[PeterDonis]

If you just take the Schwarzschild metric, which has a parameter $M$, and you replace that parameter by $-M$, that will still be a solution to Einstein's field equations, right?

Yes, but it's not the same solution as the one with $M$.

the resulting metric would be equivalent to the Schwarzschild metric under the coordinate transformation ##r \rightarrow -r##.

Yes, because you have now done a second transformation which happens to undo the first one.

 

[WannabeNewton]

If you just take the Schwarzschild metric, which has a parameter $M$, and you replace that parameter by $-M$, that will still be a solution to Einstein's field equations, right?

As Peter explained these are not the same solution up to gauge. They are physically distinguishable solutions. Furthermore the change in sign has a very clear physical effect that is easy to compute. If we consider two initially radially separated test particles with infinitesimal connecting vector $\xi^{\mu}$ then from the geodesic deviation equation we have initially $\ddot{\xi^r} = \frac{M}{r^3}\xi^r$ so $M \rightarrow -M$ will change the divergence ($\nabla_{\mu}\xi^{\mu} > 0$) of $\xi^r$ due to the attractive non-uniform gravitational field into a convergence ($\nabla_{\mu}\xi^{\mu} < 0$) coming from a non-uniform repulsion.

 

[stevendaryl]

Yes, but it's not the same solution as the one with $M$.
Yes, because you have now done a second transformation which happens to undo the first one.

Well, the second transformation is just a coordinate transformation, while the first seems physically meaningful. What I don't remember (or never knew in the first place) is whether the Schwarzschild solution can be analytically continued to the region r < 0 to get to the white hole spacetime on the "other side" of the singularity. I've only seen the extended solution done in Kruskal-Szekeres coordinates.

 

[PeterDonis]

the second transformation is just a coordinate transformation, while the first seems physically meaningful.

The physically meaningful parameter is really $M / r$, so changing the sign of either $M$ or $r$ is physically meaningful.

whether the Schwarzschild solution can be analytically continued to the region r < 0 to get to the white hole spacetime on the "other side" of the singularity.

There is no "other side of the singularity". There are two singularities, the future one (future boundary of the black hole region) and the past one (past boundary of the white hole). Both singularities have $r = 0$, and there is no continuation beyond them (because curvature invariants go to infinity).

 

[PAllen]

I propose a slightly different way to look at this. The normal SC charts have M>0, and (1) r > 2M, and (2) r ∈ (0,2M). If you do R= -r as a pure coordinate transform, nothing changes about the physics because the relevant charts now have M>0 with (1) R < -2M and (2) R ∈ (-2M,0). Further, the event (t,R,θ,φ)=(t,-n,θ,φ) is taken to be the same as (t,r,θ,φ)= (t,n,θ,φ).

However, taking the SC metric as given, with M > 0, and asking about a new 'region' charted with r < 0 representing a different event than r >0, produces a physically different region (I put region in quotes because it is not possible to extend geodesics across r=0; this is not an 'extension' of the normal SC geometry). This region is physically the same as with M < 0, and r > 0. Given the semantics of r for spherical symmetry, it is simply more sensible to describe this geometry in terms of M<0, r >0. This geometry is not the same the white hole region of the KS analytic extension of an M>0 SC chart. It is simply a physically different geometry.

 

[stevendaryl]

The physically meaningful parameter is really $M / r$, so changing the sign of either $M$ or $r$ is physically meaningful.
There is no "other side of the singularity". There are two singularities, the future one (future boundary of the black hole region) and the past one (past boundary of the white hole). Both singularities have $r = 0$, and there is no continuation beyond them (because curvature invariants go to infinity).

With a little Googling, I found a paper considering the analytic continuation of the Kerr metric (rotating black hole) to negative values of r:

http://leverett.harvard.edu/w/media/1/14/Reyes-blackholes.pdf

 

[PeterDonis]

I found a paper considering the analytic continuation of the Kerr metric (rotating black hole) to negative values of r

Yes, but this doesn't apply to the Schwarzschild metric. The presence of rotation (making the singularity at $r = 0$ a ring rather than a point) makes a difference.