# On a modified Schwarzschild metric

• I

## Main Question or Discussion Point

Hello.
I am looking for help in establishing all the consequences of a modified Scwazschild metric where the length contraction is removed.
ds^2=(1-rs/r)c^2dt^2-dr^2-r^2(... )
Thanks

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PeroK
Homework Helper
Gold Member
Hello.
I am looking for help in establishing all the consequences of a modified Scwazschild metric where the length contraction is removed.
ds^2=(1-rs/r)c^2dt^2-dr^2-r^2(... )
Thanks
Why don't you try to work out the geodesics?

But can a black hole arise in a new form from this metric ?

PeroK
Homework Helper
Gold Member
But can a black hole arise in a new form from this metric ?
A spherical mass causes the Schwarzschild geometry, which is the only static, spherically symmetric solution to the Einstein field equations. Your metric, therefore, cannot be a solution and is not physically viable.

... not physically viable ...

Ibix
Here's the mixed-index Einstein tensor in $c=G=1$ units:
$$\pmatrix{0&0&0&0\cr 0&{{2m}\over{r^3-2mr^2}}&0&0\cr 0&0&{{-m(r-m)}\over{r^2 (r-2m)^2}}&0\cr 0&0&0&{{-m (r-m)}\over{r^2 (r-2m)^2}}\cr }$$
I'm not sure that's physically meaningful.
A spherical mass causes the Schwarzschild geometry, which is the only static, spherically symmetric solution to the Einstein field equations.
It's the only such vacuum solution. The metric above can be produced by some stress-energy distribution (divide each element of the Einstein tensor by $8\pi$), but whether or not that stress-energy tensor represents anything physically possible is another matter...

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• PeterDonis and PeroK
Ibix
And the Ricci scalar is $${{2m^2}\over{r^4-4mr^3+4m^2r^2}}$$Which implies true singularities at r=0 and r=2m, so two disconnected patches of spacetime. An odd place indeed.

Edit: in fact, it's not a spacetime when 0<r<2m since the signature is ----.

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haushofer
Hello.
I am looking for help in establishing all the consequences of a modified Scwazschild metric where the length contraction is removed.
ds^2=(1-rs/r)c^2dt^2-dr^2-r^2(... )
Thanks
Isn't this effectively the Schwarzschild geometry which particles experience in the non-relativistic limit? For low speeds, in the geodesic equation the spatial speed of the particle couples to the spatial curvature, and this coupling is in the non-relativistic limit neglected.

With that, it is clear that this will not be an exact solution to the Einstein equations, but merely an approximate solution under certain conditions.

Ibix
Isn't this effectively the Schwarzschild geometry which particles experience in the non-relativistic limit?
Ah! So that's what the singularity at r=2m means - non-relativistic approximations cannot apply here. Edit: and, of course, this metric is Riemannian, not pseudo-Riemannian inside this radius, which means that this approximation doesn't work there either.
For low speeds, in the geodesic equation the spatial speed of the particle couples to the spatial curvature
I think I need to write out the geodesic equations to make sure I followed this.

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As informative as it is, is there any particular reason this thread is marked Basic? It feels to me like this should be I at a minimum

Nugatory
Mentor
As informative as it is, is there any particular reason this thread is marked Basic? It feels to me like this should be I at a minimum
Fixed

haushofer
I think I need to write out the geodesic equations to make sure I followed this.
Well, to be clear, I was talking about the Newtonian limit here. So the metric is time-independent and can be written as diagonal in a suitable coordinate system. Then

$$\Gamma^{i}_{0k} = 0$$

(one can always put this term to zero by using a time dependent rotation, as follows from its transformation law), and

$$\Gamma^{i}_{jk} = \frac{1}{2}g^{im} [\partial_{j}g_{mk} + \partial_{k}g_{mj} - \partial_{m} g_{jk}]$$

We have made a foliation such that we can regard $g_{ij}$ as the metric on spatial hypersurfaces.

This last connection coefficient couples to the spatial velocities. If the spatial curvature perturbations are regarded as "order epsilon" and the spatial velocities also, then this whole term disappears (when you expand around Minkowski; for (a)dS it's a different story of course). Effectively the particle thus only experiences $$\Gamma^{i}_{00}$$ (it's the only surviving term in the geodesic equation).

This is a subtlety: it doesn't mean that the spatial curvature is zero in the Newtonian limit; it only says that its coupling to the particle's velocity is a higher order epsilon term in your expansion and hence is neglected. Of course, in full fledged Newtonian gravity the spatial curvature is zero. That corresponds to your metric.

PAllen
2019 Award
Here's the mixed-index Einstein tensor in $c=G=1$ units:
$$\pmatrix{0&0&0&0\cr 0&{{2m}\over{r^3-2mr^2}}&0&0\cr 0&0&{{-m(r-m)}\over{r^2 (r-2m)^2}}&0\cr 0&0&0&{{-m (r-m)}\over{r^2 (r-2m)^2}}\cr }$$
I'm not sure that's physically meaningful.
It's the only such vacuum solution. The metric above can be produced by some stress-energy distribution (divide each element of the Einstein tensor by $8\pi$), but whether or not that stress-energy tensor represents anything physically possible is another matter...
This is interesting. It says that there is pressure but no energy density throughout the spacetime. It trivially violates the dominant energy condition, so it is a universe of all exotic ‘matter’, scare quotes because ther is no mass/energy density given this Einstein tensor, so exotic matter sounds funny. Exotic essence?

pervect
Staff Emeritus
Yes, except perhaps for sign the components of $G^{\mu}{}_{\nu}$ in a coordinate basis are the same as the components of $G_{\hat{\mu}\hat{\nu}}$ in an orthonormal basis.
The magnitude of the presssures approach infinity close to the event horizo at r=2m, far away though, for large r, the pressures can be small, on the order of $m/r^2$.