# Plane Schwarzschild geometry

1. Oct 22, 2006

### lalbatros

Dear all,

I would like to know about a possible plane Schwarzschild geometry.
This would be the analog of a uniform and infinite gravitational field, or a uniform acceleration.
I would like to compare it with the solution of the uniformly accelerated motion in SR.

Thanks,

2. Oct 23, 2006

### pervect

Staff Emeritus
What you are looking for is probably the Rinlder metric. This is the metric associated with a uniformly accelerating observer. Note that this metric has an event horizon, quite similar to the event horizon of a black hole.

There may be some better web references out there. For a textook reference there is always Misner, Thorne, Wheeler "Gravitation".

I assume you know the Schwarzschild metric, one of the usual form of the Rindler metric is

ds^2 = (1 + gz/c^2)^2 c^2 dt^2 - dx^2 - dy^2 - dz^2

You can see that the event horizon occurs at gz/c^2 = -1 by analogy with the Schwarzschild metric, where the event horizon also occurs when the coefficient of dt^2 goes to zero.

3. Nov 19, 2006

### Chris Hillman

A plane symmetric static vacuum solution?

Hi again, Michel,

This is suprisingly tricky!

The quick answer is that the so-called Levi-Civita A3 vacuum (c. 1917), aka the Taub plane symmetric vacuum, aka the plane symmetric case of the Kasner vacuum (1921), aka the Taub plane symmetric vacuum (1953), is a static vacuum solution whose hyperslices (orthogonal to static observers) have planar symmetry. It often turns up in various coordinate charts, such as the Levi-Civita chart
$ds^2 = -\frac{dt^2}{z} + \left( dx^2 + dy^2 \right) + z \, dz^2, -\infty < t, x, y < \infty, 0 < z < \infty$
the isotropic chart
$ds^2 = -dT^2/dZ^2 + Z^4 \; \left( dX^2 + dY^2 + dZ^2 \right)$
the Kasner chart
$ds^2 = -\zeta^{-2/3} \, d\tau^2 + \zeta^{4/3} \, (dx^2 + dy^2) + d\zeta^2$
and the Weyl canonical chart (I'm running out of letters)
$ds^2 = -\frac{dt^2}{z+\sqrt{z^2+r^2}} + \frac{ 2 z^2+r^2 + 2 z\, \sqrt{z^2+r^2}}{2 \sqrt{z^2 +r^2} \; \left(dz^2+dr^2) + (z + \sqrt{z^2+r^2} ) \, r^2 \, d\phi^2$

A longer and possibly more interesting answer might begin by noting that the Weyl family of all static axisymmetric vacuum solutions shows that they all arise from axisymmetric harmonic functions:
$ds^2 = -\exp(2 u) \, dt^2 + \exp(-2u) \left( \exp(2 v) ( dz^2+dr^2) + r^2 \, d\phi^2,$
$u_{zz} + u_{rr} + \frac{u_r}{r} = 0$
$v_z = 2 \, u_z \, u_r, \; \; v_r = r ( u_z^2 - u_r^2)$
Here, the two first order equations for v(z,r) in terms of u(z,r) are consistent exactly in case the second order equation for u(z,r) alone is satisfied. This happens to be the Laplace equation (for axisymmetric functions $$u(z,r,\phi)$$.) So given any axisymmetric harmonic equation u, we can compute v by quadratures (integrals) and obtain the metric of the corresponding Weyl vacuum solution. But of course the Laplace equation is Newton's vacuum field equation, and expanding the line element to first order in u,v you can see that the in the weak-field approximation, the gravitational field is completely determined by u, so we can think of u as being analogous to the Newtonian potential.

For an excellent discussion of the Kasner vacuums, see Hawking and Ellis, The Large Scale Structure of Space-Time. The plane symmetric Kasner vacuum also arises in the study of colliding plane wave (CPW) solutions of the Einstein equation; see Griffiths, Colliding Plane Waves in General Relativity. And it arises when we use the ideas of inverse scattering (as in the solution of the KdV equation) to obtain (vacuum) solutions of the Einstein equation; see Belinsky and Verdaguer, Gravitational Solitons.

Now the surprise: the correspondence between Newtonian potentials and Weyl vacuum solution is not at all straightforward in the strong field situation. For example, a constant potential u gives the cartesian cart for the Minkowski vacuum, but the potential of an infinite uniform density ray (with linear density 1/2) gives the Rindler chart! The potential $$u(z,r)=m/r$$ does not give the Schwarzschild vacuum, but a distinct solution known as the Chazy-Curzon vacuum, which is not even spherically symmetric! The Schwarzschild vacuum is in fact obtained from the potential of a uniform density thin rod of length 2m and density 1/2. And to come to the point, the potential of a uniform density thin plate, $$u(z,r) = mz$$, does not really correspond (except in the weak-field limit) to the solution you probably want! Rather, the A3 vacuum arises (see the Weyl canonical chart given above) from a uniform density ray with another (constant) density.

Pervect already noted that the "Rindler metric" is probably what you want for your second request.

Chris Hillman

Last edited: Nov 19, 2006
4. Nov 21, 2006

### Chris Hillman

As (bad) luck would have it, the forum went down while I was in the middle of changing notation in the above post--- with disastrous consequences for anyone trying to verify that the Ricci tensor vanishes, or trying to confirm that these are indeed different charts for the same vacuum solution! I am still very new to this forum, so I am not sure, but it looks like I no longer have the option of editing the post. So let me try to write out those line elements again:

Levi-Civita chart (1917):
$ds^2 = -\frac{d\tilde{t}^2}{\tilde{z}} + \tilde{z}^2 \, \left( d\tilde{x}^2 + d\tilde{y}^2 \right) + \tilde{z} \, d\tilde{z}^2 ,$
$-\infty < \tilde{t}, \tilde{x}, \tilde{y} < \infty, \; \; 0 < \tilde{z} < \infty$

Kasner chart (1921):
$ds^2 = -\frac{dT^2}{Z^{2/3}} + Z^{4/3} \, \left( dX^2 + dY^2 \right) + dZ^2$

Taub (1951):
$ds^2 = -\frac{d\bar{t}^2}{\sqrt{\bar{z}}} + z \, \left( d\bar{x}^2 + d\bar{y}^2 \right) + \frac{d\bar{z}^2}{\sqrt{\bar{z}}}$

Weyl canonical chart:
$ds^2 = -\frac{dt^2}{z+\sqrt{z^2+r^2}} + \left( 1 + \frac{2z^2+r^2}{2 \sqrt{z^2+r^2}} \right) \, \left( dz^2 + dr^2 \right) + r^2 \, \left( z + \sqrt{z^2+r^2} \right) d\phi^2$

... hmm... that turns out to be harder to read than I hoped... those letters are respectively adorned with tildes, capitalized, adorned with bars, and unadorned.

For anyone who wants to try to rediscover the coordinate transformations, try computing the Kretschmann scalar $$R_{abcd} \, R^{abcd}$$, which gives a unique geometrical meaning to the scalar quantities called above $$\tilde{z}, \, Z, \, \bar{z}, \, z[/itex] respectively, which should get you started. Some points I wasn't able to include before the crash: The Lie algebra of Killing vector fields has the structure [tex]R \oplus e(2)[/itex]. That is, the group of self-isometries of this spacetime consists of time translation plus the group of isometries of the euclidean plane acting on hyperslices [tex]t=t_0$$ in the obvious way. Hence the designation "Taub plane-symmetric vacuum", one of the many names which has been bestowed upon it.

The Kasner vacuum mentioned above is not one of the cosmological models, but a static vacuum obtained by Kasner in 1921 using the same simple Ansatz he used to obtain the better known cosmological solutions.

And to amplify the "surprise" I mentioned: the "other density" is -1/2! That is, the source in the A3 vacuum can be understood as a plane symmetric distribution of matter with -negative- mass. This kind of nonsense (negative mass matter has some very strange properties!) is also permitted by the field equation of Newtonian gravitation, of course, and in both gtr and Newtonian gravitation, is probably best parsed using the slogan "garbage in, garbage out!" That is, a vacuum solution whose source seems to consist of a matter with negative mass should probably be regarded, at least provisionally, as the result of imposing unphysical boundary conditions.

There is a further interesting point lurking here: the notion of an infinite thin plate (consisting of positive mass matter) is viable (modulo issues of realism and stability) in Newtonian gravitation, but not, it seems, in gtr. One way to understand this kind of phenomenon is that inventing "a better theory" (better than Newtonian gravitation, in this case) can come with a price: a more stringent theory may no longer admit some idealizations familiar from prior experience.

Chris Hillman

Last edited: Nov 21, 2006