- #1
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- TL;DR
- Flat slabs are weird in GR.
Off the back of a recently closed thread where there was some discussion about the gravitational field of an infinite flat slab, I decided to have a play at investigating that. I've found a few interesting things.
It's fairly straightforward to solve for this situation. You use Cartesian-esque coordinates to reflect the symmetry of the situation and let the slab lie in the ##xy## plane. The metric is diagonal due to reversal symmetry in the ##x##, ##y##, and ##t## directions, and can only depend on ##z##. Since the resulting Ricci tensor turns out to contain only terms like ##\frac 1{g_{00}}\frac{d g_{00}}{dz}=\frac{d}{dz}\ln g_{00}##, it's convenient to write the line element as $$ds^2=e^{f_t(z)}dt^2-e^{f_{xy}(z)}(dx^2+dy^2)-e^{f_z(z)}dz^2$$I shan't grind through the process explicitly here - have a look at Fulling et al sections 1-4 if you want to see the gory details. Suffice it to say that there are no constraints on the second derivative of ##f_z##, so we "gauge" it to zero, and that ##R_{tt}+R_{zz}-2R_{xx}## turns out to be
$$\left(\frac{df_{xy}}{dz}\right)^2+2\frac{df_t}{dz}\frac{df_{xy}}{dz}=0$$which tells you that either ##\frac{df_{xy}}{dz}=0## or ##\frac{df_{xy}}{dz}=-2\frac{df_t}{dz}##. The first turns out to be flat spacetime (in Minkowski or somewhat disguised Rindler coordinates depending on how you play it). The latter leads to $$ds^2=\frac{B}{\left(z-A\right)^{\frac 23}}dt^2-C\left(z-A\right)^{\frac 43}\left(dx^2+dy^2\right)-dz^2$$where ##A##, ##B## and ##C## are constants to be determined. It's easy to calculate the Kretschmann scalar ##R^{abcd}R_{abcd}=\frac{64}{27}(z-A)^{-4}##, and the presence of a curvature singularity at ##z=A## hints what ##A## should be - since we insisted that the solution be vacuum everywhere but we need a source of curvature, we've got a singularity as a source of curvature. So ##A## is the ##z## coordinate of our plate and can be set to zero without loss of generality.
Making that substitution we have$$ds^2=\frac{B}{z^{\frac 23}}dt^2-Cz^{\frac 43}\left(dx^2+dy^2\right)-dz^2$$which yields the following non-zero Christoffel symbols$$\begin{eqnarray*}
\Gamma^z_{tt}&=&−\frac{B}{3z^{5/3}}\\
\Gamma^t_{tz}=\Gamma^t_{zt}&=&−\frac 1{3z}\\
\Gamma^z_{xx}=\Gamma^z_{yy}&=&-\frac 23z^{1/3}C\\
\Gamma^x_{xz}=\Gamma^x_{zx}=\Gamma^y_{yz}=\Gamma^y_{zy}&=&\frac 2{3z}
\end{eqnarray*}$$
Here's where things get a bit strange. I wanted to find the constants ##B## and ##C##, so I thought I'd look at the Newtonian limit. I planned to write down the proper acceleration of a hovering observer, ##U^\mu\nabla_\mu U^\nu##, where ##U^t=1/\sqrt{g_{tt}}## and ##U^x=U^y=U^z=0##. The only non-zero component of the four acceleration is ##A^z=-\frac 1{3z}##. This is strange on two counts. Firstly, the negative sign suggests that the plate is repulsive, rather than attractive. Second, there is no regime comparable to the Newtonian behaviour, which produces a constant acceleration at all ##z##. The negative sign stems from the fact that that only non-zero component of ##\nabla_\mu U^\nu## turns out to be ##\nabla_t U^z=\Gamma^z_{tt}U^t## (it might look like ##\nabla_z U^t## ought to be non-zero, but ##\partial_zU^t=-\Gamma^t_{zt}U^t## and they cancel). I consistently get a minus sign there.
So I wrote down the geodesic equations, using that ##\partial_t##, ##\partial_x##, and ##\partial_y## are manifestly Killing vectors and the symmetry of ##x## and ##y## to write without loss of generality$$U^\mu=\left(\frac{E}{B}z^{2/3},\frac{p}{C}\frac 1{z^{4/3}},0,\frac{\partial z}{\partial \lambda}\right)$$where ##E## and ##p## are Killing constants that I've labelled (in hope) as an energy and momentum. That yields$$
\left(\frac{\partial z}{\partial\lambda}\right)^2=\frac{CE^2z^2-\epsilon BCz^{4/3}-p^2B}{BCz^{4/3}}$$where ##\epsilon=g_{\mu\nu}U^\mu U^\nu## is 1 for timelike geodesics and 0 for null geodesics. There is a solution for ##\lambda(z)## in terms of hypergeometric functions, but that's not very helpful. But we can make a few observations. For example, if I drop an object from rest ##p=0##, which implies ##\partial x/\partial\lambda=0## so its ##x## coordinate is unchanging (ditto its ##y## coordinate). But that means that if I drop two objects from rest a short distance apart at the same ##z##, the unchanging ##x## and ##y## coordinates means that the distance between them falls as ##z^{2/3}##. So the singularity is one of distance - all points on it have zero distance between them.
I'm still trying to make sense of sections 5 and onwards of the paper I linked above. Some of it kind of makes sense to me, but the bits about "matching Rindler and Minkowski spaces" is boggling the mind a bit. I thought the above was interesting, anyway.
It's fairly straightforward to solve for this situation. You use Cartesian-esque coordinates to reflect the symmetry of the situation and let the slab lie in the ##xy## plane. The metric is diagonal due to reversal symmetry in the ##x##, ##y##, and ##t## directions, and can only depend on ##z##. Since the resulting Ricci tensor turns out to contain only terms like ##\frac 1{g_{00}}\frac{d g_{00}}{dz}=\frac{d}{dz}\ln g_{00}##, it's convenient to write the line element as $$ds^2=e^{f_t(z)}dt^2-e^{f_{xy}(z)}(dx^2+dy^2)-e^{f_z(z)}dz^2$$I shan't grind through the process explicitly here - have a look at Fulling et al sections 1-4 if you want to see the gory details. Suffice it to say that there are no constraints on the second derivative of ##f_z##, so we "gauge" it to zero, and that ##R_{tt}+R_{zz}-2R_{xx}## turns out to be
$$\left(\frac{df_{xy}}{dz}\right)^2+2\frac{df_t}{dz}\frac{df_{xy}}{dz}=0$$which tells you that either ##\frac{df_{xy}}{dz}=0## or ##\frac{df_{xy}}{dz}=-2\frac{df_t}{dz}##. The first turns out to be flat spacetime (in Minkowski or somewhat disguised Rindler coordinates depending on how you play it). The latter leads to $$ds^2=\frac{B}{\left(z-A\right)^{\frac 23}}dt^2-C\left(z-A\right)^{\frac 43}\left(dx^2+dy^2\right)-dz^2$$where ##A##, ##B## and ##C## are constants to be determined. It's easy to calculate the Kretschmann scalar ##R^{abcd}R_{abcd}=\frac{64}{27}(z-A)^{-4}##, and the presence of a curvature singularity at ##z=A## hints what ##A## should be - since we insisted that the solution be vacuum everywhere but we need a source of curvature, we've got a singularity as a source of curvature. So ##A## is the ##z## coordinate of our plate and can be set to zero without loss of generality.
Making that substitution we have$$ds^2=\frac{B}{z^{\frac 23}}dt^2-Cz^{\frac 43}\left(dx^2+dy^2\right)-dz^2$$which yields the following non-zero Christoffel symbols$$\begin{eqnarray*}
\Gamma^z_{tt}&=&−\frac{B}{3z^{5/3}}\\
\Gamma^t_{tz}=\Gamma^t_{zt}&=&−\frac 1{3z}\\
\Gamma^z_{xx}=\Gamma^z_{yy}&=&-\frac 23z^{1/3}C\\
\Gamma^x_{xz}=\Gamma^x_{zx}=\Gamma^y_{yz}=\Gamma^y_{zy}&=&\frac 2{3z}
\end{eqnarray*}$$
Here's where things get a bit strange. I wanted to find the constants ##B## and ##C##, so I thought I'd look at the Newtonian limit. I planned to write down the proper acceleration of a hovering observer, ##U^\mu\nabla_\mu U^\nu##, where ##U^t=1/\sqrt{g_{tt}}## and ##U^x=U^y=U^z=0##. The only non-zero component of the four acceleration is ##A^z=-\frac 1{3z}##. This is strange on two counts. Firstly, the negative sign suggests that the plate is repulsive, rather than attractive. Second, there is no regime comparable to the Newtonian behaviour, which produces a constant acceleration at all ##z##. The negative sign stems from the fact that that only non-zero component of ##\nabla_\mu U^\nu## turns out to be ##\nabla_t U^z=\Gamma^z_{tt}U^t## (it might look like ##\nabla_z U^t## ought to be non-zero, but ##\partial_zU^t=-\Gamma^t_{zt}U^t## and they cancel). I consistently get a minus sign there.
So I wrote down the geodesic equations, using that ##\partial_t##, ##\partial_x##, and ##\partial_y## are manifestly Killing vectors and the symmetry of ##x## and ##y## to write without loss of generality$$U^\mu=\left(\frac{E}{B}z^{2/3},\frac{p}{C}\frac 1{z^{4/3}},0,\frac{\partial z}{\partial \lambda}\right)$$where ##E## and ##p## are Killing constants that I've labelled (in hope) as an energy and momentum. That yields$$
\left(\frac{\partial z}{\partial\lambda}\right)^2=\frac{CE^2z^2-\epsilon BCz^{4/3}-p^2B}{BCz^{4/3}}$$where ##\epsilon=g_{\mu\nu}U^\mu U^\nu## is 1 for timelike geodesics and 0 for null geodesics. There is a solution for ##\lambda(z)## in terms of hypergeometric functions, but that's not very helpful. But we can make a few observations. For example, if I drop an object from rest ##p=0##, which implies ##\partial x/\partial\lambda=0## so its ##x## coordinate is unchanging (ditto its ##y## coordinate). But that means that if I drop two objects from rest a short distance apart at the same ##z##, the unchanging ##x## and ##y## coordinates means that the distance between them falls as ##z^{2/3}##. So the singularity is one of distance - all points on it have zero distance between them.
I'm still trying to make sense of sections 5 and onwards of the paper I linked above. Some of it kind of makes sense to me, but the bits about "matching Rindler and Minkowski spaces" is boggling the mind a bit. I thought the above was interesting, anyway.
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But in any event, I had hoped to take a limiting case, substituting a finite g for an infinite M hoping that would work better. But as