Gravitational field of an infinite flat slab

  • #1
Ibix
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TL;DR Summary
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.
 
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  • #2
Vanadium 50
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Infinities always scare me. Two thoughts come to mind:

1. If instead of an infinite slab, you make a disk of radius R and let terms of order rho/R go to zero, what happens? (I am assuming you will have some bare z's left and not z/R)

2. What is this slab made of? If you say, "it doesn't matter, so make it dust", but my next comment will be "what keeps the dust from collapsing". If you introduce a pressure, that's got its own infinities. I think it all cancels and it just acts as a mass, but would want to look at it more carefully.
 
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  • #3
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  • #4
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What is this slab made of?

The way to find that out is to compute the Einstein tensor of the metric; then you divide that by ##8 \pi## to get the stress-energy tensor. The properties of the SET then tell you what the slab is made of.

If you say, "it doesn't matter, so make it dust", but my next comment will be "what keeps the dust from collapsing". If you introduce a pressure, that's got its own infinities.

None of these things can be done if you already know the metric. If you already know the metric, you have already specified what the slab is made of, since the Einstein tensor is computed from the metric. So you can't "make" it anything; you can only figure out what it is by computation.
 
  • #5
Vanadium 50
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I agree that if you start with the metric, that tells you the source. I was thinking we're coming at it the other way.

I also wonder whether we can look at this as a spherical solution, very far away:

[tex]ds^2 = -\left( 1 - \frac{2GM}{r} \right)dt^2 + \frac{dr^2}{\left( 1 - \frac{2GM}{r} \right)} + r^2\left(d\theta^2 + \sin^2\theta d\phi^2 \right)[/tex]

Replace G and M with g,

[tex]ds^2 = -\left( 1 + 2gr\right)dt^2 + \frac{dr^2}{\left( 1 + 2gr \right)} + r^2\left(d\theta^2 + \sin^2\theta d\phi^2 \right)[/tex]

Replace r with z and replace angular pieces with flat Cartesian pieces. (We're far away)

[tex]ds^2 = -\left( 1 + 2gz\right)dt^2 + \frac{dz^2}{\left( 1 + 2gz \right)} + dz^2 + dy^2[/tex]

Doesn't this work? It has the right limiting behavior.
 
  • #6
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I was thinking we're coming at it the other way.

The OP has a metric in it.

I also wonder whether we can look at this as a spherical solution

Not if the slab is infinite. A finite slab, perhaps; but none of the references I have seen (including those referenced in this thread) treat this case.
 
  • #7
DrGreg
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[tex]ds^2 = -\left( 1 + 2gz\right)dt^2 + \frac{dz^2}{\left( 1 + 2gz \right)} + dx^2 + dy^2[/tex]
(I've corrected an obvious typo in the penultimate term in the quote above.)

If you perform the substitution$$
Z = \frac{1}{g} \sqrt{1+2gz}
$$so that$$
dZ = \frac{dz}{\sqrt{1+2gz}}
$$you get (if I haven't made a mistake)$$
ds^2 = -g^2 Z^2 dt^2 + dZ^2 + dx^2 + dy^2
$$i.e. Rindler coordinates. This represents an infinite slab that is made of nothing!

(I remember Rindler did a similar substitution in one of his textbooks.)
 
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  • #8
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This represents an infinite slab that is made of nothing!

Actually, it doesn't quite even represent that, since at ##Z = 0##, which is where the "slab" is supposed to be, the line element with all spatial coordinates constant (i.e., only ##dt## is nonzero) is null, not timelike.

I notice, btw, that the Fulling paper referenced in the OP appears to me to be claiming that you can have a Rindler metric in the vacuum region "above" the slab but still have an actual slab, not vacuum, at ##Z = 0## (or the equivalent in whatever coordinates are adopted). I don't see how that can be true. (See in particular the discussion in section 4 and Fig. 1 of the paper.)
 
  • #9
Vanadium 50
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The OP has a metric in it.

And so it does. Doh!

Not if the slab is infinite. A finite slab, perhaps

How do you tell if the slab is infinite if parts of it are in your future light cone? :wink: 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 @DrGreg points out, it works too well - one more step and there's nothing left. So it's not a useful approach.
 
  • #10
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How do you tell if the slab is infinite if parts of it are in your future light cone?

An observer stuck in the spacetime can't tell for sure that the slab is infinite. But we, who know the exact metric describing the spacetime, can.
 
  • #11
vanhees71
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Isn't it a priori clear that an "infinite slab" is simply unphysical? It's as in classical electrodynamics, where you consider an infinite charged plane. What's really meant is a finite plane (e.g., a rectangle) and approximately calculating the field at points not too far way from the plane such that you can approximately use the field of the simplifying translational symmetry of the infinite plane.
 
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  • #12
Ibix
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I notice, btw, that the Fulling paper referenced in the OP appears to me to be claiming that you can have a Rindler metric in the vacuum region "above" the slab but still have an actual slab, not vacuum, at Z=0 (or the equivalent in whatever coordinates are adopted). I don't see how that can be true. (See in particular the discussion in section 4 and Fig. 1 of the paper.)
That's where I started to have problems with that paper. I think this particular claim is related to its discussion of the electrostatic analogy in 5.1. Their research is related to the Casimir effect, and I think they're arguing that there's nothing in the field equations that prohibits asymmetric solutions. Intuitively, two parallel plates might well have a flat spacetime between them, and I think this is an argument in support of that, but looking at one plate (or an approximation thereto) in isolation.

Thanks for the links, also - I'll have a read (the paper is reference 8 in Fulling, by the way, but I hadn't read it yet).
Isn't it a priori clear that an "infinite slab" is simply unphysical? It's as in classical electrodynamics, where you consider an infinite charged plane. What's really meant is a finite plane (e.g., a rectangle) and approximately calculating the field at points not too far way from the plane such that you can approximately use the field of the simplifying translational symmetry of the infinite plane.
Of course it's unphysical. But it doesn't seem to be anything like a Newtonian infinite flat slab, which is interesting. In particular, you seem to have to accelerate towards it to stay at constant ##z##. Unless I've messed up the maths (but note that section 5.5 in Fulling seems to me to imply that a thin flat plane would have to be made of exotic matter to have the metric in #1 on both sides, which is a partial response to @Vanadium 50's point 2 in #2 as well).
 
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Isn't it a priori clear that an "infinite slab" is simply unphysical?

Sure, an actual infinite slab would be, but (as you appear to agree) it can often be a useful approximate model for cases where a finite slab is a lot larger than the region we are interested in and we want to avoid having to deal with complications like edge effects.
 
  • #14
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So what's the upshot in terms of experienced effects?

I think I recall reading something about a thousand years ago that the net effect is an attraction to the plane that is constant, independent of height/ distance from it. ie, if it were 1g at the surface, it will be 1g at a thousand light years. Yes?
 
  • #15
Ibix
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I think I recall reading something about a thousand years ago that the net effect is an attraction to the plane that is constant, independent of height/ distance from it. ie, if it were 1g at the surface, it will be 1g at a thousand light years. Yes?
That's the Newtonian result, yes, a fairly simple result of applying Gauss' theorem. I think the Jones et al paper Peter linked in #3 does conclude that you can indeed construct such a thing in GR, but it isn't as simple as it looks - you have to add transverse stresses and/or cosmological constants.

I'm still not quite sure how that matches up with Fulling et al, which I linked in my #1 and which appears to be the same as the solution I discovered myself. Although, admittedly, that's because this got pushed down my priority list a way and I haven't spent much time on it.
 
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