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I am speaking about GR with classical fields and energy. One question, spread over three increasingly strict situations:

Is there an energy density limit in GR? (literally, can the energy density have an arbitrarily large value at some point in space at some point in time)

Is there an energy density limit beyond which a blackhole will always form?

Let's choose a small volume, for here I'll just choose the Planck volume. Is there an average energy density limit over this volume beyond which a blackhole will always form?

Clarification:

In light of http://en.wikipedia.org/wiki/Mass_in_general_relativity , can those that are answering that the energy density is limited and referring to a mass $M$ in some equations please specifically state how you are defining the $M$ in terms of the energy density, or defining $M$ in terms of $T^{\mu\nu}$ the stress-energy tensor. Does your $M$ depend on coordinate system choice?

Also, reading some comments, it sounds like there is confusion on what energy density means. Based on wikipedia http://en.wikipedia.org/wiki/File:StressEnergyTensor.svg , it sounds like we can consider energy density = $T^{00}$ of the stress-energy tensor. If you feel this is not correct terminology, please explain and I'll edit the question if necessary.

John
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  • There used to be another nice looking answer which had positive votes, and now it is gone. What happened? Is a moderator censoring answers? – John Mar 30 '11 at 02:47
  • @John, I doubt it. Moderators don't willy-nilly delete answers AFAIK. Must have been the poster who had second thoughts about his answer. –  Mar 30 '11 at 03:34
  • Dear John, yes and no. The answer was deleted by a moderator - in fact, the main moderator of this server. However, the answer was also deleted by its author. It's because the author of the answer and the main moderator on this server is the same person, David Z. It was deleted after another user found a flaw in the argument related to the coordinate freedom. – Luboš Motl Mar 30 '11 at 07:41
  • Roy had written a nice little discussion up. And now that answer has been removed too! I was hoping during this last week Deepak would also add his own answer since apparently Edward is oversimplifying something. Instead of more discussion, almost literally a negative amount. The very opposite of what I hoped the bounty would do. :( ... I give up. – John Apr 07 '11 at 02:56
  • I did have an answer and an extended answer written up, but I have deleted them due to the fact that there are still too many uncertainties around this question. What tipped the balance for me was the last minute realisation that the MM book does not necessarily equate "energy density" with $T^{00}$ either. The MM book discusses proper energy density, but the book is weak on Tensor-based explanations of its terms. So another uncertainty is whether the MM argument already accepts Edward's answer, so to speak. Combined with the Hoop and Mass definition uncertainties there was [cont.] – Roy Simpson Apr 08 '11 at 11:34
  • ... not enough definitive to give an answer different from a large question mark. Nevertheless Lawrence's answer (and similar comments) are to be found around the literature (not just the MM book), but they are not related to $T^{00}$ itself, perhaps not even to $T^{ab}$. I dont know whether they are "wrong" or whether it matters that $T^{00}$ is not used in these arbuments. I am sure research will continue on these topics. – Roy Simpson Apr 08 '11 at 11:40
  • @Roy Mass definitions and the hoop conjecture are indeed some interesting follow ups to this question. However I'm not entirely sure the MM book is even talking about 'proper' energy density since he uses it to claim a limit on the electric field. Even in SR I don't think it is meaningful to try to define a proper density for an electromagnetic plane wave as there is no inertial frame in which it is at rest. Considering his "derivation" of GR, it is probably best to just ignore the Motion Mountain book whenever it invokes GR. – Edward Apr 09 '11 at 02:48
  • @Edward, "I don't think it is meaningful to try to define a proper density..." - the equivalent concept for EM would be $E^2+B^2$ wouldnt it? I see this question as about invariant physical aspects of energy density generally (matter mostly) - and whether a BH can be proven to form - not about the variations and interpretations possible via coordinate freedoms. I could be wrong about that and maybe John just wanted to know what numbers could potentially appear inside $T^{ab}$? – Roy Simpson Apr 10 '11 at 13:29
  • @Roy $E^2 + B^2$ is just $T^{00}$ for an electromagnetic field, and therefore is just as coordinate dependent. The usual way of defining a 'proper density' for an object or fluid is to state the density in its rest frame. We don't have that luxury with a electromagnetic plane wave for example. So I don't think it is meaningful to try to define a proper density in that case. There could be a clever way to redefine the terminology that isn't immediately obvious, but without referring to other vectors like a velocity, the only scalar invariant is the trace $T$ which is 0 for electrodynamics. – Edward Apr 10 '11 at 21:29

3 Answers3

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The answer is NO. There is no energy density limit (for all three questions).

The easiest way to see this is that the energy density is just the $T^{00}$ component of the stress energy tensor. The solution in GR depends on the full stress energy tensor, so it is not enough to just talk about the energy density. Furthermore, because the energy density is just a component of a tensor, it is a coordinate system dependent quantity. So starting from a solution that doesn't become a blackhole, and has some energy somewhere, we can always choose the coordinate system to make the energy density arbitrarily large.

More clearly stated: Local Lorentz symmetry alone is enough to show that the energy density is not limited in GR. And furthermore since there exist non-zero energy solutions that don't become blackholes, this also answers your second question.

To make the answer to the third question more clear, let's discuss an exact solution. Consider the Robertson-Walker solution with a perfect fluid. Here's an example stress energy tensor for a perfect fluid in the comoving frame:

$T^{ab} =\left( \begin{matrix} \rho & 0 & 0 & 0 \\ 0 & p & 0 & 0 \\ 0 & 0 & p & 0 \\ 0 & 0 & 0 & p \end{matrix} \right)$

Now if we change to a different coordinate system, using the coordinate transformation: $\Lambda^{\mu}{}_{\nu} =\left( \begin{matrix} \gamma &-\beta \gamma & 0 & 0 \\ -\beta \gamma&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{matrix} \right)$

We see the energy density will transform as: $\rho' = \gamma^2 \rho + p \beta^2 \gamma^2 = \gamma^2 (\rho + p \beta^2)$

So not only can the energy density be arbitrarily large, but even over a finite volume.

Edward
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  • How exactly do you go from the saying the energy density transforms as ... to the energy density can be arbitrarily large, but even over a finite volume? –  Mar 30 '11 at 22:38
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    @Deepak I don't understand your and Roy's comments on this. Doesn't that coordinate transformation show the energy density will be increased at more than just a point? For example, what do you claim the new stress energy tensor will look like? You continually disagree with Edward and make insults below like "amateur hand-waving" or worse in the other thread. Provide your own answer if you disagree with Edward. Please show us the correct answer. – John Mar 30 '11 at 23:36
  • @John it doesn't matter what transformation you choose to apply to the stress-energy tensor. The fact is you are dealing with a solution to Einstein's equations here. No matter what any given observer might experience it is not going to change the equations of motion - the Friedmann equations in this case. Consequently for an FLRW metric there is such a thing as a critical density $\rho_c$ (ref: Wald) which determines whether the cosmology is open (hyperbolic), closed (spherical) or flat. In particular there is a fiducial choice of observers - those comoving with the metric. –  Mar 30 '11 at 23:42
  • For a trivial solution such as Minkowski there is no such fiducial choice and Edward's reasoning might go through. But this is not going to happen for a general time-dependent metric. As for amateur hand-waving and insults - you're right. I'm overstepping the bounds of polite conversation. But then again, Edward is trying to provide an overly simplistic answer to a complex question for which no such simple answer exists. And pointing out that this is the case seems to have no effect. Anyhow, my apologies @Edward for referring to you as an amateur. I'm sure you not. Cheers! –  Mar 30 '11 at 23:47
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    @Deepak Please provide your own answer. It would be nice to see your full reasoning and math shown like Edward and Roy gave. – John Mar 31 '11 at 05:49
  • @Deepak You write "it doesn't matter what transformation you choose to apply to the stress-energy tensor". Do you agree with the following: (1) $T^{00}$ depends on the coordinate system. (2) $T^{00}$ can be arbitrarily large over a finite region in space as shown by Edward. (3) regardless of coordinate choice $T^{00}$ is the energy density. ... I'm having trouble understanding, since you seem to keep responding to a question different than asked. So a direct yes/no for those three points would help clear things up. It sounds like you are saying (1)yes,(2)yes,(3)no. – John Mar 31 '11 at 05:51
  • since you seem to keep responding to a question different than asked ... yes, I think that is a birth defect in my case :). Please provide your own answer ... I will, in my own sweet time. I don't claim to have all the answers. This is a difficult question and I will reply as and when I feel I have a physically reasonable response. I hope you can live with that. –  Mar 31 '11 at 06:03
  • In the meantime @Roy's and @Lawrence's answers are most representative of my stand. I hope I've also made that clear. Also this is not a multiple choice exam and good questions hardly ever have cut and dried responses of the sort you seem to be looking for. –  Mar 31 '11 at 06:05
  • @Deepak "I will, in my own sweet time. ... I hope you can live with that." Of course, take your time, and I look forward to seeing your own answer. Because currently I don't understand how Edward's answer is 'overly simplistic', so I don't see what is limiting his answer and thus how to reconcile his comments with yours. I look forward to seeing the details of your answer on this subject. – John Mar 31 '11 at 06:32
  • @Edward, So just to be clear, are you saying that in any spacetime there is no limit to the value of physical energy density (at a point or region - and with any type of matter/radiation)? Why is this different from saying that (matter-based) Black Holes cannot form? Phrased in terms of the $\rho$ above one could ask for the proof that it can be unlimited in a GR (with matter) solution. I (now) think the original question was always about GR physics and not about coordinate freedom. – Roy Simpson Apr 10 '11 at 12:55
  • @Roy In my opinion, it sounds like John is still struggling with the concept that only invariant, not coordinate dependent, quantities have any real meaning. So this leads to sloppy associations like 'large amounts of matter can collapse to a black hole' -> 'E=mc^2' -> 'energy is limited by black hole formation' -> 'if I move to fast I become a blackhole'. This is clearly wrong, but it is a common question/confusion students have. – Edward Apr 10 '11 at 21:38
  • @Roy I'm not sure how to answer your question there. If you could clearly define what you mean by "physical energy density", maybe you should ask your question in the last comment as a follow up to John's question. It would likely be much more interesting than discussion on limits of the coordinate dependent $T^{00}$. – Edward Apr 10 '11 at 22:03
  • @Edward, the physical value of a quantity is its measurement in a lab. So physical mass density would be its measurement in a lab. Use E=mc^2 to convert to physical energy density measured (E=m if c=1). The question then is whether this can become arbitrarily high without a BH forming (in the Lab). – Roy Simpson Apr 11 '11 at 23:27
  • This latter question isnt my question; it is the original question: Is there an average energy density limit over this volume beyond which a blackhole will always form? I am just emphasising physical to get away from references to tensors, which has been distracting. – Roy Simpson Apr 11 '11 at 23:28
  • @Roy Again, the issue is this quantity is coordinate system dependent. The scientist can use whatever coordinate system he wishes to describe a lab experiment. Coordinate system choices are not "physical" which is the very reason the physical laws are independent of them. So we need to cast the question into invariant terms. You instead seem to be saying $T^{00}$ is already physical since it can be measured with a given coordinate system ... if you take that stance on "physical" then the question is not interesting and already answered. – Edward Apr 11 '11 at 23:36
  • @Roy Due to confusion, John chose a definition for energy density for the question. If you can appropriately rephrase this into discussing an invariant quantity instead, the question would be different (and much more interesting) so I encourage you to start a new question. – Edward Apr 11 '11 at 23:38
  • @Edward, I did not mention coordinate systems, because I agree they are arbitrary too. The Lab will have a fixed length say. A tensor-coordinate question here is about the range of representations of that length. A physical question would be "can an an object of unlimited length enter and be placed in the lab"? I agree that there are at least two types of questions around here: ones on tensor-coordinate properties, and ones on actual physical properties. The challenge of this question has been to address the phyisical question: I dont think that it is a tutorial on tensors. – Roy Simpson Apr 12 '11 at 08:41
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    Furthermore it was not John, but you who introduced $T^{00}$ into this question - only to prove that $T^{00}$ was coordinate dependendent and non-invariant. Great! So the original question still stands unanswered. – Roy Simpson Apr 12 '11 at 09:03
  • Here is a scientific discussion of density with SI units and some actual values: http://en.wikipedia.org/wiki/Density. So the claim that you appear to make that this concept (density - mass density $\rho$) has no invariant or physical or laboratory meaning doesnt make sense to me. – Roy Simpson Apr 12 '11 at 09:25
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There is an energy density limit in physics (imagine a Hubble volume full of photons with a wavelength reduced to a plank length packed 1 per planck volume, collapsing into a singularity). However general relativity has no fundamental quantities besides c and thus no such constraint.

John
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The Schwarzschild radius is $R~=~2GM/c^2$, where if you pack a mass $M$ into a volume with a radius $R$ you get a black hole. The term mass-energy limit is not standard language usage, but if you push sufficient mass into a volume it will becomes a black hole and causally sealed off from the outside world.

Lawrence B. Crowell
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  • As a clarification for John, what Lawrence is essentially referring to here is the hoop conjecture ( http://en.wikipedia.org/wiki/Hoop_Conjecture ). However the mass referred to here can't simply be related to the energy density, otherwise the hoop conjecture is trivially wrong. We can always change coordinates to add as much kinetic energy as we want and it clearly doesn't change whether a black hole forms or not. So if speaking of only the energy density -- no there is not a limit. Mass in general relativity can be a tricky concept http://en.wikipedia.org/wiki/Mass_in_general_relativity – Edward Mar 30 '11 at 03:33
  • You never mention energy density in your answer. So while it appears you are saying 'yes' there is an energy density limit, I can't figure out how this meshes with the 'no' answer from Edward. Can you provide some connecting detail from your statement to a 'yes'/'no' for the title question? – John Apr 02 '11 at 00:44
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    -1 This doesn't seem to address the question. – Ginsberg Apr 05 '11 at 22:51