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?


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.

  • $\begingroup$ There used to be another nice looking answer which had positive votes, and now it is gone. What happened? Is a moderator censoring answers? $\endgroup$
    – John
    Mar 30 '11 at 2:47
  • $\begingroup$ @John, I doubt it. Moderators don't willy-nilly delete answers AFAIK. Must have been the poster who had second thoughts about his answer. $\endgroup$
    – user346
    Mar 30 '11 at 3:34
  • $\begingroup$ 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. $\endgroup$ Mar 30 '11 at 7:41
  • $\begingroup$ 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. $\endgroup$
    – John
    Apr 7 '11 at 2:56
  • $\begingroup$ 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.] $\endgroup$ Apr 8 '11 at 11:34

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.

  • $\begingroup$ 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? $\endgroup$
    – user346
    Mar 30 '11 at 22:38
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    $\begingroup$ @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. $\endgroup$
    – John
    Mar 30 '11 at 23:36
  • $\begingroup$ @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. $\endgroup$
    – user346
    Mar 30 '11 at 23:42
  • 3
    $\begingroup$ @Deepak Please provide your own answer. It would be nice to see your full reasoning and math shown like Edward and Roy gave. $\endgroup$
    – John
    Mar 31 '11 at 5:49
  • 1
    $\begingroup$ 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. $\endgroup$ Apr 12 '11 at 9:03

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.


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.

  • $\begingroup$ As a clarification for John, what Lawrence is essentially referring to here is the hoop conjecture ( 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 en.wikipedia.org/wiki/Mass_in_general_relativity $\endgroup$
    – Edward
    Mar 30 '11 at 3:33
  • $\begingroup$ 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? $\endgroup$
    – John
    Apr 2 '11 at 0:44
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    $\begingroup$ -1 This doesn't seem to address the question. $\endgroup$
    – Ginsberg
    Apr 5 '11 at 22:51

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