How much energy can be added to a small volume of space? Perhaps like the focal point of a very high power femtosecond laser for a short time, or are there other examples like the insides of neutron stars that might be the highest possible energy density? Is there any fundamental limit?
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1$\begingroup$ Even prior to formation of black holes very strong radiation/fields facilitate creation of virtual particles thus loosing energy. So the practical limit is somewhat lower, as determined by en.wikipedia.org/wiki/Schwinger_limit $\endgroup$– oakadJun 9, 2022 at 11:53
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$\begingroup$ @oakad Thanks good link! $\endgroup$– UVphotonJun 9, 2022 at 15:28
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2 Answers
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There is a limit to how much energy that can be contained in a finite volume, after which the energy density becomes so high that the region collapses into a black hole.
We also know that matter and energy are equivalent according to the Einstein equation $$E=mc^2\tag1$$ So if we can determine the greatest amount of matter that can fit into a volume just before it collapses into a black hole, the corresponding energy should also indicate the greatest energy confined in the volume just before it becomes a black hole.
The maximum amount of matter, mass $M$, that can be contained in a given volume before it collapses into a black hole, is given by the Schwarzschild radius $$r_s=\frac{2GM}{c^2}\tag2$$ Using (1) we can then write $$M=\frac{E}{c^2}$$ so that equation (2) becomes $$r_s=\frac{2GE}{c^4}$$ or $$E=\frac{r_sc^4}{2G}$$
Note that this is still energy, and to get to energy density we need to define the volume which is of course $$V=\frac 43 \pi r_s^3$$ so that the energy density is $$\epsilon =\frac{3c^4}{8\pi Gr_s^2}$$
This computation is based on not much more than the equivalence of matter and energy. It represents a bound on the maximum amount of matter, and therefore energy, in a spherical volume of radius $r_s$ before the volume containing the matter collapses into a singularity, which of course has no properly defined volume.
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2$\begingroup$ It may be helpful to add the numerical value of that constant in front of the 1/r^2 in some common unit(s), eg J/cm and whatever would be used for this in imperial units. $\endgroup$– GraipherJun 8, 2022 at 12:36
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$\begingroup$ Thanks, would this also be true for electromagnetic fields, if matter wasn’t present? $\endgroup$– UVphotonJun 8, 2022 at 13:48
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2$\begingroup$ yet, the Schwartzschild radious can't be associated easily to density, as the volume of a sphere in the interior of the BH is NOT $\frac{4\pi r_s^3}{3}$, distances inside a massive body depend on the distribution of matter... so, the association you make between density and the Schwartzschild radious is incorrect $\endgroup$ Jun 8, 2022 at 15:18
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2$\begingroup$ @UVphoton Assuming spherical symmetry and asymptotic flatness (these are good assumptions for a highly compressed isolated system), yes, this numerical bound holds regardless of the form the energy takes, even if some is present in an extended electric field (as in the Reissner-Nordstrom solution), provided you make the iffy identification of the black hole's "volume". Outside of asymptotic flatness, one can violate the bound given here: indeed, every nontrivial FLRW metric violates this bound at each cosmological time beyond some length scale. $\endgroup$– jawheeleJun 8, 2022 at 15:25
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1$\begingroup$ I think the cosmological point is an important caveat to make in this answer. This is not at all a fundamental upper bound to the energy density that can be contained within a given radius. Since the given density goes to zero as $r_s \to \infty$, and since the observable universe looks quite homogeneous on large scales (i.e., on large scales it has constant energy density in the standard spatial slices of cosmology), this bound is always violated on a sufficiently large scale. And the farther back in the universe's history we go, the smaller the violation scale, going to zero at the big bang. $\endgroup$– jawheeleJun 8, 2022 at 15:28
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This limit is given by the Bekenstein Bound. It is proportional to the surface area of the surface enclosing a given volume. Exceeding this bound would create a black hole.
answered Jun 8, 2022 at 0:10
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1$\begingroup$ The Bekenstein bound is technically an upper limit on the entropy in a region. Also, please do not use mobile device friendly links. I've edited the link to the desktop kind. Thanks. $\endgroup$– joseph hJun 8, 2022 at 1:23
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