As far as I have seen it is assumed that the universe's energy is conserved or at least every reference agrees that pair production doesn't violate the conservation of energy as there is, ultimately, some reckoning to account for it.

Now, consider the following example which seems to avoid this reckoning and appears to violate the energy conservation law.

Suppose that near a black hole horizon there is pair production due to vacuum fluctuations such that the formed entities end up on either side of the horizon. Now, it is clearly impossible for them to ever result in pair annihilation. So it appears that a particle seemingly came out of nowhere flouting the energy conservation law. Furthermore, the only way to reconcile things (at least as far as I can think) would be to have some random correlation which absolutely forces some amount of energy to suddenly disappear and this correlation could very well be a very very long range correlation but I don't think there is any evidence to show for it.

This clearly depends on the frequency of pair production near horizon of the kind mentioned above but it doesn't look like it should be so rare as to go undetected till now.

Could someone please shed some light on this issue?

Edit : For clarification -

Suppose that the energies before pair production were $x$ for BH and $y$ for the rest of the universe. Now, pair production (with energy of pair being $p$) from vacuum fluctuations resulted in the total energy $x+y+p$ which in normal case, due to pair annihilation, would have again become $x+y+p-p=x+y$ but due to the separation because of horizon does not come about immediately. It might happen that after the BH radiates some energy and that vacuum fluctuations of that energy eventually take care of the $p$ energy excess but it could very well take a really long time (which might even be corresponding to $\mathcal{I}^{+}$ in a Penrose Diagram). But as far as I have read such energy conservation violations are only allowed for very short times. Anyhow, the question is : if there is no fault in my argument then is is possible to detect such energy conservation violations?

  • 1
    $\begingroup$ As a black hole radiates Hawking radiation, its mass decreases. The energy of the radiation doesn’t come from nowhere. $\endgroup$
    – G. Smith
    Commented Dec 23, 2020 at 1:45
  • $\begingroup$ @G.Smith Yes but where did the energy to produce the pair come from in the first place? Had they not been separated by the horizon they would have annihilated and the energy would be preserved. Since they have been separated by the horizon that is no longer possible and hence the apparent conservation violation. $\endgroup$ Commented Dec 23, 2020 at 2:05
  • 2
    $\begingroup$ There is no conservation violation: the reduction in black hole mass compensates for the the energy of the (now real) radiated particle. $\endgroup$
    – bapowell
    Commented Dec 23, 2020 at 2:32
  • $\begingroup$ @bapowell Sorry I am still having trouble understanding. I edited the question to include some explanation of what I am thinking. Can you please take a second look? $\endgroup$ Commented Dec 23, 2020 at 3:48

1 Answer 1


As far a Feynman diagrams go, the best I could do is the Schwinger Effect (pair production in a strong electric field):

enter image description here

If you focus on the far left diagram, which is the one-loop pair production and annihilation, returning back to the vacuum, what makes you think energy conservation is violated? Four momentum is conserved in Feynman diagrams, and all possible 4-momentum must be summed over. What ever $p^{\mu}$ is running forward over the top is balanced by $-p^{\mu}$ running back on the bottom.

  • $\begingroup$ This holds true for regular cases but here we have a horizon which separates the two produced entities (say each with momentum $p^{\mu}$). Since the horizon separates the two before they could recombine (and preventing the $-p^{\mu}$s from occurring) we do not have a closed loop. Furthermore, this closed loop exists for an infinitesimally small amount of time (right?) but in case of horizon separation this is apparently no longer true (if my argument is correct). Regular QFT can't deal with horizons as far as I know that's why the F-diagrams you have drawn are not so pertinent to the question. $\endgroup$ Commented Dec 23, 2020 at 9:37
  • $\begingroup$ you are completely missing the point. Diagrams that violate $\Delta E$ for some short time because of HUP are old fashioned perturbation theory, not Feynman diagrams. $\endgroup$
    – JEB
    Commented Dec 23, 2020 at 14:12
  • $\begingroup$ But still the Feynman diagrams you have drawn are in flat Minkowski space and cannot deal with the presence of a horizon so I cannot see how your argument holds. The whole point on which my doubt rests is the presence of a horizon. $\endgroup$ Commented Dec 26, 2020 at 7:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.