2
$\begingroup$

I was reading this post Black holes and positive/negative-energy particles because I was wondering how Hawking radiation works and why always the anti-particle falls into a black hole, which is nicely explained there.

However, I feel that I have an understanding problem what vacuum fluctuation really is. Is it a pair of particle-anti-particle, e.g. electron and positron (as I assumed...) or positive-negative-mass-particle? If the latter is true, what is the negative particle then, surely not a positron...?

Matter/antimatter or matter/negative-matter is not the same and it's important when understanding Hawking radiation (as in the post mentioned above). That is because anti-particles like positrons have still positive mass, thus falling into the black hole would just grow the hole. (Or in other words: If we would produce antimatter in particle accelerators and pour it into a black hole, it would grow, correct?)

Also, if vacuum fluctuations consist of positive and negative mass (or energy), the annihilation would produce zero energy, whereas I thought the explanation for vacuum fluctuations is some remaining "zero-point-energy" in quantum mechanics... thus I assumed real antimatter.

Improve this question
$\endgroup$
11
  • $\begingroup$ What is negative matter? The particle that falls into the black hole has negative energy - it falls into the black hole thereby decreasing the total energy/mass of the black hole. The particle that leaves the black hole has positive energy. $\endgroup$
    – Prahar
    Commented May 16, 2021 at 12:05
  • $\begingroup$ @Prahar Mitra, Well, that's exactly my question: Is there a difference between antimatter and matter of negative energy? I think yes, because the annihilation of a particle with its anti-particle (as produced in accelerators) produces a lot of energy, which is not possible if the anti-particle would have negative energy. $\endgroup$
    – Charles Tucker 3
    Commented May 16, 2021 at 12:13
  • $\begingroup$ The particle that enters the black hole must have negative energy, but it could be a matter particle or an anti-matter particle. There is no correlation between the two. Annihilation of two matter and anti-matter particles with both positive energy produces a lot of energy, but if one of them has negative energy, then they don't. $\endgroup$
    – Prahar
    Commented May 16, 2021 at 12:15
  • $\begingroup$ Note that its crucial that we're in a BH background and in a region near the horizon of a black hole. In flat spacetime, particles cannot have negative energy so the annihilation process always produces large amounts of energy. However, there is a region outside the horizon of a black hole known as the ergosphere where particles may have negative energy. Negative energy particles are forever confined within the ergosphere and eventually fall into the black hole whereas positive energy particles can escape the ergosphere to a far away observer where it would be observed as Hawking Radiation. $\endgroup$
    – Prahar
    Commented May 16, 2021 at 12:16
  • $\begingroup$ @Prahar Mitra, that's interesting! So the negative energy is not associated with a negative mass (because of E=m*c^2)? $\endgroup$
    – Charles Tucker 3
    Commented May 16, 2021 at 13:05

1 Answer 1

Reset to default
2
$\begingroup$

However, I feel that I have an understanding problem what vacuum fluctuation really is.

It is hypothetical loops or particle-antiparticle pairs. Hypothetical because if there is no real particle with a fourvector from which the loop can get some energy, there is no way to detect it. Loops of particle-antiparticle are in higher order Feynman diagrams and are used to get better calculations for observables as crossections and decays.

vacuum loops

However, I feel that I have an understanding problem what vacuum fluctuation really is. Is it a pair of particle-anti-particle, e.g. electron and positron (as I assumed...) or positive-negative-mass-particle? If the latter is true, what is the negative particle then, surely not a positron...?

The invariant mass of elementary particles that can be in particle antiparticle pairs is always larger than zero, neutrinos, electron-positron, quark-antiquark....

That is because anti-particles like positrons have still positive mass, thus falling into the black hole would just grow the hole

Not true. The hypothetical loop pair of electron-positron takes its four vector (energy/momentum) from the fields of the horizon. When the electron is absorbed because its momentum was pointing down to the black hole, the positron escapes with its part of the energy/momentum, (vice verso for positron electron)and thus the black hole loses mass, not gains.

If we would produce antimatter in particle accelerators and pour it into a black hole, it would grow, correct?)

That antimatter would not have any connection with the black hole, and by falling in would increase the mass of the black hole. Hawking radiation takes energy/momentum out of the black hole, that is why micro black holes from the Big Bang are supposed to have evaporated by now.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Many thanks for the comprehensive answer! But I always thougt mass and energy are the same according to E=m*c^2. So when the anti-particle has negative energy... then it must have a negative mass as well? $\endgroup$
    – Charles Tucker 3
    Commented May 16, 2021 at 12:33
  • $\begingroup$ The m in the formula is misleading. It is not the invariant mass but the inertial mass of a particle moving at relativistic energies. the invariant and this m coincide only when the particles are at rest . The negative in energy comes from balancing energy content, it is a component of a fourvector, not an invariant.. see hyperphysics.phy-astr.gsu.edu/hbase/Relativ/vec4.html $\endgroup$
    – anna v
    Commented May 16, 2021 at 13:09
  • $\begingroup$ to conclude, vacuum fluctuations produce particle-antiparticle pairs, e.g. electron-positron, both having positive mass (in our everyday meaning of what is mass, and there is no negative mass anyways). But one of it, when falling into a black hole, behaves like having negative energy, no matter if that's the electron or positron. Is that correct? $\endgroup$
    – Charles Tucker 3
    Commented May 16, 2021 at 13:29
  • $\begingroup$ not negative energy per se. One has to do the subtraction from the combined four momentum vector of the pair, the four momentum vector of the one escaping in order to know the four momentum of the absorbed. $\endgroup$
    – anna v
    Commented May 16, 2021 at 13:44
  • 1
    $\begingroup$ @safesphere yes, the loop part has taken over in illustrations the mostly photon part. It is a similar mechanicsm as black body radiation , a charged particle falls into the horizon becoming part of the black hole, but there is a probability of interacting with other charges and fields and some of those interactions have a probability to have a photon in a direction that allows it to escape the inevitable attraction of the black hole, the four vector of the escaped photon diminishing the mass of the black hole. $\endgroup$
    – anna v
    Commented May 16, 2021 at 15:46

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.