1
$\begingroup$

Here, the following explanation for Hawking radiation in canonical quantum gravity is given:

The local energy density is well-defined as the 00 component of the stress-energy tensor. It is frame dependent but in locally strong fields it is locally large in every frame. In the quantum version, a strong gravitational field is like a strong electromagnetic field, described not by an empty vacuum state but by a state full of energy (as defined by the energy-stress tensor).

The space-time inhomogeneity due to gravitation is described by a massless tensor field called the gravitational field (or the metric, in a geometric view that doesn't survive quantization). In canonical quantum field theory, which must be used in order to describe particle production, space-time is just a smooth manifold without predefined metric. The gravitational field (i.e., the quantized metric) is now described by a massless quantum field tensor operator that gives rise in the usual way to creation and annihilation operators for gravitons.

Just as particle pair production from strong electromagnetic fields is inevitable through processes such as $2\gamma\to e^-+e^+$, where $\gamma$ denotes a photon, so particle production from strong gravitational fields is inevitable: If one looks at the S-matrix in the tree approximation of canonical quantum gravity + QED, one gets processes such as $2g\to 2\gamma$ and $2g\to e^-+e^+$, where $g$ denotes a graviton. The first process happens at any positive energy since both sides are massless; the second process happens once the local energy concentration exceeds the energy equivalent of two electron masses.

The "inevitable" particle pair production in electromagnetic fields refers to the Schwinger effect, as far as I understand.

My question. Let's say we have a quantum field theory with a single massive particle.

  • This massive particle exhibits gravity, right?
  • So the "local energy density" (even though I don't quite understand how this is defined exactly since the stress-energy tensor is an operator, not a number, right?) should be bigger than the rest mass of the particle times the speed of light squared. (Is this also correct?)
  • If that is so, then, according to the paragraph above, there would be pair production of electrons at the cost of the energy of the gravitational field caused by the massive particle whenever the mass of the massive particle is bigger than twice the mass of an electron.
  • But what can loose energy here? The massive particle can't have any energy below its rest mass energy by definition.
  • Furthermore, if electrons are emitted, they also exhibit gravity, correct? Where do the corresponding gravitons come from?
$\endgroup$
2
$\begingroup$

Hawking radiation is derived using a fixed black hole background, which is then used to define a vacuum and quantize a quantum field.

Since the quantum effects produce particles in the field, it is correct that the correct approach would be then to go and create a new background that factors in the created particles, which would then create a different matter distribution, which would then require a new background, etc.

This process gets computationally complex pretty quickly (and to my knowledge, I don't think anyone has proven that it converges). But that is how one would deal with computing what the effects of the "gravity of the created particle" are.

Also worth considering: in General Relativity, you have a globally conserved energy of the whole spacetime (for asymptotically flat spacetimes), but the energy isn't really "located" anywhere. That you can't define a local energy density in a coordinate invariant way is intertwined with the equivalence principle.

$\endgroup$
1
  • $\begingroup$ I'll also say that I don't like the "pair production at the horizon" picture for much else than justifying why you have Hawking radiation at all. It is way to easy to let it confuse you. At the end of the day, the math is "1.use Schwarzschild/Kerr instead of Minkowski, do some QFT with this different metric, 2.take the resultant stress-energy tensor, solve ensintein's equation, 3. repeat" $\endgroup$ Oct 21 '21 at 16:41
1
$\begingroup$

Don't forget that in quantum physics a 'particle' is not a particle in the classical sense. The mass and charge are never concentrated at a point; they are concentrated in a region whose dimensions depend on the motional state. I think the source of the energy for the pair production here is whatever process managed to confine the charged particle sufficiently that the electric field around it got to a high enough value to cause pair creation. To get the particle's wavefunction confined to a region of size $\Delta r$ one has to provide the field energy and the kinetic energy which are implied by small values of $\Delta r$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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