Trans-Planckian Problem for Black Holes: Do Infalling Particles Become BHs?

  • Thread starter Thread starter T S Bailey
  • Start date Start date
  • Tags Tags
    Black holes Holes
T S Bailey
Messages
26
Reaction score
0
I have heard that, given the energy of a quantum of Hawking radiation, we can extrapolate backward in time to its 'creation' near the event horizon. When we do this we find that, because of time dilation and conservation of energy, the wavelength of the emitted particle becomes smaller than the Planck length. Could we then say that at some point before reaching the event horizon the wavelength of any of infalling particle will become smaller than its Schwarzschild radius? Do infalling particles become black holes themselves?
 
Physics news on Phys.org
T S Bailey said:
I have heard that, given the energy of a quantum of Hawking radiation, we can extrapolate backward in time to its 'creation' near the event horizon. When we do this we find that, because of time dilation and conservation of energy, the wavelength of the emitted particle becomes smaller than the Planck length.

Reference, please?

T S Bailey said:
Could we then say that at some point before reaching the event horizon the wavelength of any of infalling particle will become smaller than its Schwarzschild radius?

No. Infalling particles are not the same as outgoing particles, and ordinary matter or radiation falling in is not the same as Hawking radiation emitted out.

I can't really give any more specifics unless I see what particular reference you got this from.
 
PeterDonis said:
Reference, please?
No. Infalling particles are not the same as outgoing particles, and ordinary matter or radiation falling in is not the same as Hawking radiation emitted out.

I can't really give any more specifics unless I see what particular reference you got this from.
Jacobson, T. (1991). "Black-hole evaporation and ultra short distances." Physical Review D
 
T S Bailey said:
Jacobson, T. (1991). "Black-hole evaporation and ultra short distances." Physical Review D

This paper is behind a paywall so I can only read the abstract, which is here. Nothing I can see in the abstract changes my answer in post #2. (Note that the paper, at least from what I can see in the abstract, is actually proposing a way to avoid the "trans-Planckian" problem in the derivation of Hawking radiation; so if its proposal is correct, there wouldn't be such a problem even for Hawking radiation, let alone for an ordinary object falling into the hole.)
 
I haven't read the paper I cited though I have read a few different proposals on how to avoid the problem, none of which have explained how conservation of energy is maintained. If I drop a photon into the black hole I would expect its wavelength to be inversely proportional to the gravitational time dilation it experiences on its way to the horizon. If time dilation goes to infinity (as measured by an external observer) as one approaches the horizon then shouldn't we expect field modes with arbitrarily short wavelengths to exist there simply by assuming conservation of energy?
 
T S Bailey said:
none of which have explained how conservation of energy is maintained.

Conservation of energy has to be defined very carefully in curved spacetime. The only notion of "energy" that is conserved in the spacetime around a black hole is what is called "energy at infinity", which is a constant of the motion for any object in free fall. But this is not the same as the energy that would be measured locally by an observer seeing the free-falling object go past him; "energy" defined that way is simply not conserved, in the sense that different observers at different altitudes will measure different energies for the free-falling object, and there's nothing "compensating" the change to keep anything conserved.

T S Bailey said:
If I drop a photon into the black hole I would expect its wavelength to be inversely proportional to the gravitational time dilation it experiences on its way to the horizon.

The concept of "gravitational time dilation" doesn't apply to a photon; in fact it doesn't really apply to anything that is free-falling into the hole. It only applies to things that are "hovering" at a constant altitude above the hole.

Also, the photon's wavelength is not an invariant; it depends on the state of motion of whatever is measuring the wavelength.

T S Bailey said:
If time dilation goes to infinity (as measured by an external observer) as one approaches the horizon then shouldn't we expect field modes with arbitrarily short wavelengths to exist there simply by assuming conservation of energy?

No. See above.
 
  • Like
Likes T S Bailey
Thread 'Can this experiment break Lorentz symmetry?'
1. The Big Idea: According to Einstein’s relativity, all motion is relative. You can’t tell if you’re moving at a constant velocity without looking outside. But what if there is a universal “rest frame” (like the old idea of the “ether”)? This experiment tries to find out by looking for tiny, directional differences in how objects move inside a sealed box. 2. How It Works: The Two-Stage Process Imagine a perfectly isolated spacecraft (our lab) moving through space at some unknown speed V...
Does the speed of light change in a gravitational field depending on whether the direction of travel is parallel to the field, or perpendicular to the field? And is it the same in both directions at each orientation? This question could be answered experimentally to some degree of accuracy. Experiment design: Place two identical clocks A and B on the circumference of a wheel at opposite ends of the diameter of length L. The wheel is positioned upright, i.e., perpendicular to the ground...
In Philippe G. Ciarlet's book 'An introduction to differential geometry', He gives the integrability conditions of the differential equations like this: $$ \partial_{i} F_{lj}=L^p_{ij} F_{lp},\,\,\,F_{ij}(x_0)=F^0_{ij}. $$ The integrability conditions for the existence of a global solution ##F_{lj}## is: $$ R^i_{jkl}\equiv\partial_k L^i_{jl}-\partial_l L^i_{jk}+L^h_{jl} L^i_{hk}-L^h_{jk} L^i_{hl}=0 $$ Then from the equation: $$\nabla_b e_a= \Gamma^c_{ab} e_c$$ Using cartesian basis ## e_I...

Similar threads

Back
Top