So I have seen an animation about Stephen Hawking (after his recent study state universes claim) that Hawking evaporation is due to negative mass; But how is this possible? I mean, there is no such thing as negative mass!
Asked
Active
Viewed 383 times
2 Answers
5
You do not state the level of understanding you have in physics.
Negative mass is a mathematical construct necessary to explain Hawking radiation, and is due to what are called virtual particles.
These are lines in Feynman diagrams which diagrams are a shorthand in generating the integrals necessary for the calculation of crossections and lifetimes in elementary particles.
This is a Feynman diagram of electron electron scattering or electron positron annihilation, depending on the axis taken for time:
The photon is virtual because it is not on the mass shell: In the diagram the incoming from the left have at least 1MeV invariant mass and the same is true for the outgoing on the right. A real photon has mass zero, so the photon line is called virtual because it is off its mass shell, unreal. In virtual particle exchanges the "particle" has all the quantum numbers of its name except the four vector describing it is off mass shell, positive or negative.
Hawking radiation involves loops of virtual particles, i.e. particles off mass shell, over the event horizon. When one becomes real , then the other one falls into the black hole, mathematically having negative mass, as it is virtual. The energy for the whole interaction is taken out of the gravitational field of the black hole. See also my answer to a similar question here.
-
Looking at it this way emphasizes a key point that I've seen intelligent people miss - Hawking radiation violates the weak energy condition in General Relativity. If you're going to consider spacetimes where Hawking Radiation is important, you're going to have to also abandon a whole lot of the intuition built about black holes that rest on this assumption about the matter distribution. – Zo the Relativist Feb 05 '14 at 21:16
-
@JerrySchirmer when you use quantum mechanics in a semi classical form , as in Hawking radiation, paradoxes are inevitable when enlarging the scope. Only a TOE which includes the particle SM and a quantized gravity have to be consistent in all aspects, and we are not there, though there are string theory based papers doing calculations (google) At the moment these are interesting hand waves, in a sense. – anna v Feb 06 '14 at 05:06
-
@JerrySchirmer : I find this explanation by prof Steve Carlip, very pleasant. There is no negative energy for the incoming particle, but a negative momentum, because the nature of the Schwarszchlid time and radial coordinates is changing at the horizon. – Trimok Feb 06 '14 at 11:48
-
@annav: sure. I disagree with none of that. I should note that the bulk of what is going on in the string theory literature is specialized to black holes with $Q=M$, which would correspond to a zero classical temperature (there are exceptions), so I would argue that even that is halting and of limited physical applicability. I don't think that realistic models would require a full TOE, though. Something with quantized gravity and a quantized scalar field would be enough to get at least a picture of the qualitative behavior. – Zo the Relativist Feb 06 '14 at 14:15
-
I wish I had the citation of a particularly clever paper that took a Vaidya spacetime, calculated the Hawking radiation, and then nonperturbatively matched this to the spherically symmetric modes of a quantized photon field. Doing this got them different decay modes for the black hole than the naïve ${\dot M} \propto T^{4}$ models. – Zo the Relativist Feb 06 '14 at 14:18
-
@Trimok: I don't see any discussion about momentum/energy distinctions there. I personally am not a big fan of disuccsions that generate distinctions between virtual and real particles. I would rather ground my intiution of QFT in the S-Matrix and transformations between in and out states, since so many discussions about virtual particles are deceptively simple. But smart people disagree with me there, certainly. – Zo the Relativist Feb 06 '14 at 14:33
0
That's a way to interpret it...and I agree that it is a confusing one in my view. To be honest, the way I see it is as if the black hole was in an excited state so high in energy that it can emit massive particles.
The emission would be of the type
$BH^* \rightarrow BH + p$
where $BH$ stands for black hole and $p$ for particle. During this reaction, you have to conserve momentum and energy and the fact that it is a black hole doesn't change that fact. Charge has to be conserved as well so I would imagine that if the black hole radiates a charged particle, it becomes charged as well.
Some other stuff are usually conserved in such reactions (e.g. leptonic or baryonic numbers etc..) but I don't know what happens to them in this case.
Maybe this betrays the limitations of my naive point of view.
gatsu
- 7,202