How are general relativity and quantum mechanics united in mainstream theory?
Established Member
How are general relativity and quantum mechanics united in mainstream theory?
Administrator
They are united by the Grand Unified Field Theory... which doesn't exist yet.
Forming opinions as we speak
Order of Kilopi
They are also united by Quantum Electrodynamics (QED) - which I don't even begin to understand at a level beyond the Feynman book of that title.
Order of Kilopi
Ah, you might want to look into that a bit. QED only applies to electrons and photons(positrons are considered time reversed electrons). There are currently no generalized Gravitational equations that incorporate quantum effects, and work. Both Superstring Theory and Loop Quantum Gravity (LQG) are both trying, but have yet to succeed. Superstring theory is trying to quantize gravity as a force, much as the rest of the forces. LQG is trying to quantize spacetime itself and leave gravity as geometry.Originally Posted by Strange
Order of Kilopi
Well, maybe "united" is too strong a word: they "meet" in QED might be more accurate
Established Member
QED is compatible with Special Relativity but not uniquely so, it can be made compatible with many relativity theories. GR is still out on its own.
Order of Kilopi
Ah. Good point. Thank you. I was over-generalizing
Member
They're united only in the sense that one can do quantum mechanics on a fixed spacetime background. Which isn't really much of a unification at all--how to apply quantum mechanics to spacetime itself isn't something that's found in mainstream theory. Yet.
We have lots of guts. Toes are another matter.
Established Member
Will there be a theory of quantum gravity within the next 50 years ? If no, why not ?
Far away is close at hand in images of elsewhere...
Established Member
Certainly hope so. Since a fair share of theoretical physicists have been working on the issue for 50 years already it would be nice to think their efforts aren't going to waste. Of course there is always the possibility that the two theories are fundamentally incompatible at all levels such that unification is impossible. That would be really annoying.
Member
I wouldn't say the efforts would go to waste, though it would be very disappointing. String theory's AdS/CFT correspondence is applicable to parts of condensed matter physics, including superconductors, regardless of whether it is the correct path to quantum gravity.
Established Member
I might feel better about that if I could understand it!
Moderator
For the record, Leonov's initial post in this thread was a duplicate of one he made elsewhere which was moved to ATM. I have moved his post here, tusenfem's admonishment, and Leonov's response to that thread in ATM.
Never attribute to malice that which can be adequately explained by ignorance or stupidity.
Isaac Asimov
Moderation will be in purple.
Rules for Posting to This Board
Member
I thought Hawking radiation was an example if GR and QM. Basically at the Schwarzchild radius there is a probability of an annihilating pair being created, separating and recombining within the Heisenberg limit. Except one may get captured by the black hole (or tunnel beyond the event horizon) and the other then escapes. Not sure if this unites them but they co-exist quite nicely if true.
Member
That's true. However, that's a case of a quantum field on a classical spacetime background. In other words, it's a semi-classical treatment of the problem. There's a difference between that and a true unification, in which spacetime itself should be (somehow) be described by quantum mechanics.
Established Member
That is not the GR explanation of Hawking Radiation though. In GR it is a form of Unruh radiation, which can be thought of as horizon radiation. The particles being captured picture is not what is actually hypothesised to happen, it was more a QM sketch of how a horizon could be considered to radiate. The maths for it basically don't quite hold up as I understand it. Hawking radiation is a GR effect, the Unruh effect, not a QM one.
Member
The maths for the Unruh radiation do hold up. I'm not nearly as sure about Hawking radiation, but since near the horizon the Schwarzschild black hole is locally very well-approximated by a Rindler chart on Minkowski spacetime, I wouldn't expect there to be any fundamental problems (this does make the comparison of Hawking to Unruh valid, though they're still different effects). The captured-particles picture is a crude description of the field(s), but it isn't wrong.
For a Schwarzschild black hole, ∂t is a Killing field corresponding to conservation of energy outside the horizon, but inside, it is spacelike and thus locally corresponds to momentum instead. Energy conservation requires a virtual particle-antiparticle pair to annihilate again, but if they're separated by a horizon, then this 'energy conservation' actually adds energy of one particle and the momentum of another, and there's no requirement that real particles have positive momentum (this gets interpreted as negative energy by observers outside the horizon). So GR can explain why it's possible, but QM is still needed to make it go.
Established Member
OK that was not my understanding. Can you explain how the particle emissions approximate a thermal emission spectrum consistent with a temperature? There seems to be no reason that you could characterise the BH to have a temperature in the QM case.
As for the maths - I meant with the QM version. Unruh holds up very well, I thought that the basic mechanism of particle capture was rather hand-waving. I may be wrong - but I was always taught that the ripping apart of virtual pairs was a convenient 'this is how a horizon could emit something' without being rigorous.
Is it your understanding that Hawking radiation and Unruh radiation are different phenomena with respect the a BH?
Member
It only has a temperature because of quantum effects. Classically, it has no temperature and does not radiate anything.
An intuitive way to think about it if one accepts the Unruh effect as a given is that static observers are accelerated, so if the freefalling observers observes a vacuum of no temperature, the static one should measure a positive temperature. Making this slightly more exact, a Schwarzschild black hole can be approximated by a Rindler chart of Minkowksi spacetime under the substitution,
,
, which shows that the black hole has surface gravity of 1/4M and acts locally just like the Rindler horizon of an accelerated observer in flat spacetime. Therefore, Hawking temperature of a black hole should match the Unruh temperature with acceleration replaced by this surface gravity. And since redshifting thermal radiation just gives a thermal radiation with different temperature, the end result is thermal as well. This is the correct conclusion that Hawking derived by more rigorous means.
It is an explanation of 'how a horizon could emit something' (and a somewhat more rigorous formulation of the same thing in terms of energy conservation can be found in the previous post), but it isn't that bad of a description. It's alright to say that the Hawking effect is like the Unruh effect and make analogies between them, but there's also something very intrinsically different going on. The Rindler horizon can't turn virtual particles into real ones, but a Schwarzschild horizon can, because is a real horizon. Unruh predicts that an accelerated observer in a vacuum observes a thermal bath. Things act as if they were in a radiation bath, but there's still no actual energy flux going on. That's different for actual black holes.
Particle capture is just a heuristic way of describing the mechanism. The particles are pulled apart by tidal forces, and the horizon is one-way. To make this more precise, one would have to propagate both of them in the curved spacetime, and energy conservation can be fulfilled without annihilation if one of them goes inside the horizon (see previous post for basic reason as to why). That's the key point here that makes 'particle capture' a working description. It's certainly not the only heuristic description, but it is a valid one. Alternatively, one can view the particles as tunneling out of the event horizon.
Not exactly. Technically, Unruh does not apply to a BH at all. But locally, they are the same effect. However, there are also important non-local differences between them because black hole horizons are real and Rindler horizons are not, which makes the Hawking radiation have a genuine particle flux.
Order of Kilopi
I have a question about this distinction. We can define both absolute and apparent horizons. It happens to be the case that in the schwarzschild solution those coincide, but they don't have to. For example during stellar collapse, or a general perturbation, the two will be distinct. You say there is a distinction that the schwarzschild case is a real horizon, by this i presume you mean it's an absolute horizon (both Rindler and standard Schwarzschild coordinates have an apparent horizon)?
The question is as follows, if the distinction between Hawking and Unruh radiation is due to them being associated with an absolute and an apparent horizon (as seems to be suggested by your "because it is a real horizon"), how can we calculate what happens when the black hole is perturbed and we can't use schwarzschild's apparent horizon as a substitute for the absolute horizon? How can we derive Hawking radiation in this general case? Does it not present a problem that we need to know the entire future evolution of our entire spacetime, and that it must be asymptotically flat, in order to locate the absolute horizon and thus derive any Hawking radiation? If that is not a problem, does it not present a problem that in, what seems to be, our FLRW spacetime we don't even have a future null infinity to define our absolute horizon with?
If the above is not correct, and we're still defining Hawking radiation wrt Schwarzschild's apparent horizon, in what way exactly is it different from Unruh radiation?
Member
I've been using 'real horizon' as a deliberately vague contrast to the usual acceleration horizon, because the issue is a bit more involved than just the coincidence of absolute and apparent horizons. For the Schwarzschild black hole, take a look and the Killing field ξ = ∂t we started with, which is timelike with norm 1 at infinity and turns null on the horizon, and the gradient of ξ·ξ there is -2κξ, where κ is the surface gravity. The Killing field connects the near static (accelerated) observers with the very far static (inertial) observer in a way that preserves energy, so you will have radiation even in there.
The above generalizes to every stationary black hole solution, with their absolute horizons also being Killing horizons, and the connection to spacelike infinity being necessary to set the scale of the Killing field at the horizon.
The freefalling observers detect no extra radiation, whether it's a perturbation or the formation in the first place; every observer accelerated relatively to a colocated freefalling one measures Unruh-type radiation. But it's the Killing field that defines, roughly, which particles are 'real radiation' and which were measurements of local quantum fluctuations. If you don't have a Killing field, I'm not sure what sense can be made of that distinction, and it's essential to the difference between Unruh and Hawking, because it's that radiation that needs to be accounted for in the black hole's mass.
I'm far from clear on the quantum-mechanical details of what happens when there is no Killing field in general (in some cases, the distinction simply loses meaning: e.g., in stellar collapse, the scales of time and energy of the process are such that they skirt the uncertainty principle limit anyway) or when it's not everywhere timelike even outside the horizon, but there's a classical approximation that I understand models the situation fairly well, at least for slowly rotating black holes and weak perturbations.
It builds yet another horizon, on the order of a Planck length over the true one. The stretched horizon acts like a membrane of viscous fluid that's electrically charged and conductive, with finite uniform temperature and entropy, but no heat conduction. As matter falls onto this membrane, it grows correspondingly. Pretty much any response to an external electromagnetic field or gravitational perturbation can be analyzed in this formalism. And in addition to being a good approximation (or so I'm assured by people who actually work on this stuff), it also naturally enables several completely off-the-wall ideas, like using black holes in circuits or electric motors. As if black holes weren't crazy enough on their own.
Problem in what sense? Remember we're mixing two different concepts that are only locally equivalent for BH spacetimes...
Unruh: an accelerated observer measures a positive vacuum temperature where an inertial one does not
Hawking: a black hole produces radiation
So we need a global definition to even make sense of the latter.
Exploring the black hole thermodynamics in FRW spacetimes has been a long-standing problem. What the recent state of the art on that is, I've no idea. There are methods to extend the Hawking radiation results in terms of apparent horizons only in hopes to deal with just such cases, but I'm far from familiar enough to comment even semi-competently.
Order of Kilopi
In Rindler space we also have a Killing field, which we can, under a suitable coordinate transformation
, make timelike with norm 1 at infinity and null on the horizon. However Rindler's Killing horizon is degenerate, as the surface gravity vanishes (the gradient of
at the horizon).
Am i then right that the distinction lies in wether the Killing horizon is degenerate or non-degenerate? So by a "real" horizon you mean a non-degenerate Killing horizon?
As an aside, does the existence of a non-degenerate Killing horizon imply the existence of an absolute horizon? I thought this was not necessarily true, and if it isn't, what happens when you have a non-degenerate Killing horizon without an absolute horizon? It seems like you'd get Hawking radiation without having a black hole, so there must still be something missing from my understanding.
Posting Permissions
Reply With Quote