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QFT with respect to general relativity

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tom.stoer
#19
Dec25-11, 05:49 PM
Sci Advisor
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As I said, we expect the geometry to be quantized for several reasons - mainly consistency reasons. Quantum effects would then be small far away from the Planck scale, i.e. quantum gravity would be the UV completion of an effective QFT on smooth classical spacetime (however there are proposals for so-called fuzzball blackholes in string theory which indicate deviations from classical metric even far away from the Planck sale)
suprised
#20
Dec26-11, 06:08 AM
P: 407
Actually to be fair, there are some proposals out that challenge the conventional wisdom. One is classicalization and self-completeness. This posits that if one tries to probe the Planck scale, eg by an energetic scattering process, then one creates black holes before one ever enters into the quantum gravity regime. These are classical objects, so in this sense one never would be able to probe quantum gravity near the Planck scale: the theory protects itself. Pumping in more energy just makes the black holes larger and even more classical.

This is not undisputed, however, but some version of this may be true, perhaps only in particular kinematical regimes; see the ref. in my previous post. The key point is unitarity, not renormalizeability.

Nevertheless, for consistency, the whole theory needs to be quantum mechanical. This is independent of whether one can probe the Planck scale by scattering experiments or not.
martinbn
#21
Dec26-11, 08:18 AM
P: 353
Quote Quote by tom.stoer View Post
Einstein equations couple gravity to matter - and we know that matter is described by QFT. So how do you want to change the equation and couple gravity to non-quantized matter?
I don't want that, just saying that may be the equation can be changed so that gravity need not be quantized. You, yourself, say that the equation has to be changed (the whole theory), but you use the equation (that has to be changed) as the reason why gravity should be quantized. I am only saying that that is not very convincing.
Dickfore
#22
Dec26-11, 10:28 AM
P: 3,014
I think you guys are running in circles with your discussion. Would you please define what you mean when you say an equation is "quantized", and similar terms. What is "classical" then?
martinbn
#23
Dec26-11, 10:32 AM
P: 353
Quote Quote by Dickfore View Post
Would you please define what you mean when you say an equation is "quantized", and similar terms.
Where was that said?
Dickfore
#24
Dec26-11, 10:35 AM
P: 3,014
Quote Quote by martinbn View Post
Where was that said?
Quote Quote by tom.stoer View Post
First reason: the Einstein equations read "metric-dependent terms = matter-dependent term"; b/c the r.h.s. is quantized, the l.h.s. should be quantized, too.
Also, if you do a search of this thread for "quantized", you will see it is applied very liberally for various concepts. Could you define what you mean by "quantized" before you start discussing?
martinbn
#25
Dec26-11, 10:47 AM
P: 353
First reason: the Einstein equations read "metric-dependent terms = matter-dependent term"; b/c the r.h.s. is quantized, the l.h.s. should be quantized, too.
Ah, but it does NOT say anything about an equation being quantized, right?

Quote Quote by Dickfore View Post
Also, if you do a search of this thread for "quantized", you will see it is applied very liberally for various concepts. Could you define what you mean by "quantized" before you start discussing?
I could.
Dickfore
#26
Dec26-11, 10:48 AM
P: 3,014
Quote Quote by martinbn View Post
Ah, but it does NOT say anything about an equation being quantized, right?
What does r.h.s or l.h.s. stand for?!

Quote Quote by martinbn View Post
I could.
Please do.
suprised
#27
Dec26-11, 10:59 AM
P: 407
There seems a lot of confusion. So let's do a little thought experiment. Just scatter two electrons - one from the left, the other coming from the right, in some rest frame.

Quantum mechanics is used to describe the scattering matrix. This is like a black box which tells you what comes out from this scattering process, given the incoming particles. And you want to have unitary scattering, so that probabilities do not exceed one. So far so good, I guess nobody objects that QM is the right concept here.

To make things easier, the electrons have an offset, or impact parameter, which is large, say 1km. Ordinarily one wouldnt expect that something would be peculiar or problematic.

But I didnt tell you that the kinetic energy of the electrons equals to the mass of a large star. A star with such a mass would form a black hole. So what's going to happen is that when the electrons are still, say 2km apart, a large black hole forms. But you dont really want to know the details now; all that matters is the "black box", or S-Matrix, and the question is, without caring about the details of what happens in the black box, what are the final states? Is the scattering unitary? This is obviously a quantum mechanical question. And if the scattering is unitary, this implies that the black hole must be able to decay. So Hawking radiation must necessarily occur, if quantum mechanics is supposed to be valid.

Note that this involves quantum mechanics and gravity, and ultra-plankian energies, but still these questions are insensitive to the Planck scale: small distances are not relevant here. So we talk about highly non-perturbative non-local effects.

Related problems occur when considering loops of virtual black holes; do these induce non-unitary scattering for low-energy particle physics? Better not!

Obviously one needs to describe gravity and quantum mechanics in one single coherent framework, in order to address this kind of questions. AFAIK a suitable framework to describe this quantitatively is still lacking. Although I know of some attempts using AdS/CFT.
friend
#28
Dec26-11, 12:35 PM
P: 969
Quote Quote by tom.stoer View Post
As I said, we expect the geometry to be quantized for several reasons - mainly consistency reasons. Quantum effects would then be small far away from the Planck scale, ...
Planck scale this and Planck scale that... How can we be sure that any of the constants of nature and thus the planck scale should remain the same as we approach ever more tightly curled up spacetimes? I mean, if we cannot see inside a black hole or cannot see the big bang, then it seems we are just guessing.
Fra
#29
Dec26-11, 12:44 PM
Fra's Avatar
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Quote Quote by suprised View Post
There seems a lot of confusion. So let's do a little thought experiment. Just scatter two electrons - one from the left, the other coming from the right, in some rest frame.

Quantum mechanics is used to describe the scattering matrix. This is like a black box which tells you what comes out from this scattering process, given the incoming particles. And you want to have unitary scattering, so that probabilities do not exceed one. So far so good, I guess nobody objects that QM is the right concept here.
I see a big problem with this closed black box view - it is valid ONLY when the scattering picture is which is when you have an inert observer that can make observations as well as preparations from a distance where the coupling to the black box is is weak/controlled in the sense that the observer itself (which is a generalized "background") does not severly deform during the interaction.

The other problem is that it also only makes sense when ensembles can be realized.

In cosmological pictures, where the observer is strongly coupled, the observers entire ENVIRONMENT (ie remainder of the universe) is the effective "black box", and here most of the premises in the scattering picture fails. Also an inside observer can hardly encode arbitrary amounts of inforamtion - someting that is usually not cared about in a good way in the scattering pictures as I see it.

It's no news that my own view is that QM formalism as it stands is unlikely to be sufficient here. That's not to say the scattering matrix is interesting, it is. But I think it's a good abstraction of observed reality only in limiting/special case.

/Fredrik
suprised
#30
Dec29-11, 07:25 AM
P: 407
Quote Quote by Fra View Post
In cosmological pictures, where the observer is strongly coupled, the observers entire ENVIRONMENT (ie remainder of the universe) is the effective "black box", and here most of the premises in the scattering picture fails. Also an inside observer can hardly encode arbitrary amounts of inforamtion - someting that is usually not cared about in a good way in the scattering pictures as I see it.

It's no news that my own view is that QM formalism as it stands is unlikely to be sufficient here. That's not to say the scattering matrix is interesting, it is. But I think it's a good abstraction of observed reality only in limiting/special case.
Well I am not talking about cosmological pictures, but just a transplanckian scattering experiment, if you wish with asymptotic oberservers. So what is the S-Matrix for this scattering? It should have a concrete answer, and better be unitary.

If you dispute the valitidy of QM and the S-Matrix - well QM has been proven to be extremely robust against deformations and so far no one, AFIAK, was able to replace it by something else. It is very common (because cheap) to say "according to my opinion QM needs somehow be modified", but very difficult to actually do it ...
suprised
#31
Dec29-11, 07:27 AM
P: 407
Quote Quote by Dickfore View Post
I think you guys are running in circles with your discussion. Would you please define what you mean when you say an equation is "quantized", and similar terms. What is "classical" then?
Why circles? This just means it's an equation involving operators. And this makes sense only if the complete equation, and not just part of it, becomes operator valued.
Dickfore
#32
Dec29-11, 12:47 PM
P: 3,014
Quote Quote by suprised View Post
Why circles? This just means it's an equation involving operators. And this makes sense only if the complete equation, and not just part of it, becomes operator valued.


So, what is the meaning of the operators [itex]g_{\mu \nu}[/itex], and [itex]R_{\mu \nu}[/itex]?
friend
#33
Dec29-11, 04:55 PM
P: 969
Quote Quote by Dickfore View Post


So, what is the meaning of the operators [itex]g_{\mu \nu}[/itex], and [itex]R_{\mu \nu}[/itex]?
It turns them into probability distributions instead of absolute values.
Dickfore
#34
Dec29-11, 05:00 PM
P: 3,014
Quote Quote by friend View Post
It turns them into probability distributions instead of absolute values.
Please elaborate. Are you saying the metric tensor becomes a probability distribution? If yes, whose random variable it is a distribution of? Or, is the metric tensor a (multivariate) random variable. In this case, what determines its distribution?
tom.stoer
#35
Dec30-11, 12:52 AM
Sci Advisor
P: 5,366
In canonically quantized GR g and R are field operators with a huge gauge symmetry and therefore w/o a direct physical meaning.
Dickfore
#36
Dec30-11, 12:48 PM
P: 3,014
Quote Quote by tom.stoer View Post
In canonically quantized GR g and R are field operators with a huge gauge symmetry and therefore w/o a direct physical meaning.
What do you mean by "gauge symmetry" of GR?

Also, if you canonically quantize the gravitational field, what are the canonical commutation relations?


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