Quantum Gravity

1. Feb 13, 2006

Sumo

I have often heard about how when the quantum equations dealing with the other forces are applied to gravity they produce infinities. Can anyone point me to these equations? What exactly are they?

2. Feb 13, 2006

f-h

Oh boy...
Fix a background eta_mu,nu, write GR as a perturbation around it: h_mu,nu. Write a path integral for h_mu,nu read off the feynman rules, and study the perturbative quantum field theory (QFT) you get for a couple of decades and you will see that unlike the infinities in the Standard Model, the infinities in the resulting perturbative QFT can not be removed perturbatively, and that is the only thing about the quantum theory you have defined, by adding a finite set of parameters. Therefore you can not perturbatively define a theory with the GR QFT as its low energy limit.

To understand this more deeply pick any book on QFT (I recommend to get started with Zee: QFT in a Nutshell), QFT has no simple fundamental formula (except arguably for the path integral).

3. Feb 13, 2006

marcus

nice concise explanation. I hope to see more such posts.

4. Feb 13, 2006

Sumo

I dont. :P Well actually thats not true, what I was really looking for was where you pointed me in the last paragraph, a few books which could get me started to understanding these ideas.

One other question. What are these equations attempting to solve for? Are they trying to find the properties of particles (mass ect)? So when they are applied to gravity they are solving the properties of a graviton?

5. Feb 13, 2006

marcus

Sumo might be glad if someone would expand f-h discussion out, and make it the long version instead of the short version

Sumo, please assure us that you know what a power series is. Just an ordinary freshman calculus power series. this may sound dumb, but we get questions from people of all levels

If you know about power series then you may know that in conventional QED calculations involve expanding stuff in a complicated powerseries in the terms of the finestructure constant ALPHA.

there are successive terms involving higher and higher powers of alpha----a linear term, a term in alpha2, a term in alpha3, and so on.

at least theoretically----it gets harder to construct the higher power terms

BUT ALPHA IS A SMALL DIMENSIONLESS NUMBER, and powers of it get small quickly------so after a few terms in the power series it is very small and the rest of the series doesnt matter.

Alpha is the electromagnetic coupling strength. The problem with gravity appears when one tries to do the ANALOGOUS THING and build a power series using powers of the GRAVITATIONAL coupling strength, which is essentially newton G

Newton G is not dimensionless, it needs you to provide an energy scale (a mass scale) to multiply it by, before it will give you somethat you can meaningfully take powers of.

taking higher and higher powers of some physical quantity doesnt always give anything physically meaningful. What does the cube of mass mean? What does the fourth power of pressure signify?

but it is OK to take higher and higher powers of a dimensionless pure number, like 1/3 or like 1/137.
Look at how successive powers of 1/137 get smaller and smaller
1/137, 1/1372, 1/1373 this is already very tiny.

So the situation with gravity is already different from QED in the very simple way that newton G is not dimensionless, whereas the ALPHA of QED is dimensionless.

G is not a pure number, it is only a number when units are attached to it. It can only appear as a number when you dress it in units-----otherwise it is a naked primitive swamp-critter that you can't do arithmetic with: a physical quantity. Like the pre-literate length that I show you with my hands.

Other people can correct me but I think, Sumo, that the root of the difficulty is that in the case of gravity the coupling strength (that you would use to expand the series in powers of) is not already dimensionless, and nobody has yet found a way to make it dimensionless and to build a powerseries where the successive terms get smaller and smaller

this reply is recklessly oversimplified----but how do we know what is appropriate to your frame of mind, Sumo. Maybe f-h explanation will work for you. But if not, think some about power series.

Last edited: Feb 13, 2006
6. Feb 13, 2006

Staff Emeritus
Let me try. A Quantum Field Theory(QFT) attempts to find the PROBABILITIES of various kinds of interactions between particles. The basic objects it deals with however are fields, more or less complicated ones, extending throughout space, and the particles are viewed as excitations ("quanta") of these fields.

Richard Feynman made two contributions to QFT: Feynmann diagrams and path integrals. Feynmann diagrams are drawings of the interactions between the particles. A very basic diagram looks like an H; two particles come up from the bottom, exhange a force particle (the crossbar of the H) and are thereby transformed into some other particles, perhaps, and the two outgoing particler exit out the top. So you break down this drawing and it has five lines (bottom two legs, crossbar, top two legs), and two vertices (the ends of the crossbar, where it meets the other legs). And Feynmann showed how to associate each of these seven items with a mathematical expression and put the expressions together to form an integral, which integral gives the basic probability for the interaction that the diagram describes. Think that over a bit, it's really not hard. Of course most particle interactions are much more complicated and the mathematics can get realy long and intricate.

A word about the perturbation approach. This means starting with the simplest diagrams, calculating the probability, then going to the next simplest, calculating and adding these two results together, and so on. You get an infinite series which hopefully, although it may not converge, gives good answers for the first few terms. You'll see the terms referred to as "n-loop" meaning a diagram in which n photons (or other particles) are emitted and then reabsorbed; this produces a loop in the diagram; and these loops are the principle complication of the diagrams, and of the probabiity calculations. The theory called Quantum Electrodynamics has been around for nearly sixty years and they are just now getting into five loop calculations.

This then gives a classical, unquantized theory, and Feynmann's other contribution, path integrals, is one of the ways to make it into a quantized theory. I am not going to describe this, becaise Feynmann has done it far better in a little book for intelligent but not mathematically sophisticated people that he calls QED. I most strongly suggest that you read it.

But I will say this, all QFTs have a problem. At very short distances, really close to the particles themselves, those probability integrals go to infinity. Physicsts have learned that this is because any QFT they have is only an "effective theory"; it correctly describes physics at some energy scale, but cannot be expected to give meaningful results at a much higher energy scale. In quantum land, small distance means high energy, so that, the physicsts tell themselves, is the reason for the infinities.

This gives them a strategy for dealing with the infinities; they apply a smooth function which turns off the integral at very short distances; this is called a Regulator and after a little work it shows up as an undetermined constant in the integral. Depending on the theory you may have several of these constants. After you quantize and use the integrals to calculate probabilities, you will have various quantities like the masses of the particles that the theory doesn't predict and you can combine the undetermined constants into these "counterterms" formally and get rid of them. Then you just plug the measured values of those quantities into the integrals. This is called Renormalization. Then you have a set of integrals that give finite probabilities, and the ones we know best are very very good at that.

Now we come to gravity. When you do all this with general relativity, you get an infinite number of undetermined constants, and you don't have an infinite number of counterterms to absorb them. So we say that general relativity cannot be renormalized, and therefore a valid QFT of it cannot be derived.

Last edited: Feb 13, 2006
7. Feb 13, 2006

Sumo

That was actually exactly what I was looking for. Thank you for taking the time to respond Marcus, and selfAdjoint.

8. Feb 13, 2006

Staff Emeritus
9. Feb 13, 2006

duke_nemmerle

Wow! Really nice thread, I love both of the explanations. I might have to get my Feynman QED book out and look it over again. I hadn't thought to do such a thing as it was pretty out of my league in high school. I haven't looked at it in a long time, may have been before high school even. Knowing basic Calc might make it more comprehensible, it certainly made what Marcus and SelfAdjoint said work. Honestly, replies like these make these forums great for me, keep it up guys :)

Last edited: Feb 13, 2006
10. Feb 14, 2006

f-h

Sumo, sorry if that was to quick, but it's really difficult to give a correct and good laymans answer to these sort of questions based on only half correct popular physics statements, marcus and sa have done a good job there indeed, but if you want to grog what's truely going on you really need to study GR, QFT and RG.

For example you can't directly apply the perturbation theory of QFT to Gravity since it assumes at it's very foundation that there is no gravity (fixed background). Therefore what you apply them to is an approximation of GR compatible with this assumption.

And another point I want to emphasize here: Nonrenormalizability means we can not define the theory perturbatively, it does not mean that it can not be defined and it does not mean it doesn't have a well defined perturbative expansion once defined nonperturbatively.
An example would be 2+1 gravity.

It also really helps if you give your level of expertise with these questions.

11. Feb 14, 2006

George Jones

Staff Emeritus
I juist want elaboate on one point in your very nice post.

After regularization and renormalization, the individual terms in a perturbation series are all finite, but a typical perturbation series still diverges. Some people think that this is a problem, but some don't.

Regards,
George

12. Feb 14, 2006

marcus

fh, in the "group mind" stew that is PF
a comment that is OVER SUMO'S HEAD will often be just right for
someone else. this is how it works. So thanks to Sumo for asking a good question! And everybody can answer at all different levels of sophistication and it all seems to work in a synergistic way---each person uses what is right for him.

I suspect you are a QG grad student somewhere with a good background in QFT and probably string as well, and I am wanting very much to see more posts from you----at whatever mathematics level. It's all good. thanks.

You get high marks for this one!

Last edited: Feb 14, 2006
13. Feb 15, 2006

f-h

Thanks, but you're slightly of the mark. I certainly hope to be all that in a few years though ;)

14. Feb 15, 2006

Staff Emeritus
Yes this is a very important point. Perturbation and renormalization are strategies for getting approximate answers, not deep ontological criteria. What may be true in the end is not clear from the current state of QFT.

One of the brags of string physics is that it doesn't require renormalization, but it still is heavily, heavily dependent on the perturbative strategy.

15. Feb 16, 2006

Careful

Who says you have to care about taking the continuum limit (ie. UV limit) at all ? Concerning the non perturbative approaches (actually none exists yet beyond 2+1 - but in these dimensions we do not have gravitons and the theory is also perturbatively renormalizable), all evidence points in the direction that we shall encounter an infinite number of ambiguities again. This presumed non perturbative renormalizability of gravity is comparable to looking for a needle in a haystack.

Cheers,

Careful

16. Feb 16, 2006

f-h

Careful, this is a widely held assumption, that is known to fail for some models. Quoting from Ashtekar:
http://www.gravity.psu.edu/people/Ashtekar/articles/final.pdf

"There exist quantum Field theories (such as the Gross-Neveu model in three dimensions) in which the standard perturbation expansion is not renormalizable although the theory is exactly soluble!"

17. Feb 16, 2006

marcus

Hi f-h, I found the passage you quoted on page 13 of "Gravity and the Quantum". The Gross-Neveu example may not satisfy Careful because apparently it is 3D instead of 4D. I will leave it to Careful to decide if he is happy with it. But I think it is still worth thinking about to see what one can conclude. I will quote the passage more fully:

"...Are there any situations, outside loop quantum gravity, where such physical expectations are borne out in detail mathematically? The answer is in the affirmative. There exist quantum field theories (such as the Gross-Neveu model in three dimensions) in which the standard perturbation expansion is not renormalizable although the theory is exactly soluble! Failure of the standard perturbation expansion can occur because one insists on perturbing around the trivial, Gaussian point rather than the more physical, non-trivial fixed point of the renormalization group flow. Interestingly, thanks to recent work by Lauscher, Reuter, Percacci, Perini and others there is now non-trivial and growing evidence that situation may be similar in Euclidean quantum gravity. Impressive calculations have shown that pure Einstein theory may also admit a non-trivial fixed point. Furthermore, the requirement that the fixed point should continue to exist in presence of matter constrains the couplings in non-trivial and interesting ways [52]. However, as indicated in the Introduction, even if quantum general relativity did exist as a mathematically consistent theory, there is no a priori reason to assume that it would be the ‘final’ theory of all known physics..."

18. Feb 17, 2006

Careful

A small remark, as far as I know gravitation is not *perturbatively* renormalizable, meaning that you have to add an infinite number of counterterms in the action to make the formal power series well defined (the resulting series won't converge even then I guess - but nobody really knows that). A theory which is exactly soluble always has a formal power series expansion, the solution is just of C infinity class in the coupling constant and not analytic. So, the Gross-Neveu model would still be perturbatively renormalizable around a value of the coupling constant in which the exact solution is of C infinity class, while gravity isn't unless you can rigorously show that perturbation theory is formally well defined around a nontrivial background (but I do not believe it to be the case).

Anyway, a theory of quantum gravity would still require one to dynamically determine such background (and many ambiguities would creep in there too).

Cheers,

Careful

Last edited: Feb 17, 2006