Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

In Hartle-Hawking "no boundary" proposal it is proposed that Riemannian spacetimes rather than Lorentzian dominated the path integral near the big bang.

Moments after the big bang however spacetimes with Lorentzian metrics started to dominate over the Riemannian. The dominance of the Riemannian spacetimes is characterized by positive definite metrics obtained by applying Wick rotation.

Now to compute the path integral some approximations are made and the process of analytic continuation is applied. What bothers me is analytic continuation of an approximate holomorphic function is not guaranteed to remain analytic in another region.

Why then this unreliable process applied—can one rely on this scheme which seems to be mathematically dubious?

share|improve this question
3  
I find your question tendentious and the wording misleading. The HH wave function is just a mathematical expression to determine a preferred state of the Universe on a three-sphere. It uses maths of a Euclidean spacetime because it is natural and the Ansätze are allowed to use any maths: you may still view the Euclideaniation as a math trick. Moreover, contrary to your implicit assertions, quantum field theory in a Euclidean spacetime is more mathematically well-behaved than quantum field theory (or QG) in the Lorentzian space, so the Euclideanization makes things better, not worse. –  Luboš Motl Apr 10 '11 at 5:45
2  
What I want to say is that HH were formulating a hypothesis about a state, and one can use any mathematical machinery to formulate a hypothesis. They used what they considered the most natural one, one which solved the WDW equation etc., and I tend to agree. Testing against evidence is what decides about the validity of a scientific hypothesis. I don't understand your problem. In the second part of the question, you assume that the expression in the Minkowski space are always "more true" than those in the Euclidean space. You have no evidence for that and in fact, I believe you're wrong. –  Luboš Motl Apr 10 '11 at 5:53
4  
Otherwise, approximating the HH state by a minisuperspace approximation etc. is always a problem and can lead to wrongness by itself. It's surely true. However, it may also be a good approximation for many problems. At any rate, this problem has nothing to do with the Euclideanization: the approximations done in the Euclidean signature are likely to be more accurate, much like in other cases where the Wick rotation is used. It is being used for a good reason - it makes the maths more convergent etc. E.g. path integrals only "rigorously" exist as a measure in the Euclidean signature. –  Luboš Motl Apr 10 '11 at 5:57
    
It uses maths of a Euclidean spacetime because it is natural and the Ansätze are allowed to use any maths: Still doesn't explain what mechanism would cause the signature of the space-time manifold to change from Lorentzian to Riemannian. I think that is @sb1's original question - or at least that's my question. –  user346 Apr 10 '11 at 7:21
    
@Sklivvz Reposting comments is not helping and is seemingly considered controversial, so I deleted your answer. Yet, @Luboš, please don't post long texts as comments, it is just hard to read. –  mbq Apr 11 '11 at 9:00

1 Answer 1

What is bothering you here is not Hartle-Hawking per-se, but the process of analytic continuation inside a path integral. This can be mystifying, because the actual functions which dominate the integral are never analytic, they are hardly ever continuous or even bounded. Usually they are distributional, so that they fluctuate in terrible ways at short distances. How can you analytically continue them?

But you don't ever continue the functions, you continue the path integral. The functions which you integrate over in imaginary time are of no relation to the functions which you integrate over for real times. The reason you can continue the path integral is because of positive energy--- the contribution of high-energy intermediate states decay quickly in positive imaginary time, and there are no negative energy contributions.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.