How does string theory avoids IR divergences

[zetafunction]

since the strings are no longer points, the UV divergences are avoided from calculations, but what happens with IR divergences ? ,

 

[suprised]

They are there, fortunately.

 

[Demystifier]

They are there, fortunately.

Why fortunately?

 

[zetafunction]

perhaps they are easier to compute rather than UV divergences , for example

$$f(a)= \int_{0}^{a+1}dx (a^{2}+x^{2})^{-3}$$

is well defined for every real 'a' let us make an analytic continuation to the complex plane $$a \rightarrow -ia+i\epsilon$$

so $$f(i\epsilon-ia) = \int_{0}^{a+1}dx (-a^{2}+x^{2})^{-3}$$

and we have AVOIDED the IR divergence at the point x=a

 

[genneth]

Why fortunately?

IR divergence means phase change / parametric instability. Considering how many symmetries string theory must "spontaneously" break to approach known phenomenology, there must be a good cascade of IR divergences :-)

 

[suprised]

Why fortunately?

Well IR divergences must be there, otherwise field theory results wouldn't be recovered at low energies. In particular the running of the renormalized gauge coupling constants is due to IR divergences in string theory, and if they wouln't be there it would be quite a disaster, isn't it.

No one wants a completely finite theory!

Edit: ... and those who want to compute eg the fine structure constant out of a miracoulus formula should also take this running into account, to make any sense at all.

 

[MTd2]

I still don't get it. In the universe, right after the big bang, strings would be essentially a gas . So, we have, at that time, an effective theory that is divergent!

 

[suprised]

Look there are different kinds of divergences and if one is bothered one should first of all specify which ones one means. Some IR divergences (tachyonic ones) indeed signal a vacuum instability, other ones take care about the logarithmic running of gauge couplings, still other ones arise from massless particles being on-shell, etc. Most of them are pretty harmless.

 

[MTd2]

I don't get it why getting an infinity would be harmless. You mean, they are renormalizable right?

 

[humanino]

I don't get it why getting an infinity would be harmless. You mean, they are renormalizable right?

I think what is meant is that some IR divergencies are physical. The infinity spitted out is not due to the theory being sick, but the question asked being meaningless, or at least ill-posed (does not correspond to feasible measurement). Take QED IR divergencies. Simplifying a little bit, if you ask the question "how many photons collinear and with zero energy propagate along a charged particle ?", you do get infinity. In real life, detector resolutions always save the day. If you improve your detector resolution, you do count more collinear photons. One worries about these infinities only before one realizes that the question asked was ill-posed. Of course, it is not obvious at first. This is what lead to the concept of jet in QCD for instance.

 

[MTd2]

But, if strings are fundamental, there shouldn't be an ill posed question to anything!

 

[suprised]

humanino said it already well. Every S-Matrix has all sorts of singularities; the moment an internal leg of an amplitude sits on the mass-shell, you get a divergence. String scattering (Veneziano amplitude) even has infinitely many poles. There is nothing wrong with that. This is an idealized situation and in reality such singularities don't occur because of a finite line width of the intermediate states; generically IR singularities tend to disappear in finite volume anyway. This issue is essentially a non-problem!

 

[MTd2]

Alright. So, these IR happens just in approximative calculations.

 

[Demystifier]

genneth and suprised,
If the size of the Universe is finite (say 100 billions light years or more), then IR divergences are absent. I don't see how that could be a problem for particle phenomenology.

 

[zetafunction]

in general demystifier this would be the main reason the fact that there is no INFINITE wavelength in nature

in many cases (if not all) the IR divergences appear as poles inside the integrand and can be evaluated by using Cauchy's Residue theorem

 

[lpetrich]

If one can get the Standard Model out of string theory, then that's a proof that string theory can have infrared divergences, because the SM has such divergences. Bremsstrahlung processes produce an infinite number of low-energy soft photons, and such divergences will appear in any theory with massless particles, at least in 4 space-time dimensions.

 

[Demystifier]

Bremsstrahlung processes produce an infinite number of low-energy soft photons, and such divergences will appear in any theory with massless particles, at least in 4 space-time dimensions.

... provided that the IR cutoff is absent. But in a finite universe, it is not absent.

 

[humanino]

... provided that the IR cutoff is absent. But in a finite universe, it is not absent.

I must say, I am not sure I fully understand that argument. What precisely prevents a wavelength substantially larger than the radius of a ball to exist on the ball ? Would not it simply show as a slowly rotating breathing mode of the ball ?

Besides, is it supposed to be a valid argument in a negative cosmological constant universe ? The cosmos would expand faster than it takes the wave to go around it.

 

[Demystifier]

I must say, I am not sure I fully understand that argument. What precisely prevents a wavelength substantially larger than the radius of a ball to exist on the ball ?

I am talking about closed finite universe, which can be thought of as a periodic universe. A Fourier expansion of a function on such a universe involves the biggest possible wavelength.

 

[genneth]

Demystifier: I agree, but I don't think that was the point of the OP.

 

[humanino]

Well I still do not agree. The argument works on a compact spacetime. There, periodic boundary condition on the hyperplane modulo the compactification group amounts to single-valuness on the compact spacetime. But with a negative cosmological constant, although you can have a compact space, you do not have a compact spacetime anymore, and you do not have periodic boundary conditions in the time direction anymore. The reason I am complaining is that apparently our universe has a negative cosmological constant.