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zetafunction
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since the strings are no longer points, the UV divergences are avoided from calculations, but what happens with IR divergences ? ,
Why fortunately?suprised said:They are there, fortunately.
Demystifier said:Why fortunately?
Demystifier said:Why fortunately?
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 said:I don't get it why getting an infinity would be harmless. You mean, they are renormalizable right?
... provided that the IR cutoff is absent. But in a finite universe, it is not absent.lpetrich said: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.
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 ?Demystifier said:... provided that the IR cutoff is absent. But in a finite universe, it is not absent.
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.humanino said: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 ?
An IR (infrared) divergence is a mathematical issue that arises in string theory when calculating certain quantities, such as scattering amplitudes. It occurs when the momentum of particles involved in the calculation approaches zero, resulting in an infinite value. This is a problem because it means the theory cannot accurately predict physical phenomena at low energies.
String theory addresses IR divergences by introducing the concept of extended objects, or strings, instead of point particles. These strings have a finite length and therefore do not experience IR divergences in the same way that point particles do. This allows string theory to make predictions at low energies without encountering infinite values.
Supersymmetry is a fundamental symmetry of string theory that relates bosons (particles with integer spin) and fermions (particles with half-integer spin). This symmetry plays a crucial role in avoiding IR divergences by canceling out the contributions from bosons and fermions in calculations, resulting in finite values.
No, string theory does not completely eliminate IR divergences. While it greatly reduces the occurrence of these divergences, they can still appear in certain calculations. However, string theory provides a more consistent and reliable framework for dealing with IR divergences compared to traditional quantum field theory.
Yes, there are challenges in incorporating string theory into other areas of physics due to IR divergences. For example, in cosmology, string theory can face difficulties in predicting the behavior of the early universe due to IR divergences. However, ongoing research and developments in the field are working towards addressing these challenges and furthering our understanding of string theory.