Thursday 28 February 2019

quantum electrodynamics - Where and how exactly does string theory and Q.E.D. use zeta function regularization?


In the video they mention it being used in many fields of physics inclusing String and QED theory.


https://www.youtube.com/watch?v=w-I6XTVZXww


But I remember reading somewhere that 1+2+3..=-1/12 is obviously a "mathematical trick" (something about stupidly equating incompatible sets), and if so, how does this turn out to be true for things that are real (like QED)?



Answer



Zeta function regularization is used in other fields, and even in pure mathematics to obtain finite answers from otherwise divergent integrals. In bosonic string theory, the mass of states in lightcone gauge is,


$$M^2 = \frac{4}{\alpha'} \left[ \sum_{n>0} \alpha^{i}_{-n}\alpha^{i}_n + \frac{D-2}{2}\left( \sum_{n>0} n\right) \right]$$


where $\alpha'$ is the universal Regge slope, $D$ is the spacetime dimension, and $\alpha^{i}_n$ may be interpreted as Fourier coefficients of the expanded form of the embedding functions $X^{\mu}(\sigma)$ in the Polyakov action which provide a map from the worldsheet to the target space. We use the fact that


$$\sum_{n>0} n = 1+2+3+...=\zeta(-1)=-\frac{1}{12}$$



to write the expression for the mass of states as,


$$M^2 = \frac{4}{\alpha'} \left(N - \frac{D-2}{24} \right)$$


If we look at the ground state, corresponding to $N=0$, we see


$$M^2 = -\frac{1}{\alpha'}\frac{D-2}{6}$$


which corresponds to a particle with an imaginary mass, known as a tachyon. The demand that we preserve $SO(1,D-1)$ Lorentz symmetry forces us to choose that the first excited state $(N=1)$ be massless, and so we must choose the spacetime to be $D=26$. In other string theories, the critical dimension of the string may be lower, e.g. $10$ or $11$. For further details, I recommend Prof. Tong's lectures notes on string theory available at: http://www.damtp.cam.ac.uk/user/tong/string.html.


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...