Recent content by Trifis

Does anybody can give me a hint on how to reach S_0(x_t) = \frac{3}{2}\frac{x_t^3}{(1-x_t)^3} \ln{x_t} - [\frac{1}{4}+\frac{9}{4}\frac{1}{1-x_t}-\frac{3}{2}\frac{1}{(1-x_t)^2}] x_t from A(x_i,\,x_j) = \frac{J(x_i)-J(x_j)}{x_i-x_j} in the limit of large mt?

You are right, I was thinking about p as the integral variable, but it was M' instead. Thanks for clearing this out!

I've made some typing errors. The integration after introducing the new variable t=\ln{(p/M)} is: \frac{de}{dt}=-\frac{e^3}{12\pi^2} \int_{e(p)}^{e(M)} \frac{ de}{e^3} = -\frac{1}{12\pi^2} \int_{\ln{(p/M)}}^0 dt which yields the final result e(p)^2 = \frac{ e(M)^2}{1 + \frac{e(M)^2}{6\pi^2}...

So there is always one e_0 factor, which we trade for its expansion in e. The \delta_3 contains only e prefactors. In the end we get a beta-function depending only on e, which one can use to calculate the running coupling.

The beta function is defined as: \beta(\lambda)=M\frac{d}{dM}\lambda If we make the substitution t=ln(p/M) the above equation becomes: \beta(\lambda)=-\frac{d}{dt}\lambda Now if we use e.g. the QED beta function \beta(e)=\frac{e^3}{12\pi^3} and for e(p=M)=e_0 the result is...

So what happens at higher loop orders? Do we get a e_0 ⋅ e^n factor and then solve e_0 for e and replace? What about the \beta(e) = \mu \frac{d}{d\mu} e \bigg|_{e_0} equation (that I just noticed on pg. 417) ? Setting e_0=e at any order would yield a convenient result. But then again, I don't...

If this is the case, then when I try to compute the correct QED beta function I get: \beta(e) = \mu \frac{d}{d\mu} e = \mu \frac{d}{d\mu} [(1+\frac{\delta_3}{2})e_0] = \frac{e_0 e^2}{12\pi^2} and not the e_0^3 factor (or e^3 factor, I no longer know what is right). PS: In QED Z_1=Z_2, so...

Now that we are at it, I think they have the coefficients of the \delta terms also wrong. For example the \Pi_2 diagram was calculated before the normalization and thus it has to contain the bare electric charge e_0 and not the renormalized charge e. Another way to see this is the definition of...

I've spent a day searching for alternative formulations. Totally forgot about the ERRATA, so I guess it was my fault. Thank you very much for the quick reply.

I think I have found a mistake/wrong formulation at Peskin’s, when he discusses the renormalization of QED. In particular, he defines the 1PI of the electron’s self-energy on page 331 as: –i\Sigma( \displaystyle{\not}p ) and the corresponding counterterm on page 332 as: i(...

I would be more than interested in a continuation of this discussion. Monopoles symmetrizing the Maxwell equations seem to cause a hell of a lot of problems as far as spacetime and Lagrangian formulation of electromagnetism is considered. We practically have to invent a new, highly non-trivial...

@fzero Excellent reply, many thanks.

It is not clear to me, why textbooks do not mention an upper bound for the e-foldings of the basic inflation theory. To my knowledge, in order to deal with the flatness problem, we require: \frac{Ω^{-1}(t_0)-1}{Ω^{-1}(t_i)-1} = \frac{Ω^{-1}(t_0)-1}{Ω^{-1}(t_e)-1}...

Yes, and the fact that operators are no longer observables in this many-body QM context, makes me think that the basic Dirac-von Neumann axioms do no longer have to hold. I've done some reading on the proof of the spin-statistics theorem. It seems that the basic assumption is indeed the...

Ok, so field operators after the second quantization do not follow the basic axioms of 1-particle QM. There must be a lot of mathematical subtleties in this one, but let us not go off-topic.