Recent content by Maurice7510

That makes a lot of sense, thanks. I was just thinking about it in a general sense and honestly just failed to consider that, but I now realize that was a foolish oversight. Thanks a lot!

I just came across this and it was extremely useful but I'm having an issue when the number of indices on the two tensors is the same. For example, (in the integral) \begin{align*} \sqrt{g}A^{\mu}\nabla_{\nu}B^{\rho} &= \sqrt{g}A^{\mu}\partial_{\nu}B^{\rho} +...

Thanks a lot! I was looking up the Kiefer book when I happened to come across one by Percacci, the latter of which covers a ton of extremely relevant topics from our research, so this is ideal for us

I'm currently doing my thesis in QG and there's a distinct gap between where QFT and GR left off and QG begins, and as I'm sure most of you know, in a thesis you're sort of just thrown right into the deep end. As such, I was hoping someone could recommend a decent textbook that gives a solid...

I'll start by saying I'm posting this in Beyond the SM just because we have no elementary spin-3/2 particles in the SM as far as we know, though I was also considering posting it elsewhere. If you feel it's more appropriate in another area just let me know. As for the question itself, I'd like...

I'm not sure how that would be the case though; the indices on the creation operator are ##i## (or ##a##) and ##-1##. The lower index is the state number (i.e. ##\alpha_{-1}## creates a one particle state) and the upper indicates indicate whether the BCs are NN or DD.

In string theory, if we have NN BCs along ##X^i, i = 1, \ldots, n-1## and DD BCs along ##X^a, a = n, \ldots, 25## then you get, from ##\alpha^{i,a}_{-1}|0,p\rangle ##, ##n## massless vectors and ##24-n## massless scalars. I understand that for the first excited level, ##M^2=0## and so we have...

Thanks a lot! The thing I'm trying to show in the end is that the integral is proportional to ##e^{-mx}##, which clearly this is.

Yes, the factor of ##i## isincluded in the derivations I've seen, and I see now how it works. How might you recommend trying to solve this now? With Cauchy's integral formula, having $$\oint \frac{f(z)}{z-im}$$ would mean ##f(z)## would have a factor of ##\sqrt{z-im}## and would thus be zero...

Now I've seen an extremely similar argument before (the same integral with just ##r^2+m^2## in the denominator is commonly approached) but they never include the ##Im##; perhaps this is just a result of physicists lacking proper mathematical rigour, but I've seen it in a few sources and was...

Oh sorry, ##\theta## was from ##0## to ##\pi## and ##r## from ##0## to ##\infty##

Homework Statement The integral I want to solve is $$ D(x) = \frac{-i}{8\pi^2}\int dr\,d\theta \frac{e^{-irx\cos\theta}}{\sqrt{r^2+m^2}}r^2\sin\theta$$ which I've reduced to $$ D(x) = \frac{-i}{4\pi x}\int dr \frac{r\sin(rx)}{\sqrt{r^2+m^2}} $$ by integrating over ##\theta##. However, I...

I now realize I made a mistake in the spherical coordinate substitution, the integral should be $$ D(x) = \frac{-i}{8\pi^2} int dr\,d\theta \frac{e^{-irx\cos\theta}}{\sqrt{r^2+m^2}} r^2\sin\theta $$ The integral in ##\theta## is fairly straight forward at this point, and I got $$ D(x) =...

That is curious, though I wrote exactly what he has in the book. Initially, we have $$ D(x) = -i \int \frac{d^3k}{(2\pi)^32\omega_k}[e^{-i(\omega_kt - \boldsymbol{k}\cdot\boldsymbol{k})}\theta(x^0)+e^{i(\omega_kt - \boldsymbol{k}\cdot\boldsymbol{k})}\theta(-x^0)] $$ which he reduces to the...

Homework Statement I'm working through Zee for some self study and I'm trying to do all the problems, which is understandably challenging. Problem 1.3.1 is where I'm currently stuck: Verify that D(x) decays exponentially for spacelike separation. Homework Equations The propagator in question...