Recent content by Markus Kahn

Assume that I have the Lagrangian $$\mathcal{L}_{UV} =\frac{1}{2}\left[\left(\partial_{\mu} \phi\right)^{2}-m_{L}^{2} \phi^{2}+\left(\partial_{\mu} H\right)^{2}-M^{2} H^{2}\right] -\frac{\lambda_{0}}{4 !} \phi^{4}-\frac{\lambda_{2}}{4} \phi^{2} H^{2},$$ where ##\phi## is a light scalar field...

My attempt at this: From the general result $$\int \frac{d^Dl}{(2\pi)^D} \frac{1}{(l^2+m^2)^n} = \frac{im^{D-2n}}{(4\pi)^{D/2}} \frac{\Gamma(n-D/2)}{\Gamma(n)},$$ we get by setting ##D=4##, ##n=1##, ##m^2=-\sigma^2## $$-\frac{\lambda^4}{M^4}U_S \int\frac{d^4k}{(2\pi)^4} \frac{1}{k^2-\sigma^2} =...

This seems rather straight forward, but I can't figure out the details... Generally speaking and ignoring prefactors, the Fourier transformation of a (nicely behaved) function ##f## is given by $$f(x)= \int_{\mathbb{R}^{d+1}} d^{d+1}p\, \hat{f}(p) e^{ip\cdot x} \quad\Longleftrightarrow \quad...

@vanhees71 Thanks a lot for the explanations and I will be sure to check out your lecture notes! Just as a quick check, the issue is that I basically conflated the following, right? i.e. I assumed that ##\phi^4## has this one extra loop diagram that appears due to a ##\phi^3## interaction...

Alright, this makes sense. Then we have $$m_{\text{ren}}^2=m^2[1+I(m_{\text{ren}}^2)] \approx m^2[1+I(m^2)].$$ When exactly did that happen? Where in post #1 did I make a mistake so that I ended up in ##\phi^3## theory?

I'm sorry, but I don't understand how to do that... What I have tried (thought about) so far: $$ \frac{1}{p^{2}-m^{2}-m^2I(p^2)} \approx \frac{1}{p^2-m^2} + \frac{1}{p^2-m^2}m^2I(p^2)\frac{1}{p^2-m^2}.$$ Can we use this maybe like this: $$\frac{i Z...

Thank you very much for the response! I hope you mean the ##\log## that will eventually show up in ##I(p^2)##, if not, I'm not really sure what you mean. I just went back to my QFT1 lecture notes (Chp. 11.2) one more time to check, and my Prof. got for this integral two different expression...

Before I start, let me say that I have looked into textbooks and I know this is a standard problem, but I just can't get the result right... My attempt goes as follows: We notice that the amplitude of this diagram is given by $$\begin{align*}K_2(p) &= \frac{i(-i...

My attempt: Realize we can work in whatever coordinate system we want, therefore we might as well work in the rest frame of the fluid. In this case ##u^a=(c,\vec{0})##. The conservation law reads ##\nabla^a T_{ab}=0##. Let us pick the Levi-Civita connection so that we don't have to worry about...

Perfect, thanks a lot for checking and looking up the references!

1) We know that for a given Killing vector ##K^\mu## the quantity ##g_{\mu\nu}K^\mu \dot q^\nu## is conserved along the geodesic ##q^k##, ##k\in\{t,r,x,y\}## . Therefore we find, with the three given Killing vectors ##\delta^t_0, \delta^x_0## and ##\delta^y_0## the conserved quantities $$Q^t :=...

Thanks for spotting the typo. I'm rather new to this entire GR-formalism, i.e. the covariant derivatives, etc., so I was just a bit unsure if I'm really doing operations that are permitted. Also, ##C=1## seemed a bit odd in the first moment, but if you think this works, then I'm happy!

My try: $$ \begin{align*} \nabla^a T_{ab} &= \nabla^a \left(\nabla_{a} \phi \nabla_{b} \phi-\frac{C}{2} g_{a b} \nabla_{c} \phi \nabla^{c} \phi\right)\\ &\overset{(1)}{=} \underbrace{(\nabla^a\nabla_{a} \phi)}_{=0} \nabla_{b} \phi + \nabla_{a} \phi (\nabla^a\nabla_{b} \phi)-\frac{C}{2}...

@PeroK Could you elaborate? I'm asking because I don't see how what you write adds up with the very next line in my professors reasoning, i.e. $$\begin{aligned} \left(\gamma^{\mu} \partial_{\mu}^\prime-m\right) \psi^{\prime}(x^\prime)\neq\left(\gamma^{\mu} \partial_{\mu}-m\right) S \psi(\Lambda...

The problem is given in the summary. My attempt: Assume that ##\psi^\prime (x^\prime)## is a solution of the Dirac equation in the primed frame, given the transformation ##x\mapsto x^\prime :=\Lambda^{-1}x## and ##\psi^\prime (x^\prime)=S\psi(x)##, we have $$ \begin{align*} 0&=(\gamma^\mu...