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...
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
$$Q^t :=...
Thanks for spotting the typo. I'm rather new to this entire GR-formalism, i.e. the covariant derivatives, etc., so I was just a bit unsure if I'm really doing operations that are permitted. Also, ##C=1## seemed a bit odd in the first moment, but if you think this works, then I'm happy!
@PeroK Could you elaborate? I'm asking because I don't see how what you write adds up with the very next line in my professors reasoning, i.e.
$$\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...