I was reading about the renormalization of ##\phi^4## theory and it was mentioned that in order to renormalize the 2-point function ##\Gamma^{(2)}(p)## we add the counterterm :
\delta \mathcal{L}_1 = -\dfrac{gm^2}{32\pi \epsilon^2}\phi^2
to the Lagrangian, which should give rise to a...
Sorry, I assumed the concepts and notation were well known. Here's how Hobson puts it:
"A maximally symmetric space is specified by just one number - the curvature K, which is independent of the coordinates. Such constant curvature spaces must clearly be homogeneous and isotropic."
So it's...
Why does the constraint:
$$R_{ijkl}=K(g_{ik} g_{jl} - g_{il}g_{jk})$$
Imply that the resulting space is maximally symmetric? The GR book I'm using takes this relation more or less as a definition, what is the idea behind here?
I had a vague idea of how things could work out, let me see if it's what you're suggesting. If we bombard an atom with muon antineutrinos we ought to observe electrons being scattered out of it, which would give an indirect confirmation for the existence of the ##Z^0## boson. Is that it? No hope...
In decay processes where no mixing between quark families is present, the mediator of the weak force is the neutral ##Z^0## boson. If that is the case, how is it experimentally possible to detect neutral currents in processes such as: $$\bar{\nu}_\mu + e \rightarrow \bar{\nu}_\mu + e$$ What...
I see now. I just had to use the product rule and the fact that:
$$\frac{\delta J(y)}{\delta J(x)} = \delta^4(x-y)$$
What really confused me was the ##\frac{Z[J]}{Z_0}## term appearing out of nowhere. But on second look that's just:
$$\frac{Z[J]}{Z_0} = \text{exp}[-\frac{1}{2}\int d^4 x \; d^4 y...
I don't understand a step in the derivation of the propagator of a scalar field as presented in page 291 of Peskin and Schroeder. How do we go from:
$$-\frac{\delta}{\delta J(x_1)} \frac{\delta}{\delta J(x_2)} \text{exp}[-\frac{1}{2} \int d^4 x \; d^4 y \; J(x) D_F (x-y) J(y)]|_{J=0}$$
To...
Sorry, I'm still confused. If I do something like: $$\frac{d}{dt}(-P_a \xi^a) = \frac{1}{c}\frac{d}{d\tau}(-P_a\xi^a)=-\frac{1}{c}\left(\frac{dP_a}{d\tau}\xi^a + P_a\frac{d\xi^a}{d\tau}\right)$$
It vanishes because the derivative of the killing vector is zero by definition and we are imposing...
Hi, sorry for not providing the reference. I read it on Wald (page 287).
Can we be more precise about this statement? If I take the derivative with respect to time of ##E = -P_a \xi^a## it should yield zero, right? But how exactly do I calculate this derivative?
I'm struggling to get the hang of killing vectors. I ran across a statement that said energy in special relativity with respect to a time translation Killing field ##\xi^{a}## is: $$E = -P_a\xi^{a}$$ What exactly does that mean? Can someone clarify to me?
Nevermind, you're right. It's the same thing as the expression I wanted, since: $$\delta^3(p_A-p_2)=\delta^3(-(p_2-p_A) )= \frac{\delta^3(p_2-p_A)}{|-1|}=\delta^3(p_2-p_A)$$
Everything is fine.
I did the same thing that I had done with ##a_2a_{A}^{\dagger}##, but this time for ##a_2 a_{B}^{\dagger}##:
$$a_2 a_{B}^{\dagger} = [a_2,a_{B}^{\dagger}] + a_{B}^{\dagger}a_2$$
I did this to ##a_1 a_{B}^{\dagger}## and ##a_1 a_{A}^{\dagger}## too. The result is pretty close to what I wanted...
Can I bother you a little more? I got a bit further on the deduction of the formula, but now I'm stuck at:
$$ \int \bar{dk_{A}^{z}}\delta(\sqrt{ \bar{k_{A}^{2}}+m_{A}^{2}}+\sqrt{ \bar{k_{B}^{2}}+m_{B}^{2}}-\sum E_f)|_{\bar{k_{B}^{z}}=\sum p_{f}^{z}-\bar{k_{A}^{z}}} $$
$$ =...