Recent content by Sonny Liston

Nevermind -- got it sorted.

...Help?

This isn't a homework problem, but something from a set of notes that I'd like to better understand. My confusion starts on page 23 here: http://isites.harvard.edu/fs/docs/icb.topic1146665.files/III-9-RenormalizationGroup.pdf. I'm having trouble reproducing his calculation for the...

If by "the first term" you mean the \pi^2 term, then yes, the \phi derivative will kill that term; reason being \phi doesn't appear in that term, only its time derivative. Edit: Sorry, I understand what you meant by "first term" now. The first term vanishes at infinity so you can ignore it.

That was my original thought -- you just couldn't write down sensible diagrams for the process. The wording was what then got me all tangled up thinking diagrams should cancel and whatnot. Thanks for all your help!

Right, that's part of what I was trying to get at (and expressing poorly) when I said I was having trouble writing down diagrams that made sense. Following the rules for electron/positron arrows produces a diagram that violates the rule concerning arrows flowing into and out of vertices. So I...

I think I've got my arrows right. I tried to type out the diagrams themselves, but couldn't get it to work so I'll just describe them. Starting from the left: Diagram 1: incoming electron, momentum P, with an arrow to the right flowing toward the first vertex. A scalar line with momentum K...

Homework Statement I'm working with the Yukawa theory, where the interaction term in the Lagrangian density is g\varphi\overline{\psi}\psi. As an exercise for getting used to using the Feynman rules for the theory, I'm asked to show explicitly (i.e. I'm not allowed to invoke charge...

Btw, I don't think I understand why the magnetic flux is a function of the energy here.

I can't thank you enough for all your help.

B\int x(t) v_{y} (t) = B \frac{-(v_{0})^2}{\omega} \int (sin(\omega t)^2) Using (sin(\omega t)^2) = \frac{1 - cos (2\omega t)}{2} and doing the integral, I get \Phi = B\pi(- \frac{v_{0}}{\omega})^2

So there is nothing wrong in my reasoning and the flux actually is zero? It seems odd to ask me to evaluate it "as a function of the energy" then.

In my notes, I have that it is given by \frac{1}{2\pi}\oint A dx, and the baby E&M book I am looking at says that it can be given by the Area of the surface dotted with the B field times cos(theta), for theta the angle between the B field and the surface. But since the B field is homogeneous...

Assuming that's satisfactory, may I ask you one last quick question (you are a saint, by the way. I have been unbelievably useless throughout this.) How would I go about evaluating the magnetic flux through the classical trajectory as a function of the energy? In case you're wondering about my...

1 - cos (2 \omega \frac{(2\pi}{\omega}) = 0, so it is appropriately periodic.