# Tag Info

2

First note that $$\frac{\partial}{\partial m} \frac{i}{\not p - m} = \frac{i}{(\not p-m)^2}$$ Now the exact answer for finite $q$ for that diagram is $$\mathcal{A} = \left( \frac{im}{v} \right) \frac{i}{\not p-m} \frac{i}{\not p + \not q - m} G(q)$$ where $G(q)$ is the propagator for whatever ...

1

We have to change the renormalization scale to a typical scale of the process we are considering because otherwise perturbation theory becomes rather useless, since process far away from the renormalization scale get large higher order corrections due to the "large log" $\ln(-p^2/\mu^2)$, where $p$ is the process scale and $\mu$ the renormalization scale. ...

-1

Forgive my ignorance but it seems to me that the other 3 fundamental forces can be described in electrical terms (hence unification I presume). We use nuclear forces even in power stations. Yet gravity seems to me to be an entirely different entity, it can be seen to be a distortion in space-time. If I picture in my mind a future time where some sort of ...

1

Every phase transition has an order parameter: something that vanishes above the transition temperature and is finite below. In superconductors, the order parameter is a complex quantity related to the superconducting gap: $\Delta = |\Delta| e^{i \phi}$. In BCS theory, there is a self-consistent equation for the gap: $\Delta_k = -\sum_q V_{kq} ... 0 Renormalization group is usually used because of some divergence your are avoiding in perturbation theory(such as the divergence of your greens function). In perturbation theory your still expanding in terms of the small parameter before you apply the renormalization group action. Therefore your expansion is still valid. Also just because your parameter ... 1 I am not sure whether the standard model has a Landau pole at$10^{34}$GeV but there are two obstacles to providing a definite answer to the question: (1) perturbation theory is no longer valid when the coupling constants get large, and (2)$10^{34}\$ GeV is well beyond the Planck scale, so that ignoring the effects of (quantum) gravity is not valid. If ...

Top 50 recent answers are included