DJBunk
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Starting with the Lagrangian for a massive $U(1)$ vector boson $A_\mu$ which like you said has 3 DOF: $$\mathcal{L} = - \frac{1}{4 e^2} F^{\mu \nu} F_{\mu \nu} - m^2 A^\mu A_\mu$$ now if we change ...

The anomalous dimension for the field strength is defined as (eqn 12.63 Peskin): $\gamma = \frac{1}{2} \frac{M}{Z} \frac{\partial Z}{\partial M} = \frac{1}{2} \frac{\partial \log Z}{\partial \log M} ... View answer 2 votes Unless I am missing something the relation is trivial since starting with $$\epsilon^{klm}\sigma^m = \sigma^m \epsilon^{mkl}$$ and permuting the$m$past the$l$gives a factor of -1 $$(-1)\... View answer 5 votes I personally recommend Georgi's book with a particular focus on SU(3). And there is also Ramond's book, which is along the same lines as Georgi's textbook. Also online there are some notes ... View answer 1 votes Building on DavePhD's answer, you need to have some sort of experiment in order to measure/resolve such a length, and this would be done with an interference experiment. In order to have any meaning, ... View answer Accepted answer 5 votes Inductors resist changes in current. So in a circuit like the one you describe, for short times after the switch is closed, the inductor acts like a broken wire. This is consistent with the statement ... View answer 2 votes The fine tuning you describe isn't present in the \phi^4 model. You need some some other heavy fields around to see it. For example, couple your \phi to a heavy fermion of mass M, Then when you ... View answer 7 votes If you are getting used to 'natural' units I think its best to think of it like this: we are basically defining a new time variable t' \equiv c t to work in. t ' has units of distance. We can ... View answer 1 votes The time dilation you speak of is a description of the apparent time an observer outside witnesses someone falling into a black hole. That is, if you are standing outside the black hole (some distance ... View answer 3 votes First of all, by electromagnetic radiation we basically mean photons. That is particles are radiated by a black hole. Hawking radiation in general can be any type of particles, as in the same way in ... View answer Accepted answer 1 votes Any term in the action that does not transform under any symmetries is allowed. This means for a scalar field you can have any power \phi^n in your Lagrangian. In equation (92.2) Scrednicki just ... View answer Accepted answer 3 votes The first one is probably correct, and you are likely not counting correctly in the second method. You should compare you result \ddot{t}+ 2 \frac{\partial U}{\partial x^{i}} \dot{x}_i \dot{t}=0 ... View answer 4 votes You can find this by noting the the photon potential A_\mu couples to the electromagnetic current J_\mu in the form \mathcal{L}_{int} = A^\mu J_\mu. Where J_\mu obeys the continuity ... View answer 3 votes Picture this. Take your desk and put in top of the tallest building on earth. Put a penny on the desk and flick it with your finger. (The point of this is that we are only giving the penny a ... View answer Accepted answer 4 votes I think you are asking for a finite well of width L that is from -L/2< x< L/2. Why do we only use \psi(x) = A e^{+\kappa x} for x<-L/2 and \psi(x) = B e^{-\kappa x} for ... View answer Accepted answer 1 votes I can't see where this has any utility at all. The point of having any equation, differential or algebraic, is to put constraints on a system. We then solve these equations to obtain an unknown ... View answer 1 votes Your \mu^\epsilon is still there, it's just that you have expanded in small \epsilon so you got$$ \mu^\epsilon \approx 1 + \epsilon \log \mu = 1 - \epsilon \log \frac{1}{\mu}$$The$\log \...

First of all just to refine your notion of force and acceleration by `force = mass X acceleration' what we mean is the relationship between the vectors: $\vec{F}_{net} = m \vec{a}$ that is, its the ...

As far as where you put things like the $2 \pi$ and the $\hbar$ in the Fourier transform or Inverse Fourier transform, it doesn't really matter. What really matters is that the operations are the ...

If you take your Lagrangian, including the $A^\alpha A_\alpha$ and vary it with respect to $A^\alpha$, you will get the classical equation of motion: $\partial_\beta \partial^\beta A^\alpha + \mu^2 ... View answer -1 votes This: Which I borrowed from: enter link description here The lines go off to infinity and never terminate. EDIT: As per your comment, you seem to be asking about the difference between an electric ... View answer Accepted answer 1 votes The truck is indeed moving up the hill, and the tires are not slipping. There are a couple of ways to see why the friction points in the direction of the motion of the truck. One way is to keep in ... View answer 3 votes The massive verse massless cases are different. Massive vector bosons are a bit more 'honest' in their representation of the Lorentz group in that they have all 3 DOF implied by the$j=1$... View answer Accepted answer 5 votes First of all, this is just a change of basis, which is up to us to make. Furthermore we should always choose a basis that makes our calculations easier, and hopefully makes things more intuitive. For ... View answer Accepted answer 3 votes Your first interaction term is bilinear in 2 gauge fields. Terms of this form indicate that you need to diagonalize your mass matrix. So you could work out the Feynman rules for your first interaction,... View answer 7 votes There is an aspect to this question that nobody seems to have addressed and that is, although the higgs (the 'radial' component of the field) is neutral, and therefore doesn't interact with the photon ... View answer 4 votes If you want to see this from a straightforward implementation of the Feynman rules: You can always calculate the diagram for$e^- e^+ \rightarrow \gamma$and for arbitrary momentum it will be nonzero. ... View answer 7 votes If I understand your question correctly its just a matter of what you are calculating whether you put the external particles on shell or not. If you are, for example, calculating an amplitude to use ... View answer 2 votes Either way is fine:$\delta(x-x')= \langle x| x' \rangle= \delta(x' - x) =\langle x'| x \rangle$. You can see this either from the fact that in the limiting definition$\delta (x-x') = \lim_{\...