DJBunk
• Member for 9 years, 9 months
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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 ...

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_{\... View answer Accepted answer 2 votes In your post when you say you 'know' the position and momentum of a single photon you really don't know anything, you are just making a prediction, not making a measurement. In your head you are ... View answer 2 votes All your equations are correct. $$H\Psi(x,t) = i\hbar\frac{\partial}{\partial t}\Psi(x,t)$$ says that we will differentiate wrt to time, keeping x constant. $$H|\Psi\rangle = i\hbar\frac{d}{dt}|\... View answer 2 votes As far as renormalizability of gauge bosons, a nonabelian gauge theory with massive gauge bosons is non-renormalizable as I describe here: What evidence is there for the electroweak higgs mechanism? . ... View answer Accepted answer 2 votes For a static situation (i.e. no charges moving or current flowing) the net electric field is always zero inside of a conductor. Where by 'inside' I mean actually inside the material itself, not a ... View answer Accepted answer 2 votes In the case of quantum field theory: First of all for a massless gauge field the most general form of the effective action will contain the renormalizable term \mathcal{L} = -\frac{1}{4 g^2} F^{\... View answer 1 votes The anomalous dimension for the field strength is defined as (eqn 12.63 Peskin): \gamma = \frac{1}{2} \frac{M}{Z} \frac{\partial Z}{\partial M} = \frac{1}{2} \frac{\partial \log Z}{\partial \log M} ... View answer 1 votes Building on DavePhD's answer, you need to have some sort of experiment in order to measure/resolve such a length, and this would be done with an interference experiment. In order to have any meaning, ... View answer 1 votes The time dilation you speak of is a description of the apparent time an observer outside witnesses someone falling into a black hole. That is, if you are standing outside the black hole (some distance ... View answer Accepted answer 1 votes Any term in the action that does not transform under any symmetries is allowed. This means for a scalar field you can have any power \phi^n in your Lagrangian. In equation (92.2) Scrednicki just ... View answer Accepted answer 1 votes I can't see where this has any utility at all. The point of having any equation, differential or algebraic, is to put constraints on a system. We then solve these equations to obtain an unknown ... View answer 1 votes Your \mu^\epsilon is still there, it's just that you have expanded in small \epsilon so you got$$ \mu^\epsilon \approx 1 + \epsilon \log \mu = 1 - \epsilon \log \frac{1}{\mu}$$The$\log \...
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: $\partial_\beta \partial^\beta A^\alpha + \mu^2 ... View answer Accepted answer 1 votes The truck is indeed moving up the hill, and the tires are not slipping. There are a couple of ways to see why the friction points in the direction of the motion of the truck. One way is to keep in ... View answer 1 votes OK just to be clear the lagrangian for a scalar field theory for scalar field is written as$ \mathcal{L} = \frac{1}{2} \partial^\mu \phi \partial_\mu \phi -\frac{1}{2} m^2 \phi^2 $sometimes this ... View answer 1 votes Electromagnetic fields, which include static electric and magnetic fields, are indeed made of photons. From a particle physics perspective the Quantum Electrodynamics as a model of particles carrying ... View answer 1 votes From a practical standpoint it depends on your field and motivation whether relativity and the Dirac equation are necessary tools. If you are working with things like the Aharonov–Bohm effect or in ... View answer 0 votes First of all just to refine your notion of force and acceleration by `force = mass X acceleration' what we mean is the relationship between the vectors:$\vec{F}_{net} = m \vec{a}$that is, its the ... View answer 0 votes Colin did a good job answering this question but I can't help fill in some other things. As for the utility of force, momentum and energy: First of all, when you are trying to solve a physics ... View answer 0 votes I am not an expert either, but I think the answer is still simpler than the one that James provided. First of all, the Coriolis effect is only mysterious if you think that your spot on the earth is ... View answer -1 votes This: Which I borrowed from: enter link description here The lines go off to infinity and never terminate. EDIT: As per your comment, you seem to be asking about the difference between an electric ... View answer -1 votes Nima gives a better explanation in this video I will be able to in written format here. Relevant portion starts at about 20 min in http://streamer.perimeterinstitute.ca/Flash/c53bd5c4-4108-489c-b43b-... View answer -4 votes All we can ascribe to a process like you are describing ($\gamma + p \rightarrow products\$) is a probability. The sum over all process will give us %100. I couldn't find a plot of anything like the ...