It should depend on what the Resume is for, i.e. if you want to show to show (i.e. think that the most important thing for whoever this is for) that you have experience in coding the first version sounds better, if you want to stress your experience with molecular dynamics the second one, if...
Well, what is possible is to turn around my smartphone very quickly, which evidently has no effect. So I would think it should not affect at least our every day electronics.
Maybe, but if we lived in a world with such a very strong magnetic field we would we would I assume not have the...
Even if it happened over night, why should if affect electronics (apart from, as mentioned, devices that are constructed to use the magnetic field to tell directions)? I would think things like my laptop shouldn’t care since I can use them in an arbitrary direction relative to the earth’s...
Well, the reason that certain nuclei decay in general could could be seen as the fact that other configurations are energetically favorable, yes.
But for a single nucleus, the decay is inherently random, e.g. it will decay at some random time, without any further cause. You can predict how many...
Maybe I am misunderstanding your notation, but just ##G_{a\bar{b}}G^{a\bar{b}}## written out is not the term you are replacing in the lagrangian (even ignoring the different number of terms that would be summed), so how are you supposing this should work?
Ok, in that case you still need ghosts. From a pragmatic point of view, there are contributions form unphysical states, and you need to subtract them out, and adding ghost particles is the way to do that. Without introducing the gauge fixing term, you would not even get to the stage where one...
I am not sure what you are asking.
First, with your reference to gauge theories it makes most sense you are talking about Faddeev-Popov ghosts, but then the wiki page you linked lists a bunch of other possible meanings of ghost, so which one are you asking about?
Second, assuming you are...
I don't think there is a real theoretical limit for photon energies from particle collisions, other than something like (half) the energy available in the universe. Practically of course, our ability to build larger and more powerful accelerators is very much limited, but that is just an...
In addition to what was said before:
I think you are even misinterpreting what the Wikipedia article is doing. The calculation there is an estimate for ##a_C##, assuming that you can approximate ##Z(1-Z)## by ##Z^2##. So they start out with a spherical symmetrical charge distribution, for which...
Ok, so this is supposed to be an cartesian inner product over spherical coordinates, i.e.
##\langle\Psi|\Phi\rangle = \int \mathop{d\phi}\mathop{d\theta}\mathop{dr} \Psi^*(r,\theta,\phi)\Phi(r,\theta,\phi)##
that then is somehow only defined on the subspace where the integral converges? What...
Which are exactly the expressions I wrote above. My point is that
##\langle\Psi|\Phi\rangle = \int \mathop{dx}\mathop{dy}\mathop{dz} \Psi^*(x,y,z)\Phi(x,y,z) = \int \mathop{d\phi}\mathop{d\theta}\mathop{dr} r^2 \sin\theta \Psi^*(r,\theta,\phi)\Phi(r,\theta,\phi)##
so what sense does it make...
Yes, but ##w(x)## is different in different coordinates systems in a way that the integral value stays the same, right? I was assuming this is implicitly contained in the notation ##\mathop{d^3 x}##, i.e. in cartesian coordinates
##\mathop{d^3 x} = \mathop{dx} \mathop{dy} \mathop{dz}##
whereas...
What is a “carthesian inner product”? The usual inner product should be
##\langle\Psi|\Phi\rangle = \int \mathop{d^3 x} \Psi^*\Phi##
and it should not matter what coordinates I choose in practice.
Sure, I understand what you were saying there, but I do not think it is enough to just wave your hands and say “at infinity everything will be fine” without deriving what the terms at infinity really would be. Anyway, I was pointing the the last equation in #7. Do you agree that, if we just...
Sure, but in this specific case everything is real anyway.
I know, and I think we have arrived at these conclusions already multiple times during this thread. With my latest posts I was trying to understand @Hans de Vries comments about ##p^2##, since I am not familiar with this “getting...
Where are you getting your definition of what a photon is from?
If we start with quantum theory, there is a quantum field for the electromagnetic force, and its excitation are photons. Charged particles are (by definition of the word charged) the particles this quantum field couples to, and...
I am having the feeling we are just discussing nomenclature here. Is there an actual physics question this is building up to? Do you want to as about the use of the word stability in a particular context? Then it would be nice to provide some reference.
Basically yes (as @PeterDonis pointed out, a bound state of charged particles can still interact even if its net charge is zero). There is really no notion of a photon required here. What would your answer be to “Why does the electromagnetic field only influence charged particles?”
Ok, I guess this is where the problem is. Photons are the particles that mediate the electromagnetic interaction, they appear after quantizing the em field. There are many particles, all have Energy and may or may not be charged. But coupling to charged particles is really what makes a photon a...
What do you mean by that?
I don’t think there is a real answer to why a particular particle has charge 0 or not. Of course proton and neutron are themselves bound states of quarks, so their charge is give by the charge of the quarks they contain. But that is not really an answer since it just...
ok, I see what you did but not really how it applies here. I would have assumed that, to get a matrix representation, I would have to calculate something like
## (p^2)_{mn} = \langle\Psi_m|p^2|\Psi_n\rangle##
and this would be symmetric ##(p^2)_{mn} = (p^2)_{nm}##?
I would say it is the other way around, we call those particles charged that the photon interacts with (with the magnitude of the charge giving the strength of the interaction).
I mean a process where 1 particle decays into several (usually 2). That is what the word decay means to me. Fusion starts with two particles (at least).
I would say a decay is a ##1\to n## process. Of course this is just a label that has nothing to do with physics, so I am not sure what the question is actually about.
Can you give an example?
I guess it somewhat depends on what you mean by included. I have not seen it included in general treatments of the SM, other than saying somewhere at the beginning that gravity is too weak to play a role for what we are going to do anyway.