12:30 AM
Hi all
I'm having trouble with the Lorentz Transformations
@Jim It would be great if anyone could help.
1:09 AM
If anyone can help, that would be great
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I've spent the last couple of hours trying to derive the Lorentz Transformation from Maxwell's Equations. What I ended up with is L_{\nu}^{-1}=\left(\begin{array}{ll} \frac{1}{\sqrt{1-v^{2}}} & \frac{-v}{\sqrt{1-v^{2}}} \\ \frac{-v}{\sqrt{1-v^{2}}} & \frac{1}{\sqrt{1-v^{2}}} \end{array}\right)$... 4 hours later… 4:42 AM @JohnRennie Sir can you tell me a more precise definition of Refraction (of light)? I'm not sure exactly how the term refraction is defined. In practice it's the speed change of light as it enters a medium. 5:01 AM @JohnRennie Everywhere It is defined as bending of light which I found vague. The bending is just a result of the speed change. 5:35 AM This comment seems a bit confused to me. Can anyone who knows more about QFT than me say if there is any validity to it? A hadron is not a Fock state but it can still be expressed as a linear combination of Fock states because the Fock states form a basis for the Hilbert space (up to UV issues). And if so, the question is a good one -- how can some linear combination of Fock states be an energy eigenstate? — Eric David Kramer 9 hours ago 5:56 AM By "Fock state" does he mean a specific number state Those are indeed not physical states I suspect he doesn't know what he means. A quick glance through his previous answers reveals no great knowledge of QFT. I guess what he means is Number states are a basis of the Hilbert space and they are all eigenstates of the Hamiltonian But bound states are also eigenstates of the Hamiltonian? Yet are not part of the basis Are the number states a basis for the Hamiltonian in an interacting field? I thought not. Well we pretend they are somewhat That's the point though. We approximate field states as a combination of Fock states, but it's only an approximation as they cannot actually be written as a linear combination of Fock states. 6:11 AM IIRC the real Hilbert space for an interacting theory is thought to be more the direct sum of one particle states and bound states Can't really have just two particles with energy$2m$in an interacting theory So obviously the Fock state$|0,0,1,0,\ldots\rangle$isn't gonna work out quite the same as the free theory 1 hour later… 7:28 AM @JohnRennie It depends on what you mean by "hadron" and at what energy scale we are :P A proton at low energies compared to the proton mass. At ordinary energies, a "real hadron state" like the proton is very far from being a linear combination of free quark states (cf. the "99% of a hadron's mass is from the sea") But could you write it as a combination of free particles states where you'd have to include all the particles that interact with quarks. Basically just gluons I guess, though in principle you'd need all the other fields too. 7:46 AM Gluons can produce quark-antiquark pairs, which can produce photon pairs, which can produce electron-positron pairs!!! Probably gravitons are produced in there! 7:58 AM [sorry for the pause I'm also working :P] but in HEP scatterings at high energy - i.e. where QCD is weak - you can approximate a hadron in (or out) state pretty well as a collection of 3 quarks 8:27 AM @JohnRennie@ACuriousMind Sir I have doubt. Are the electrons of a conductor motionless when conductor is not connected to a power source? That's a complicated question because the electrons in a conductor are in a delocalised state i.e. they are spread out over a large volume of the metal. To a good approximation they are in momentum eigenstates so any individual electron has a non-zero and actually very large velocity. But on average the electron velocity is zero when no current is flowing. 8:56 AM @JohnRennie Now I think that doesn't really work - the interacting Hilbert space is distinct from the free Hilbert space of Fock space (Haag's theorem), there aren't any free states in it. In the regime where there's gluons in the hadrons, the free state approximation is poor, and in the regime where the free state approximation is good, there's no gluon sea. Aha! So the free particle states do not form a basis for the Hilbert state of an interacting field? No, the two spaces are non-isomorphic We often kinda pretend they do, but it's controversial to what extent that's a valid idea some think this is a big problem, others think te mathematicians are being too pedantic :P 9:28 AM IIRC people usually theorize that the Hilbert space is probably like vacuum sector + one particle states + bound states + multiparticle states or something like that At least as far as decomposing it by energy eigenstates yeah, that's essentially what the Kallen-Lehmann representation says 1 hour later… 10:50 AM When we're looking for the representations of the Lorentz group, is this because the representation space (a vector space) can be turned into a vector field which has Lorentz symmetry? as in, the objects that live in the representation space can be used to construct vector fields with Lorentz symmetry so when arrive at the "spinor representation", we're interested in using these objects to construct a spinor field which we use to reach the dirac equation @Charlie I wouldn't talk about " a vector field with Lorentz symmetry". We're generically interested in "fields", something that assigns a value to a point on spacetime. For this to be a consistent notion, we need to know how these values transform under the transformations of spacetime we're allowing Objects in physics are assumed to be invariant under the Lorentz group If you have such objects, they can always be decomposed into sums of irreducible representations So the fields need to take values in a representation of the Lorentz group. Which are overall easier to work with than the raw objects you may deal with @Slereah Either you mean covariant, or you don't mean "objects". :P "invariant" would be the trivial representation 10:55 AM @ACuriousMind so what you're saying here is that at each point in our field we want an object that lives in a representation space of the lorentz group @ACuriousMind You know what I mean :p @Charlie we want the value of the field at each point to be an element of a rep of the Lorentz group yes, so that we can do Lorentz transformations and know what the value of the field will seem to be in the new frame ahh I see ok nice once we have a representation of the Lorentz group, is it non-trivial to find a corresponding representation of the poincare group? Poincaré group isn't too hard to find yeah ok that would make sense since the lorentz representation is essentially a tighter restriction than the poincare 10:58 AM @Charlie There is no unique "corresponding" representation. You can trivially get a representation of the Poincaré group from one of the Lorentz group by letting translations do nothing. It's slightly non-trivial because the composition of two Poincaré transformations will rotate the translation vector yeah I realised "corresponding" was a bad word to use there @Slereah even if the two transformations are just translations? @Charlie Well sure, but then that rotation is just the identity :p ah lol ok for a field, you don't need to think about translations acting on its value at a point - the space of fields with values in a Lorentz representation$(V,\rho)$automatically gets an induced representation of the Poincaré group where$\phi(x) \mapsto \rho(\Lambda)\phi(\Lambda x + a)$11:02 AM and just out of curiosity, why is it particularly useful to have an irreducible representation of the group? but recall that we've already talked about there being two different representations in the QFT context - the one on the fields and the one on the states @Charlie because (for nice groups) you can build all the reducible ones out of them, i.e. by understanding the behaviour in the irreducible cases you understand the behaviour in all cases ahh I see also if you use reducible reps it's gonna be a bit weird for some things ie trying to compute the spin of a reducible field It will not be an eigenstate @Slereah Nobody's stopping you from saying the standard model is one big field with values in some large reducible representation of all groups involved :P @ACuriousMind It's just some$\mathbb{R}^{17}$field 11:04 AM it's just not a very useful point of view Imagine the rotation matrix for the standard model as a single field I tried making that field once Spinor field with all the flavor and group indices The fact that the states are related to the Poincare group, while the fields are only related to the Lorentz group, is vaguely one of the motivations for susy 11:53 AM I can't seem to get any of the suggestions online to work, is there an easy way to diagonally strikethrough symbols in mathjax? for the shorthand for contraction with the gamma matrices$\not \gamma$? yeah there's a thread about it on the latex Stackexchange We went through this here before,$\,\,\,/\gamma $or something ah got it ty @Slereah it's \cancel{symbol} but the usepackage thing was missing on the links I checked 11:56 AM$k\!\!\!/ = \gamma^\mu k_\mu$7 Small tip to enter the Dirac slash: k\!\!\!/ = \gamma^\mu k_\mu 12:12 PM Found the first diagram on n-lab: is it helpful? I recall that one of the instanton pages had weird diagrams Well they took it from some book so... :p Did the author write the book at least Or is it theft I would say inspiration 12:52 PM "First quantization is a mystery, but second quantization is a functor" Doesn't that mean that QFT can be done in a 'first quantization' formalism like normal QM D: "I did not invent category theory to talk about functors. I invented it to talk about natural transformations." 1:24 PM Does third quantization actually make sense 1:34 PM Dunno 2 hours later… 3:06 PM Given a basis of a lie algebra, if I can find a set of matrices that satisfy the lie bracket (in the form of the commutator) does that guarantee the existence of a matrix representation for the rest of the algebra? (and by extension the elements of the lie group connected to the identity) @Charlie That depends on what you mean by "a set of matrices that satisfy the Lie bracket" (but I'm being nitpicky again and the answer to the spirit of your question is likely yes ;) ) :P as in the Lie bracket equation involving the structure constants, with the Lie bracket being replaced by the matrix commutator It seemed a bit strange to me that in the search for a representation of the Lorentz group we need only define the Dirac gamma matrices, but if what I've said is correct then this is sufficient information in other words we define the gamma matrices from the clifford algebra, we use them to define matrices for the generators of the Lorentz group, and from there that generates the entire Lorentz algebra *generates a representation of the entire Lorentz algebra 3:22 PM Yes, e.g. the$M_{\mu \nu}$generators are the basis elements of a Lie algebra, knowing their commutation relations means you know the Lie algebra because all elements are linear combinations of the basis generators as one always has in a vector space. When seeking representations of this algebra, one needs generators which satisfy the same commutation relations, the Gamma matrices can be used to contruct anti-symmetric tensors built from the Gamma's which obey the same commutation relations @Charlie Well, the formal statement that is true is that, for a Lie algebra$\mathfrak{g}$with generators$T^a$and structure constants$f^{ab}_c$, if you have a subalgebra$A \subset \mathbb{R}^{n\times n}$such that for a basis of matrices$M^a$of$A$you have$[M^a, M^b] = f^{ab}_c M^c$, then the linear map$\mathfrak{g}\to A$defined by$T^a\mapsto M^a\$ defines an isomorphism of Lie algebras.
ok had to think about it for a minute but it's clear now ty guys
(My nitpick is that "set of matrices" is too weak - you need to require that they are linearly independent, even though that probably wouldn't be an issue in practice)
ah ok I see what you mean that's fine
3:54 PM
Does anyone know a more comprehensive explanation to the following question?
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I am unable to find images of pure crystalline graphite with high confidence, but based on various sources I believe that it should actually be both black and shiny, in the sense that it reflects much less visible light than a white piece of paper, and yet has a much more metallic sheen than pape...

There is also a bounty on the question from @rob. I like Ruslan's answer, and intend to accept it within the bounty period if nobody else can give a better answer, especially one that addresses our specific interest:
@user21820 I don't know of the microscopic mechanism that results in so much lower refractive index of extraordinary ray. But it's definitely not always so: e.g. in calcite (ordinary, extraordinary rays) both refractive indices in the visible range are 1) purely real, 2) similar in magnitude. — Ruslan yesterday
Thanks for any help! =)
Alternatively, you can help to upvote Ruslan's answer if you think it's good enough. =)
4:11 PM
@user21820 whenever light falls on a surface where the refractive index changes some of the light is reflected. The reflection coefficient is given by the Fresnel equations and it's simply related to the refractive indices of the two phases.
Shininess is just specular reflection i.e. a reflection from a smooth surface, so in this sense everything is shiny. Things only look dull if the surface is rough so the reflection is diffuse rather than specular.
The eye is very good at picking up this specular reflection even if the percentage of the light reflected is very small. For example freshly washed glasses look shiny even though they reflect only about 5% of the light and transmit the rest.
@JohnRennie Are you saying Ruslan's explanation for why crystalline graphite is not as shiny as crystalline silicon is not the main reason?
If so, you should post another answer explaining what you think is the main contributing reasons.
I don't doubt that a smoother surface will look more shiny.
So your graphite crystal looks shiny because it has a smooth surface which reflects a few percent of the light, and the eye is very good at detecting this specular reflection.
I think what you're saying is not consistent with what Ruslan's post conveys. As I said, you should post another answer if that is really the case.
@user21820 I suspect that if you polished graphite and silicon to equal smoothness they would look very similar.
@JohnRennie That's what I thought when I posted the question, but Ruslan's answer says "no".
4:19 PM
I think he is wrong but I can't prove it because I don't have polished graphite and silicon surfaces to hand to test. Since I can't make any concrete statements I couldn't post a convincing answer. You'd need to do the experiment to answer this with any confidence. Without the experimental evidence it seems like post rationalisation to me.
Then yes I hope someone else has crystalline graphite to test this out somewhere because I definitely don't have any and can't find any images online.
But I suspect it's not as shiny because of what I said in my post:
> I also noticed that if you take a graphite pencil and shade an oval completely (yes MCQ), it looks black under usual lighting unless it is at an angle to catch the light from an overhead lamp at which point it appears to be very shiny.
Even at that very shiny point, it hardly looks like a metal shininess.
Though one might argue that is not crystalline graphite.
At least, not one big crystal.
@ACuriousMind : And your Phys.SE question is mike stone's ref. 1! Cool!
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2 hours later…
6:48 PM
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You may think of this as an application of Mach's Principle. This represents an observed fact of physics and cosmology, not derivable from some other principle. The local, non-rotating frame appears to be determined by matter, mostly distant matter. General Relativity, partly inspired by this ide...

3 hours later…
9:36 PM
question: is it true to say: any rotation in a system f(x,y,z) can be written as a parity in another system g(x',y',z')? what do you think about it
9:57 PM
@2physics Are the systems f and g related by a rotation, or is there a parity inversion in the transformation between them?