"It appears that the square of the index of refraction is equal to the product of the specific dielectric capacity and the specific magnetic capacity." -Maxwell
I found this quote at Quora:
In March 1916 Ramanujan graduated from Cambridge with a Bachelor of Science by Research (This degree was later renamed as Ph.D. from 1920) for his work on Highly composite numbers.
Given how important his paper on Highly composite numbers was it appears that Cam...
Bear in mind that at Hardy's time mathematics in the UK was rather stagnant. All the big advances were being made on mainland Europe. Hardy was instrumental in the resurgence of UK mathematics.
@JohnRennie I don't know about relativity. I was studying from my chemistry book imgur.com/a/ns1wxMh I got that magnetic quantum number and that zeeman effect written there. I have a doubt that while drawing electronic config. diagrams for atoms, we can draw the first electron in any p orbital in a particular subshell as all the 3 have the same energy.Doesn't the earth's magnetic field break the degeneracy and change the energy of all the three p orbitals?
@user8718165 Yes, the Earth's field will cause a splitting, but the Zeeman effect is very small unless the magnetic fields are strong. In practice the Earth's magnetic field is too small to make any noticeable difference.
Since there are like an almost endless number of sources of EM fields in the universe there are quite an incalculable number of splitting, but as they are all very small it basically just translates as a tiny width to the energy level
Amazon are currently offering £5 off any order above £25. 20% off is a useful discount. But ... I don't want anything from Amazon at the moment. Life is wretched sometimes.
due to my prolonged habit of reading on web or downloaded articles, I feel a little awkward when reading physical books. For example, when reading on the compter screen, I can search for some words easily, but I can't do that on physical books or paper.
the dream I made today I can remember now is that I was arguing with a clerk. It's like a scenario in which I can't find the bonus attached with the merchandise I bought. so I requested to withdraw it.
@JohnRennie neither do I. Reading on trembling bus would make me get eyeache, but reading on train seems not to have this problem. But I really hate to waste time, so I don't like to have long-time commute every day.
I usually watch scenery outside the window when on vehicle.
yesterday dream is that I went to visit a university.
@students but settled is not a noun. I want to say something like "I am in a limbo while he is in a xxx." in which xxx is a place or status where his work is supported, concerned, regarded, appreciated, etc.
@All I was wondering if this is possible. So we all know that light has mass but the mass is really really small. And we all know the speed of light. But is it possible that there is something with no mass that would be able to go faster than light? The only thing I can think of is the speed of darkness.
But yet darkness is only there when light disappears. So that would mean darkness goes the speed of light
"A 2.50-kg fireworks shell is fired straight up from a mortar and reaches a height of 110 m. Neglecting air resistance, calculate the shell’s velocity when it leaves the mortar."
How do you calculate that with only mass and distance?
The unitary infinite dimensional reps of Poincare are usually labelled by the eigenvalues of the P^2 and W^2 operator, i.e. the mass squared and the square of the Pauli Lubanski vector.
However, finite dimensional non-unitary reps are based on finite dimensional reps of Lorentz. Thus, in addition to P^2, one would need 2 labels to properly pin down the rep of Lorentz... that means the unitary irreps need one less label...
Of course while the projectile is in the mortar it will experience an acceleration opposite to gravity but the question is simply asking about projected velocity
I could not find this post I've been looking for it.
@knzhou I like the infinite-dimensional irreps of Lorentz...
would you have a reference to the procedure you give as part of your answer?
BTW I have Tung's book and it is indeed nice; this chapter in particular is indeed quite nice. I have been reading Bogolubov, N. N., Logunov, A. A., Oksak, A. I., & Todorov, I. (2012). General principles of quantum field theory (Vol. 10). Springer Science & Business Media., which I find to be very clear.
@ZeroTheHero I'm not sure if there's an ideal reference for this stuff, this is what I got trying to match up Wu-Ki Tung section 10.5 with what I learned in regular QFT class.
Suppose I have a finite dimensional irrep of Poincare. How many labels do I need to completely identify it?
Certainly I can see that I need the eigenvalue of $P^\mu P_\mu$.
and given that the finite-dims of Lorentz have 2 labels one would naively think that Poincare would require 3 labels: one for the translation part and "two" for the Lorentz subgroup part.
@ScientistSmithYT physicists don't say light has a mass. I'm a physicist and I'm saying light has zero mass. A quick search through the main site will find lots of statements that photons have zero mass.
@ZeroTheHero The reason you need to labels for Lorentz is that the Lorentz algebra has two casimirs, the two $su(2)$ casimirs from $so(1,3) \sim su(2) \times su(2)$. These are no longer casimirs for Poincare, so there is no 3-label thing going on - for Poincare we have $P^2$ and $W^2$ as casimirs. What you think should be the 'Lorentz subgroup part' is actually the 'little group' part, e.g. in the massive case it's $SO(3)$
As an explanation of why a large gravitational field (such as a black hole) can bend light, I have heard that light has momentum. This is given as a solution to the problem of only massive objects being affected by gravity. However, momentum is the product of mass and velocity, so, by this defini...
@bolbteppa actually what is trowing a spanner in my wheel is Mathews, P. M. "Covariant fields: Poincaré group representations and metric structure in the space of quantum states." Pramana 3.4 (1974): 261-276., available here, link.springer.com/content/pdf/10.1007/BF02872244.pdf, where 2 "Lorentz" indices are present.
@ScientistSmithYT all particles are fundamentally massless, and they get their masses from the Higgs field. (Actually it's possible this isn't true for neutrinos, but that's not relevant here.)
The four vector bosons, the photon, Z, W+ and W-, interact with the Higgs field but the interaction gives only three of the four particles a mass. The photon is the one that misses out and stays massless while the other three get masses.
@ZeroTheHero @knzhou The sentence "the Poincare action on the fields induces a Poincare action on the Hilbert space; we identify particles with Poincare irreps on that Hilbert space" in knzhou's answer is one I would like to see more formally substantiated, in particular with respect to the uniqueness of how the unitary representation is "induced", since it would answer this question of mine
@ScientistSmithYT But, just to be confusing, while photons are massless they can add mass to a bound state. This is probably where a lot of the confusion arises.
For example consider a hydrogen atom in its 1s ground state. Suppose we shine a photon of energy 10.2eV onto this atom, then the atom will absorb the photon and it will be excited to the 2p state.
The mass of the 2p state is greater than the mass of the 1s state, and it's greater by 10.2eV/c^2 i.e. when the atom absorbed the photon its mass increased.
@ACuriousMind Ah, I don't have a proof for that and don't think I've ever even see it explicitly stated. It just worked for the examples that first came to mind.
@ACuriousMind with reference to your question why are you choosing to induce from $s_1+s_2$, i.e. the largest $j$ value, rather than $\vert s_1-s_2\vert$, i.e the smallest.
@ZeroTheHero Because the actual rotation algebra in 3d sits diagonally in the $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$, meaning the "true 3d spin" of $(s_1,s_2)$ is $s_1+s_2$.
so just to be clear (as we must with topic): in your su2+su2 one of them is like (L+iK) and the other is (L-iK) where L is angular momentum and K is boost.
You're actually pointing at something important: It's not obvious from the mere representations that there are no d.o.f. of the field corresponding to e.g. $s_1 -s_2$
I.e. no one guarantees us that a single field corresponds to a single particle irrep, and indeed, for e.g. photons this is not the case - one massless field gives us two helicity irreps
I might have to read up on your point about s_1+s_2 a little more. In my notes I have that the angular momentum contents of the finite dimensional Lorentz (s1,s2) is from Abs[s1-s2] to s1+s2.
But mysteriously, the physical fields always have funny constraints on them such that only the $s_1+s_2$ actually survives (a massive vector field does not induce a scalar particle!)
However, these constraints are dynamical, i.e. imposed on the level of the Lagrangian, so it's impossible to deduce them merely representation-theoretic
E.g. For a massive four-vector field with canonical Lagrangian you have that the $A^0$-component (the singlet under 3d rotations) is non-dynamical
And for a massless four-vector field you have a gauge theory the eliminates two d.o.f., leaving us with two, corresponding to the two 1d helicity irreps
But this is all rather explicitly linked to the dynamical constraints of these fields and not solely to their mass and finite Lorentz rep
I.e. one could imagine fields with non-standard kinetic terms that produce very different particles, and I think this has been done e.g. for the massless continuous spin representations
This means, I guess, that I believe the answer to my question is "particle reps are determined by actually quantizing a specific theory, not purely from classical field representations"
@ZeroTheHero This is also all in agreement with what Wu-Ki Tung says about relativistic wave equations (e.g. 10.5.4), though he doesn't really say it that directly.
I’m not the greatest fan of L&L as a standalone learning tool: it’s very Russian in style in that they assume the reader will fill in the blanks - sometimes considerable blanks - by themselves in between bouts of intuition.
On the other hand, as a companion to a more pedagogical text, the intuitive nature of all these books, and the hypnotic manner in which the material is presented, is second to none.
The volumes on stat mech and on class. mech. are absolutely representative of this: brilliant brilliant brilliant but they require a lot of academic maturity to use as standalone texts.
@knzhou @ACuriousMind thanks for the inputs. I have enough reading to do for a short while but I’m interacting closer to the exposition I need.
If you want to be pointlessly but technically accurate, the opposite of "limbo" is "heaven" :P (limbo is originally a subdivision of hell in Christian mythology)
@students yea, in some sense. If you are in a wonderland, you are greatly favored and people appreciate and recognize your achievements and thus respect you; if you are in a wasteland, nobody regards your existence because nobody can appreciate or recognize your achievements.
@ACuriousMind but I just saw limbo means "(often cap.) a region on the border of hell or heaven in Roman Catholic teaching, serving as the abode after death of unbaptized infants and of the righteous who died before the coming of Christ."
@CaptainBohemian A dictionary may not fully grasp the finer points of medieval theology ;) The much more extensive Wikipedia article is rather clear about Limbo not being part of heaven (as only those redeemed of original sin are admitted to heaven), but part of the underworld/an outer circle of hell.
@ACuriousMind the free dictionary doesn't say limbo is part of heaven, but a region on the border of hell or heaven. I think that means a region characteristic of uncertain fate.
@students thank you for your information; this webpage gives me more comprehensive idea of what limbo means.
when one says "an infinite-dimensional representation of a Lie group", what does the "infinite-dimensional" therein mean? I read somewhere that an infinite-dimensional Lie group means the manifold realizing the Lie group is infinite-dimensional? Based on this idea, does "an infinite-dimensional representation of a Lie group" means the bundle realizing the representation is infinite-dimensional?
I just saw @ZeroTheHero refers to the term (infinite-)dimensional representation in the above. I have learnt this term for long but keep not knowing its meaning. The reference above makes me think of to ask.
@JohnRennie If the photon adds mass to the whole thing doesn’t that mean that the photon in that state has mass, but then how come it doesn’t have mass in any other situation yet when it’s in the same form?
@CaptainBohemian the Hilbert space of states that is closed under the action of the generators is irreducible AND infinite-dimensional.
The simplest example of course is the "standard" representation of the Heisenberg-Wey algebra (or group) where the Hilbert space is spanned by $\{ \vert n\rangle, n=0,1,\ldots \infty\}$
Another slightly less trivial example is the sp(2,R) algebra. The Hilbert for the representation is the same as that generated by the complex conbinations $a^\dagger a^\dagger$, $a^2$ and $a^\dagger a+ a a^\dagger$ of the observables $x^2$, $p^2$ and $xp+px$
For a single particle there are two irreps usually denoted by k=1/4 and k=3/4. The k=1/4 contains the H.O. states $\{\vert 2n\rangle, n=0,2,\ldots\}$ and the $k=3/4$ the odd-numbers Fock states $\{\vert 2n+1 \rangle, n=0,2,\ldots\}$
I found this quote at Quora:
In March 1916 Ramanujan graduated from Cambridge with a Bachelor of Science by Research (This degree was later renamed as Ph.D. from 1920) for his work on Highly composite numbers.
Given how important his paper on Highly composite numbers was it appears that Cam...
