The h Bar

ACuriousMind

8:00 PM

@EmilioPisanty Could work if you're lucky, but there's no guarantee simple trimming will produce anything useful.

bolbteppa

Q.S. Zheng, “Theory of representations for tensor functions - A unified invariant approach to constitute equations”

Emilio Pisanty

@bolbteppa I'll have a loook

@ACuriousMind wait, seriously?

ACuriousMind

@bolbteppa That the existence of such invariants is a more general fact doesn't mean you can't use calculus when you're using them on a vector space over the reals.

Emilio Pisanty

@bolbteppa sure, but I'm working over $\mathbb R$

bolbteppa

If you could use calculus in abstract algebra it wouldn't be so hard haha

Emilio Pisanty

8:01 PM

true that

I'm also going wtf, shouldn't this stuff have been ironed out in the sixties?

I mean, how are there still papers in 2018 about how to parameterize tensor spaces in isotropic ways?

ACuriousMind

@EmilioPisanty Yeah, seriously. The invariant basis has no reason to be chosen in a way that that would work. I'm thinking e.g. about the characteristic polynomial of a matrix, which is an invariant under conjugation by another matrix. The coefficients of the polynomial "parametrize" the similarity classes of matrices, but you're not able to restrict them such that only a subset of them varies on a small neighbourhood of matrices - varying the matrix slightly varies all the coefficients.

Emilio Pisanty

@ACuriousMind that's not what I was saying, though

what I want is a set of coefficients that captures enough of the variation so that it is locally invertible

I don't care if I'm dropping coefficients that also vary if I vary the matrix

ACuriousMind

@EmilioPisanty Also, I want to caution against your feeling that the result should be 7-dimensional in some sense. Consider doing the same just for the $\mathrm{R}^3$, not for its tensors. The orbits are clearly spheres, and the invariant parameter is the radius, that is, the "isotropic space" is one-dimensional (a half-line, in fact), even though $\mathrm{SO}(3)$ is 3-dimensional so the result should 0-dimensional by your logic.

Emilio Pisanty

8:16 PM

@ACuriousMind I'm aware of that

but that's because points are too symmetric

some of the tensors will be like that

so it'll have some ugly orbifold behaviour in parts

but the bulk of the tensors are asymmetric enough that any rotation will change them

ACuriousMind

@EmilioPisanty Right. I still have a feeling that taking the invariants and dropping a few will not be directly lead to useful charts, but I have no actual argument against it

Emilio Pisanty

I'm OK with not having useful charts, I think

Emilio Pisanty

8:37 PM

@ACuriousMind hmmmmm, OK. A random sampling on that space suggests that that Jacobian (almost always) has rank eight instead of seven

which is.... odd

well, eight is even, but the result is odd

to say the least

bolbteppa

8:53 PM

If I understand right, the paper is saying that given a third-order 3-D symmetric tensor $\mathbf{A}$ which takes the form $\mathbf{A} = A_{i_1 i_2 i_3}$ in an orthonormal basis $\{\mathbf{e}_i \}$, and defining any polynomial $f$ which satisfies $f(A_{i_1 i_2 i_3}) = f(Q_{i_1 j_1} Q_{i_2 j_2} Q_{i_3 j_3} A_{j_1 j_2 j_3})$ with the $Q$'s orthogonal as an isotropic polynomial invariant,

and defining any scalar-valued function $f$ which satisfies $f(A_{i_1 i_2 i_3}) = f(Q_{i_1 j_1} Q_{i_2 j_2} Q_{i_3 j_3} A_{j_1 j_2 j_3})$ with the $Q$'s orthogonal as an isotropic function,

and using the Olive 2014 result that any isotropic polynomial invariant can be expressed as a polynomial in 13 isotropic polynomial invariants,

one can show that by removing two of the 13 one can express any isotropic function as a function of the 11. What does it meant to split up the space of third order 3-D symmetric tensors $\mathbf{A}$ into '"shape" and "orientation" parameters', how do Euler angles have anything to do with anything other than the $O(3)$ matrices which are irrelevant since $f$ is left invariant under them? How can differentiation do anything but change the 11/13 functions you were using?

ACuriousMind

@bolbteppa I'm not sure what you're on about. The definition of the invariants means that the invariants are functions on the orbits of $\mathrm{SO}(3)$ in $\mathrm{Sym}^3 \mathbb{R}^3$. Emilio is hoping to use these functions to parametrize the orbits, and then parametrize points in each orbit through some equivalent of the Euler angles, just like for $\mathbb{R}^3$ the radius parametrizes an orbit (a sphere) and then a set of Euler angles identifies a point on that sphere.

@EmilioPisanty that would imply that "but the bulk of the tensors are asymmetric enough that any rotation will change them" isn't true, right?

Emilio Pisanty

@ACuriousMind so it would seem

bolbteppa

The only things the paper is doing is abstract algebra, forming syzygy's e.g. expressing the $K_6$ on page 5 in terms of the other 11 invariants, trying to use anything about $SO(3)$ at all to talk about the 11 or 13 "invariant" functions makes no sense since they are invariant under the behaviour of $SO(3)$ :\

Emilio Pisanty

@bolbteppa 'shape' parameters should be isotropic invariants in the language of Chen and Olive

@ACuriousMind yep, that's exactly what I'm shooting for

@ACuriousMind I still don't know what that means, though

ACuriousMind

9:09 PM

I found a paper that does what you want to do for elasticity tensors $\mathrm{Sym}^2(\mathbb{R}^6)$ instead of 3d 3-tensors.

ACuriousMind

9:27 PM

@EmilioPisanty Could you do the same random sampling of the Jacobian for the second-order symmetric tensors and the invariants of trace, sum of principal minors and determinant?

Emilio Pisanty

@ACuriousMind probably

Semiclassical

@ACuriousMind This sounds like random matrix stuff.

Emilio Pisanty

lemme see how much hassle it'll be to change the code

Semiclassical

(That doesn't mean it is, of course)

Emilio Pisanty

@Semiclassical quite far from it

I'm just doing some random sampling to do some heuristic checks

ACuriousMind

9:29 PM

Because here's the things: These three invariants - the coefficients of the characteristic polynomial - are the analogue of your invariants for second-order symmetric tensors. The second-order symmetric tensors are 5-dimensional, and really most of them have 3d orbits

Emilio Pisanty

@ACuriousMind goodness

ACuriousMind

Their orbits are degenerate precisely if they (as matrices) do not have three distinct eigenvalues.

Emilio Pisanty

why $\mathbb R^6$?

wait, got it

> As stated above, elasticity tensors can be viewed as fourth-rank tensors in $\mathbb R^3$ with intrinsic symmetries or as a second-rank symmetric tensor in $\mathbb R^6$

ACuriousMind

So we do expect "most" of the space of orbits to be "2d", really. If we don't get rank 2 here, then that means something is fundamentally wrong with the approach of sampling the Jacobian rank.

Emilio Pisanty

weird

ACuriousMind

9:33 PM

Ah, shoot, I forgot we're not in the traceless case, so the 2nd order symmetric tensors have dimension 6

Emilio Pisanty

@ACuriousMind I was just going to ask you to do a harmonic decomposition on those invariants

i.e. to re-express them in terms of the trace and the traceless (quadrupole) component

ACuriousMind

Wait, maybe you already know this anyway: Do you have shape parameters for second-order tensors?

If yes, how many are there, and do you know if they are also a "function basis" for isotropic invariants in the sense of these other papers?

Emilio Pisanty

7

Q: How many truly different multipolar charge distributions are there?

Dipolar charge distributions are essentially all the same: regardless of how one adds up a combination of the form $$ \sigma(\theta,\varphi) = \operatorname{Re}\left[\sum_{m=-1}^1 a_m Y_{1m}(\theta,\varphi)\right], $$ you only ever get a rotated version of the canonical $Y_{10}(\theta,\varphi) \p...

ACuriousMind

@EmilioPisanty Ah, yes. So, do you know whether $Q_0, Q_2$ together with the trace are a "function basis" for isotropic invariants of 2nd order symmetric tensors?

Emilio Pisanty

@ACuriousMind I'm not a big fan of that notation in this context though

it requires you to move to a principal frame and I'd rather work with explicit polynomials

I just got one set working, though

ACuriousMind

9:43 PM

Or, if not, if taking such a function basis and dropping stuff from it would produce parameters equivalent to these three?

Emilio Pisanty

$\mathrm{Tr}(T)$, $\det(Q)$ for $Q=T-\frac14 \mathrm{Tr}(T) \mathbb I$, and $Q_{ij}Q_{ij}$

random sampling of the jacobian gives rank 3

ACuriousMind

Hmm...so what one would expect. Alright.

Emilio Pisanty

@ACuriousMind I just tried the explicit characteristic-polynomial coefficients

also rank 3 generically

ACuriousMind

Yeah, okay, so the rank thingy works correctly for things that we know are proper parameters

Remains to test it for a "function basis" of invariants for this case, though

Emilio Pisanty

@ACuriousMind ugh

aren't the characteristic-polynomial coefficients a functional basis?

ACuriousMind

9:51 PM

@EmilioPisanty I would naively say so, but then I would naively have said that there should also only be 7 independent invariants on the rank 3 tensors :P

Emilio Pisanty

@ACuriousMind indeed

well

you can separate the trace from the other two

so if you want irreducibility you just need a traceless tensor with zero determinant with nonzero $Q^2$

which is pretty easy

and potentially also something with vanishing $Q^2$ and nonzero determinant?

that'll be much harder though

danielunderwood

10:11 PM

What's the difference in having a system where only certain states are allowed and where any state is allowed, but some have probability 0?

Here's the footnote that made me ask in case that helps explain what I'm asking

> For our discussion it is not important that certain values of a, b, or c might be excluded by quantum mechanics but not by classical mechanics. For simplicity, assume the values are the same for both but that the probability of certain values may be zero.

ACuriousMind

@danielunderwood Continuous probabilities behave very counter-intuitively: Saying an event is not allowed is much stronger than saying it has probability 0! Consider the case of drawing a random number from the unit interval with uniform probability density: The probability to draw any specific number is zero (because the density assign non-zero probabilities only to subintervals), yet you will get a number if to draw one: Having probability 0 is very different from being not allowed.

However, in the case of the footnote you cite, I suspect the author is trying to make a much simpler (and less interesting :P) point: In order not to have to change the space of event when switching between classical and quantum mechanics, he's including all classically possible results in the space of possible events in quantum mechanics, even if they're things like "measure angular momentum $\hbar/3$" which can't occur in quantum mechanics.

This is done for convenience of always dealing with the same event space, not because it means anything important.

danielunderwood

10:28 PM

For the first part, what about when you're looking at an entire region? Say you have an interval a,b where the probability is uniformly non-negative and b,c where it is 0. What would be the difference there besides saying it's allowed in a,b and not in b,c? The probability of a certain point anywhere is 0, but the whole picture changes when you back out and look at regions, right?

ACuriousMind

@danielunderwood The difference would be that in the first case your theory still assigns a probability to the interval [a+b/2,b+c/2], while in the second case it does not.

danielunderwood

So they would be equivalent from a physical perspective, but make the math not have to handle the smaller space? I think that's what you may have been getting at in the second part of your response.

bolbteppa

10:54 PM

4

A: Can Schroedinger equation be derived from the unitary representation of Galilean group?

Yes, you can derive the Klein-Gordon, (free) Dirac, (free) Maxwell, linearized Einstein vacuum, etc., equations from the representation theory of the Poincaré group. Yes, you can derive the ordinary non-relativistic (free) Schrödinger equation from the representation theory of the Galilei group ...

"Yes, you can derive the ordinary non-relativistic (free) Schrödinger equation from the representation theory of the Galilei group"

@Semiclassical any idea if there's a Bohm link to this perspective

This kind of stuff:

Representation theory of the Galilean group

In nonrelativistic quantum mechanics, an account can be given of the existence of mass and spin (normally explained in Wigner's classification of relativistic mechanics) in terms of the representation theory of the Galilean group, which is the spacetime symmetry group of nonrelativistic quantum mechanics. In 3 + 1 dimensions, this is the subgroup of the affine group on (t, x, y, z), whose linear part leaves invariant both the metric (gμν = diag(1, 0, 0, 0)) and the (independent) dual metric (gμν = diag(0, 1, 1, 1)). A similar definition applies for n + 1 dimensions. We are interested in projective...

Semiclassical

I don't know, no. there is one source I could imagine, though. lemme look

bolbteppa

@Semiclassical this paper projecteuclid.org/euclid.cmp/1103840281 referenced there using these ideas gets a non-relativistic Dirac equation that actually predicts the correct intrinsic magnetic moment without any relativity

"The ''extraordinary" value of the spin gyromagnetic ratio, like the very existence of the spin, is nothing of a "relativistic effect" as is plainly or tacitly stated in many textbooks"

Semiclassical

@bolbteppa this is what I had in mind: books.google.com/…

(a pity that's a google book source and not a pdf)

@bolbteppa Yeah

It reminds me of one fact which rather bothers me about the standard account of deBB: Replacing the probability current $\vec{j}:=\dfrac{\hbar}{m}\text{Im}\left[\Psi^\dagger \nabla\Psi \right]$ with $\vec{j}\to \vec{j}+\nabla f$ would leave the continuity equation unchanged

bolbteppa

"The absence of this term in the galilean case means that the Maxwell equations introduce relativity (under the Poincare group) in a quite literal sense. In physical situations where the displacement current is of negligible importance, the predictions of the theory are in perfect agreement with galilean relativity. The whole of pre-Maxwellian electromagnetism (laws of Faraday, Ampere, Biot-Savart etc. . . .)

is simultaneously exact and consistent with the old Newtonian conception of space-time. But as soon as one takes into consideration the essentially Maxwellian equation (62), one obtains specifically relativistic phenomena such as the propagation of electromagnetic waves with constant velocity etc. This is the one equation which definitely ruins the old Galilean relativity, introducing the Einsteinian one."

Pretty cool

Semiclassical

And that's a problem, since in deBB you want to be able to interpret $\vec{j}/\rho$, evaluated at a given spacetime point, as the velocity a particle which happens to be at that spacetime point would have

bolbteppa

11:05 PM

"It is commonly stated that electric fields and magnetic fields are two separate entities in nonrelativistic theory, mixing only under the effect of Lorentz transformations, which make them to appear as two aspects of a same fundamental quantity, the electromagnetic field. The galilean transformation properties (71) show the necessity of revising this opinion: ..."

Hmm

Semiclassical

So there seems a charge of arbitrariness, i.e. that you could modify $\vec{j}$ (and therefore the allowed velocity field and thus the allowed trajectories) by a term $\nabla f$ without changing the physical predictions

The best explanation I've seen for this is that, if you go to the Dirac equation, then the Dirac current is uniquely determined. you then take the non-relativistic limit and find that the usual $\vec{j}$ is what pops out

I've never really liked that argument, though. I'd much prefer an argument which is self-contained within non-relavistic physics.

(And this is one of the objections which iirc is labelled as reasonable in the Towler lectures, so it's not a new concern)

Semiclassical

11:32 PM

@bolbteppa one thing I've found, reading the L-L paper, is that they rather annoyingly never actually write down the equivalent of the time-dependent Schrodinger equation. The wave equation he obtains in equation (25) is the time-independent case.

presumably one just substitutes $E\mapsto i\hbar\partial_t$ but it'd be nice if he actually had said so

bolbteppa

Ah, equation (14) he does it, missed it myself

Semiclassical

equation 14 or equation 34?

bolbteppa

(14), page 292

Semiclassical

that's the schrodinger equation, sure

I meant the time-dependent equivalent of the Levy-Leblond equation

but, coincidentally, that does show up as equation (34)

a pity that the time-independent Levy-Leblond equation is equation (25) and not equation (24)

bolbteppa

11:51 PM

Hmm the wiki on Galilean reps seems to break down to only the vacuum for massless reps, so photon states from a Galilean representation theory perspective don't seem to show up, but are fine for Poincare group reps

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