Ah, I see

I'm looking at Arnol'd's classical mechanics text, pg 257. He gives the following problem: solve the Cauchy problem for the Hamilton-Jacobi equation $S(q, t_0) = q^2/2$, $\partial_t S + H(\partial_q S, q, t) = 0$ with $H(p, q, t) = p^2/2$, and draw the graphs of the multivalued function $S(q, t)$ for some fixed time $t$. I get first order nonlinear PDE $\partial_t S + (\partial_q S)^2/2 = 0$ with initial $S(q, 0) = q^2/2$ out of this, which I do not know how to solve explicitly.

Is the intent here to interpret the Hamilton-Jacobi equation for $H = p^2/2$ as the geodesic flow on $T\Bbb R^2$, with initial wavefront being the graph of $S = q^2/2$ (a parabola), and then use Huygen's principle to draw a sketch of the evolution? It looks like a parabola developing a pair of cusps, as drawn in pg 257.

Or is there some trick which solves the PDE that I'm too topology-pilled to see?

This is my sketch, with a closely related initial curve which gets the idea across. Just draw wavelets and sketch. $S_0$ is the blue curve, which evolves to the red one and then to the green one.

@BalarkaSen wait a moment...is $H = p^2$ not just the free particle?

see also physics.stackexchange.com/q/420922/50583 and the answers Qmechanic links there

Apparently some authors have a different definition of a Cartan connection

Although I think the nuance is that a Cartan geometry is a Cartan connection on some natural bundle

or maybe not idk

it is hard to parse

Never noticed that before, "Cartan" is two letter swaps away from "arctan". Wow, I must be tired.

two letter swaps, so we do not know if it fermion or boson

both give the same positive sign

lol

I had enough signs chasing for today ^_^

<tiredus maximilius/>... ;) see you later!

> Don't advertise your recent questions. If you just posted something on the main site, give it some time, and don't tell people to go there and look at it. Particularly by pinging people. That's rude. Those who can answer are already watching the queue on the main site!

taps the sign

If $\Psi\in(1/2,0)\oplus(0,1/2)$, then in what representation is $\overline{\Psi}$? $\Psi\in(0,1/2)\oplus(1/2,0)$?

@Mr.Feynman yes

@ACuriousMind and considering $\Psi\in(0,1/2)\oplus(1/2,0)$ how do I see that they transform with the inverse matrix with representation theory arguments (i.e. not conjugating $\Psi$ etc.)? Because for $\Psi$ we have the matrix \begin{pmatrix} \Lambda_L & 0 \\ 0 & \Lambda_R\end{pmatrix}

Where those are the matrices under which left and right spinors transform

In the conjugate case the matrix should be \begin{pmatrix} \Lambda_R & 0 \\ 0 & \Lambda_L\end{pmatrix} which is... not the inverse?

that who transforms with the inverse matrix?

also, note that $(0,1/2)\oplus(1/2,0)$ and $(1/2,0)\oplus(0,1/2)$ are isomorphic - there is no meaningful distinction between them

also, both conjugation and the transpose exist abstractly in representation theory

@ACuriousMind and what is the relationship between those two (apart from the fact they are isomorphic), both conjugate and transpose?

@Mr.Feynman The representation $(s_1,s_2)$ is the conjugate representation of $(s_2,s_1)$

@ACuriousMind oh ok I was just confused with the direct sum

My problem is that I want to understand in rigorous terms what's going on in the naive calculation $$\Psi'=D(\Lambda)\Psi\implies\overline{\Psi}'=\overline{\Psi}[D(\Lambda)]^{-1}$$

@Mr.Feynman That's not a "calculation", that's just the definition of the conjugate representation

@ACuriousMind I called it "calculation" because I've seen people deriving that conjugating the first equality. Anyways, shouldn't the conjugate rep give $$\overline{\Psi}'=\overline{\Psi}[D(\Lambda)]^{*}$$ instead?

@SillyGoose honk back atcha

@Mr.Feynman Okay, let's carefully unpack the definition: The definition of the conjugate I linked doesn't have any row/column distinction, so it would say $\bar\Psi' = D(\Lambda)^\ast \bar\Psi$. Now switch to row vectors for $\Psi$, you get $\bar\Psi {D(\Lambda)^\ast}^T = \bar\Psi D(\Lambda)^\dagger = \bar\Psi D(\Lambda)^{-1}$ because the representation is unitary

@ACuriousMind Unitary? Isn't $(0,1/2)\oplus(1/2,0)$ finite dimensional? Because that's my main concern

@Mr.Feynman Ah, well, not exactly unitary (I was thinking in terms of the $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$).

The swapping iso $(1/2,0)\oplus (0,1/2)\to (0,1/2)\oplus (1/2,0)$ probably makes this work but I'm too lazy to work it out :P

@ACuriousMind Yes. Admittedly I don’t understand how the answers there can be applied to my situation; they seem to do away with the initial value of S at t = 0 while writing Hamilton’s principal function.

Plugging t = 0 in the principal function gives a linear initial value whereas my initial value is S(q, 0) = q^2/2

@BalarkaSen Your $S$ is not the principal function, it's what Qmechanic calls the on-shell action

and you get it directly from what Arnold writes: The solution to the Cauchy problem is eq. (3): $S(q,t) = \frac{q_0}{2} + \int_{q_0,t_0}^{q,t} L(q,\dot{q},t)$. You get $L$ by Legendre transforming $H$ and then just computing that integral, no need to solve any differential equation explicitly

@ACuriousMind Alright, so it's not completely straightforward. I'll think about it as soon as I have less workload

also, @BalarkaSen I think you got confused by the terrible typesetting: The task of drawing the graph of $S(q)$ is not supposed to be for the solutions for the free particle. It's a separate problem where you are supposed to look at figure 201, not any equations

is the fact that the lenz vector is conserved under rotations in $SO(4)$ useful at any point?

@ACuriousMind Am I supposed to get $S(q, t) = 1/2 \cdot q^2/t$ from there? That blows up at $t = 0$.

The only thing that seems to match the initial value is $S(q, t) = 1/2 \cdot q^2/(1 - t)$. This blows up at $t = 1$, now.

@ACuriousMind Figure 201 seems to be a schematic to me, not a picture of anything particular. What are the $H$ and $S_0$ if not the ones given in the previous problem?

@ACuriousMind I think you're wrong about this, and I am right. Take the multi-valued function $S = S(q, t)$ defined as follows. Introduce an auxiliary variable $z$, and set $S(z, t) = z^4/4 + t z^2/2$ and $q(z, t) = z^3/3 + t z$. Then solve for $z$ in terms of $q$ by inverting the second equation, and plug in $z = z(q)$ to get $S = S(q, t)$.

This satisfies $\partial_t S + 1/2 (\partial_q S)^2 = 0$

And the picture for $S = S(q)$ is exactly the swallowtail catastrophe

The initial conditions are still weird, somehow. Hmm.

But isn't it true that the Hamiltonian flow for $H = p^2/2$ is exactly the geodesic flow on $T\Bbb R^2$? So solution to $\partial_t S + H(\partial_q S, q, t) = 0$ should get you from $S = S(q, t_i)$ to $S = S(q, t_f)$ by moving the points of the former graph along the geodesic flow in the normal directions to the graph?

I'll appreciate it if anyone wants to comment on whether I have badly botched up this answer here especially because it was accepted by the OP... I am not partial to leaving inaccurate / misleading information on the site when it's an accepted answer <_< (note, comments were made before the last edit...)

@Relativisticcucumber Maybe. ;) See johncarlosbaez.wordpress.com/2015/03/17/… & math.ucr.edu/home/baez/hydrogen/4d

@Amit The name is written opposite. It is under-determined, not over. Over means that there are too many data points, so unique answer is impossible. Under-determined is the case where you have extra free parameters to play with.

That's the thing, I think that the technical jargon is sticking to "number of unknowns" vs. "number of equations" regardless of other kinds of information, like boundary conditions, compatibility, etc. so in that sense I think I am just sticking to the (maybe confusing) standard terminology, aren't I?

an almost identical example where over determined is used as the term...

Never wanted to go down the PDE rabbit hole that deeply when I wrote this :) I won't be able to climb out, lol

@BalarkaSen It's possible, I only half-remember HJ theory

I am not sure though, because the initial conditions for the solutions I wrote down are still off.

@naturallyInconsistent Consider a perhaps silly example... $\frac{\partial{f}}{\partial{x}} = \frac{\partial{f}}{\partial{y}}$ and $\frac{\partial{f}^2}{\partial{x}^2} = 0$ . Two equations and one unknown function right? Yet, we still have infinitely many solutions of the form $f(x,y)=mx+my$

Thanks, good save, lol

$\partial f^2$ on the numerator???

grrr sorry, was meant to be $\partial^2 f$

I would not consider it as two equations and one unknown. The 2nd order x derivative means that it should be 2 unknowns just for x alone, and then the 1st order in y means there there is yet one more unknown for the y dependence once that is done.

but the unknown is $f$... again this is just terminology apparently...

It seems to be one of those things that aren't agreed widely enough so it's 50% chance you'll be using the opposite terminology from someone else :) it now also explains why I saw this is a point of confusion in many SE posts...

@naturallyInconsistent I suppose I can add a remark that because there are in fact (as someone said in a comment) $100$(!) unknown functions if we consider each first and second order partial derivative of the metric... it does somewhat explain where the extra free parameters come from and why the metric isn't unique... but I am concerned it would just further confuse things :)

@ACuriousMind Riddle me this Batman

If I have a curve with the pullback of the frame bundle on it

Is there always a section of it on it?

I'm guessing 1) if it has loops, maybe not 2) If it has a nice tubular neighbourhood, definately

but what if it doesn't have a nice tubular neighbourhood

What if it's the nasty curve on the irrational torus

What if the curve passes arbitrarily close to itself in a spacetime that doesn't admit global frames

@Slereah You mean a global section? Since the frame bundle is a principal bundle, this is the same question as asking whether the pullback bundle is trivial

I'm guessing you're probably already in hot water if the spacetime isn't time orientable, but let's assume it is

[this question was secretly about spacetime the entire time]

:O

I guess the pullback would be trivial if there's no self intersection since it's over a contractible space?

It feels a little weird if you consider the case I considered, though, idk

Think of a Klein bottle, and $\gamma$ be a "longitude" of it. The pullback of the frame bundle is Mobius.

@Slereah yes

So it does not admit a section

This is for a closed curve, though

What if on a closed curve, it isn't trivial, but on an open curve that passes arbitrarily close to the points considered, it is?

On open curves, the frame bundle will always trivialize, because an open interval is contractible.

Not sure if there's a specific case for this I can think up but it makes me wonder

Torus is parallelizable so I can't use that example unfortunately

Is there an irrational moebius strip

what do you mean by that

A non-orientable equivalent of the irrational torus

what's the irrational torus

a donut with unhinged opinions

Apparently the proper term is the "irrational winding of a torus"

The spacetime version being the Carter spacetime

The only possible issue I can see is that the map between the manifolds has to be continuous

Is that true for such curves

I would guess so but I amwary

i don't really like donuts

the donut holes are far superior

@SillyGoose That's what algebraic topology is all about

Schopenhauer had a whole diatribe about why math was fake and a swindle and sometimes I agree

"On the polarization of light the Frenchmen have nothing but nonsensical theories on undulations and homogeneous light, besides computations which are not based upon anything. They are constantly in a haste to measure and to calculate ; they consider this as the main thing, and their slogan is le calcul! le calcul! But I say, Où le calcul commence, l'intelligence des phénomènes cesse: he who has only numbers in his head cannot find the trace of the connective cause."

You understand things are getting serious with SE when you are scrolling reddit and you find a thread about the new policies in the Math sub

@Slereah I wouldn't go so far as to saying it's a swindle. But it is true that we're like the drunk guy searching for his keys under the streetlight. We only look at the physics that's illuminated by mathematics. But maybe some core stuff isn't covered by mathematics that we have developed, or it's covered by maths that's too complex for us to comprehend.

@Amit No, the more I think about it, and the more I look up, the less I think you can even back up the claim that it is "aren't agreed widely enough", let alone "50% chance".

Overdetermination means you have to do some statistical inference and throwing some data to make the fit better.

This is definitely the opposite case.

@naturallyInconsistent Careful, there are (at least) two meanings of "overdetermination": The one you're using and the one @Amit's using

Note that it is possible for $\Gamma$ to overdetermine $g$ without causing problems, because what that means is that $\Gamma$ is not completely free. Once there are plenty of internal constraints, then the actual amount of degrees of freedom of $\Gamma$ could be lowered enough for it to work.

@ACuriousMind No, I was using the latter too.

Amit is correct that the equations relating $R$ and $g$, when viewed as equations for $g$ given $R$, is an overdetermined system of equations in spacetime dimension 4 or higher

@ACuriousMind It was not about $R$, but rather about $\Gamma$. Also, the fact that $R$ would be too much, is also why Einstein had to labour to get $G$

@naturallyInconsistent why about $\Gamma$? The question is about getting $g$ from $R$.

Plenty of g's have the same R

Even not isometric ones

You can look at the whole family of Schwarzschild metrics parametrized by mass and they're all Ricci flat

Oh wait the question is about the Riemann tensor

@ACuriousMind He linked to another post talking about that instead. Sorry

Thanks for taking an interest everyone. To avoid this confusion I just removed mention of overdetermination completely from the answer.. frankly due to my noobiness about such complex PDEs I feel more comfortable about the answer that way, and it still addresses the question

All I know is, integration is mighty complicated, ooga booga

@Slereah makes it a bit complicated? :) From here: "In dimensions 4 and higher, there are more equations than unknown functions and therefore there are no solutions unless further conditions are imposed on the curvature-like tensor. These conditions are poorly understood." , that's where I also poorly understand what is meant 🤣

I am quite sure however that by "no solutions" he meant "not possible to determine solutions"

how do i compute how many ways there are to swap N slots D:

Swap N distinct slots?

You mean # of arrangements?

given $N$ slots, how many ways are there to swap only two of the slots with each other

Oh, should be choose 2 from N... i think...

oh the word is transposition

yes i think it is $(N 2)$ as you say :D

great

Cos you choose and you swap

I tried swapping before choosing one time , store owner called the police

xD

Are you swapping fermions?

Btw that turns out to be $N(N-1)/2$ right? Same like summing an arithmetic series of $1,2,3..$ cos when you add a slot it forms an additional possible $N-1$ pairs

yes it does turn out to be that i think at least for the first few Ns

im swapping tensor factors in a hamiltonian

do yall know of any circumstance where analytically computing a derivative of a function is easier to do by hand than by computer?

For those functions that have a tonne of identities, it is sometimes the case that the derivative can be immediately stated by hand, but the computer algorithm might return a complicated version that would need a lot of identities work to realise is the same as the version we would pick by hand.