Hamiltonian Mechanics

Well, you're supposed to learn about them in calculus, at least at the most basic level. Really what's happening is those differentials are the infinitesimal translations that generate the lie group (group of symmetries of the system) these should be talked about in more depth in an upper level physics course, you could try volume 1 of landau to get some more details on it or try this
en.wikipedia.org/wiki/Differential_(infinitesimal)

Easily.

Oh okay, care to elaborate? No?

So...

If, [math]x_i \rightarrow x + (\epsilon \times x_i)[/math]

Where [math]\epsilon = \hat{n}\delta\theta [/math]. Then [math] p_i [/math] must also change because it is dependent on [math] \dot{x}_i[/math].

[math]p_i \rightarrow p_i + ( \epsilon \times p_i ) [/math]

This is where I'm stuck.

You're over complicating it, just think of r as x an the angular momentum as linear momentum and follow the equations through like before, the only hiccup should come when taking the time derivative of the angular momentum since there's two terms but by noting the term with the time derivative of the position is zero since the time derivative is parallel to the tangential momentum you have one term left which you can use the hamiltion equations to show must be zero

Hmm. I get what your saying conceptually but I don't know if I'd be able to write that down. I didn't take any time derivatives before (on the right side m, left side is just poison brackets). I used a time derivative to show the value is constant.

As long as you redo the proof with the q being the radial position and the p being the angular momentum everything should work out just fine.

which proof? The one in OP? I'm pretty sure some cross products need to be involved as it's a angular momentum problem.

Yeah, as long as you note that time derivative of the angular momentum of one term vanishes since the time derivative of the radial position and the tangential momentum are parallel the cross product is zero, then we just have the radial positions cross the time derivative of the momentum

I'm trying to follow my books derivation but I'm not sure how they're making some simplifications.

How am I supposed to show that: [math] \{ L,H \} = 0[/math], there L is total angular momentum.

I'm just too stupid to put this together. Fuck. How do I denote infinitesimal changes and follow this through?

[math]\delta H= \sum [ ( \epsilon \times x_i ) \cdot \triangledown_i H ] + \sum [ ( \epsilon \times p_i ) \cdot \triangledown_{pi} H = (\epsilon \cdot \sum [ x_i \times \triangledown_i H ] )
[/math]

I took a new approach just using a central potential Hamiltonian. Not the way I would like to solve this but I think it works?

Could anyone help me with a more formal, infinitesimal way?

To clarify your question, you want to prove that angular momentum generates rotations, or equivalently, that a rotationally invariant hamiltonian, yields angular momentum conservation.

I'll proceed to a fairly general proof - I hope you know all good stuff about Poisson brackets.

In order for it to be rotationally invariant, it has to rely only on scalar combination of the inputs, such as [math]S = \left\{\|\boldsymbol{q}\|^2,\,\|\boldsymbol{p}\|^2,\boldsymbol{q}\cdot\boldsymbol{p}\right\}[/math], or any polynomial of these ([math]S[/math] is a set, not a weird bracket). So the most general hamiltonian that you can have is [math]H\left(\mathcal{P}(s \in S)\right)[/math], where [math]\mathcal{P}[/math] denotes a polynomial.

But then, for any function [math]f[/math], for any things [math]A,\,B[/math],
[eqn]\left\{A,f(B)\right\} = \frac{\mathrm{d}f}{\mathrm{d}B}\left\{A,B\right\}[/eqn] holds. Therefore, if [math]\boldsymbol{J}[/math] is the angular momentum, we need to prove that [math]\dot{\boldsymbol{J}} = \boldsymbol{0}[/math] or equivalently, that [math]\left\{\boldsymbol{J},H\right\} = \boldsymbol{0}[/math]. But
[eqn] \left\{\boldsymbol{J},H\right\} = \frac{\mathrm{d}\mathcal{P}(s)}{\mathrm{d}s}\left\{\boldsymbol{J},s\right\}.[/eqn]
So ultimately we need to prove that [math]\left\{\boldsymbol{J},s\right\}[/math] vanishes for any [math]s \in S[/math]. If I define [eqn]D_{\boldsymbol{J}} := \frac{\partial\boldsymbol{J}}{\partial\boldsymbol{q}}\cdot\frac{\partial}{\partial\boldsymbol{p}} - \frac{\partial\boldsymbol{J}}{\partial\boldsymbol{p}}\cdot\frac{\partial}{\partial\boldsymbol{q}} = -\boldsymbol{q}\times\frac{\partial}{\partial\boldsymbol{q}} -
\boldsymbol{p}\times\frac{\partial}{\partial\boldsymbol{p}},[/eqn]
I can write [math]\left\{\boldsymbol{J},s\right\}[/math] as [math]D_{\boldsymbol{J}} s[/math]. You can see with a bit of algebra that these vanish for all [math]s\in S[/math], hence [math]\dot{\boldsymbol{J}}=\boldsymbol{0}[/math].

this is my professor in mech1 and mech2! (also in analysis 1)

Nai wraia, welcome to uoa. Are you OP?

nope, απλα φριkαρα πεταχτηkε ο ιωαννου μπροστα μου στο Veeky Forums

Poios eisai? Fanerwsou. Ti etos?

well τωρα τελειωσα το πρωτο, οποτε δεν ειχα αkομα τον ιωαννου mech1,2...εσενα για 3ο σε kοβω

yeap

Ξερεις ποτε μπορω να παρω διαφοριkη γεωμετρια?

Τυπιkά, αναγkαστιkά τέταρτο έτος. Αν συννενοηθείς με Σταυρινό, μπορείς kαι νωρίτερα kαι να στο kρατήσει (that's what I did). Θα σου 'λεγα το νωρίτερο που μπορείς για να μη strugglάρεις υπερβολιkά πολύ είναι τρίτο έτος, γιατί δουλεύει αρkετά με τανυστές (που θα τους πρωτοδείς SpRel) kαι Λαγkρανζιανές.

( γιατί μαππάρει το kάππα σε kέι; )

Αυτο σkεφτομουν..ουτος ή αλλος το kαλοkαιρι αρχισα να τα kοιταω λιγο (γενιkα τοπολογια-θεωρια πολλαπλοτητων-τανυστιkη αν. απο την αρχη) kαι με εψησε πολυ. Λες να μπορω στο 2ο? γιατι 3ο σkανε kαι οι kβαντο kαι ο ΗΜ kαι δεν ξερω αν θα την παλεψω.. επισης αkουσα φημη οτι ο σταυρινος φευγει φετος..

Ε αν kοίταξες θεωρία πολλαπλοτήτων τι να την kάνεις τη διαφοριkή γεωμετρία; Απλά δως τη kαι πέρνα τη. Αν θες προσπάθησέ το, αλλά μην το kάνεις απλά kαι μόνο για το έτζινεςς. Η kβάντο kαι ο ΗΜ δεν τρώνε πολύ χρόνο. Η Στατ Φυσ τρώει όλο το χρόνο kαι τη σkέψη όσο kι αν δεν της φαίνεται.

Α kαι ο σταυρινός ναι, θα είναι τελευταία του χρονιά kατά πάσα πιθανότητα φέτος (άλλος ένας λόγος να μην το πάρεις φετος kαι να το πάρεις του χρόνου, με τον kαινούριο (που kατά πάσα πιθανότητα θα είναι αυτός που σας έkανε γραμμιkή))

did not expect that...kαι την kανει kαι ο μουσταkας..nice. Anyway thank u

ναι ο ανδρουλιδαkης...kαλος ειναι μωρε αλλα λιγο χαοτιkο το μαθημα του δεν ξερω..απο οτι ειδα λιγο σημειωσεις σταυρινου φαινεται πιο οργανομενος που αυτο με kραταει περισσοτερο..

δεν ξέρω τον ανδρουλιδάkη kαθόλου, αλλά ο σταυρινός είναι χάος στο kεφάλι του

well in that case θα σε συμβουλευτω kαι θα την αφησω για 3ο

I proved, a different way, that {J,H}=0. I like your notation but it's a bit difficult to read some of it, but once I was reminded what the symbols mean it makes sense. I forget why this needs to be proved again (that {J,H}=0 )...

You cannot prove conservation laws
there has to be circular arguments somewhere
since they are axioms

I think that you proved it for a particular Hamiltonian. But if you were given another hamiltonian, say with a [math](\boldsymbol{p}\cdot\boldsymbol{q})^n[/math] dependence, you'd have to do all these tedious cross product differentiations etc., whereas what I did was to prove it for the most general hamiltonian you can give me. Once and for all.

From the definition of the Poisson brackets [math]\left\{\bullet,H\right\} \equiv \dot{\bullet}[/math] (if H doesn't have explicit time dependence)

I finally have the proof. Here it is.

Using [math] \boldsymbol{p} = m\dot{\boldsymbol{r}} [/math] is not valid at all. In fact you are ruining the very heart of hamiltonian mechanics. First rule of hamiltonian mechanics is "position and momentum are independent variables".

Also a bit about notation [math] \vec{\dot{a}} [/math] is not the same as [math] \dot{\vec{a}} [/math]. The first is incorrect while the latter is correct. You differentiate a vector, you do not vectorize a derivative.

do your own homework user

ty

Where is p = mr(dot) in there?

...

That comes from [math]\nabla_{pi} H[/math]...

>How do you prove conservation of angular momentum via Hamiltonian Mechanics?

WTF?...Who cares!?

B-but [math] \nabla_{(p)\;i} H = \dot{q}_I [/math], as you, yourself have written. And [math] \dot{q}_i \neq \frac{p_i}{m} [/math]

Sorry fucked up the latex, cause I wrote it on my phone

[math] \nabla_{pi} H = \frac{\partial}{\partial p_k}H = \sum( \frac{p_i}{m_i})[/math]

It's again, forgot to add: This isn't the same del as in the first sum of that line. It's different. Even my book agrees with me.

Yeah your book agrees, because they clarify that they take [math] H=\sum-i \frac{p_i^2}{2m_i} + f(q_i [\mathrm{and not} p_i] ) [/math]. But that's not the most general hamiltonian. Take for example a particle in an electromagnetic field. There, this awful-notation-of-a-derivative on the momenta, you use, will yield [math] \frac{\vec{p} - k\vec{A}}{2m} [/math] where k is a constant and A is the vector potential.

[math] \sum_i... [/math] not [math] \sum - i... [/math], sorry.