Timeline for Help with geometric view of conjugate momenta and Legendre transformation
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jun 7, 2023 at 19:23 | comment | added | peek-a-boo | Actually, I said it again in Why is the Legendre transform (of vector bundles) a smooth morphism $FL:E\to E^*$. Look at the local formula for the Fiber derivative; that says precisely that (letting $(q,\dot{q})$ be coordinates on $TQ$ and $(q,p)$ coordinates on $T^*Q$ (keeping in mind the abuse of notation regarding $q$ I mentioned in my answer here)) $p_i\circ \mathbf{F}L=\frac{\partial L}{\partial \dot{q}^i}$. If you want to avoid confusion, call one of them $p_i$ and the other $\pi_{L,i}$, then $p_i\circ \mathbf{F}L=\pi_{L,i}$. | |
| Jun 7, 2023 at 19:17 | comment | added | peek-a-boo | but $\frac{\partial L}{\partial \dot{q}^i}$ is a smooth function on $TU$, not $T^*U$. The symbol $p_i$ has no meaning until you define it for me. I can define, just for notation sake, $p_i=\frac{\partial L}{\partial \dot{q}^i}$; this is done in Lagrangian formalish. But, this of course is not equal to the adapted coordinate on the cotangent bundle (which is also commonly denoted $p_i$, and is what’s there in the Hamiltonian formulation). I suggest re-reading the first link once again. | |
| Jun 7, 2023 at 19:15 | comment | added | peek-a-boo | @JPeterson yes, the symbol $p_i$ is very overloaded, as I mentioned in my very first link. | |
| Jun 7, 2023 at 19:12 | comment | added | J Peterson | From your response and links, I understand this much better now except one detail: the relation $p_i=\frac{\partial L}{\partial\dot{q}^i}$ is still given in every source I've seen, even the more mathematically rigorous ones. But if $p_i\in\mathcal{F}(T^*Q)$ and $\frac{\partial L}{\partial\dot{q}^i}\in\mathcal{F}(TQ)$, then they cannot be equal. Am I correct in thinking that this too is an abuse of notation and what it really means is that $p_i\circ \mathbf{F}L(v) = \frac{\partial L}{\partial\dot{q}^i}(v)$ for $v\in TQ$? | |
| Jan 1, 2023 at 21:13 | comment | added | peek-a-boo | Now, just as $\gamma’:I\to TQ$ is a “lift” of the base curve $\gamma$ (i.e so we say a velocity vector field of the particle, along its trajectory), the mapping $(\Bbb{F}L)\circ \gamma’:I\to T^*Q$ is also a lift of $\gamma$, so you could just as well call it a momentum field along the particle’s trajectory. But, you should not confuse this with the momentum 1-form $\mu_L$. | |
| Jan 1, 2023 at 21:07 | comment | added | peek-a-boo | @JPeterson let $L:TQ\to\Bbb{R}$ be a Lagrangian, and suppose $\gamma:I\subset\Bbb{R}\to Q$ satisfies the Euler-Lagrange equations (think motion of a particle as it moves in $Q$). For each $t\in I$, this has a velocity vector $\gamma’(t)\in T_{\gamma(t)}Q$. Using the fiber derivative (see first bullet point of section 3) we thus get a covector $(\Bbb{F}L)_{\gamma’(t)}\in T_{\gamma(t)}^*Q$, which you would be justified in calling the momentum of the particle. But again, this is not quite the same as the “momentum 1-form” $\mu_L$, which is a 1-form on $TQ$, not Q. | |
| Jan 1, 2023 at 19:39 | vote | accept | J Peterson | ||
| Jan 1, 2023 at 19:37 | comment | added | J Peterson | ... But I'm sure I'll be able to answer this from the links you provided | |
| Jan 1, 2023 at 19:35 | comment | added | J Peterson | wow. This is massively informative and I'm not close to finished reading. Your comments about coordinates on $Q$, $TQ$, and $T^*Q$ clears up a lot on its own. So is their no notion of a "momentum" 1-form on $Q$ itself? only on $TQ$ or $T^*Q$? I had thought a "particle" moving on $Q$ has a velocity vector $\mathfrak{X}(Q)\ni \mathbf{v}:x\mapsto \mathbf{v}_x\in T_xQ$ and momenta 1-form $\Omega^1(Q)\ni \mathbf{p}:x\mapsto \mathbf{p}_x \in T_x^* Q$, which I thought was related (but not the same) to $ \boldsymbol{\theta} \in \Omega^1(T^*Q)$ and $\boldsymbol{\mu}_L\in\Omega^1(TQ)$... (cont.) | |
| Jan 1, 2023 at 13:15 | history | answered | peek-a-boo | CC BY-SA 4.0 |