Timeline for Help with geometric view of conjugate momenta and Legendre transformation
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 2, 2023 at 14:03 | comment | added | Leonid | @peek-a-boo: You're right, my mistake. | |
| Jan 2, 2023 at 13:59 | comment | added | Leonid | @JPeterson: Regarding the first point: Initially you have coordinates $q^i,v^i$ on $TM$; Then when you are trying to describe a curve on $M$ (which has coordinates $q^i(t)$) it naturally induces a curve on $TM$ (called the lifted curve) with coordinates $\big(q^i(t), \dfrac{dq^i}{dt}(t)\big)$. This should be reminiscient of what you do in classical mechanics: You first take the derivative of the lagrangian with respect to velocities (treated as independent variables) then you evaluate the derivatives of the lagrangian at the lifted curve (when you plug them in the Euler Lagrange equations). | |
| Jan 1, 2023 at 13:32 | comment | added | peek-a-boo | most of the answer is fine, but your first sentence isn’t. Equation (3), with $d\dot{q}^i$ is wrong; it should be $dq^i$. Note that $\frac{\partial L}{\partial \dot{q}^i}\,d\dot{q}^i$ doesn’t have the right transformation law; if you change coordinates, you get something like $\frac{\partial L}{\partial \dot{r}^i}\,d\dot{r}^i+\frac{\partial L}{\partial \dot{r}^i}\frac{\partial r^i}{\partial q^j}\frac{\partial \dot{q}^j}{\partial r^l}\,dr^l$, and the second term cannot be made to vanish. | |
| Dec 31, 2022 at 16:05 | comment | added | J Peterson | The $\frac{d\pmb{q}}{dt}\neq \dot{\pmb{q}}$ point is something I have never fully understood and need to dwell on more. So let's re-name $\dot{\pmb{q}}$ as $\pmb{v}$ so that some $(\text{x},\mathbf{v})\in TQ$ has coordinates $(\pmb{q},\pmb{v})=(q^i,\dots , q^n , v^1,\dots, v^n)\in\mathbb{R}^{2n}$. Are you saying the conjugate momenta is then $\mathbf{p}= p_i \mathbf{d}v^i = \frac{\partial L}{\partial u^i}\mathbf{d}v^i\in \Omega^1(Q)$? are the $\mathbf{d}v^i\in\Omega^1(Q)$ then some basis for each $T^*_{\text{x}} Q$? What is the transformation $\mathbf{d}q^i\mapsto \mathbf{d}v^i$? | |
| Dec 30, 2022 at 19:31 | history | answered | Leonid | CC BY-SA 4.0 |