From the Legendre transform we can deduce that $$\frac{\partial H(q,p)}{\partial q}=-\frac{\partial L(q,\dot{q})}{\partial q}.$$
Similarly can we prove $$\frac{\partial H(q,p)}{\partial p}= \frac{\partial L(q,\dot{q})}{\partial p}~ ?$$
Asked
Active
Viewed 118 times
1
-
Possible duplicate: https://physics.stackexchange.com/q/105912/2451 – Qmechanic Aug 30 '22 at 16:28
-
I'm pretty sure these are not true, but it is true that $\frac{\partial H}{\partial t}= - \frac{\partial L}{\partial t}$. – agaminon Aug 30 '22 at 18:17
-
The question/title doesn't make sense because it's trivially untrue. Do you mean "equivalence"? Or maybe "equation involving"? – Sean E. Lake Sep 01 '22 at 18:49
-
Technically $\partial L/\partial p=0$ if you define partial derivatives of $L(q,,\dot{q})$ the obvious way. – J.G. Sep 01 '22 at 19:08
2 Answers
1
Unless you made a typo, this is not true. In the general case of the Legendre transform: let $F$ be a function of $x$ and $G$ a function of $y$ its Legendre transform defined by: $$ y = \frac{dF}{dx} \\ G = xy-F $$ you are asking whether: $$ \frac{dG}{dy} = \frac{dF}{dy} $$ You rather have: $$ \frac{dG}{dy} = x \\ \frac{dF}{dy} = y\frac{dx}{dy} $$ so you have equality iff $x$ is linear in $y$ i.e. $F$ (or equivalently $G$) is quadratic. While this is false in general, since in a lot of cases you have a quadratic $L$ (or $H$) this is often true (and probably the reason why you thought it was true in the first place perhaps).
Hope this helps.
-
But after differentiating legendre transform w.r.t 'y' shouldn't we get $\frac{\partial F} {\partial y} $=x i.e $\frac{\partial F} {\partial y} $= $\frac{\partial G} {\partial y} $ isn't that right ? – Keshav Shrestha Aug 30 '22 at 15:48
-
No as the above derivation just proved. You can convince yourself with an example, such as $F=x^4$ or for a physically motivated example $F = \sqrt{1-x^2}$. I think you are rather looking for the relation: $$\frac{dF}{dx}=y \ \frac{dG}{dy}=x$$ – LPZ Aug 30 '22 at 15:54
1
Similarly can we prove $$\frac{\partial H(q,p)}{\partial p}= \frac{\partial L(q,\dot{q})}{\partial p}~ ?$$
No, not in general.
But the above equation is true when the velocity is a linear function of the canonical momentum $p$, as shown below.
To see this, you must remember that $L$ is not a normally a function of $p$. This means we can only make sense of the above equation by assuming you have made a substitution for $\dot q$ with a different function, say, $v(q,p)$.
Sometimes it is helpful to use a different letter for the expression for $\dot q$ as a function of $q$ and $p$. So, I write: $$ \dot q = v(q,p)\;, $$ where the function $v$ comes from inverting: $$ \frac{\partial L}{\partial \dot q} = p $$
For example, for a free particle: $$ \frac{\partial L_{free}}{\partial \dot q} = m\dot q = p $$ $$ \to v_{free}(q,p) = p/m $$
In the above example, $v_{free}$ actually depends only on $p$ and not on $q$. But in general $v$ can depend on both.
Given the above definitions, the Hamiltonian is: $$ H = pv(q,p) - L(q,v(q,p)) $$
Therefore: $$ \frac{\partial H}{\partial p} = v(q,p) + p\frac{\partial v}{\partial p} - \frac{\partial L}{\partial \dot q}\frac{\partial v}{\partial q} = v(q,p) $$
Whereas: $$ \frac{\partial L(q,v(q,p))}{\partial p} = \frac{\partial L}{\partial \dot q}\frac{\partial v}{\partial p} = p\frac{\partial v}{\partial p} $$
These are only equal if $$ v(q,p) = p\frac{\partial v}{\partial p}\;, $$ which is not true in general.
However, it is true, for example, for a free particle where $v_{free}$ is linearly related to $p$. It's also true for any Lagrangian of the form: $$ L = \frac{1}{2}m{\dot q}^2 - U(q) $$
hft
- 19,536