While doing some calculation I came across the following term [$G=G(x,x'),$ $x$ is independent of $x'$] $$K=\frac{\partial G(x,x')}{\partial (\frac{\partial G}{\partial x'})}$$ I tried to think of it as $G=\int_0^{x'}\frac{\partial G(x,y)}{\partial y}dy$ and then i got something like $$K= \int_0^{x'}\frac{\frac{\partial G}{\partial y}}{\frac{\partial G}{\partial x'}}dy$$ $$K=\int_0^{x'}\delta(y,x')dy$$ Is this correct so far, if yes is there any further simplifcation that I can do? Is there any other way to do it? to give further context I was evaluating Poisson bracket $\dot{x'}=\{x', H' \}$ and $H'(x',p')= H(x,p) + \frac{\partial G(x,x')}{\partial t}$. In addition to that we have the relations $p=\frac{\partial G}{\partial x}$ and $p'=-\frac{\partial G}{\partial x'}$
4
-
$\begingroup$ You need to be a lot clearer. For example, is $x^\prime=\dfrac{\mathrm dx}{\mathrm dt}$ or something like that? Also, if you want the integral form, you have to have it be $$G(x,x^\prime)=G(x,0)+\int_0^{x^\prime}\frac{\partial G(x,y)}{\partial y}\mathrm dy$$ $\endgroup$naturallyInconsistent– naturallyInconsistent2025-12-07 09:04:44 +00:00Commented yesterday
-
1$\begingroup$ I think you need to rewrite $G$ as a function such that one of its argument is $\dfrac{\partial G}{\partial x'}$. There is nothing like $\dfrac{\partial G(x,x')}{\partial(\frac{\partial G}{\partial x'})}$. $\endgroup$JC Q– JC Q2025-12-07 09:23:37 +00:00Commented yesterday
-
$\begingroup$ $x'$ is just another variable independent of $x$ $\endgroup$neutrino_cuber– neutrino_cuber2025-12-07 12:00:16 +00:00Commented 23 hours ago
-
$\begingroup$ i have added some more info, i hope it clears things out $\endgroup$neutrino_cuber– neutrino_cuber2025-12-07 12:12:49 +00:00Commented 23 hours ago
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Actually, you face an implicit change of variable. Indeed, the definition of the variable $y = \frac{\partial G}{\partial x'} = f(x,x')$ establishes a functional relation between $x$, $x'$ and $y$. You might keep $x$ as an independent variable and invert the said functional relation, hence $x' = \phi(x,y)$ and finally $$ K = \frac{\partial}{\partial y}G(x,x') = \left.\frac{\partial G}{\partial x'}\right|_{x'=\phi(x,y)} \cdot \frac{\partial\phi}{\partial y} $$ by the chain rule.
