Monday, 25 February 2019

classical mechanics - Does the variation of the Lagrangian satisfy the product rule and chain rule of the derivative?


I have seen wikipedia use the product rule and maybe the chain rule for the variation of the Langragin as follows:


\begin{align} \dfrac{\delta [f(g(x,\dot{x}))h(x,\dot{x})] } {\delta x} = \left( \dfrac{\delta [f(g)] } {\delta g} \dfrac{\delta [g(x,\dot{x})] } {\delta x} \right) h(x,\dot{x}) + f(g(x,\dot{x})) \dfrac{\delta [h(x,\dot{x})] } {\delta x} \end{align} where the variation of the Lagrangian is defined \begin{align} \dfrac{\delta \mathcal{L} } {\delta x} = \dfrac{\partial \mathcal{L} } {\partial x} - \dfrac{d}{d \tau} \dfrac{\partial \mathcal{L} } {\partial \dot{x}} \end{align} and $\mathcal{L}=f(g(x,\dot{x}))h(x,\dot{x})$.


Does the variation of the Lagrangian satisfy the product rule and chain rule of the derivative?




Answer





  1. OP considers the 'same-time' functional derivative (FD) $$\tag{1} \frac{\delta f(t)}{\delta x(t)}~:=~\frac{\partial f(t)}{\partial x(t)} - \frac{d}{dt} \frac{\partial f(t)}{\partial \dot{x}(t)} +\ldots. $$ Here $f(t)$ is shorthand for the function $f(x(t), \dot{x}(t), \ldots;t)$. Although the 'same-time' FD (1) can be notationally useful, it has various fallacies, cf. my Phys.SE answer here.




  2. The Leibniz rule $$\tag{2} \frac{\delta (f(t)g(t))}{\delta x(t)} ~=~\frac{\delta f(t)}{\delta x(t)} g(t) +f(t)\frac{\delta g(t)}{\delta x(t)}\qquad(\leftarrow \text{Wrong!}) $$ for the 'same-time' FD (1) does not hold. Counterexample: Take $f(t)=g(t)=\dot{x}(t)$.




  3. The chain rule $$\tag{3} \frac{\delta f(t)}{\delta x(t)} ~=~\frac{\delta f(t)}{\delta y(t)}\frac{\delta y(t)}{\delta x(t)}\qquad\qquad(\leftarrow \text{Wrong!}) $$ for the 'same-time' FD (1) does not hold. Counterexample: Take $f(t)=y(t)^2$ and $y(t)=\dot{x}(t)$.





  4. However, the usual FD $\frac{\delta F}{\delta x(t)}$ (where $F[x]$ is a functional) does satisfy a Leibniz rule $$\tag{4} \frac{\delta (FG)}{\delta x(t)} ~=~\frac{\delta F}{\delta x(t)} G +F\frac{\delta G}{\delta x(t)}, $$ and a chain rule $$\tag{5} \frac{\delta F}{\delta x(t)}~=~ \int dt^{\prime} ~\frac{\delta F}{\delta y(t^{\prime})}\frac{\delta y(t^{\prime})}{\delta x(t)}.$$




No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...