Calculus of Variations. Lagrangian Hamiltonian Mechanics Mathpages.

Question

Over at http://www.mathpages.com/home/kmath523/kmath523.htm is an article about Lagrangian and Hamiltonian Mechanics with a derivation of the Euler-Lagrange equations of motion.

Mid-way through is this statement:

"Variations in x,y,z and X at constant t are independent of t (since each of these variables is strictly a function of t), so we have"

$ \frac{\partial x} {\partial X} = \frac {\partial \dot x} {\partial \dot X} $ ; $ \frac{\partial y} {\partial X} = \frac {\partial \dot y} {\partial \dot X} $ ; $ \frac{\partial z} {\partial X} = \frac {\partial \dot z} {\partial \dot X} $

Can someone please explain why this is true. To me it seems wrong because by applying the chain rule I get:

$ \frac {\partial \dot x} { \partial \dot X} = \frac {\partial (\frac {dx} {dt})} {\partial (\frac {dX} {dt})} $ $ = \frac {\partial (\frac {\partial x} {\partial X}\frac {dX} {dt} + \frac {\partial x} {\partial Y}\frac {dY} {dt}) } {\partial (\frac {dX} {dt})} $ $ = \frac {\partial (\frac {\partial x}{\partial X}\frac{dX}{dt})} { \partial (\frac {dX} {dt})} + \frac {\partial (\frac {\partial x} {\partial Y}\frac {dY} {dt}) } {\partial (\frac {dX} {dt})} $ $ = \frac {\partial x} { \partial X} + \frac {\frac {\partial x} {\partial Y} \dot Y } {\dot X} $ $ \ne \frac {\partial x} { \partial X} $

Thanks

Answer 1

I think I answered my own question after writing this in MathJax. If I had

$\frac {\partial{(Ax + By)}} {\partial x} = A$

Then

$ \frac {\partial \dot x} { \partial \dot X} = \frac {\partial (\frac {dx} {dt})} {\partial (\frac {dX} {dt})} $ $ = \frac {\partial (\frac {\partial x} {\partial X}\frac {dX} {dt} + \frac {\partial x} {\partial Y}\frac {dY} {dt}) } {\partial (\frac {dX} {dt})} $ $ = \frac {\partial (\frac {\partial x}{\partial X}\frac{dX}{dt})} { \partial (\frac {dX} {dt})} + 0 $ $ = \frac {\partial x} { \partial X} $

Since the differentiation is with respect to $\frac {dX}{dt}$, which does not appear in the terms involving Y.

Answer 2

It is confusing because there is some serious abuse of notation going on in classical mechanics and it results in unclarity when taking partial derivatives. Let's see what's going on:

We have two coordinate systems $(x,y), (X,Y)$ related by a transformation $(x,y)=(x(X,Y),y(X,Y))$ and we have a curve $\displaystyle (X,Y)=(X(t),Y(t))$ (see I have already abused notation), so we have a curve $\displaystyle (x,y)=(x(X(t),Y(t)),y(X(t),Y(t)))\,.$ Using chain rule we get $\displaystyle \dot{x}=\frac{\partial x}{\partial X}\dot{X}+\frac{\partial x}{\partial Y}\dot{Y}.$ There has been some abuse of notation so far, but I can make sense of everything, and in fact this notation is unfortunately standard. Now doing the obvious thing gives you what you want.

But the last part is one which I find confusing; what does $\displaystyle \frac{\partial \dot{x}}{\partial\dot{X}}$ really mean? Is it a directional derivative of some sort? I think this part was never really clear to me.