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$\newcommand{\dd}[2]{\frac{\partial {#1}}{\partial {#2}}}$ $\newcommand{\DD}[2]{\frac{d {#1}}{d {#2}}}$

Begin with the Euler-Lagrange equation $\dd Ly = \DD \ x [\dd L {y'}]$ where $L=L(y,y',x)$, (implicit dependence on $x$, so $\dd L x= 0$).

We get $\DD \ x L = \dd L {y'} y'' + \dd L y y' + [\dd L x = 0] \\= \DD \ x [\dd L {y'} y']$

so $$L - \dd L{y'}y'=Const.$$

This is a well known result.

My problem follows in this math:

$\dd L y = \dd \ y (\dd L {y'} y') =\dd { (\dd L {y'} )} y y' *$ From the equation just derived

but it also equals from the E-L equations $\dd L y=\DD \ x \dd L {y'} = \dd {\dd L {y'}} y y' + \dd {\dd L {y'}} {y'} y'' + [\dd {\dd L {y'}} x = 0] *$

Setting the two starred equations (*) equal to each other, this gives $$ \dd {\dd L {y'}} {y'} y'' =0$$

which is clearly wrong. What was the step that messed things up here?

Qmechanic
  • 201,751

1 Answers1

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The EL equations and the Beltrami identity are on-shell relations. There are no reason why they should still hold after a partial differentiation wrt. a dependent variable $y$ or $y^{\prime}$.

Example: Let $L=\frac{1}{2}y^2$. Then $y\approx 0$ on-shell. However we are not allowed to differentiate the relation $y\approx 0$ wrt. $y$. That would lead to the contradiction $1\approx 0$.

Qmechanic
  • 201,751
  • I havent studied field theory and don't really know what on-shell means. Can you elaborate please? Thanks – Mondo Duke Jan 23 '21 at 04:15
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    @Mondo Duke: I added links. – Qmechanic Jan 23 '21 at 04:16
  • I see, doing this, I'm effectively using the 'wrong' lagrangian in the E-L equations. Next time, you might want to be more straightforward and just say that, but I guess it's good to get it in a bigger context. Thanks for the answers. – Mondo Duke Jan 23 '21 at 04:33