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The Euler Lagrange equation $\frac{\partial F}{\partial q}-\frac{d \frac{\partial F}{\partial \dot{q}}}{d t}=0$ can also be put in the form $\frac{\partial F}{\partial t}-\frac{d (F- \dot{q}\frac{\partial F}{\partial \dot{q}})}{d t}=0$,

How is the second form of the equation arrived at mathematically, and does equating them and rearranging (as above) lead to any simplifications of the equations? I'm probably missing some elementary rearrangement.

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migrated from Dec 24 '12 at 19:20

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It looks like you are making some confusion with partial and total derivatives. But the second form writen as:$$\frac{\partial F}{\partial t}-\frac{d }{d t}[F-\dot q \frac{\partial F}{\partial \dot q}]=0,$$

is nothing but the conservation of energy, since $\dot q \frac{\partial F}{\partial \dot q}-F$ is defined as the Hamiltonian of the system, or by other words the energy.

The way to derive this form is only to expand the total derivative of the function (in physics: Langrangean) $F$ in order to it's explicit dependencies in $q$, $\dot q$ and $t$ and make use of the Euler-Langrange equations to re-write the term that involves the partial derivative in order to $q$.

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Wouldn't $\frac{d}{dt}(F-\dot{q}\frac{\partial F}{\partial \dot{q}})=0$ be the conservation of energy, or is $\frac{\partial F}{\partial t}=0$? – Alyosha Dec 24 '12 at 17:09
Conservation of energy is verified when there is no explicit time dependency of the Lagrangean and so one get's the expression you have writen in the comment. If there is then the system is losing or gaining energy and that is expressed by that term. – PML Dec 24 '12 at 17:56
By 'expand the total derivative of the function $F$' do you mean taking the differential: $\frac{dF}{dt}=\frac{d}{dt}(F_q dq+ F_{\dot{q}} d\dot{q}+ F_t dt)$? – Alyosha Dec 24 '12 at 18:27
Well, yes but be careful when you write that because what you wanted to write was $dF=F_q dq + F_{\dot q} d\dot q + F_t dt$ and not $F=...$. – PML Dec 25 '12 at 12:01

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