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If $L$ is a Lagrangian for a system of n degrees of freedom satisfying Lagrange's equations, show by direct substitution that $$L' = L + \frac{\mathrm{d}F(q_1,\dots,q_n,t)}{\mathrm{d}t}$$ also satisfies Lagrange's equation where $F$ is any arbitrarily, but differentiable function of its arguments.

So I think I got everything correct except I don't know why the last step vanishes?

\begin{align} \frac{d}{dt}&\frac{\partial M}{\partial \dot{q_j}} - \frac{\partial M}{\partial q_j} = \frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial(L + \mathrm{d}F(q_1,\dots,q_n,t)/\mathrm{d}t)}{\partial \dot q_j} - \frac{\partial(L + \mathrm{d}F(q_1,\dots,q_n,t)/\mathrm{d}t)}{\partial \dot q_j}\\ &= \left(\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot q_j} - \frac{\partial L}{\partial q_j}\right) +\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial}{\partial \dot q_j}\frac{\mathrm{d}F(q_1,\dots,q_n,t)}{\mathrm{d}t}\right) \\ &\qquad-\left(\frac{\partial}{\partial q_j}\frac{\mathrm{d}F(q_1,\dots,q_n,t)}{\mathrm{d}t}\right)\tag{1} \end{align}

By chain rule we have $$\frac{\mathrm{d}F}{\mathrm{d}t} = \Sigma_i \frac{\partial F}{\partial q_i}\dot{q_i} \ + \frac{\partial F}{\partial t}$$

Notice since $L$ satisfies Euler-Lagrange equation the first two components of equation (1) vanishes. So equation (1) becomes

$$\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial}{\partial \dot q_j}\left(\Sigma_i \frac{\partial F}{\partial q_i}\dot{q_i} \ + \frac{\partial F}{\partial t}\right)\right) -\frac{\partial}{\partial q_j}\left(\Sigma_i \frac{\partial F}{\partial q_i}\dot{q_i} \ + \frac{\partial F}{\partial t}\right) \tag{2}$$

So why does Eq (2) vanish? Can someone explain?

Qmechanic
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user99741
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  • Possible duplicates: http://physics.stackexchange.com/q/174137/2451 , http://physics.stackexchange.com/q/87628/2451 and links therein. – Qmechanic Dec 13 '15 at 07:59

1 Answers1

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Hint: Keeping in mind that $F$ has no explicit dependence on $\dot{q}_i$, work out the first term in the last equation you wrote. By showing the following, the road to the answer becomes very clear.

\begin{align*} \frac{d}{dt}\left(\frac{\partial}{\partial \dot{q}_j}\left(\sum_i\frac{\partial F}{\partial q_i}\dot{q}_i+\frac{\partial F}{\partial t}\right)\right)&=\frac{d}{dt}\left(\frac{\partial F}{\partial q_j}\right) \end{align*}

Arturo don Juan
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  • Yeah I got that, but why is the second term equal to $\frac{d}{dt}(\frac{\partial F}{\partial q_j})$? – user99741 Oct 14 '15 at 03:47
  • Because in this case, the derivatives commute. The second term is, from the outset, $\frac{\partial}{\partial q_j}\left(\frac{dF}{dt}\right)$. I strongly suggest you don't take my word for it and expand the first and second terms out. – Arturo don Juan Oct 14 '15 at 04:01
  • Alright if I expand the second term out and since partial derivatives commute we get $\frac{\partial F}{\partial q_j}\frac{\partial}{\partial q_j}\frac{dq_j}{dt} + \frac{\partial}{\partial q_j}*\frac{\partial F}{\partial t}$ – user99741 Oct 14 '15 at 04:08
  • but this doesn't simplify to the above or does it ? – user99741 Oct 14 '15 at 04:09
  • The expansion you wrote in that comment isn't quite right. Remember that there's the sum to take into account (which you did quite well in your original question statement). – Arturo don Juan Oct 14 '15 at 04:15
  • And also, once you do that expansion of the second term, do it also for what I wrote in my original answer (above), and compare them. – Arturo don Juan Oct 14 '15 at 04:16
  • Alright I did the expansion as follows $\Sigma_i \ \frac{\partial F}{\partial q_i}\frac{\partial}{\partial q_j}\dot q_i + \frac{\partial}{\partial q_j}\frac{\partial F}{\partial t}$, but it doesn't simplify further ? – user99741 Oct 14 '15 at 04:23
  • Let us continue this discussion in chat. – Arturo don Juan Oct 14 '15 at 04:24