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In page 30 of Goldstein's CM there is this line that says, if we are dealing with a system where the Euler-Lagrange equation with $L = T - V$ is satisfied, so does the $L' = L + \frac{dF}{dt}$ where $F(q,t)$ is a differentiable function in $q$ and $t$.

So, to check this I wrote the equation and the result was that:$$ \frac{d}{dt} \frac{\partial }{\partial \dot{q}}(\frac{\partial F}{\partial q}\dot{q} + \frac{\partial F}{\partial t}) - \frac{\partial}{\partial q}(\frac{\partial F}{\partial q}\dot{q} + \frac{\partial F}{\partial t})= \frac{\partial F}{\partial q}\frac{\partial \dot{q}}{\partial q} $$

I am relatively new to Lagrangians, so I assume that $\frac{\partial \dot{q}}{\partial q} = 0$ in the sense that they are independent of each other. But I don't understand what does the function $\dot{q}$ mean in the sense that if $q$ also depends on $t $, can't there be a relation between $q$ and $q'$? For example $q$ can be $e^t$ so that equation wise partial derivative is not zero, i.e $\frac{\partial e^t}{\partial e^t} = 1$.

I can't make sense that there is no relation between $q$ and $q'$ so that each are independent of each other but on the other hand $\dot{q} = \frac{dq}{dt}$ is already a relation between them so they are not independent, at least $q$ implies $\dot{q}$ and other direction holds upto a constant.

Qmechanic
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Mahammad Yusifov
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1 Answers1

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The easiest way to see is to notice that $q = q(t)$, i.e $q$ is only a function of time. Therefore,

$$ \dot{q} = \frac{\partial q}{\partial t} = \frac{dq}{dt}$$

We therefore have

$$ \frac{\partial \dot{q}}{\partial q} = \frac{\partial ^2q}{\partial q \partial t} = \frac{\partial}{\partial t} \frac{\partial q}{\partial q} = \frac{\partial}{\partial t}1 = 0$$

More philosophically, this reflects the fact that you use 2 variables to characterise your system. In Hamiltonian Mechanics you use $p$ instead of $\dot{q}$. Hope this helps.

emir sezik
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  • Yes it is helpful but not satisfactory yet, because I am expecting a answer that uses the definition of partial derivative eventually. For example, for the above if q(t) = e^t, assume d here is $\partial$ then 0 = $\partial$q'/$\partial$q = 1 as q = q'. – Mahammad Yusifov Jul 01 '23 at 22:09
  • Also, could you elaborate what does you use 2 variables mean. – Mahammad Yusifov Jul 01 '23 at 22:10
  • The problem is that when you take a partial derivative with respect to q, you assume it is an independent variable. – emir sezik Jul 01 '23 at 22:33