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I was reading through Lagrangian Mechanics and noticed this sort of question popping up everywhere. To illustrate, the 3d x coordinate was considered to be a function of X, Y (degrees of freedom). Now in the differentiating mess, the authors considered both x and X as functions of time. But I thought x was a function of X and hence the confusion.

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  • $\begingroup$ Yes, velocity can be considered as a function of position rather than as a function of time. $\endgroup$ – G. Smith Jul 4 '19 at 17:37
  • $\begingroup$ @G.Smith Stated this way I am afraid that there could be some problem with counting the number of independent degrees of freedom. $\endgroup$ – GiorgioP Jul 4 '19 at 17:53
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    $\begingroup$ I don't know if this is what you're getting at, but velocity is not in general a function of position, because an object can have two different velocities at the same position, if it returns to the same point at two times. $\endgroup$ – user4552 Jul 4 '19 at 17:58
  • $\begingroup$ @ Ben Crowell Thanks for making that point. Lemme edit the question $\endgroup$ – Eesh Starryn Jul 4 '19 at 18:01
  • $\begingroup$ Maybe you could show the math you are talking about? $\endgroup$ – user234190 Jul 4 '19 at 19:34
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If Lagrangian is written as $L=L(x(t),v(t),t)$, $L$ is time-dependent explicity through $t$ and implicitely through $x$ and $v$. If $L=L(x(t),v(t))$, then $L$ is not explicitly dependent on time but implicitly dependent on time through $x$ and $v$. This means $L$ can change if and only if $x$ and $v$ changes with time and can't change with time without a change in $x$ and $v$.

Same explanation goes for all the functions depending explicitly and implicitly on time.

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