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enter image description here This problem comes from Goldstein.

What does $s=e^{\gamma t}q$ mean? Do I just put $q=e^{-\gamma t}s$ into the Lagrangian?

But I don't know what that means.

I think the point transformation may relate canonical transformation from wikipedia, but this is not enough to understand it. Could you give me an advice to learn about this?

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In this context, it is a change of variables. The variable in the original Lagrangian is $q$, and Goldstein is asking you to use another variable $s$, which is related to the original $q$ via the "transformation": $$s = \exp(\gamma t) \ q$$ and later on, make sense of it (with the later questions). Point transformation in this context refers merely to this co-ordinate transformation. If you are seeking to understand from the point of view of canonical transformations, these point transformations are transformations of the adopted generalized co-ordinates. These would form a subset of canonical transformations, since this change of co-ordinates $q \rightarrow s$ would be accompanied by a corresponding change in the generalized momenta (via the Legendre transformation), and hence, will not affect the Hamilton's equations in this context.

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  • $\begingroup$ So, if one can think of any transformation that gives a corresponding change in the generalized momenta, then he can use it without restrictions? And thus, it is just a mathematical trick that just does not change Hamilton's equations? $\endgroup$
    – TheQuantumMan
    Commented Nov 13, 2015 at 1:55
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    $\begingroup$ @LandosAdam - Sure. But it matters whether you are looking at a transformation which ends up simplifying the problem. If not, this transformation hardly has any virtue. e.g. consider this detour: In integral transforms, you transform a real space equation into a momentum space one, but while the procedure works for every physically meaningful real space equation, doing this is not beneficial for all cases. It is useful only when doing this simplifies the problem, otherwise we were better off solving the original equations themselves! Same idea over here too. $\endgroup$
    – 299792458
    Commented Nov 13, 2015 at 11:24

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