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Legendre Transformation

The above image shows the Legendre Transformation in the context of an introduction to the Hamiltonian formalism.

My question is in 4.6, wherein $u(x, y)$ has been defined; what is the guarantee that we would be able to invert this relation in order to get $x(u, y)$? Or in other words, what is the guarantee that we would be able to write $\dot{q_{i}}$ as $\dot{q_{i}}(p_{i}, q_{i}, t)?$. And doesn't that affect the invertibility of the Legendre transformation as a whole as written in the lines following equation 4.9?

Another question: Is time supposed to be a spectator variable when it comes to a Legendre Transformation in Hamiltonian dynamics?

Could somebody possibly dwell on this rigorously? Thanks.

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  • $\begingroup$ Please write equations yourself, rather than posting an image. The image cannot be searched and cannot be edited. $\endgroup$
    – DanielSank
    Commented Apr 27, 2020 at 16:40
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    $\begingroup$ Here's the simple answer without full derivation: in order for a Legendre transform $f(v)\rightarrow f(p)$ to be well-defined and invertible, you need $df/dv$ to be strictly concave in $v$. For your second question, yes, time is a spectator variable in classical Hamiltonian dynamics. $\endgroup$
    – Arturo don Juan
    Commented Apr 27, 2020 at 16:46
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    $\begingroup$ Related : A mathematically illogical argument in the derivation of Hamilton's equation in Goldstein. $\endgroup$
    – Frobenius
    Commented May 1, 2020 at 17:55

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