Could a Legendre transform be equivalent to evaluation in a retarded time? We have the following function:
$$U = U(r(t), p(t)) = (α/r)·\sqrt{1 - p^2/c^2}$$
being $α$ and $c$ positive constants, $t$ a real variable $0$ or positive, and $p = dr/dt$.
I would like to show that if we define
$t'= t – r(t)/c$
it is true that
$U' = U - p·(∂U/∂p)$
where the prime ' stands for evaluated at $t'$.
The origin of the problem is electrodynamics.
I try to show that the Legendre transform of the electrodynamic potential is equivalent to a retarded potential evaluated at instant $t'$.
I have tried to prove it from:
$\Delta U=\left( \frac{\partial U}{\partial r} \right)\cdot \Delta r+\left( \frac{\partial U}{\partial p} \right)\cdot \Delta p$
being
$\Delta U={U}'-U$,
$\Delta r={r}'-r$,
$\Delta p={p}'-p$
and $p={dr}/{dt}\;$
Then
${U}'=U+\left( \frac{\partial U}{\partial r} \right)\cdot \Delta r+\left( \frac{\partial U}{\partial p} \right)\cdot {p}'-\left( \frac{\partial U}{\partial p} \right)\cdot p$
In order to prove my equation, then
$0=\left( \frac{\partial U}{\partial r} \right)\cdot \Delta r+\left( \frac{\partial U}{\partial p} \right)\cdot {p}'$
We can assume that (for $\Delta t\ll 0$, and taken into account that ${t}'=t-r(t)/c$
$\Delta r=p\cdot \Delta t=p\cdot \left( {-r}/{c}\; \right)$
And $p’$ should be
${p}'=c\cdot \left( 1-\frac{{{p}^{2}}}{{{c}^{2}}} \right)$
But I can't think of how to prove the latter.
I think the approach I have followed is inappropriate. That is why I am asking for help.
 A: As far as I understand the comments you look at the
time transformation $$t'=t-r(t)/c$$ and evaluate
$U(r(t),p(t))$ at $t=t'$ which gives
$$\tag{A}
U\Big(r(t'),p(t')\Big)=U\Big(r\big(t-r(t)/c\big),p\big(t-r(t)/c\big)\Big)\,.
$$
(I do not think we have to denote this by $U'\,.$) Then you ask if (A) is of the form
$$\tag{B}
U\Big(r(t),p(t)\Big)-p(t)\frac{\partial U}{\partial p}\Big(r(t),p(t)\Big)\,.
$$
I find this very unlikely. Taking the $t$ derivative of (A) and (B) leads
to a nonlinear system of first order ODEs for $r(t)$ and $p(t)\,.$ Only when these functions satisfy that system we have equality of (A) and (B).
Some background on Legendre transforms:
For simplicity I assume $r>0$ and $p=\dot r>0\,.$ From
$$
\frac{\partial U}{\partial p}=\frac{\alpha}{rc^2}\frac{-p}{\sqrt{1-\frac{p^2}{c^2}}}=\frac{-\alpha}{rc^2\sqrt{\frac{1}{p^2}-\frac{1}{c^2}}}\,,\quad \frac{\partial^2U}{\partial p^2}=\frac{\alpha}{2rc^2}\frac{-2p^{-3}}{\Big(\frac{1}{p^2}-\frac{1}{c^2}\Big)^{3/2}}
$$
it follows that $U$ is strictly concave in $p\,.$ This allows to express the Legendre transform of $U$ as a nice function of $\partial U/\partial p\,.$ The convex
conjugate of $p\mapsto U(r,p)$ is the function
$$\tag{1}
U^*(r,q)=\sup_{p>0}\{qp-U(r,p)\}
$$
which is also called Fenchel-Legendre transform. Since $p\mapsto U(r,p)$
is strictly concave the supremum in (1) is attained for the unique $p$ that satisfies
$$\tag{2}
q=\frac{\partial U}{\partial p}(r,p)\,.
$$
This allows to write
$$\tag{3}
U^*(r,q)=qp-U(r,p)\,.
$$
Note that

*

*$p$ in (3) is a function of $q\,.$


*Therefore, $U^*$ is not a function of $p$ but instead of the conjugate variable $q=\frac{\partial U}{\partial p}\,.$


*In the physics literature a Legendre transform (3)  is often written quite sloppily as
$$\tag{4}
U^*=p\frac{\partial U}{\partial p}-U\,.
$$
This ignores the fact that $U^*$ is not a function of $p\,$.
For further details, see Zia et.al..


*Up to a minus sign, your $U'$ (my (B)) is the sloppy $U^*$ from (4) but this would be a Legendre transform and not a time transform.
