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I explain all this below because I want to exactly see the problem in my flow and would need an explanation depending on this example that I'm bringing. Please don't try to bring $p$ as generalized variable, let's stick with $ds/dt$.

We have $s(t) = t^2$ function which is distance-time graph.

  • t=1, s = 1
  • t=2, s = 4
  • t=3, s=9

Let's do Legendre transformation. We know that legendre is $y$ interception of each of tangent line of our curve. So we're looking for the equation of $b$ from $kx +b$.

$b = \frac{ds}{dt}t - s(t)$

By getting rid of $t$ from the equation in order to fully transform to different variable other than t, we end up with: $\frac{1}{4}(\frac{ds}{dt})^2$. Now, if we plug in velocities, we will get the distance again - e.x when car had velocity = 4, distance he had travelled at that moment was 4. which is correct.

So here, distance was exactly $b$ as in $b(v)$ which gets us the distance.

Doing the same for $s = t^3$, I end up:

$$b = \sqrt{\frac{1}{3}\frac{ds}{dt}} \frac{ds}{dt} - \frac{1}{3}\frac{ds}{dt}\sqrt{\frac{1}{3}\frac{ds}{dt}}$$ (Specifically don't mention $p$ generalization as I want to understand this both physically and mathematically. Note now that at $t=2$, $s = 2^3=8$. By looking at $s' = 3t^2$, we see that at $t=2$, $v = 12$. Now, if we plug this for $\frac{ds}{dt}$ in the derived legrende , we get $b = 2*12 - 4*2 = 16$, but our distance was 8. So I don't get it, I plugged in velocity, it gave me 16.

I know that it would be more confusing if it always had matched the distance. I don't get what $16$ truly means for this case. Should It always be matching such as $s(t) = b(v)$ where pluggined in $v$ is velocity at time $t$ which we plug in $s(t)$ ?

Qmechanic
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Giorgi
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1 Answers1

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For any function $s(t)$ the Legendre transform is defined as $$ s^*(p)=\inf_{t\in\mathbb R}\{p\,t-s(t)\}\,. $$ $p$ is the new conjugate variable. For nice enough functions $s(t)$ that infimum is attained for $t=t(p)$ satisfying $\dot s(t)=p\,.$ Then, $$\tag{1} s^*(p)=t(p)\,p-s(t(p))\,,\quad \dot s(t(p))=p\,. $$ In your case, $s(t)=t^2\,,\; \dot s(t)=2\,t\,,\; t(p)=p/2\,,$ so that $$ \boxed{\phantom{\Bigg|}\quad s^*(p)=\frac{p^2}{2}-\Big(\frac{p}{2}\Big)^2=\frac{p^2}{4}\,.\quad} $$ When $s(t)=t^2$ is the distance as a function of time $t$ then this expression for $s^*(p)$ is obviously the distance as a function of velocity $p=\dot s(t)\,.$

  • It seems like this coincidence in the case $s(t)=t^2$ makes you think that $s^*(p)$ is always the distance travelled as a function of velocity. But this is not the case and your own example $s(t)=t^3$ is a counter example.

  • It is true that $s^*(p)$ has the units of a distance and is a function of velocity and contains the same information as the original function $s(t)\,.$ For further details see [1].

For $s(t)=t^3$ we get $\dot s(t)=3\,t^2\,,\;t(p)=\sqrt{p/3}\,,$ so that now $$ \boxed{\phantom{\Bigg|}\quad s^*(p)=p\sqrt{\frac{p}{3}}-\Big(\frac{p}{3}\Big)^{3/2}=2\Big(\frac{p}{3}\Big)^{3/2}\,.\quad} $$ In this case $$ s(t(p))=\Big(\frac{p}{3}\Big)^{3/2}\not=2\Big(\frac{p}{3}\Big)^{3/2}=s^*(p)\,. $$

[1] R. K.P. Zia, E.F. Redish, S.R. McKay. Making Sense of the Legendre Transform. https://arxiv.org/abs/0806.1147.

The following edit has the only purpose to simplify a discussion which tries to draw a connection between Legendre transform an integration-by-parts:

From (1)
$$ s^*(p)+s(t(p))=t(p)\,p\,. $$ Differentiating w.r.t. $p$ gives

$$ \dot s^*(p)+\dot s(t(p))\,\dot t(p)=\dot t(p)\,p+t(p)\,. $$ Using $\dot s(t(p))=p$ leads to $$ \dot s^*(p)=t(p)\,. $$ In short: When (1) holds it is (with some sloppy notation) always true that $$\boxed{\quad\phantom{\Big|} \frac{d}{dt}s(t)=p\,,\quad\frac{d}{dp}s^*(p)=t\quad} $$ hold.

To draw a connection to integration-by-parts I think we have to assume that the variable $p$ is a function of another variable $x$ and write with some abuse of notation $t(x)=t(p(x))\,.$ Then $$ s^*(p(x))+s(t(x))=t(x)\,p(x)=\int_0^x \dot t(y)\,p(y)\,dy+\int_0^x t(y)\,\dot p(y)\,dy\,. $$

Kurt G.
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