How do I set up this Legendre Transform for Hamiltonian

[13Nike]

Homework Statement

Im trying to understand the Legendre transform from Lagrange to Hamiltonian but I don't get it. This pdf was good but when compared to wolfram alphas example they're slightly different even when accounting for variables. I think one of them is wrong. I trust wolfram over the pdf but I like the pdf approach better. How does the wolfram formula relate to Lagrange formula given by wolfram? for example

• f = L
• x = ?
• y = ?
• u = (p maybe ?)
• v = ?
• g = H

also in the original pdf it states H is negative when H = L - p qdot. Why? Mathematically and physically why is this, and why does the Hamiltonian equations H = p qdot- L (in one dimension) equal to just the derivative of H in respect to q and p?

Homework Equations

Equation 6 in the pdf says that its negative? It likes like they're saying g(arbitrary) is -H = (L-p qdot) therefore H = (p qdot - L) I am confused as to why it can be -H.

The Attempt at a Solution

I attempted to do the matching above but I don't think I'm doing it right.

 

[vela]

Why don't you first identify what variables the Lagrangian and Hamiltonian depend on. Then you can identify which variable is being replaced by the another.

Both the PDF and MathWorld are saying the same thing. I'm not sure why you think there's an inconsistency.

As far as the sign of H goes, if you do the plain old Legendre transformation on the Lagrangian, you end up with a quantity that's equal to the negative of the total energy. We define H to be the negative of that quantity so we can identify it with the energy of the system rather than keeping the pesky negative sign around.

 

[13Nike]

Why don't you first identify what variables the Lagrangian and Hamiltonian depend on. Then you can identify which variable is being replaced by the another.

Both the PDF and MathWorld are saying the same thing. I'm not sure why you think there's an inconsistency.

As far as the sign of H goes, if you do the plain old Legendre transformation on the Lagrangian, you end up with a quantity that's equal to the negative of the total energy. We define H to be the negative of that quantity so we can identify it with the energy of the system rather than keeping the pesky negative sign around.

Okay thanks I understand the negative part but in terms of deriving it I would like to wolfram's formal definition but I don't get how it relates to Lagrangian. Can you help me where I'm wrong. L is a function of q and qdot, H is a function of q and p. Therefor qdot is being replaced with p. In other words, q and qdot are the independent variables. Meaning that qdot and p are the conjugate pairs? The equivalent table using the wolfram definition would be

• f = L
• x = q
• y = qdot
• u = u
• v = p
• g = H

?

 

[vela]

I think you meant $u=q$. Yes, $q$ and $\dot{q}$ are independent variables, and $\dot{q}$ and $p$, which is defined as $p=\partial{L}/\partial{\dot{q}}$, are the conjugate pair.

 

[13Nike]

I think you meant $u=q$. Yes, $q$ and $\dot{q}$ are independent variables, and $\dot{q}$ and $p$, which is defined as $p=\partial{L}/\partial{\dot{q}}$, are the conjugate pair.

I don't think that's correct for the equivalency? What is x then?

 

[vela]

$x$ is $q$. It's the variable that remains. That's why $u=q$ as well.