MHB Derivation of Euler-Lagrange equations w/ Hamilton's equations

AI Thread Summary
The discussion focuses on deriving the Euler-Lagrange equations using Hamilton's equations and the Hamiltonian's definition in terms of the Lagrangian. The initial attempt involved manipulating the Hamiltonian to express the Lagrangian, but the user encountered a roadblock when deriving the necessary relationships. A key insight shared is that the momentum \( p \) is a function of \( \dot{q} \), which affects the partial derivatives in the calculations. This oversight is crucial for correctly applying Hamilton's equations to arrive at the Euler-Lagrange equations. The user plans to rework their approach with this understanding in mind.
skate_nerd
Messages
174
Reaction score
0
I've got a problem that asks us to derive the Euler-Lagrange equations by only using Hamilton's equations and the definition of the Hamiltonian in terms of the Lagrangian. Here's what I tried:

The Hamiltonian is defined as
\begin{align*}
\mathcal{H} = \dot{q}_ip_i - \mathcal{L}
\end{align*}
(where the summation convention is implied), and solving for $\mathcal{L}$, we have
\begin{align*}
\mathcal{L} = \dot{q}_ip_i - \mathcal{H}
\end{align*}
Taking the partial derivative with respect to $\dot{q}_i$ on both sides of the above equation, we have
\begin{align*}
\frac{\partial\mathcal{L}}{\partial\dot{q}_i} &= \frac{\partial}{\partial\dot{q}_i}\left[\dot{q}_ip_i - \mathcal{H}\right] \\
&= p_i - \frac{\partial\mathcal{H}}{\partial\dot{q}_i}
\end{align*}
We are given that
\begin{align*}
\frac{\partial\mathcal{L}}{\partial\dot{q}_i} = p_i
\end{align*}
so going back to our definition for the Hamiltonian, we have
\begin{align*}
p_i = p_i - \frac{\partial\mathcal{H}}{\partial\dot{q}_i}
\end{align*}
So we find that
\begin{align*}
\frac{\partial\mathcal{H}}{\partial\dot{q}_i} = 0
\end{align*}

Clearly what I have tried is going nowhere, but the professor gave a hint where he says to start with the definition of the Hamiltonian and invert it to solve for the Lagrangian, which is exactly what I did. I feel like I'm at a bit of a roadblock, so any hints would be appreciated. Thanks everybody
 
Mathematics news on Phys.org
Hi skatenerd,

This is a nice question.

skatenerd said:
I've got a problem that asks us to derive the Euler-Lagrange equations by only using Hamilton's equations and the definition of the Hamiltonian in terms of the Lagrangian. Here's what I tried:

The Hamiltonian is defined as
\begin{align*}
\mathcal{H} = \dot{q}_ip_i - \mathcal{L}
\end{align*}
(where the summation convention is implied), and solving for $\mathcal{L}$, we have
\begin{align*}
\mathcal{L} = \dot{q}_ip_i - \mathcal{H}
\end{align*}
Taking the partial derivative with respect to $\dot{q}_i$ on both sides of the above equation, we have
\begin{align*}
\frac{\partial\mathcal{L}}{\partial\dot{q}_i} &= \frac{\partial}{\partial\dot{q}_i}\left[\dot{q}_ip_i - \mathcal{H}\right] \\
&= p_i - \frac{\partial\mathcal{H}}{\partial\dot{q}_i}
\end{align*}
We are given that
\begin{align*}
\frac{\partial\mathcal{L}}{\partial\dot{q}_i} = p_i
\end{align*}
so going back to our definition for the Hamiltonian, we have
\begin{align*}
p_i = p_i - \frac{\partial\mathcal{H}}{\partial\dot{q}_i}
\end{align*}
So we find that
\begin{align*}
\frac{\partial\mathcal{H}}{\partial\dot{q}_i} = 0
\end{align*}

Clearly what I have tried is going nowhere, but the professor gave a hint where he says to start with the definition of the Hamiltonian and invert it to solve for the Lagrangian, which is exactly what I did. I feel like I'm at a bit of a roadblock, so any hints would be appreciated. Thanks everybody

Your attempt was good and was in the right direction. Given your calculation, I imagine you've overlooked the same thing I did when I first encountered the relationship between Hamiltonian and Lagrangian mechanics. To simplify things, I will present things in one generalized coordinate dimension and let you work out how to extend things to the case of several variables (i.e. I won't have any $i$ indices anywhere)

Note that via the Legendre transformation, $p=p(q,\dot{q})$ and so its partial derivative with respect to $\dot{q}$ isn't zero in general. Furthermore, you can use this fact to further expand the partial derivative of the Hamiltonian in your calculation. From here you should be able to apply Hamilton's equations to derive the Euler-Lagrange equations.

Hopefully this can help you make some sense out of things this time around.
 
GJA said:
Hi skatenerd,

This is a nice question.
Your attempt was good and was in the right direction. Given your calculation, I imagine you've overlooked the same thing I did when I first encountered the relationship between Hamiltonian and Lagrangian mechanics. To simplify things, I will present things in one generalized coordinate dimension and let you work out how to extend things to the case of several variables (i.e. I won't have any $i$ indices anywhere)

Note that via the Legendre transformation, $p=p(q,\dot{q})$ and so its partial derivative with respect to $\dot{q}$ isn't zero in general. Furthermore, you can use this fact to further expand the partial derivative of the Hamiltonian in your calculation. From here you should be able to apply Hamilton's equations to derive the Euler-Lagrange equations.

Hopefully this can help you make some sense out of things this time around.

Thanks for the response! I think I see what you mean. When I took the partial derivative with respect to $\dot{q}_i$ I neglected the fact that $p_i$ is a function of $\dot{q}_i$. I'll rework this keeping that in mind.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top