Linked Questions

2
votes
0answers
84 views

Is there a physical connection between Lagrangian mechanics and Hamilton mechanics [duplicate]

I know that the Hamilton's equations results from the Legendre transformation of the Lagrange equation. It's also well-known that the Hamiltonian is equal to the energy of a system if the system is ...
0
votes
0answers
37 views

Why are so many (all?) Key concepts in thermodynamics derived from a legendre transformation? [duplicate]

Given a function $f(x,y)$, a legendre transform w.r.t. $x$ is $f^*(p,y)=p x - f(x,y) |_{p=\frac {\partial f(x,y)}{\partial x}}$. E.g. , the various free energies, enthalpy, etc are all legendre ...
11
votes
5answers
3k views

A mathematically illogical argument in the derivation of Hamilton's equation in Goldstein

In the book of Goldstein, at page 337, while deriving the Hamilton's equations (canonical equations), he argues that The canonical momentum was defined in Eq. (2.44) as $p_i = \partial L / \partial \...
16
votes
2answers
1k views

Why exactly do we say $L = L(q, \dot{q})$ and $H = H(q, p)$?

In classical mechanics, we perform a Legendre transform to switch from $L(q, \dot{q})$ to $H(q, p)$. This has always been confusing to me, because we can always write $L$ in terms of $q$ and $p$ by ...
9
votes
5answers
534 views

Why can't we obtain a Hamiltonian by substituting?

This question may sound a bit dumb. Why can't we obtain the Hamiltonian of a system simply by finding $\dot{q}$ in terms of $p$ and then evaluating the Lagrangian with $\dot{q} = \dot{q}(p)$? Wouldn't ...
5
votes
3answers
1k views

Writing $\dot{q}$ in terms of $p$ in the Hamiltonian formulation

In the Hamiltonian formulation, we make a Legendre transformation of the Lagrangian and it should be written in terms of the coordinates $q$ and momentum $p$. Can we always write $dq/dt$ in terms of $...
7
votes
1answer
2k views

Why is the Hamiltonian the Legendre transform of the Lagrangian?

So, as the title says, why is the Hamiltonian the Legendre transform of the Lagrangian? I know that from quantum mechanics, one can start with the Hamiltonian defined as the generator of time ...
6
votes
2answers
1k views

Legendre transformation: non-convex/non-convave functions

The Legendre transformation is used to derive the Hamiltonian from the Lagrangian, and it finds many applications in thermodynamics to convert between the different potentials. $ f(x) \rightarrow g(u)...
11
votes
1answer
789 views

Physical Interpretation of the Graph of the Legendre Transform?

See Making Sense of the Legendre Transform and Legendre Transforms for Dummies. Look at the following diagram from the first link: I was trying to think of the simplest example to interpret this ...
2
votes
1answer
573 views

Motivating the Legendre Transform Mathematically

If I begin with a functional of the form $$J[y] = \int_a^b f(x,y,y')dx$$ and find its Euler-Lagrange equations $$\frac{\partial f}{\partial y} - \frac{d}{dx}\frac{\partial f}{\partial y'} = 0 = \...
-1
votes
1answer
214 views

Apply the Heisenberg Equation to the Hamiltonian [closed]

$\frac{d}{dt}$$\hat{H}$ = $\frac{i}{\hbar}$$[\hat{H},\hat{H}]$ +$\frac{\partial{\hat{H}}}{\partial{t}}$ That's as far as I've got. I do not know much about the Heisenberg equation or even what it ...
1
vote
2answers
441 views

Is this a valid derivation of the Legendre transformation from the Euler-Lagrange condition

E-L condition: $$\frac{d p}{dt}=\frac{\partial L}{\partial q}$$ Where $p=\frac{\partial L}{\partial \dot{q}}$ Are the following steps valid: $$\frac{\partial q}{dt} dp=\partial L$$ $$\dot{q} \: ...
3
votes
1answer
147 views

Conceptual difficulty with the Legendre transformation

I feel like I have a fairly muddled understanding of function transformations, and so I'm hoping someone can clarify things for me a bit. I think this might be in part to do with confusion surrounding ...
7
votes
1answer
84 views

What properties, specifically, make the Legendre transform so useful in physics?

The Legendre transform plays a pivotal role in physics in its connecting Lagrangian and Hamiltonian formalisms. This is well-known and has been discussed at length in this site (related threads are e....
1
vote
0answers
65 views

What exactly is the Legendre transformation? [duplicate]

Goldstein et al's Classical Mechanics states that: The Hamiltonian $H(q,p,t)$ is generated by the Legendre transformation $$ H(q,p,t) = \dot{q}_i p_i - L(q,\dot{q},t). \tag{8.15} $$ But I don't ...

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