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8 votes
1 answer
2k views

If the Lagrangian depends explicitly on time then the Hamiltonian is not conserved?

Why is the Hamiltonian not conserved when the Lagrangian has an explicit time dependence? What I mean is that it is very obvious to argue that if the Lagrangian has no an explicit time dependence $L=L(...
Spectree's user avatar
  • 227
1 vote
0 answers
83 views

Independence of variables in Lagrangain and Hamiltonian mechanics - Rigorous Mathematical approach

I am trying to self-learn the Hamiltonian and Lagrangian mechanics and I came across thoughts to which I could not find an answer therefore I would like to try and ask them here. My questions are as ...
Gauss_fan's user avatar
0 votes
0 answers
64 views

What would make the Legendre transformation interesting, from the graphical point of view?

The term "transformation" is often used in physics and mathematics for functions to denote a different (and useful) way of encoding the information in a function. In this question, I want to ...
Jbag1212's user avatar
  • 2,599
2 votes
1 answer
861 views

Proof that Hamiltonian is constant if Lagrangian doesn't depend explicitly on time

I know that on solutions of motion we have $\frac{dH}{dt}=\frac{\partial H}{\partial t} $ and i understand the proof for this fact. Then, we have that $$\frac{\partial H}{\partial t}=-\frac{\partial L}...
abc's user avatar
  • 49
0 votes
2 answers
42 views

Is there a way to understand which variable is more influential in the dynamics of a system?

Is there any known way to identify which variable has the most impact in the dynamics of a system given its lagrangian or hamiltonian formulation? Let's say i have a system with 3 variables, two ...
francesco pagano's user avatar
2 votes
2 answers
156 views

How to derive the fact that $p\sim d/dx$ and $H\sim d/dt$ from classical mechanics?

I am trying to understand Noether's conserved quantities to shifts in time and or position. I have seen the derivation of the operators for Schrodinger's equation but not for classical mechanics. Is ...
nemui's user avatar
  • 381
0 votes
1 answer
240 views

A relationship between Lagrangian formalism and Hamiltonian formalism

In the Lagrangian formalism, The Lagrangian $$L = T\text{(kinetic energy)} - V\text{(potential energy)}$$ The equations of motion for a given system is given by minimizing the action functional which ...
yo-yos's user avatar
  • 11
20 votes
3 answers
881 views

What properties 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....
glS's user avatar
  • 14.8k
2 votes
1 answer
86 views

I don't get this "derivation" of canonical transformation

Given a transformation $$(q, p, t)\to (Q(q, p, t), P(q, p, t), t),$$ the modified Hamiltonian, $K$ is related to the original one, $H$, as $$H(q, p, t) = K(Q(q, p, t), P(q, p, t), t).$$ Now, what I've ...
Atom's user avatar
  • 1,951
0 votes
1 answer
237 views

Why is the energy function not always equal to total energy? [duplicate]

Why is the energy function $h = \dot{q_i}\frac{\partial L}{\partial \dot{q_i}} - L $ not always equal to total energy $E = T + V$? Here $T$ is Kinetic Energy and $V$ is Potential Energy. I've read ...
dimes's user avatar
  • 75
3 votes
3 answers
704 views

Why is the Legendre transformation the correct way to change variables from $(q,\dot{q},t)\to (q,p,t)$?

I always found Legendre transformation kind of mysterious. Given a Lagrangian $L(q,\dot{q},t)$, we can define a new function, the Hamiltonian, $$H(q,p,t)=p\dot{q}(p)-L(q,\dot{q}(q,p,t),t)$$ where $p=\...
Solidification's user avatar
7 votes
3 answers
324 views

Hamiltonian of non-regular Lagrangian is well-defined on phase space

In section 1.1.3 of Quantization of Gauge Systems by Henneaux and Teitelboim, it is stated that the Hamiltonian $$H=\dot{q}^np_n-L,\tag{1.8}$$ although trivially a function of $q$ and $\dot{q}$, can ...
Ivan Burbano's user avatar
  • 3,915
0 votes
0 answers
69 views

When the hamiltonian isn't equal to energy? [duplicate]

I have the following hamiltonian: $$H = \frac{p_1^2}{2}+\frac{(p_2-k\;q_2)^2}{2} ,\qquad k\in\mathbb{R}.$$ I know that the hamiltonian isnt explicitly dependent on time so $H$ is a motion ...
Ezequiel Saidman's user avatar
1 vote
1 answer
139 views

What if we set Hamilton-Jacobi equation as an axiom?

We usually postulate the principle of least action. Then we can get Lagrangian mechanics. After that we can get Hamiltonian mechanics either from postulate or from the equivalent Lagrangian mechanics. ...
Semyon  Yurchenko's user avatar
1 vote
0 answers
30 views

How to calculate Hamiltonian when Lagrangian has higher order derivatives? [duplicate]

If we have a Lagrangian density $\mathcal{L}$ for a scalar field $\phi$ depending on $\phi$, $\partial _{\mu} \phi$, and $\partial _{\mu} \partial _{\nu} \phi$, what is the Hamiltonian? Additionally, ...
Joshua Tilley's user avatar
2 votes
1 answer
1k views

Difference between the energy and the Hamiltonian in a specific example

The problem is the following: Consider a particle of mass $m$ confined in a long and thin hollow pipe, which rotates in the $xy$ plane with constant angular velocity $\omega$. The rotation axis ...
Slayer147's user avatar
  • 1,045
2 votes
4 answers
1k views

The definition of the hamiltonian in lagrangian mechanics

So going through the "Analytical Mechanics by Hand and Finch". In section 1.10 of the book, the Hamiltonian $H$ is defined as: $$H = \sum_k{\dot{q_k}\frac{\partial L}{\partial \dot{q_k}} -L}.\tag{1.65}...
user avatar
3 votes
1 answer
1k views

General Form for Kinetic Energy Given Velocity Independent Potential such that $\mathcal{H}=E$

Suppose the potential energy is independent of $\dot{q},$ i.e $\frac{\partial V}{\partial\dot{q}}=0$. What is the most general form of the kinetic energy such that the Hamiltonian is the total energy? ...
garserdt216's user avatar
19 votes
1 answer
1k views

Why are Hamiltonian Mechanics well-defined?

I have encountered a problem while re-reading the formalism of Hamiltonian mechanics, and it lies in a very simple remark. Indeed, if I am not mistaken, when we want to do mechanics using the ...
Frotaur's user avatar
  • 2,296
4 votes
2 answers
801 views

Liouville's theorem for systems with dissipation described by a single hamiltonian

Following this link, one can treat dissipation by using a factor $e^{\frac{t \beta}{ m}}$ in addition to the Lagrangian $L_0$ of a system without disspation: $$ L[q, \dot{q}, t] = e^{\frac{t \beta}{ m}...
Quantumwhisp's user avatar
  • 6,763
6 votes
1 answer
924 views

Independence of generalised coordinates and momenta in Hamiltonian mechanics [duplicate]

I am told that in Hamiltonian mechanics, we put the generalised coordinates $q_i$ and generalised momenta $p_i$ on equal footing, and treat them as being independent from one another. But I'm ...
jl2's user avatar
  • 379
4 votes
1 answer
806 views

Do time-invariant Hamiltonians define closed systems?

In classical mechanics, every time-invariant Hamiltonian represents a closed dynamical system? Can every closed dynamical system be represented as a time-invariant Hamiltonian? Or are there closed ...
Ricardo's user avatar
  • 155
25 votes
3 answers
29k views

Constructing Lagrangian from the Hamiltonian

Given the Lagrangian $L$ for a system, we can construct the Hamiltonian $H$ using the definition $H=\sum\limits_{i}p_i\dot{q}_i-L$ where $p_i=\frac{\partial L}{\partial \dot{q}_i}$. Therefore, to ...
SRS's user avatar
  • 26.8k
3 votes
0 answers
222 views

Does the additivity property of Integrals of motion and Lagrangians valid in all situations?

I would like to know if the additivity property of an integral (constant) of motion valid in all situations ? It works for energy but does it work for all other integrals of motion in all kinds of ...
singularity's user avatar
8 votes
5 answers
716 views

Why can't we obtain a Hamiltonian from the Lagrangian by only 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 ...
carllacan's user avatar
  • 590
10 votes
3 answers
4k views

Is there a valid Lagrangian formulation for all classical systems?

Can one use the Lagrangian formalism for all classical systems, i.e. systems with a set of trajectories $\vec{x}_i(t)$ describing paths? On the wikipedia page of Lagrangian mechanics, there is an ...
Nikolaj-K's user avatar
  • 8,523