Linked Questions

4
votes
4answers
380 views

Looking for a Basic ( or “for Dummies” ) Explanation of the Lagrangian - Hamiltonian Relationship. ( Mathematician ) [duplicate]

(Mathematician here - first time physics.stack poster). I'm basically looking for as simple as possible explanation of the Hamiltonian - Lagrangian relationship. $\textbf{My understanding :}$ $\textbf{...
3
votes
1answer
556 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 ...
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 \...
26
votes
3answers
4k views

Are the Hamiltonian and Lagrangian always convex functions?

The Hamiltonian and Lagrangian are related by a Legendre transform: $$ H(\mathbf{q}, \mathbf{p}, t) = \sum_i \dot q_i p_i - \mathcal{L}(\mathbf{q}, \mathbf{\dot q}, t). $$ For this to be a Legendre ...
15
votes
5answers
1k views

Does a Lagrangian always generate a unique Hamiltonian?

Hamiltonian is related to Lagrangian with the equation: $$H= p\dot{q}- L(q,\dot{q},t) $$ Now, $H$ is function of $p,q,t$ so the Hamiltonian to be unique, $\dot{q}$ must be expressed uniquely using $...
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 ...
18
votes
2answers
1k views

Can one write down a Hamiltonian in the absence of a Lagrangian?

How can I define the Hamiltonian independent of the Lagrangian? For instance, let's assume that i have a set of field equations that cannot be integrated to an action. Is there any prescription to ...
10
votes
2answers
2k views

What do the derivatives in these Hamilton equations mean?

I have a Hamiltonian: $$H=\dot qp - L = \frac 1 2 m\dot q^2+kq^2\frac 1 2 - aq$$ In a system with one coordinate $q$ (where $L$ is the Lagrangian). One of the Hamilton equations is: $$\dot q =-\...
12
votes
3answers
2k views

Constraints in classical mechanics

I am self-studying classical mechanics and I have a couple of questions about constraints. Goldstein in his book Classical Mechanics writes in the beginning of Section 1.3 that: It is an overly ...
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 $...
3
votes
4answers
874 views

Help understanding what the Hamiltonian signifies for the action compared with the Euler-Lagrange equations for the Lagrangian?

Consider the Lagrangian for a simple harmonic oscillator \begin{equation} L (x,\dot{x}) = \frac{1}{2}m\dot{x}^2 - \frac{1}{2}kx^2 \end{equation} Obviously we have \begin{align} \frac{\partial L}{\...
17
votes
1answer
779 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 ...
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

Finding geodesics: Lagrangian vs Hamiltonian

I have a question referring to how to compute geodesics of a given spacetime (say, Kerr). I know that the direct way is via the geodesic equation $$\frac{d^{2}x^{\mu}}{d\lambda^{2}}+\Gamma^{\mu}_{\...
12
votes
2answers
1k views

Why do the Lagrangian and Hamiltonian formulations give the same conserved quantities for the same symmetries?

The connection between symmetries and conservation laws can be viewed through the lens of both Lagrangian and Hamiltonian mechanics. In the Lagrangian picture we have Noether's theorem. In the ...

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