Linked Questions

3
votes
0answers
48 views

Hamiltonian definitions in the presence of boundary term [duplicate]

Consider a Lagrangian of the form \begin{equation} L(q,\dot{q})=L_1(q,\dot{q})+\frac{d L_2(q,\dot{q})}{dt} \end{equation} I understand that $\dot{L_2}$ does not modify the equations of motion, ...
73
votes
10answers
8k views

Why are differential equations for fields in physics of order two?

What is the reason for the observation that across the board fields in physics are generally governed by second order (partial) differential equations? If someone on the street would flat out ask me ...
13
votes
3answers
1k views

Are Maxwell's equations “physical”?

The canonical Maxwell's equations are derivable from the Lagrangian $${\cal L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} $$ by solving the Euler-Lagrange equations. However: The Lagrangian above is ...
14
votes
2answers
2k views

Functional derivative in Lagrangian field theory

The following functional derivative holds: \begin{align} \frac{\delta q(t)}{\delta q(t')} ~=~ \delta(t-t') \end{align} and \begin{align} \frac{\delta \dot{q}(t)}{\delta q(t')} ~=~ \delta'(t-t') \end{...
5
votes
3answers
3k views

Does a four-divergence extra term in a Lagrangian density matter to the field equations?

Greiner in his book "Field Quantization" page 173, eq.(7.11) did this calculation: ${\mathcal L}^\prime=-\frac{1}{2}\partial_\mu A_\nu\partial^\mu A^\nu+\frac{1}{2}\partial_\mu A_\nu\partial^\nu A^\...
5
votes
2answers
1k views

Accounting for metric tensor derivatives in Einstein-Hilbert action

I'm puzzling over the canonical derivation of GR from the Einstein-Hilbert action; getting the derivation to gel with an explicit treatment of the functional derivative isn't working out. So the ...
6
votes
2answers
1k views

Lagrangian $L' = L + \frac{df}{dt}$ gives the same equations of motion

It is well known that when a Lagrangian $L$ is incremented by the total time derivative of a function $f$ that does not depend on the time derivatives of the generalized coordinates, the same ...
7
votes
2answers
1k views

Invariance of canonical Hamiltonian equation when adding the total time derivative of a function of $q_i$ and $t$ to the Lagrangian

The following is exercise 8.2 in 3rd edition (and exercise 8.19 in 2nd edition) of Goldstein's Classical Mechanics. Adding the total time derivative of a function of $q_i$ and t to the Lagrangian ...
5
votes
1answer
496 views

Expanding electromagnetic field Lagrangian in terms of gauge field

The electromagnetic field tensor is given by $F_{\mu \nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}$, and it appears in the Lagrangian as $$L = -\frac{1}{4}F_{\mu\nu}^2 - A_{\mu}J_{\mu}.$$ ...
4
votes
1answer
391 views

Gauge invariance of the Hamiltonian

Consider a Lagrangian $L(x,\dot x,t)$ and a corresponding Hamiltonian $H=\dot xp-L$ where $p=\partial L/\partial \dot x$ which satisfies Hamilton's equations $$\frac{\partial H}{\partial x}=-\dot p$$ $...
2
votes
1answer
339 views

Hamiltonian and Lagrangian classical mechanics

Is the following logic correct?: If you have an Hamiltonian, that has time has a variable explicitly, and you get the Lagrangian $L$ from it, and then you get an equivalent $L'$, since $L$ has the ...
3
votes
3answers
139 views

Why is dependence on derivatives not a problem in the definition of canonical energy-momentum tensor?

Let $\mathcal L(\phi,\partial\phi)$ be a Lagrangian for a field $\phi$. It is known that the Lagrangian $\mathcal L$ and the Lagrangian $\mathcal L+\partial_\mu K^\mu$ produces the same physics, ...
1
vote
1answer
92 views

Why is there ambiguity of the electromagnetic field energy?

In the following question the OP asked what is meant in chapter 27-4 of Feynman's Lectures on Physics Vol.II by the ambiguity in the location of electromagnetic field energy: Why is there ambiguity of ...
2
votes
0answers
106 views

Gauge freedom in Lagrangian corresponds to canonical transformation of Hamiltonian

I want to show that the gauge transformation $$L(q,\dot{q},t)\mapsto L^\prime(q,\dot{q},t):=L(q,\dot{q},t)+\frac{d}{dt}f(q, t)$$ corresponds to a canonical transformation of the Hamiltonian $H(p, q, ...
3
votes
1answer
81 views

How does the Hamiltonian change if $L\to L + \frac{dF}{dt}$? [duplicate]

The Hamiltonian is defined as the Legendre transform of the Lagrangian $$H = p\dot{q} -L .$$ In the Lagrangian formalism we are free to add the total derivative of an arbitrary function $F=F(q,t)$ to ...

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