Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with lagrangian-formalism
Search options not deleted
user 59215
For questions involving the Lagrangian formulation of a dynamical system. Namely, the application of an action principle to a suitably chosen Lagrangian or Lagrangian Density in order to obtain the equations of motion of the system.
18
votes
Accepted
Adding a total time derivative term to the Lagrangian
You have seen that the substitution
$$L\longrightarrow L':= L+\frac{\mathrm{d}F}{\mathrm{d}t}$$
does not change the Euler-Lagrange equations. Now, this happens because the time derivative satisfies t …
- 8,943
0
votes
Trying to understand relativistic action of a massive point particle
The action is a time integral, just as you wrote. However, it's also a (line) integral of proper distance. This form is convenient when you make the jump to GR, because $ds$ has an obvious generalizat …
- 8,943
3
votes
What can be inferred about this particle from a Lagrangian?
It is well known that adding a total time derivative to the Lagrangian does not change equations of motion.
The Lagrangian above adds a term $$-q\dot q=-\frac{1}{2}\frac{\mathrm{d}q^2}{\mathrm{d}t}$$ …
- 8,943
1
vote
Accepted
The Nambu-Goto action how do we know the Hamilton's principle applies?
The action principle holds by assumption. It is assumed that all equations of motion follow from this principle with the appropriate action.
By introducing an auxiliary tensor field $h_{\alpha\beta} …
- 8,943
3
votes
Accepted
In the context of quantum field theory, what does it mean to "couple" something?
In terms of Feynman diagrams, a "coupling" translates to a vertex factor. The Lagrangian for a free electromagnetic field is
$$\mathcal{L}=-\frac{1}{4}F^2$$
as you well know. Now suppose we have an el …
- 8,943
5
votes
Accepted
What assumptions about the action do we make or give up in transitioning from classical mech...
In NRQM, we represent particles by a localized wave packet $\psi$, called the wave function. We say roughly that the classical particle is located at the "peak" of the wave packet. We say that we cann …
- 8,943
6
votes
1
answer
494
views
Is there a Maupertuis principle for General Relativity?
The motion of a point particle in classical mechanics is given by Newton's equation, $\mathbf{F}=m\mathbf{a}$. Suppose all forces considered are conservative and we have a constant total energy $h$. L …
- 8,943
6
votes
Finding geodesics: Lagrangian vs Hamiltonian
They are all equivalent. The answer to your other question is: the Hamiltonian approach usually works best.
Geodesics can be defined a few ways, since the connection of spacetime is taken to be Levi …
- 8,943
7
votes
Accepted
Why does $\frac{d\tau}{d\sigma} = L$?
The Lagrangian you wrote is
$$L=\sqrt{-g_{\mu\nu}\frac{dx^\mu}{d\sigma}\frac{dx^\nu}{d\sigma}}$$
I'm sure you also know that
$$d\tau^2=-g_{\mu\nu}dx^\mu dx^\nu$$
Plugging this into the first equation, …
- 8,943
1
vote
Why does the classical electrodynamics Lagrangian density equation have a "field" term and a...
I know that the question specifically refers to classical electrodynamics, but I think it is helpful to look at this from a QED perspective. The term $-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$ is the kinetic …
- 8,943
6
votes
Accepted
Energy-Momentum Tensor for Electromagnetism in Curved Space
The energy momentum tensor is found by varying the metric and holding all other fields constant. Since clearly $$\frac{\partial F}{\partial g}=0\longleftrightarrow
\delta_gF=0$$
we end up with
$$\delt …
- 8,943
1
vote
Correct derivation of Einstein's equations from the Hilbert action
I think it is important to understand what exactly the functional differential $\delta$ is doing. We have a functional $S:\mathscr E\to \Bbb R$, where $\mathscr E$ is some vector space of field config …
- 8,943
5
votes
Why a timelike geodesic maximizes path length?
First we sketch a proof that a timelike geodesic is a maximum of proper time. (We exclude saddle points for now.) Let $\gamma$ be a curve satisfying the geodesic equation, i.e. it is an extremum of pr …
- 8,943
9
votes
Accepted
How to calculate explicitly the classical on-shell action for a harmonic oscillator?
I randomly had this typed up in personal notes. Was probably an exercise somewhere.
Consider a harmonic oscillator, which is described by the Hamiltonian
$$H=\frac{p^2}{2m}+\frac{1}{2}m\omega^2q^2$$
…
- 8,943
17
votes
Geodesic equation from Euler - Lagrange
Let us do the RHS first. This just gives us a derivative on the metric:
$$\frac{\partial L}{\partial x^\lambda}=\frac{1}{2}\partial_\lambda g_{\mu\nu}\dot x^\mu\dot x^\nu$$
The first derivative on the …
- 8,943