All Questions
17
questions
0
votes
3
answers
417
views
Potential energy of a particle inside a magnetic vector potential
Why is the potential energy of a particle inside a magnetic vector potential equal to $-\frac{e}{c}\cdot\vec{A}(\vec{x}(t))\cdot\dot{\vec{x}}(t)$? It appears for example inside the lagrangian of the ...
- 189
1
vote
1
answer
81
views
On generalised potential in Electrodynamics
I'm studying Lagrangian Mechanics from Goldstein's Classical Mechanics. My question concerns Section 1.5 which talks about velocity-dependent potentials.
I am actually unsure about how Equation 1-64' ...
- 1,638
0
votes
2
answers
102
views
How to understand "the potential energy in an EM field is determined by $\phi$ alone"?
Goldstein page 342,
Consider a single particle (non-relativisitic) of mass $m$ and charge $q$ moving in an electromagnetic field.
The Lagrangian is
$$ L = T-V = \frac{1}{2}mv^2-q\phi +q\vec{A}\cdot \...
- 1
1
vote
3
answers
443
views
Intuition on $E^2-B^2$ as a Lagrangian?
I recall reading that instead of $E$ and $B$ vectors, gradients of $E^2-B^2$ are fundamental. I understand that it is the Lorentz invariant Lagrangian for the EM field. The signs of the energy imply ...
- 381
0
votes
1
answer
169
views
How to get the Magnetic Force from the Electromagnetic Tensor using Hodge Decomposition?
The following notation is used below:
d: exterior derivative
$\delta$: codifferential (adjoint of d)
$\times$: skew-symmetric operator of a $\mathbb{R}^3$-vector
$\nabla\times$: Curl operator in ...
2
votes
3
answers
4k
views
Lagrangian in presence of an Electromagnetic Field
Given the following definition of the Magnetic Vector Potential $\vec{A}$:
$$\vec{A} \ \mid \ \vec{B}=\vec{\nabla}\times\vec{A}$$
We can derive (but I don't know how) that the Lagrangian in presence ...
- 4,577
-2
votes
1
answer
50
views
Expressing $E$ field and $B$ field using potentials
Why dont we have a term that corresponds to the change in E field in eq1.61b like we have in eq 1.61a?
0
votes
1
answer
157
views
Follow-up on "Derivation of Lagrangian of electromagnetic field from Lorentz force"
I have a follow-up on this post. The way I understand it, if one generally has a velocity-dependent potential $U(q, \dot q, t)$, then we can derive/define a generalized force $$Q_k = \frac{d}{dt}\frac{...
user248824
5
votes
4
answers
985
views
Deeper Meaning to the Nature of Lagrangian
Is there a more fundamental reason for the Classical Lagrangian to be $T-V$ and Electromagnetic Lagrangian to be $T-V+ qA.v$ or is it simply because we can derive Newton's Second Law and Lorentz Force ...
- 177
1
vote
2
answers
749
views
Velocity-dependent potentials and the dissipation function
From this previous question Charge, velocity-dependent potentials and Lagrangian where the citation is shown at the page 22, §1.5 of the book Classical Mechanics of Goldstein, we read that
"an ...
- 2,547
0
votes
2
answers
283
views
Lagrangian of a moving charge distribution under action of an eletromagnetic field
We all know that for a single charged particle, we can derive the Lagrangian starting from Lorentz law of force:
$$
\mathbf{F}=q(\mathbf{E}+\mathbf{v}\times\mathbf{B}).
$$
and by using the ...
1
vote
0
answers
527
views
Particle in electromagnetic field Lagrangian
Given the two definitions of $\vec E$ and $\vec B$ by scalar potential $\phi$ and vector potential $\vec A$:
$$\vec B=\vec \nabla \times \vec A$$
$$\vec E=-\vec \nabla \phi -\frac 1 c\frac {\partial \...
- 141
2
votes
1
answer
271
views
Lagrangian of a massive particle in an electromagnetic field
I am trying to find the Lagrangian of a massive particle in an electromagnetic field using the Lorentz force: $$ \vec F = q ( \vec E + \vec v \times \vec B)$$ with $$\vec E = - \nabla \phi - \frac{\...
- 43
0
votes
0
answers
118
views
From Lagrangian of Electromagnetic field to the Lorentz force? [duplicate]
The dynamics of a charged particle with velocity $\textbf{v}$ in electromagnetic field is dominated by the Lorentz force
$$\textbf{F} = q(\textbf{E}+{\textbf{v} \times \textbf{B}}), \tag{1}$$
and ...
- 1,757
0
votes
1
answer
2k
views
Lorentz force in terms of potential
The Wikipedia page on the Lorentz force states the following:
$$\boldsymbol{F}=q\left[- \boldsymbol{\nabla}(\phi-\boldsymbol{v} \cdot \boldsymbol{A})-\frac{d\boldsymbol{A}}{dt}\right]$$
which ...
- 381