All Questions
91
questions
4
votes
1
answer
70
views
Can we impose Coulomb gauge without using temporal gauge in source-free Maxwell electrodynamics?
Coulomb gauge is $$\vec{\nabla} \cdot A=0$$ Now, from expression for electric field in terms of potentials $\vec{E}=-\vec{\nabla} \phi-\frac{\partial \vec{A}}{\partial t}$ and Gauss Law $\vec{\nabla} \...
- 43
0
votes
0
answers
60
views
Degree of freedom - Lorentz transfomation reduces it? [duplicate]
I am having a real difficult to counting degree of freedom. In fact, I notice that sometimes I am confused about what exactly we count as DoF, and what we do not count.
See, for example, the ...
- 980
1
vote
0
answers
48
views
Gauge redundancy and Gauge fixing
Take any gauge invariant theory, for instance QED. The QED Lagrangian is invariant under
$$A_{\mu}(x)\rightarrow A'_{\mu}(x)=A_{\mu}(x)+\partial_{\mu}. \alpha(x)$$
I have chosen a local gauge ...
- 3,982
1
vote
1
answer
76
views
Magnetic vector potential in 1+1 spacetime dimensions
In the theory of electromagnetism in 1+1 spacetime dimensions (one temporal and one spatial coordinate), one can define the 2-potential vector (analogous to the 4-potential vector in 3+1 spacetime ...
0
votes
1
answer
99
views
Can we express the electrodynamic potentials $V$, $\mathbf{A}$ in terms of the electrodynamic fields $\mathbf{E}$, $\mathbf{B}$?
In Griffiths' Introduction to Electrodynamics problem 10.25, I am asked to draw a "triangle diagram" illustrating the relationship between (1) the sources $\rho$, $\mathbf{J}$, (2) the ...
- 660
1
vote
0
answers
50
views
Bibliography for the Quantization of the free electromagnetic field with the Lorenz gauge
Recently I have been studying QFT and when I arrived at the Gauge theory I learned that one can quantize the electromagnetic field with the Coulomb gauge and the Lorenz gauge.
Regarding the Coulomb, I ...
13
votes
2
answers
2k
views
Trouble reconciling these two views on gauge theory
Very generally speaking, I view gauge theory as asking what local symmetries leave our theory invariant and then seeing the consequences. Thus, taking a look at the Lagrangian for electromagnetism, we ...
- 3,350
3
votes
3
answers
233
views
When we solve the Maxwell equations for $(\phi,{\bf A})$ in a gauge, will the solution $(\phi,{\bf A})$ automatically obey the gauge condition?
As the title of the question suggest; how you could determine if a gauge fixing is a condition or a requirement. Let me explain.
Imagine you are working with Maxwell's Equations. By the definition of ...
- 1,515
3
votes
2
answers
573
views
Quantum Theory of Radiation Enrico Fermi 1932
I was reading Fermi's review on Dirac's "Quantum Theory of Radiation", which he published in 1932. I was unable to know why he expressed electric field as the following:
I understand that ...
2
votes
0
answers
107
views
Proving that the path integral formulation of scalar QED theory is independent of the choice of the gauge-fixing parameter $\xi$
I am considering the following scalar QED lagrangian:
$$L = −\frac{1}{4}F_{\mu\nu}^2 + |D_{\mu\varphi}|^2 − m^2|\varphi|^2− \frac{1}{2\xi}(\partial_\mu A^\mu)^2.$$
Where I want to show that the ...
- 41
1
vote
1
answer
180
views
Gauge choice and observable quantities
Assume that I have the usual $U(1)$ gauge field $A_{\mu}$. We know that observable quantities are invariant under global transformations of the form $A_{\mu}\rightarrow A_{\mu}'=A_{\mu}+\partial_{\mu}\...
- 3,982
2
votes
0
answers
56
views
$R_\xi$ gauge and degrees of freedom counting
In the standard classical Maxwell theory, we use the following arguments to claim that there are only two propagating degrees of freedom
$A_\mu$ has 4 components
$A_0$ is non-dynamical (-1)
$\...
- 669
1
vote
0
answers
154
views
Peskin and Schroeder's QFT eq.(9.56)
On Peskin and Schroeder's QFT book, page 296, the book give the functional integral formula after inserting Faddeev and Popov's trick of identity.
$$ \int \mathcal{D} A e^{i S[A]}=\operatorname{det}\...
- 1,421
2
votes
2
answers
263
views
Does Coulomb gauge imply constant density?
Say we have
$$\Box A = J$$
and
$$\nabla \cdot A = 0\;.$$
Then
$$0 = \Box (\nabla \cdot A) = \nabla \cdot J\;.$$
But,
$$\nabla \cdot J - \partial_t \rho = 0\;.$$
So
$$ \partial_t \rho = 0\;.$$
Thus,
$$\...
- 184
2
votes
0
answers
269
views
Coulomb gauge choice: Does $A_0=0$ imply that we also need to choose $\nabla \cdot \vec{A} =0$ from the EOM of $A_0$?
How to justify the Coulomb gauge fixing condition choice with
$$
A_0=0, \quad \nabla \cdot \vec{A} =0?
$$
Below in the text image, I find a text explaining that imposing $A_0=0$ is always possible ...
- 593