All Questions
91
questions
18
votes
2
answers
3k
views
Counting the number of propagating degrees of freedom in Lorenz Gauge Electrodynamics
How do I definitively show that there are only two propagating degrees of freedom in the Lorenz Gauge $\partial_\mu A^\mu=0$ in classical electrodynamics. I need an clear argument that
involves the ...
- 6,381
17
votes
3
answers
8k
views
What is a gauge in a gauge theory?
As I study Jackson, I am getting really confused with some of its key definitions. Here is what I am getting confused at. When we substituted the electric field and magnetic field in terms of the ...
- 1,355
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
11
votes
2
answers
12k
views
What is the physical meaning of Lorenz gauge condition? [closed]
What is the physical meaning of Lorenz gauge condition?
And what part of the solutions we throw?
- 163
10
votes
2
answers
3k
views
Showing that Coulomb and Lorenz Gauges are indeed valid Gauge Transformations?
I'm working my way through Griffith's Introduction to Electrodynamics. In Ch. 10, gauge transformations are introduced. The author shows that, given any magnetic potential $\textbf{A}_0$ and electric ...
- 103
10
votes
1
answer
3k
views
Gauge theory and eliminating unphysical degrees of freedom
In free space we can express Maxwell's equations as
\begin{align}
\varepsilon^{abcd}\partial_bF_{cd}=0 ~~\text{ and }~~ \partial_aF^{ab}=0 \tag{1}
\end{align}
where $F^{ab}=-F^{ba}$. The most general ...
- 599
10
votes
2
answers
2k
views
Question about physical degree of freedom in Maxwell Theory: Why Coulomb gauge can fix all redundant degree of freedom
Given $4$-potential $A^\mu(x)=(\phi(x),\mathbf{A}(x))$, the vacuum Maxwell equations:
$$\nabla^2\phi+\frac{\partial}{\partial t}(\nabla\cdot \mathbf{A} )=0$$
$$\nabla^2 \mathbf{A} -\frac{\partial^2 \...
user153663
9
votes
2
answers
884
views
Why the extra term $\frac{1}{2}(\partial_{\rho}A^{\rho})^2$ in the photon Lagrangian?
In my quantum field theory class we have been told to use this Lagrangian for the photon field
$$\mathcal{L}=-\frac{1}{4}F_{\alpha\beta}F^{\alpha\beta}
-\frac{1}{2}(\partial_{\rho}A^{\rho})^2.$$
but ...
- 6,067
9
votes
3
answers
1k
views
Why do we use gauges in Maxwell equation?
While solving the Maxwell's equation we often use the Lorenz or Coulomb gauge, but why is that? Are the equations unsolvable if the gauge is not fixed? Or is it just for the simplicity?
- 340
7
votes
1
answer
744
views
Why is the gauge-fixing condition squared in the QED Lagrangian?
Consider the free Maxwell Lagrangian:
$$L= -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}. $$
As we know, the gauge symmetry $A_{\mu} \rightarrow A_{\mu}+\partial_\mu \lambda$ must be fixed when quantizing the ...
- 2,296
6
votes
3
answers
781
views
Can I call additional conditions on potentials a Gauge choice?
Let's say I have an electromagnetics problem in a spatially varying medium. After I impose Maxwell's equations, the Lorenz gauge choice, boundary conditions, and the Sommerfeld radiation condition, I ...
- 555
6
votes
1
answer
1k
views
Why does Coulomb gauge condition $\partial_i A_i =0$ pick exactly one configuration from each gauge equivalence class?
There are infinitely many configurations of a vector field $A_\mu$ that describe the same physical situation. This is a result of our gauge freedom
$$ A_\mu (x_\mu) \to A'_\mu \equiv A_\mu (x_\mu) + \...
- 10.1k
6
votes
2
answers
5k
views
Landau level degeneracy in symmetry gauge, finite system
As we know, Landau level degeneracy in a finite rectangular system is $\Phi/\Phi_0$, where $\Phi=BS$ is the total magnetic flux and $\Phi_0=h/q$ is the flux quanta. This can be easily derived using ...
- 6,095
6
votes
1
answer
2k
views
Why not use the Weyl/temporal gauge?
In E&M in Minkowski space, the Lorenz and Coulomb gauges are typically used since they make things vastly simpler. On a curved background, Maxwell's equations (without sources) can be written as:
\...
- 513
5
votes
4
answers
1k
views
Why does Lorenz gauge condition $\partial_\mu A^\mu =0$ pick exactly one configuration from each gauge equivalence class?
For a vector field $A_\mu$, there are infinitely many configurations that describe the same physical situation. This is a result of our gauge freedom
$$ A_\mu (x_\mu) \to A'_\mu \equiv A_\mu (x_\mu) + ...
- 10.1k