Confused about Maxwells Equations in potential formulation


by waveandmatter
Tags: confused, equations, formulation, maxwells, potential
waveandmatter
waveandmatter is offline
#1
Nov2-12, 10:14 AM
P: 16
As i understand it, the Maxwell equations in potential form (in the Lorenz gauge) are basically 4 independent wave equations for the 3 components of A and the 1 component of Phi, with J and ρ acting as source terms:



Now from the usual formulation of the Maxwell equations and from experience we know, that an infinitesimally small oscillating current, for example Jx, generates dipole waves, which have nonzero values for all 3 field components.

I can not see, how the 4 independent wave equations of the potential formulation can ever generate Ey or Ez components, if i only plug in a Jx in one point. Contrarily, from the equations i would guess that i would get a spherical wave with only Ex component, since the equation for A works component-wise and the equation that gets me E out of A also only performs a temporal derivation.

This seems to contradict the idea that the E-vector has to stand at right angles to the k-vector (which in a spherical wave takes all kinds of directions).

The only possible explanation for me would be, that if i incorporate a J, i would also have to add the continuity equation for ρ, ∇J=d/dt ρ, which would give me a scalar potential phi as well, which in turn would generate the other field components.

Is this explanation correct? If not, how can this formulation give me proper results where E is at right angles to k etc?
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mfb
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#2
Nov2-12, 11:46 AM
Mentor
P: 10,864
To satisfy the Maxwell equations, you need charge conservation - your current Jx has to come with some charge distribution.
I don't know if that is the only source of Ey and Ez (your derivation looks fine), but it is one source.
jasonRF
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#3
Nov2-12, 03:58 PM
P: 657
I understand what you are thinking about, I think. If you only have a x component to your current density, then you only have an x component of your vector potential. However, if you take the gauge condition,
[tex]
\nabla \cdot \mathbf{A} = - \mu_0 \epsilon_0 \frac{\partial \phi}{\partial t}
[/tex]
combined this with
[tex]
\mathbf{E} = - \nabla \phi - \frac{\partial \mathbf{A}}{\partial t},
[/tex]
you can see how the [itex]\nabla \phi [/itex] term is the source of the y and z components of [itex]\mathbf{E}[/itex].

Things are simpler if you assume a time dependence that is time-harmonic [itex]e^{-i\omega t}[/itex] (allowing all vectors to be complex now, with the understanding that we take the real part at the end). The Lorenz gauge is then,
[tex]
\nabla \cdot \mathbf{A} = i \omega \mu_0 \epsilon_0 \phi
[/tex]
so the electric field can now be expressed solely as a function of [itex] \mathbf{A}[/itex],
[tex]
\mathbf{E} = i \omega \mathbf{A} -\frac{1}{i \omega \mu_0 \epsilon_0} \nabla \left(\nabla \cdot \mathbf{A} \right).
[/tex]
Clearly it is the [itex] \nabla \left(\nabla \cdot \mathbf{A} \right) [/itex] term that is giving you what you are expecting.

I hope that helped.

jason

waveandmatter
waveandmatter is offline
#4
Nov2-12, 06:34 PM
P: 16

Confused about Maxwells Equations in potential formulation


Thanks so far! If i interpret your answers correctly, my initial guess was correct and in the potential formulation it is important to model sources correctly, explicitly including charge conservation when introducing a current as a source. This will generate not merely a vector potential, but also a scalar potential, giving me the correct E-fields (including components that originally were not in the source term).

Just one more question to ensure my correct understanding:
When simulating the more popular vector-field formulation (for example with the fdtd algorithm), one can artificially introduce an (oscillating) E-field in a bounded plane (say Ex in a part of the xy plane), which will propagate (in that case in z-direction) and be correctly diffracted on the sources boundaries, creating light with other k-vectors and E-field components.
Such a model of the source would give physically incorrect results when applied to the potential formulation. Instead one would have to model the source as an oscillating current with its associated charge distribution.
Is this right?
jasonRF
jasonRF is offline
#5
Nov2-12, 08:08 PM
P: 657
Quote Quote by waveandmatter View Post
Thanks so far! If i interpret your answers correctly, my initial guess was correct and in the potential formulation it is important to model sources correctly, explicitly including charge conservation when introducing a current as a source. This will generate not merely a vector potential, but also a scalar potential, giving me the correct E-fields (including components that originally were not in the source term).
If you have a current source, then you can solve the wave equation for the vector potential, and once you have the vector potential you can compute all of the fields (see above for how to get E, B is simply curl A). If you wish, you can use the vector potential to compute the electric potential (via your gauge condition) and the charge density as well (Gauss's law). So it is perfectly fine to model a current source just as a current source.

By the way, charge conservation is implicit in Maxwells equations already. Most books show this, but in case you aren't familiar consider,
[tex]
\nabla \times \mathbf{H} = \mathbf{J} + \frac{\partial \mathbf{D}}{\partial t}.
[/tex]
Take the divergence of both sides,
[tex]
\nabla \cdot \nabla \times \mathbf{H} = \nabla \cdot \mathbf{J} + \frac{\partial \nabla \cdot \mathbf{D}}{\partial t}.
[/tex]
Now use the fact that [itex]\nabla \cdot \mathbf{D}=\rho[/itex] and [itex]\nabla \cdot \nabla \times \mathbf{Q} =0 [/itex] for any [itex]\mathbf{Q} [/itex], and you get
[tex]
0 = \nabla \cdot \mathbf{J} + \frac{\partial \rho}{\partial t}.
[/tex]
waveandmatter
waveandmatter is offline
#6
Nov3-12, 11:21 AM
P: 16
Thanks a lot! I see now, that one can calculate the vector potential from its wave-equation and then use the gauge condition to get the scalar potential from the vector potential.
I also tried this out numerically and actually got what looks like a dipole when calculating the electric fields from A and Phi!


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