- #1

intervoxel

- 195

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What has it to do with adding a gradient to A and subtract a partial time derivative from V?

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- Thread starter intervoxel
- Start date

In summary, the Coulomb gauge fixes the gauge by setting div(A)=0. This has the nice property of leaving the fields E and B unaffected. It is used to simplify expressions derived from Maxwell's equations. The two most common gauge transformations are the Coulomb and Lorentz gauges.

- #1

intervoxel

- 195

- 1

What has it to do with adding a gradient to A and subtract a partial time derivative from V?

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- #2

JANm

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Hello intervoxelintervoxel said:

What has it to do with adding a gradient to A and subtract a partial time derivative from V?

What is Coulomb gauge?

greetings Janm

- #3

intervoxel

- 195

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A gauge transformation in the electromagnetic field is the folowing pair of equations:

A' = A + grad(lambda),

V' = V - drond lambda / drond t,

where V is the scalar potential and A the vector potential. Lambda can be any function.

This has the nice property of leaving the fields E and B unnafected. It is used to simplify

expressions derived from Maxwell`s equations.

The two most common gauge transformations are the Coulomb and Lorentz gauges.

- #4

JANm

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So you have large scale fields based on the scalar potential A and vector potential V.

The elctric field based on the scalar potential and the magnetic field based on the vector potential is it not?

... and now a gauge is a local changement of overall field if I understand right.

Is Lambda a scalar function and what is drond?

greetings Janm

- #5

intervoxel

- 195

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lambda is a scalar function.

The theory says that the EM field can be derived from two potential functions through the relations:

E = -grad(V)-drond A / drond t

B = curl(A)

We may then have many pairs of potential functions generate the same EM field.

Gauge transformations extend this set even further.

Oh, come on folks, let's return to my original question: What div(A)=0 has to do with gauge transformation?

- #6

JANm

- 223

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Hello Intervoxelintervoxel said:... What div(A)=0 has to do with gauge transformation?

The chapter inhomogeneous wave equation speaks of Lorentzcondition:

divA+mu*eps*drondphi/drondt+mu*sig*phi=0,

which gives the wave equation in free space.

divA+(1/c^2)*drondphi/drondt=0

and continues with introduction of the d'Alenbertian and after that retarded potentials, so what is your problem?

greetings Janm

- #7

intervoxel

- 195

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Now I see, JANm, thank you.

The Lorentz condition is a statement of the law of conservation of charge -- It imposes a constraint between A and V. If I select a V invariant in time I get Coulomb gauge:

A' = A + grad(V)

V' = V - 0

The Lorentz condition is a statement of the law of conservation of charge -- It imposes a constraint between A and V. If I select a V invariant in time I get Coulomb gauge:

A' = A + grad(V)

V' = V - 0

Last edited:

- #8

Born2bwire

Science Advisor

Gold Member

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intervoxel said:

The Lorentz condition is a statement of the law of conservation of charge -- It imposes a constraint between A and V. If I select a V invariant in time I get Coulomb gauge:

A' = A + grad(V)

V' = V - 0

You would not want to explicitly set the scalar potential to be time invariant. The primary reason being that it restricts the physical phenomenon that can occur. For example, under the Coulomb gauge we can derive the expression that

[tex]\frac{1}{c^2}\nabla\frac{\partial^2 \Phi}{\partial t^2} = \mu_0\mathbf{J}_t[/tex]

where J_t here is the irrotational current, the curl free portion of the currents. Thus, by fixing a time invariant scalar potential you are forcing the condition that there can be no irrotational currents, among other things I am sure.

The reason for confusion is that I believe you are still trying to satisfy the Lorenz condition, which is the defining property that gives rise to the Lorenz gauge. The Lorenz gauge requires that the Lorenz condition be satisfied,

[tex]\nabla\cdot\mathbf{A}+\frac{1}{c^2}\frac{\partial \Phi}{\partial t} = 0[/tex]

This gives rise to the two decoupled inhomogeneous wave equations for \Phi and A.

The Coulomb gauge, however, satisfies the inhomogeneous wave equation,

[tex]\nabla^2\mathbf{A} - \frac{1}{c^2}\frac{\partial^2 \mathbf{A}}{\partial t^2} = -\mu_0\mathbf{J} + \frac{1}{c^2}\nabla \frac{\partial \Phi}{\partial t}[/tex]

We can see that by decomposing the current into irrotational and solenoidal components we can decouple the above wave equation as previously stated above. While this is very similar to what arises in the Lorenz gauge, it allows a different set of freedoms in the choice of our potentials. You can find a detailed explanation of this in Jackson's text.

- #9

intervoxel

- 195

- 1

With your explanation and the link below, I think I arrived where I wished.

http://www.phy.duke.edu/~rgb/Class/Electrodynamics/Electrodynamics/node31.html

Coulomb gauge fixing is a mathematical technique commonly used in quantum field theory to simplify calculations and analysis of electromagnetic fields. It involves adding a gradient term and subtracting a partial time derivative in the equations describing the field.

Coulomb gauge fixing is necessary because in certain cases, such as when studying electromagnetic fields in the presence of sources, the equations describing the field can become quite complex. By using Coulomb gauge fixing, the equations can be simplified and the physical interpretation of the results becomes clearer.

Adding a gradient term in the equations for the field helps to eliminate the longitudinal components of the field. This means that only the transverse components, which are the physically relevant ones, remain in the equations. This simplifies the analysis of the field and makes it easier to interpret the results.

No, Coulomb gauge fixing is not applicable in all cases. It is most commonly used for studying electromagnetic fields, but may not be suitable for other types of fields or in certain situations. It is important to carefully consider the physical system and the equations involved before deciding to use Coulomb gauge fixing.

Yes, there are some limitations to Coulomb gauge fixing. It is important to note that while it simplifies the equations describing the field, it does not change the physical behavior of the system. This means that in some cases, certain physical phenomena may not be accurately captured using Coulomb gauge fixing. It is important to carefully consider the applicability and limitations of this technique before using it in any analysis.

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