# Electromagnetic Potential Gauge Transformations

• Petar Mali
In summary, the given equations show the relationship between electromagnetic potentials, charge and current densities, and the constants of permittivity and permeability. The gauge transformations for the electromagnetic potentials are also discussed, with the basic formula being A_\mu\mapsto A_\mu\pm\frac{\partial f}{\partial x^\mu}. The Lorentz gauge condition is shown to be Lorentz covariant, but it is unclear if the four-potential in a different gauge also transforms like a tensor.
Petar Mali
$$\Delta\vec{A}-\frac{1}{c^2}\frac{\partial^2 \vec{A}}{\partial t^2}=-\mu_0\vec{j}$$

$$\Delta\varphi-\frac{1}{c^2}\frac{\partial^2 \varphi}{\partial t^2}=-\frac{1}{\epsilon_0}\rho$$

$$c=\frac{1}{\sqrt{\epsilon_0\mu_{0}}}$$

$$\epsilon_0=8,85\cdot 10^{-12}\frac{F}{m}$$

$$\mu_0=4\pi 10^{-7}T$$

$$div\vec{A}+\frac{1}{c^2}\frac{\partial \varphi}{\partial t}=0$$

$$\frac{\partial A_x}{\partial x}+\frac{\partial A_y}{\partial _y}+\frac{\partial A_z}{\partial z}+\frac{\partial (\frac{1}{c}\varphi)}{\partial (ct)}=0$$

$$A^{\mu}=(\vec{A},\frac{1}{c}\varphi)$$

$$A_{\mu}=g_{\mu\nu}A^{\nu}$$

$$A_{\mu}=(-\vec{A},\frac{1}{c}\varphi)$$

$$divA^{\mu}=0$$

Can I say from

$$A^{\mu}=(\vec{A},\frac{1}{c}\varphi)$$

and

$$A_{\mu}=(-\vec{A},\frac{1}{c}\varphi)$$

something more about gauge transformations of electromagnetic potentials

$$\varphi_0=\varphi-\frac{\partial f}{\partial t}$$

$$\vec{A}_0=\vec{A}+gradf$$

I think about - sign in first term $$-\frac{\partial f}{\partial t}$$ and + sign in second term $$+gradf$$

Or

$$\varphi_0=\varphi+\frac{\partial f}{\partial t}$$

$$\vec{A}_0=\vec{A}-gradf$$

$$\Delta A^{\mu}-\frac{1}{c^2}\frac{\partial^2 A^{\mu}}{\partial t^2}=-\mu_0j^{\mu}$$

$$j^{\mu}=(j_x,j_y,j_z,c\rho)$$

Last edited:
The gauge transformation of $$A_\mu$$ is

$$A_\mu\mapsto A_\mu +\frac{\partial f}{\partial x^\mu}$$

or, alternatively, which is the same replacing f by -f:

$$A_\mu\mapsto A_\mu -\frac{\partial f}{\partial x^\mu}$$

All the rest you derive from this basic formula.

Thanks!

Petar Mali said:
Can I say from

$$A^{\mu}=(\vec{A},\frac{1}{c}\varphi)$$

and

$$A_{\mu}=(-\vec{A},\frac{1}{c}\varphi)$$

something more about gauge transformations of electromagnetic potentials
I thought that only the Lorentz gauge condition was Lorentz covariant. So does the four-potential in a different gauge even transform like a tensor? I would think not.

## 1. What is an electromagnetic potential gauge transformation?

An electromagnetic potential gauge transformation is a mathematical tool used in classical electromagnetism to describe the same physical system in different ways. It involves changing the values of the electric and magnetic potentials while keeping the electric and magnetic fields constant. This transformation is based on the principle that the electric and magnetic fields are not unique physical quantities, but rather depend on the chosen gauge.

## 2. Why do we need electromagnetic potential gauge transformations?

Electromagnetic potential gauge transformations are necessary because they allow us to simplify and manipulate the equations of electromagnetism in different ways, making them easier to solve or analyze. They also help us understand the underlying physical principles of electromagnetism, such as the gauge invariance of Maxwell's equations.

## 3. What is gauge invariance in electromagnetism?

Gauge invariance in electromagnetism is the property of Maxwell's equations and the electromagnetic potentials that they are invariant (remain unchanged) under a gauge transformation. This means that different sets of electric and magnetic potentials can describe the same physical system and produce the same electric and magnetic fields.

## 4. Are there different types of electromagnetic potential gauge transformations?

Yes, there are two main types of electromagnetic potential gauge transformations: the Coulomb gauge and the Lorentz gauge. In the Coulomb gauge, the electric potential is set to zero, while in the Lorentz gauge, the divergence of the vector potential is set to zero. These two gauges are commonly used in different situations depending on the problem at hand.

## 5. Can electromagnetic potential gauge transformations be applied to other fields besides electromagnetism?

Yes, the concept of gauge transformations can be applied to other fields besides electromagnetism. In quantum mechanics, for example, gauge transformations are used to describe the behavior of particles with spin. They are also used in other areas of physics, such as in general relativity and particle physics, to simplify and understand complex equations and phenomena.

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