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If we do the following gauge trasformation for magnetic potential and electric potential :

$$\vec A(\vec r,t)'=\vec A(\vec r,t) + \nabla f(\vec r,t)$$ $$\phi((\vec r,t)'=\phi((\vec r,t) - \frac {\partial f(\vec r,t)}{\partial t}$$

Then for simplicity we do:

$$\vec A(\vec r,t)'=\vec A(\vec r,t)$$ $$\phi((\vec r,t)'=\phi((\vec r,t)$$

We simply change the name. We consider the potential with gauge our new original potentials.

Then we have:

$$(\triangle - \frac 1{c^2}\frac {\partial^2}{\partial t^2}) \vec A(\vec r,t) - \nabla[\nabla \cdot \vec A(\vec r,t) + \frac 1{c^2} \frac {\partial\phi(\vec r,t)}{\partial t}]=-\mu_0\vec j_{free}(\vec r,t)$$

$$\triangle\phi(\vec r,t) + \frac{\partial}{\partial t}\vec A(\vec r,t)=-\frac {\rho_{free}(\vec r,t)}{\epsilon_0}$$

Then we consider the Lorenz gauge:

$$\nabla \cdot \vec A(\vec r,t) + \frac 1{c^2} \frac {\partial\phi(\vec r,t)}{\partial t}=0$$

Then in order for our initial gauge to be valid in our Lorenz gauge we simply replace the potentials in the above equation:

$$\nabla \cdot \vec A(\vec r,t)' + \frac 1{c^2} \frac {\partial\phi(\vec r,t)}{\partial t}'=g(\vec r,t)$$

And in the end we find that in order for our initial gauge trasformations to be valid in the lorenz gauge we need to have:

$$(\triangle - \frac 1{c^2}\frac {\partial^2}{\partial t^2})f(\vec r,t)= g(\vec r,t)$$

And How exactly this satisfied the Lorenz gauge?

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given I perform a gauge transformation:

$(\vec{A},\phi)->(\vec{A'},\phi')$

Such that,

$\vec{A'} = \vec{A} + \nabla f$

$\phi' = \phi - \frac{\partial f}{\partial t}$

The new potentials, $\vec{A'},\phi'$, Leave the field invariant.

Performing the actual gauge transformation would change the potentials in the potential formulation of maxwell equation to be $\vec{A'}$ and $\phi'$

Suppose further, that aswell as leaving the field invariant, we would also like the new potentials $\vec{A'},\phi'$

To satisfy the condition that :

$\nabla \cdot \vec{A'} + \mu_0 \epsilon_0 \frac{\partial \phi'}{\partial t} = 0$

Thus, by substituting the definition of $\vec{A'},\phi'$ into the lorenz gauge condition. We are essentially saying this: Given the potentials $\vec{A'},\phi'$ satisfy the lorenz gauge condition, AND they are in a form that leaves the field invariant. What should the function "f" be?

If we can prove than an "f" exists given the previous 2 statements, then we have shown that:

Given the potentials $\vec{A'},\phi'$ satisfy the lorentz gauge condition, they can be written in a form that leaves the field invariant. Aka, we can prescribe the divergence of $\vec{A'}$ to follow the lorenz gauge condition. And the potentials still gives the correct E,B field.

Let's solve for the function F, to prove that F exists!

$\nabla \cdot \vec{A'} + \mu_0 \epsilon_0 \frac{\partial \phi'}{\partial t} = 0$

Sub in:

$(\vec{A'} = \vec{A} + \nabla f$), $(\phi' = \phi - \frac{\partial f}{\partial t})$

$\nabla \cdot ( \vec{A} + \nabla f ) + \mu_0 \epsilon_0 \frac{\partial }{\partial t}(\phi - \frac{\partial f}{\partial t}) = 0$

$\nabla \cdot \vec{A} + \nabla^2 f + \mu_0 \epsilon_0 \frac{\partial \phi }{\partial t}-\mu_0\epsilon_0 \frac{\partial^2 f}{\partial t^2} = 0$

$\nabla^2 f -\mu_0\epsilon_0 \frac{\partial^2 f}{\partial t^2} = -(\nabla \cdot \vec{A} + \mu_0 \epsilon_0 \frac{\partial \phi}{\partial t})$

note the function on the right isn't the same as what you have gotten.

This IS a solvable equation( "inhomogenous wave equation"). meaning F exists. Meaning the potentials can be written in a form that leaves the field invariant whilst also satisfying the lorenz gauge condition.

To answer your question more directly, this is the conditions on F such that the new potentials satisfy the lorenz gauge

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  • $\begingroup$ Its all about proving F exists for specific conditions, given the potentials satisfy the lorenz gauge, AND they can be written in a form that leaves the field invariant. IF f exists given these circumstance then A' and phi' can be written in a form that leaves the field invariant $\endgroup$ Dec 15, 2021 at 19:31
  • $\begingroup$ Thank you for your answer $\endgroup$
    – imbAF
    Dec 16, 2021 at 19:27
  • $\begingroup$ No worries :) :) $\endgroup$ Dec 16, 2021 at 22:20

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