- #1

Smacal1072

- 59

- 0

Hi All,

Thanks again to all the great mentors and contributors to this forum.

I wanted to ask a question about the Gauss's law/Ampere's law equations in Maxwell's Equations:

[itex]

\nabla \bullet \textbf{E} = \frac{\rho}{\epsilon_0}

\\

\\

\nabla \times \textbf{B} = \mu \left( \textbf{J} + \epsilon \frac{\partial\textbf{E}}{\partial t} \right)

[/itex]

For charge distributions or currents that are accelerating, I was taught that you need to consider the retarded potentials in order to derive the fields. Why are Maxwell's Equations not written like this:

[itex]

\nabla \bullet \textbf{E} = \frac{\rho_r}{\epsilon_0}

\\

\\

\nabla \times \textbf{B} = \mu \left( \textbf{J}_r + \epsilon \frac{\partial\textbf{E}}{\partial t} \right)

[/itex]

Where [itex]\textbf{J}_r \mbox{ and } \rho_r [/itex] are the retarded charge and retarded current density?

Edit: In retrospect, I should have used the integral version of the equations, in particular:

[itex]

\oint \textbf{B} \bullet dl = \mu_0 \iint \left( \textbf{J} + \epsilon_0 \frac{\partial \textbf{E}}{\partial t} \right) \bullet d\textbf{S}

[/itex]

For example, if we instantly switch on a current element at the origin at t = 0, then calculate [itex]\oint \textbf{B} \bullet dl[/itex] at a radius of a million miles, we'll get zero, even though at that instant, a current may be flowing. Unless the current density and "displacement current" cancel out, the inequality won't hold...

Thanks again to all the great mentors and contributors to this forum.

I wanted to ask a question about the Gauss's law/Ampere's law equations in Maxwell's Equations:

[itex]

\nabla \bullet \textbf{E} = \frac{\rho}{\epsilon_0}

\\

\\

\nabla \times \textbf{B} = \mu \left( \textbf{J} + \epsilon \frac{\partial\textbf{E}}{\partial t} \right)

[/itex]

For charge distributions or currents that are accelerating, I was taught that you need to consider the retarded potentials in order to derive the fields. Why are Maxwell's Equations not written like this:

[itex]

\nabla \bullet \textbf{E} = \frac{\rho_r}{\epsilon_0}

\\

\\

\nabla \times \textbf{B} = \mu \left( \textbf{J}_r + \epsilon \frac{\partial\textbf{E}}{\partial t} \right)

[/itex]

Where [itex]\textbf{J}_r \mbox{ and } \rho_r [/itex] are the retarded charge and retarded current density?

Edit: In retrospect, I should have used the integral version of the equations, in particular:

[itex]

\oint \textbf{B} \bullet dl = \mu_0 \iint \left( \textbf{J} + \epsilon_0 \frac{\partial \textbf{E}}{\partial t} \right) \bullet d\textbf{S}

[/itex]

For example, if we instantly switch on a current element at the origin at t = 0, then calculate [itex]\oint \textbf{B} \bullet dl[/itex] at a radius of a million miles, we'll get zero, even though at that instant, a current may be flowing. Unless the current density and "displacement current" cancel out, the inequality won't hold...

Last edited: