- #1

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- 0

## Main Question or Discussion Point

I understand that after writing down this:

[tex]F^{ \mu v} = \partial^{\mu} A^v- \partial^v A^{\mu} [/tex]

We can get a nice matrix connecting E and B vectors. But I just wonder what we need this matrix for? I am a little bit confused about all this relativity in electromagnetism...

And another question I have is:

How do these equations

(1)[tex]\vec{\nabla} \cdot \vec{E} = \frac{\rho}{\epsilon_0} [/tex]

(2)[tex]

\vec{\nabla} \times \vec{B} - \frac{1}{c^2} \frac{ \partial \vec{E}}{\partial t} = \mu_0 \vec{J}

[/tex]

reduce to this one:

(3)[tex]\partial_{\beta} F^{\alpha\beta} = \mu_0 J^{\alpha} [/tex]

I know that after Lorenz gauge we get:

[tex] \square A^ {\mu} = - \mu_0 j^ {\mu} [/tex]

And I think we can derive easily this Poisson equation for vector potential also from (2). But is such proof sufficient? And is it possible to get from (3) Poisson equation for scalar potential analogical like for vector one?

[tex]F^{ \mu v} = \partial^{\mu} A^v- \partial^v A^{\mu} [/tex]

We can get a nice matrix connecting E and B vectors. But I just wonder what we need this matrix for? I am a little bit confused about all this relativity in electromagnetism...

And another question I have is:

How do these equations

(1)[tex]\vec{\nabla} \cdot \vec{E} = \frac{\rho}{\epsilon_0} [/tex]

(2)[tex]

\vec{\nabla} \times \vec{B} - \frac{1}{c^2} \frac{ \partial \vec{E}}{\partial t} = \mu_0 \vec{J}

[/tex]

reduce to this one:

(3)[tex]\partial_{\beta} F^{\alpha\beta} = \mu_0 J^{\alpha} [/tex]

I know that after Lorenz gauge we get:

[tex] \square A^ {\mu} = - \mu_0 j^ {\mu} [/tex]

And I think we can derive easily this Poisson equation for vector potential also from (2). But is such proof sufficient? And is it possible to get from (3) Poisson equation for scalar potential analogical like for vector one?