Question

Hi, does anybody know how these equations are equivalent?

Answer 1

Maxwells homogeneus equations in vector form and\[F_{\mu \nu , \sigma} + F_{\nu \sigma , \mu} + F_{\sigma \mu , \nu} = 0\]

Answer 2

Does it have something to do with the wave equation??

Answer 3

and...electromagnetism? Just curious.Haha

Answer 4

No, it has to do with reformulating Maxwell's equations from ordinary vector calculus to a covariant form.

Answer 5

@Kainui

Answer 6

Where is the other equation?

Answer 7

Starting from the vector form of the equations, you reformulate in terms of the scalar and vector potentials, and then define the 4-potential \(A^\mu = (\phi,\vec A) \) (in units where c = 1). From that you can create the Electromagnetic tensor \( F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu \).

The homogeneous equations are
\[ \nabla \cdot \vec B = 0 \]
and
\[ \nabla \times \vec E + \frac{\partial }{\partial t} \vec B = 0\]
The equivalent statement using the Electromagnetic tensor is that
\[ \partial_{[\alpha} F_{\beta \gamma]} = 0\]

This can be verified by simply plugging in the definitions. Note that
\( F^{00} = F^{11} = F^{22} = F^{33}= 0\),
\(F^{i0} = \partial^iA^0 - \partial^0A^i = -\partial_i\phi -\partial_tA^i = E_i\)
and
\(F^{jk} = \partial^jA^k - \partial^k A^j = -(\partial_jA^k-\partial_k A^j) \rightarrow -\epsilon_{ijk}\partial_jA^k = -B_i\).
where \((i,j,k)\) run from 1 to 3.

For example, we can let \( \alpha = 0, \beta = j, \gamma = k\) from which we find
\[\partial_{[0}F_{jk]}=\partial_0 F_{jk} + \partial_kF_{0j} + \partial_j F_{k0} \]
\[=\partial_tF_{jk} + \partial_jE_k - \partial_kE_j \rightarrow \partial_t \epsilon_{ijk}\partial_jA^k + \epsilon_{ijk}\partial_j E_k = \partial_tB_i + (\nabla\times \vec E)_i = 0\]
So letting \(\alpha = 0, \beta = 2, \gamma = 3\) yields the x-component of the equation
\[ \nabla \times \vec E + \partial_t \vec B = 0\]
Repeating for other combinations of \((i,j)\) yield the other two components.

You can check that the cases where \(\alpha \neq 0\) correspond to the first equation, \( \nabla \cdot \vec B = 0\).

Answer 8

Verifying that the tensor form of the equations implies the vector form is fairly tedious but straightforward. Just bear in mind that the electromagnetic tensor looks like this:
\[\left(\begin{matrix}0 & E_x & E_y & E_z\\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0\end{matrix}\right)\]
And from there you should be able to move from the tensor form to the vector form (component by component, of course) without resorting back to the 4-potential in the meantime.