130_notes.dvi

~∇×B~−^1

c

∂E

∂t

=

4 π

c

J~ ∂

∂xi

Bjǫijk−

1

c

∂Ek

∂t

=

4 π

c

Jk

B~=∇×~ A~ Bj= ∂

∂xm

Anǫmnj

E~=−~∇φ−^1

c

∂A

∂t

Ek=−

∂

∂xk

φ−

1

c

∂Ak

∂t

If the fields are written in terms of potentials, then the first two Maxwell equations are automatically
satisfied. Lets verify the first equation by plugging in the B field in terms of the potential and noticing
that we can interchange the order of differentiation.

∇·~ B~= ∂

∂xi

Bi=

∂

∂xi

∂

∂xm

Anǫmni=

∂

∂xm

∂

∂xi

Anǫmni

We could also just interchange the index namesiandmthen switch those indices around in the
antisymmetric tensor.

∇·~ B~= ∂

∂xm

∂

∂xi

Anǫinm=−

∂

∂xm

∂

∂xi

Anǫmni

We have the same expression except for a minus sign which means that∇·~ B~= 0.

For the second equation, we write it out in terms of the potentials and notice that the first term
∂
∂xi

∂

∂xjφǫijk= 0 for the same reason as above.

∂

∂xi

Ejǫijk+

1

c

∂Bk

∂t

= −

∂

∂xi

(

∂

∂xj

φ+

1

c

∂Aj

∂t

)

ǫijk+

1

c

∂

∂t

∂

∂xm

Anǫmnk

=

1

c

(

−

∂

∂xi

∂Aj

∂t

ǫijk+

∂

∂xm

∂An

∂t

ǫmnk

)

=

1

c

(

−

∂

∂xi

∂Aj

∂t

ǫijk+

∂

∂xi

∂Aj

∂t

ǫijk

)

= 0

The last step was simply done by renaming dummy indices (that are summed over) so the two terms
cancel.

Similarly we may work with the Gauss’s law equation

∇·~ E~= ∂

∂xk

Ek=−

∂

∂xk

(

∂

∂xk

φ+

1

c

∂Ak

∂t

)

= 4πρ

−∇^2 φ−

1

c

∂

∂t

(~∇·A~) = 4πρ

For the fourth equation we have.

∂

∂xi

Bjǫijk−

1

c

∂Ek

∂t

=

4 π

c

Jk

∂

∂xi

∂

∂xm

Anǫmnjǫijk+

1

c

∂

∂t

(

∂

∂xk

φ+

1

c

∂Ak

∂t

)

=

4 π

c

Jk

Its easy to derive an identity for the product of two totally antisymmetric tensorsǫmnjǫijkas occurs
above. All the indices of any tensor have to be different in order to get a nonzero result. Since the