- #1

timeant

- 16

- 2

From Maxwell's equations [itex] \partial_\nu F^{\mu\nu}=J^{\mu},[/itex] one can derive charge conservation. The derivation is

[tex] 0\equiv \partial_\mu \partial_\nu F^{\mu\nu}= \partial_\mu J^{\mu} { \Rightarrow}\partial_\mu J^{\mu}=0. [/tex]

However, a circular reasoning exists in it. For the sake of better understanding, we suppose [itex] F^{kl} [/itex] is an antisymmetric

[tex] \partial_l F^{kl}= J^{k}, \qquad n=3,4,5,\cdots \qquad (\star) [/tex]

Where [itex] J^{k} [/itex] is known source. If the source is chosen as [itex] \partial_k J^{k}\neq 0 (e.g. J^{k} \propto x^k) [/itex], then the above equation(*) has no solutions. Hence, [itex] \partial_k J^{k}= 0 [/itex] is one of

Hence the charge conservation law can

[tex] 0\equiv \partial_\mu \partial_\nu F^{\mu\nu}= \partial_\mu J^{\mu} { \Rightarrow}\partial_\mu J^{\mu}=0. [/tex]

However, a circular reasoning exists in it. For the sake of better understanding, we suppose [itex] F^{kl} [/itex] is an antisymmetric

*n*-dimenstional (*n*> 2) tensor. We consider the following equation[tex] \partial_l F^{kl}= J^{k}, \qquad n=3,4,5,\cdots \qquad (\star) [/tex]

Where [itex] J^{k} [/itex] is known source. If the source is chosen as [itex] \partial_k J^{k}\neq 0 (e.g. J^{k} \propto x^k) [/itex], then the above equation(*) has no solutions. Hence, [itex] \partial_k J^{k}= 0 [/itex] is one of

**preconditions**of existence about solutions of the above equation (*). If [itex] \partial_k J^{k}= 0 [/itex] is considerd as a**corollary**of Eq.(*) ([itex] 0\equiv \partial_k \partial_l F^{kl}= \partial_k J^{k} { \Rightarrow}\partial_k J^{k}=0 [/itex]), and at the same time it is one of**preconditions**of existence about Eq.(*)'s solutions. It must involve circular reasoning. Therefore, [itex] \partial_k J^{k}= 0 [/itex] is**NOT**a corollary of Eq.(*) for any*n*. When*n*=4, Eq(*) is one of Maxwell equations.Hence the charge conservation law can

**NOT**be derived from Maxwell equations.
Last edited: