Skip to main content
I mistakenly approved an edit which had one very good point, but several other not so good things.
Source Link
user178876
user178876

In order to derive the inhomogeneous Maxwell equations, you do not need to use the antisymmetry of the field strength tensor. They follow from the equations of motion for $A_\mu$. If you start with: $$ \mathscr{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-e J^\mu A_\mu\;,$$ the Euler-Lagrange equationequations give you: $$ \partial_\mu F^{\mu\nu}=e J^\nu$$ up to a total derivative term that can be gauged or argued away. If you relabel $\mu\leftrightarrow\nu$, you get: $$ \partial_\nu F^{\nu\mu}=e J^\mu\;.$$ It might very well be that our conventions differ because I do not work with these $\mu_0$ oddities. That is, all the above assumes natural units, in which $c=1$. In any case, you do not get a sign when renaming the summation indices.

In order to derive the inhomogeneous Maxwell equations, you do not need to use the antisymmetry of the field strength tensor. They follow from the equations of motion for $A_\mu$. If you start with: $$ \mathscr{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-e J^\mu A_\mu\;,$$ the Euler-Lagrange equation give you: $$ \partial_\mu F^{\mu\nu}=e J^\nu$$ up to a total derivative term that can be gauged or argued away. If you relabel $\mu\leftrightarrow\nu$, you get: $$ \partial_\nu F^{\nu\mu}=e J^\mu\;.$$ It might very well be that our conventions differ because I do not work with these $\mu_0$ oddities. That is, all the above assumes natural units, in which $c=1$. In any case, you do not get a sign when renaming the summation indices.

In order to derive the inhomogeneous Maxwell equations, you do not need to use the antisymmetry of the field strength tensor. They follow from the equations of motion for $A_\mu$. If you start with $$ \mathscr{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-e J^\mu A_\mu\;,$$ the Euler-Lagrange equations give you $$ \partial_\mu F^{\mu\nu}=e J^\nu$$ up to a total derivative term that can be gauged or argued away. If you relabel $\mu\leftrightarrow\nu$, you get $$ \partial_\nu F^{\nu\mu}=e J^\mu\;.$$ It might very well be that our conventions differ because I do not work with these $\mu_0$ oddities. That is, all the above assumes natural units, in which $c=1$. In any case, you do not get a sign when renaming the summation indices.

In order to derive the inhomogeneous Maxwell equations, you do not need to use the antisymmetry of the field strength tensor. They follow from the equations of motion for $A_\mu$. If you start with: $$ \mathscr{L}=\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-e J^\mu A_\mu\;,$$$$ \mathscr{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-e J^\mu A_\mu\;,$$ the Euler-Lagrange equation give you: $$ \partial_\mu F^{\mu\nu}=e J^\mu$$$$ \partial_\mu F^{\mu\nu}=e J^\nu$$ up to a total derivative term that can be gauged or argued away. If you relabel $\mu\leftrightarrow\nu$, you get: $$ \partial_\nu F^{\nu\mu}=e J^\nu\;.$$$$ \partial_\nu F^{\nu\mu}=e J^\mu\;.$$ It might very well be that our conventions differ because I do not work with these $\mu_0$ oddities. That is, all the above assumes natural units, in which $c=1$. In any case, you do not get a sign when renaming the summation indices.

In order to derive the inhomogeneous Maxwell equations, you do not need to use the antisymmetry of the field strength tensor. They follow from the equations of motion for $A_\mu$. If you start with $$ \mathscr{L}=\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-e J^\mu A_\mu\;,$$ the Euler-Lagrange equation give you $$ \partial_\mu F^{\mu\nu}=e J^\mu$$ up to a total derivative term that can be gauged or argued away. If you relabel $\mu\leftrightarrow\nu$, you get $$ \partial_\nu F^{\nu\mu}=e J^\nu\;.$$ It might very well be that our conventions differ because I do not work with these $\mu_0$ oddities. That is, all the above assumes natural units, in which $c=1$. In any case, you do not get a sign when renaming the summation indices.

In order to derive the inhomogeneous Maxwell equations, you do not need to use the antisymmetry of the field strength tensor. They follow from the equations of motion for $A_\mu$. If you start with: $$ \mathscr{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-e J^\mu A_\mu\;,$$ the Euler-Lagrange equation give you: $$ \partial_\mu F^{\mu\nu}=e J^\nu$$ up to a total derivative term that can be gauged or argued away. If you relabel $\mu\leftrightarrow\nu$, you get: $$ \partial_\nu F^{\nu\mu}=e J^\mu\;.$$ It might very well be that our conventions differ because I do not work with these $\mu_0$ oddities. That is, all the above assumes natural units, in which $c=1$. In any case, you do not get a sign when renaming the summation indices.

Source Link
user178876
user178876

In order to derive the inhomogeneous Maxwell equations, you do not need to use the antisymmetry of the field strength tensor. They follow from the equations of motion for $A_\mu$. If you start with $$ \mathscr{L}=\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-e J^\mu A_\mu\;,$$ the Euler-Lagrange equation give you $$ \partial_\mu F^{\mu\nu}=e J^\mu$$ up to a total derivative term that can be gauged or argued away. If you relabel $\mu\leftrightarrow\nu$, you get $$ \partial_\nu F^{\nu\mu}=e J^\nu\;.$$ It might very well be that our conventions differ because I do not work with these $\mu_0$ oddities. That is, all the above assumes natural units, in which $c=1$. In any case, you do not get a sign when renaming the summation indices.