Why did covariant nu change sides with mu in the inhomogenous maxwell equation?

[ercagpince]

[SOLVED] Inhomogenous maxwell equation

Homework Statement

In relativistic notation , the field strenght tensor can be expressed as (A is the vector potential) as on eq.(1.1) .

The inhomogenous Maxwell equations can be written as on eq.(1.2) .
Why did covariant "nu" changed sides with "mu" and become contravariant on the second term of left hand side of the equation 1.2 when one take the derivative of eq.(1.3) by
$$\partial_{\nu}$$?

Homework Equations

$$F^{\mu\nu}=\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}$$ (1.1)

$$\partial_{\mu}\partial^{\mu}A^{\nu}-\partial^{\nu}\partial_{\mu}A^{\mu}=\frac{4\Pi}{c}J^{\nu}$$ (1.2)

$$\partial_{\mu}F^{\mu\nu}=\frac{4\Pi}{c}J^{\nu}$$ (1.3)

The Attempt at a Solution

I tried to contract all terms in the eq.(1.2) , however , I couldn't find a quantitative way to solve the problem .

 

[quantumdude]

Why did covariant "nu" changed sides with "mu" and become contravariant on the second term of left hand side of the equation 1.2 when one take the derivative of eq.(1.3) by
$$\partial_{\nu}$$?

The author didn't take the derivative $\partial_{\nu}$, he took the derivative $\partial_{\mu}$. In the second term of the left hand side he simply commuted $\partial^{\nu}$ and $\partial_{\mu}$.

 

[ercagpince]

thanks a lot!