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ercagpince
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[SOLVED] Inhomogenous maxwell equation
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}?
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)
I tried to contract all terms in the eq.(1.2) , however , I couldn't find a quantitative way to solve the problem .
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 .
