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P: 908
[QUOTE=LHS1;1714217]
 Prove that 4 vector potential does really a 4 vector?

The vector potential is not a 4-vector!

Under Lorentz transformation, the vector potential transforms according to

$$a_{\mu} \rightarrow \Lambda_{\mu}{}^{\nu} a_{\nu} + \partial_{\mu}\Omega$$

This means that $a_{\mu}$ is a 4-vector, if and only if;

$$\partial_{\mu}\Omega = 0$$

Since this is not compatible with the arbitrary nature of the gauge function, $\Omega$, we conclude that $a_{\mu}$ is not a 4-vector.

Deriving the transformation law of $a_{\mu}$ from the gauge-fixed Maxwell equation

$$\partial_{\mu}\partial^{\mu} a_{\nu} = J_{\nu}$$

is a wrong practice. The gauge-invariant Maxwell equation is

$$\partial^{\nu}f_{\mu \nu} = J_{\mu}$$

Lorentz covariance(of this gauge invariant equation) requires

$$a \rightarrow \Lambda a + \partial \Omega$$

Well, this is not how a 4-vector transforms. Is it?

Genuine 4-vectors cannot describe the two polarization states of light. So, it is not a bad thing that the vector potential is not a 4-vector.

If this does not convince you, see Weinberg's book "Quantum Field Theory" Vol I, on page 251, he says:

"The fact that $a_{0}$ vanishes in all Lorentz frames shows vividly that $a_{\mu}$ cannot be a four-vector. ........
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Although there is no ordinary four-vector field for massless particles of hilicity $\pm 1$, there is no problem in constructing an antisymmetric tensor ........
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$$f_{\mu \nu} = \partial_{\mu}a_{\nu} - \partial_{\nu}a_{\mu}$$

Note that this is a tensor even though $a_{\mu}$ is not a four-vector, ......"

See also Bjorken & Drell "Relativistic Quantum Fields", on page 73, they say this:

"Actually, under Lorentz transformation $A_{\mu}$ does not transform as a four-vector but is supplemented by an additional gauge term."