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

The vector potential is not a 4vector!
Under Lorentz transformation, the vector potential transforms according to
[tex]a_{\mu} \rightarrow \Lambda_{\mu}{}^{\nu} a_{\nu} + \partial_{\mu}\Omega[/tex]
This means that [itex]a_{\mu}[/itex] is a 4vector,
if and only if;
[tex]\partial_{\mu}\Omega = 0[/tex]
Since this is not compatible with the arbitrary nature of the gauge function, [itex]\Omega[/itex], we conclude that [itex]a_{\mu}[/itex]
is not a 4vector.
Deriving the transformation law of [itex]a_{\mu}[/itex] from the
gaugefixed Maxwell equation
[tex]\partial_{\mu}\partial^{\mu} a_{\nu} = J_{\nu}[/tex]
is a wrong practice. The
gaugeinvariant Maxwell equation is
[tex]\partial^{\nu}f_{\mu \nu} = J_{\mu}[/tex]
Lorentz covariance(of this gauge invariant equation) requires
[tex]a \rightarrow \Lambda a + \partial \Omega[/tex]
Well, this is not how a 4vector transforms. Is it?
Genuine 4vectors cannot describe the two polarization states of light. So, it is not a bad thing that the vector potential is not a 4vector.
If this does not convince you, see Weinberg's book "Quantum Field Theory" Vol I, on page 251, he says:
"The fact that [itex]a_{0}[/itex] vanishes in all Lorentz frames shows vividly that [itex]a_{\mu}[/itex] cannot be a fourvector. ........
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Although there is no ordinary fourvector field for massless particles of hilicity [itex]\pm 1[/itex], there is no problem in constructing an antisymmetric tensor ........
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......
[tex]f_{\mu \nu} = \partial_{\mu}a_{\nu}  \partial_{\nu}a_{\mu}[/tex]
Note that this is a tensor even though [itex]a_{\mu}[/itex] is not a fourvector, ......"
See also Bjorken & Drell "Relativistic Quantum Fields", on page 73, they say this:
"Actually, under Lorentz transformation [itex]A_{\mu}[/itex] does not transform as a fourvector but is supplemented by an additional gauge term."
So, my friend, you should have asked the following instead;
"prove that the vector potential is not a 4vector"
regards
sam