Let's speaking about special relativity.

A four-vector is a physical quantity with 4 components that, when passing from one inertial frame to another, transform those components applying the Lorentz matrix, as the event [itex](ct, x, y, z)[/itex] do. Now i define the 4potential as [itex](\phi, \vec{A})[/itex]. For being sure that IS a 4vector i have to change frame and show that the new 4potential is obtained applying the Lorentz matrix, but i dont know how to do it. So i put in Lorentz's gauge and write Maxwell equations that results to be:

[itex]\square A^{\mu} = j^{\mu}[/itex]

where [itex]j^{\mu}[/itex] is the 4current (that i know is a 4vec). So because the d'alembert operator is invariant, or rather, it doesnt change the algebra of the quantity applied it i obtain the 4vec nature of [itex]A^{\mu}[/itex]. But...

1) Does the 4potential is a 4vector ONLY in Lorentz gauge? Because i've seen in 2nd quantitation that it has be used as a 4vec in Coulomb's gauge.

2) How can i proof that the d'alembert operator is "invariant"?

Really thank you for attention,

Ll.

A four-vector is a physical quantity with 4 components that, when passing from one inertial frame to another, transform those components applying the Lorentz matrix, as the event [itex](ct, x, y, z)[/itex] do. Now i define the 4potential as [itex](\phi, \vec{A})[/itex]. For being sure that IS a 4vector i have to change frame and show that the new 4potential is obtained applying the Lorentz matrix, but i dont know how to do it. So i put in Lorentz's gauge and write Maxwell equations that results to be:

[itex]\square A^{\mu} = j^{\mu}[/itex]

where [itex]j^{\mu}[/itex] is the 4current (that i know is a 4vec). So because the d'alembert operator is invariant, or rather, it doesnt change the algebra of the quantity applied it i obtain the 4vec nature of [itex]A^{\mu}[/itex]. But...

1) Does the 4potential is a 4vector ONLY in Lorentz gauge? Because i've seen in 2nd quantitation that it has be used as a 4vec in Coulomb's gauge.

2) How can i proof that the d'alembert operator is "invariant"?

Really thank you for attention,

Ll.

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