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##### Definition and proof of covariance of four-current

Proving that the electromagnetic four-potential is Lorentz covariant rests on the assumption that the four-current given by

is Lorentz covariant, which can be demonstrated through charge conservation expressed in continuity equation form as

where is the four-gradient, which holds in every reference frame. We now use the fact that the gradient operator with respect to any Lorentz covariant four-vector is a Lorentz covariant four-vector which contravaries with , which was proved in [1]. Applying this to reference frame corresponding to , we have

and consequently the statement of the Lorentz invariant corresponding to charge conservation as given in implies

which holds if and only if

##### Definition and proof of covariance of four-potential

As elaborated in [2], under the Lorentz gauge condition that or

Maxwell’s equations can be simplified as, in SI units,

or even simpler than that

wherein the d’Alembertian operator

One can read in Chapter 14 of Landau-Lifshitz shorter course book 1 on classical mechanics and electromagnetism [3] the process of deriving the solution to , which is

where

is the retarded time and

is the volume element.

Given that the four-potential by definition is and , can be rewritten as

Since the prime coordinate is typically used to denote a Lorentz transformation, we replace the prime in with the overline, that is becomes . we then rewrite as

In order to demonstrate Lorentz covariance of the four-potential, we desire to show that

holds for an arbitrary Lorentz transform with the associated transformation tensor such that

Moreover, will transform to per the same rule

From now on we set in for simplicity. We will use or to denote the space coordinates. Moreover, we set

Then, substituting into both sides of gives

and

The integrals in can be viewed as a summation of infinitesimal components the corresponding infinitesimal volume elements of which, in addition to holding equal value, are disjoint. Moreover, that set of infinitesimal volume elements upon any Lorentz transformation remain disjoint. Therefore it suffices to prove that the differential forms within the integral in and are equal. By the Lorentz covariance of four-current,

which simplifies our desired sufficient condition to

We note how if we prove Lorentz covariance restricted to the simple case where only one space coordinate is transformed, we have proven Lorentz covariance in general since the Lorentz group is generated by the set of transformations where only one space coordinate is changed. This means we can without loss of generality consider only one velocity along the first coordinate, the Lorentz transform corresponding to which as derived in [1] as is

where .

Thus equates to

Moreover, length contraction, which tells us that the ratio between the length in the observer frame and proper length (also called rest length) is , upon application to the purely spatial distances occurring in . Disregarding indices ,the points by corresponds to

and thereby upon subtraction of the two equations in , we get

Similarly, through

we get

We note how by definition of rest frame, in the rest frame, the length between any two different spatial coordinates is measured to be the same even if the two spatial coordinates are measured at different times. However, in an observer frame, the measured spatial coordinate of an implicit point object moving along with the observed frame differs with respect to the time coordinate of the rest frame per

and thus, to truly and meaningfully measure the length between those two point objects in an observer frame, one must perform the two coordinate measurements at the same time of the observer frame. In , the primed coordinate corresponds the observer frame, and the unprimed coordinate corresponds to the rest frame.

In , the is the length at the same time in frame between two implicit point objects which at time are located at spatially. Moreover, we have assumed arbitrary Lorentz transformation to another frame moving relative to only along the first spatial coordinate. Note how the two point objects at time in frame are only constrained in their spatial coordinate, and there are no constraints in how their spatial coordinates vary with respect to . Then, for simplicity, we assume that both point objects move along the frame , which means by definition that , the spatial coordinates of the two respective point objects in the frame, are constant with respect to time and thus can be treated as constants . One observes that the spacetime coordinates associated with the two point objects are in , and in , they transform to the coordinates given in , which differ in the time component, which, as we just mentioned, does not affect the spatial coordinate at all. Here, is the rest frame, and is the observer frame, and by length contraction,

which is equivalent to the first equation in .

Since only one spatial coordinate changes, is equivalent to

which via substitution of is equivalent to the equality

Thus, the Minkowski norm of the electromagnetic four-potential at the same point in spacetime is the same regardless of reference frame. This shows that the four-potential is indeed Lorentz covariant.

**References**

- [1] gmachine1729：Illustrating covariance and contravariance of vectors through Lorentz transformation of four-vectors
- [2] gmachine1729：磁矢势以及经典规范场理论
- [3] Chapter 14 of Landau-Lifshitz shorter course book 1 on classical mechanics and electromagnetism (downloadable on libgenesis)
- [4] gmachine1729：How I systematically guessed hyperbolic functions and the Lorentz group from the spacetime invariant
- [5] gmachine1729：On proper time, relativistic velocity, and Lorentz covariance in Einstein notation