IMO 'covariant formulation' of electrodynamics means that the equations should remain invariant across different Lorentz frames. Now there are broadly two ways to write electrodynamics equations.

  1. using 3 vector notation $$ \frac{\partial \rho}{\partial t} + \nabla.\mathbf{J} = 0 \tag{Eq. 11.127 of J.D.Jackson} $$

  2. using 4 vector notation and tensors $$ \partial_\alpha J^\alpha = 0 \tag{Eq. 11.129 of J.D.Jackson} $$

An electrodynamics equation written in either of these two ways remains invariant (form-wise) across all Lorentz frames. Then why is it that the latter is referred to as the 'covariant formulation' but not the former?

  • 2
    $\begingroup$ This is semantics but the last one is sometimes referred to as "explicitly covariant". Also technically as written it is "invariant" rather than covariant since the rhs is a scalar. $\endgroup$ – ZeroTheHero Mar 26 '17 at 0:29
  • $\begingroup$ @ZeroTheHero Invariant is a special case of covariant; it's perfectly fine to call it covariant. $\endgroup$ – tparker Mar 26 '17 at 0:45
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    $\begingroup$ There's also the (equivalent) differential forms approach, where the above would be written as $d\star\mathbf{J} = 0$. $\endgroup$ – Joe D Mar 26 '17 at 11:44

It's true that the first equation has the same form in all Lorentz frames, but it's not obvious unless you know how $\rho$ and ${\bf J}$ individually transform. For example, the similar-looking equation

$$ \frac{\partial \rho}{\partial t} + k\, {\bf \nabla} \cdot {{\bf E}} = 0$$

(where $k$ is some constant) would not look the same in other Lorentz frames, because charge density and electric field transform differently under Lorentz boosts. The second equation, on the other hand, is clearly Lorentz-covariant, because any contraction between a covariant and contravariant index of a Lorentz tensor is automatically covariant (i.e. that contraction has the same algebraic form in any other Lorentz frame).

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