# Covariant Formulation of E&M

Can anybody explain me what does mean the "covariant formulation of electrodynamics"? What does the covariant here mean?

Invariance of Maxwell equations under Lorentz Transformations? In what way? Invariance under mathematical base change?

Does it relate to co- and contravariant derivatives from differential geometry? There the words covariant and contravariant refer to how objects transform under general coordinate transformations.

• See Wikipedia. – Qmechanic Apr 13 '18 at 12:52
• @Qmechanic: There is used the word "manifestly invariant". Does it mean that the maxwell equations are just transformating like tensors (regarding tensor derivatives) or that the shape of the equations stay the the same after a Lorentz transformation. According Ján's answer there is more require than only the "shape concervation". So I guest that "manifestly invariant" means compatible tensor behavior AND "shape concervation". Is that the definition? Or is the second require redundant, since it automatically holds if the equations behave like tensors? – KarlPeter Apr 13 '18 at 20:32

$$\frac{dp_k}{dt} = q E_k + q \epsilon_{kij} v_i B_j$$
The covariant formulation of the same law is $$\frac{d p_\mu}{d\tau} = q F_{\mu\nu} u^\nu,$$ using components of 4-vectors $p,u$ and 4-tensor $F$. This is because in other frame the equation has the same form and the quantities involved - $p,u,F$ - transform as tensors according to the Lorentz transformation.
• Hi, thank you for the answer. One question about your argument that $\frac{dp_k}{dt} = q E_k + q \epsilon_{kij} v_i B_j$ is not covariant. Do you mean by this that if I make a Lorentz trafo $E_k \to E_k'$, $B_j \to B_j'$ for all involved quantities, then it would broke the "shape" of the equation? Or do I misunderstood your explanation? – KarlPeter Apr 11 '18 at 23:13
• No, the form of the equation is the same, but the problem is that the triple $E_1,E_2,E_3$ is not a 4-vector that could be transformed in a standard way as $E'^k = \Lambda^k_l E^l$. The correct transformation of $E_k's$ is not a simple Lorentz tranformation on some quadruple of numbers, but instead, one must first form tensor $F_{\mu\nu}$ which containts $E_k'$s in one row and then transform this tensor according to the Lorentz transformation. In other words, Lorentz transformation cannot be applied directly to $E_k$'s and $B_j$'s. – Ján Lalinský Apr 11 '18 at 23:43