Even when we cast Maxwell's Equations in as coordinate-independent form as possible, $dF=0$ and $d \star F = J$, we still have to make use of the Hodge star $\star$ which is defined relative to an orientation. It doesn't look like the equations are preserved under orientation-reversal.

But does space have an orientation, or is it just an artefact? It seems unphysical. I thought all classical theories were invariant under parity.


  • $\begingroup$ I believe you are breaking parity with the external current. In the vacuum case, orientation does not change a thing. With the external current, if you don't also change it by a sign when changing orientation, you will probably just describe a different situation, however I don't yet have a stronger argument. $\endgroup$
    – ohneVal
    Dec 7, 2020 at 8:31
  • $\begingroup$ I see... so perhaps instead of "the same" current $J$ giving you the same field $F$ in altered parity, we should instead say that for every current $J$ yielding $F$ in our world there is a corresponding $-J$ that yields $F$ in opposite-parity world. Then you'd add that, observationally, we cannot detect whether we prepared $J$ in world A or prepared $-J$ in world B. I'm not quite confident in this assessment? $\endgroup$
    – DPatt
    Dec 7, 2020 at 8:37
  • 1
    $\begingroup$ I added the explanation in terms of differential forms to my answer (J flips sign, F doesn't), but the situation is imo easier to understand in the "freshman" notation (or the relativistic one). In that notation, currents yield fields (or J yields F in relativistic notation) according to exactly the same rules in the normal and opposite parity worlds (which is just another way of saying that Maxwell’s equations are true in both worlds). And since in the lab we create currents and measure fields/forces (instead of differential forms), we can't determine observationally which world we are in. $\endgroup$ Dec 7, 2020 at 9:11
  • $\begingroup$ You can find solution to rotation coordinates equations here: henrykdot.com/en/chap01a.php henrykdot.com/en/chap02a.php $\endgroup$ Dec 9, 2020 at 11:04

2 Answers 2


It doesn't look like the equations are preserved under orientation-reversal.

Short answer: Maxwell’s equations are preserved under a parity transformation, but you have to be careful to transform quantities correctly.

We have to address two questions.

  1. What happens to the quantities involved in Maxwell’s equations upon a parity transformation?

The normal vector quantities: positions, velocities, forces, and $J$ (vector current density) transform in the usual way, they all flip sign. The electric field is also defined to change sign, but the magnetic field is an odd duck, it is defined to keep its sign (and is called a "pseudovector" for that reason).

In the covariant formulation, you can deduce how both fields (and $F$) transform from just one rule: the four-potential $A$ is defined to transform as a normal four-vector (i.e. the time component stays the same, the spacial components change sign).

Finally, in the language of differential forms, $F$ is a two-form, which is invariant under the parity transformation, and $J$ is a twisted three-form, which changes sign under parity.

  1. Do Maxwell’s equations still hold if we perform a parity transformation?

Yes, if we follow the rules in the previous paragraph the equations still hold. In fact that's why we define the magnetic field (or $A$ in the covariant formulation) to transform in this way.

In the language of differential forms, $\mathrm{d}\star F = J$ stays true under parity, because both sides change sign.

At the freshman level, we can visualize why we define the magnetic field to not change sign with a simple example. Imagine a long wire with a current going through it, and the magnetic field curling around it according to the right hand rule. What happens if you reflect everything about the origin? The current will of course change direction, but now you have to decide what you will do with the magnetic field. If you decide to transform it the way normal vectors transform, by changing its sign, then the resulting picture will become unphysical, the magnetic field will stop following the right hand rule. But if you follow the "odd duck" prescription for the magnetic field, then after the transformation everything will still look great.

  • $\begingroup$ I'm not convinced that $J$ should gain a negative sign under parity transformation. The point of writing the equations in the form that I did is that they are completely coordinate independent; the $F$ that appears is a differential form, and it does not "transform" in any way whatsoever (its many coordinate representations $\phi^*F$ do, when you change $\phi$). Differential forms can be defined on non-oriented or even non-orientable manifolds, and so nothing in their definition could be sensitive to a choice of orientation. $\endgroup$
    – DPatt
    Dec 7, 2020 at 9:10
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    $\begingroup$ I'm not sure if you wrote this before or after I added the explanations in terms of differential forms (also see my comment below your original question). The 3-vector current density flips sign because it's defined in terms of velocities of moving particles, and their velocities flip sign. The differential form J flips sign because it's defined as a twisted 3-form. $\endgroup$ Dec 7, 2020 at 9:45
  • $\begingroup$ I believe I have cleared my confusion, if you'll please review the following for me: I've been taking "parity transformation" to literally mean "switch orientations" but it is not! Instead, guided by our interpretations of the mathematics*, we define "parity transformation" to literally map certain objects (like $J$) onto distinct objects ($-J$). I've been arguing that $J$ can be given a persistent identity across changes of orientation, but this is irrelevant since the meaning of "parity" is to literally remove $J$ and put $-J$ in its stead. $\endgroup$
    – DPatt
    Dec 7, 2020 at 9:52
  • $\begingroup$ *I assume something similar is done on a case-by-case basis for other physical theories, but I'm not aware of any general prescription. All that being said, if it isn't too far outside the scope here, what is a twisted differential form? $\endgroup$
    – DPatt
    Dec 7, 2020 at 9:54
  • $\begingroup$ Yes, you are correct that flipping the sign of differential form J is part of the definition of parity. The general prescription could be phrased like this: find some way to transform objects that preserve the equations, and then call that transformation by whatever name feels right :) For example, if the transformation changes the sign of ordinary vectors like positions and velocities, we call it parity. $\endgroup$ Dec 7, 2020 at 10:20

It is only necessary that the space of solutions is preserved. (And, after all, elementary particle reactions do not preserve parity.) But the physical solutions to Maxwell's equations are unchanged if you choose a different orientation.


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