According to Maxwell's equations, electric and magnetic fields (EF and MF) should have the same phase for EM waves. Also, they must be perpendicular and perpendicular to the propagation direction.

If we consider a planar wave spreading in the +x direction with an EF in the y axis, the MF would be in the z axes. At any given moment, they will have the same sign, so one can write down their expressions:

$\vec{E}= E_{max}sin(kx-\omega t) \hat j$

$\vec{B}= B_{max}sin(kx-\omega t) \hat k$

But if we consider the wave going in the -x direction, one of the fields would need to change sign if nothing else is changed. Suppose $\vec{E}$ stays the same and we change $\vec{B}$. Now we have:

$\vec{E}= E_{max}sin(kx-\omega t) \hat j$

$\vec{B}= -B_{max}sin(kx-\omega t) \hat k= B_{max}sin(kx-\omega t + \pi)\hat{k} $

So their phases are no longer the same, which is necessary. However, the wave still exists.

What am I missing? I see they will both still have the maxima and minima at a given point and time, but supposedly their phase should be equal. We haven't been shown any true proof with actual vectorial calculus, so I could be misinterpreting the "must have the same phase" condition that might be less ambiguous when seeing the process. This is clearly the case for a phase difference of $2n\pi$, as it doesn't affect anything.

  • $\begingroup$ "According to Maxwell's equations, electric and magnetic fields (EF and MF) should have the same phase." This as stated is nonsense: just take a capacitor, an inductor and a resistor and the relative phase between $V$ (electric field) and $I$ (magnetic field) can be anything. There must be restrictions for the kind of EM fields where that is true. I suggest that you rephrase your question. $\endgroup$
    – hyportnex
    Apr 26, 2023 at 15:01
  • $\begingroup$ Alright, I was thinking about EM waves but didn't write it explicitly. Thanks! $\endgroup$ Apr 26, 2023 at 15:11
  • $\begingroup$ Even for "EM waves" is not true as stated, see any, yes, any antenna and its near field or any field in a lossy medium. Maybe your question should be then if and when it is true at all, but for starters check out the Ignatowsky equations. $\endgroup$
    – hyportnex
    Apr 26, 2023 at 15:16

1 Answer 1


I'm going to assume we are talking about plane waves in vacuum.

However, the wave still exists.

It doesn't. If the original fields satisfy Maxwell's equations, if you flip the sign of the magnetic field, in general they will no longer satisfy Maxwell's equations.

The (phase) velocity of the wave is by following a point of constant phase on the wave: $$kx-\omega t=\text{const.}$$ $$\frac{dx}{dt}=\frac \omega k.$$ If your original equations describe a valid wave, to obtain a valid wave moving in the opposite direction, you also need to change the sign of $k$ (in addition to that of $B_{max}$):

$\vec{E}= E_{max}\sin(-kx-\omega t) \hat j$

$\vec{B}= -B_{max}\sin(-kx-\omega t) \hat k.$

A phase difference of $\pi$ (or $180^\circ$) just corresponds to a sign difference, so it is usually trivial and not very interesting. You can always choose components that are not strictly "in phase". Here, $E_y$ and $-B_z$ are in phase, so naturally $E_y$ and $B_z$ are $180^\circ$ out of phase (sometimes said to be "in antiphase"). This isn't unique to this new wave either: in the original wave $E_y$ and $-B_z$ are in antiphase. Two fields being "in phase" can be understood to mean the field magnitudes reach their maximum value at the same time.


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