In an exercise I am supposed to calculate the magnetic field from the electric field for a plane, harmonic wave in vacuum. $$\vec{E} = - E_0 \cdot \sin(\omega t - k z) \cdot \vec{e_y}$$

Using the law of induction $$\operatorname{rot} \vec{E} = -\mu \dfrac{\partial \vec{H}}{\partial t} $$ I end up with this solution for the $x$ component of $H$, \begin{align} \dfrac{\partial H_x}{\partial t} & = \dfrac{E_0 k}{\mu} \cos(\omega t - k z)\\ H_x(z,t)& = H_x(z,0) + \dfrac{E_0 k}{\mu \omega} (\sin(\omega t - k z) - \sin(-kz)). \end{align}

According to the provided solution this is right, except for the integration constant $H_x(z,0)$. How do I choose this constant?

Intuitively I would use the wave impedance $Z$, $$|\vec{H}| = \dfrac{1}{Z} \cdot |\vec{E}|,$$ but if I can choose the constant $H_x(z,0)$ as I want, this wave impedance formula seems to be false as well...

  • $\begingroup$ You can never calculate the magnetic from the electric field. These are partly indepenfent quantities. However in the special case of no currents and charges they are related. $\endgroup$ – my2cts Jul 24 '19 at 9:12
  • $\begingroup$ @my2cts So this exercise doesn't really make sense? $\endgroup$ – cakelover Jul 24 '19 at 9:15
  • $\begingroup$ If they are asking after the relation between the two fields for a free wave, the exercise makes sense. To say the B can be calculated from E is misleading. $\endgroup$ – my2cts Jul 24 '19 at 11:44

As my2cts mentions in the comments, the magnetic field in the presence of currents is only partially determined by the electric field. However, in vacuum they do determine each other fully.

Remember that the magnetic field is not only governed by Farday's law of induction

$$ \epsilon_0^{-1}\mathrm{rot} D = -\mu_0\partial_t B $$

but also by Ampère's law

$$ \mathrm{rot} H = \partial_t D $$

You already have evaluated the former and found the solution up to a constant of integration $H(\vec{x},0)$. If you evaluate the latter as well, you will be able to fix $H(\vec{x},0)$ up to a true constant $\vec{C}$. Maxwell's equations do not forbid that solution. However, this background field needs to vanish, because space(time) is isotropic and translational invariant. A non-vanishing constant background would violate Galilei/Lorentz-invariance.

  • $\begingroup$ Thank you! That makes kind of sense. But lets assume i choose in the example above $H_x(z, 0) = \dfrac{E_0 k}{\mu \omega} \sin(-k z) + C_2$. As fas as I think, I can still choose $C_2$ as I want to, because the rotation operator only contains derivatives for x,y and z. $\endgroup$ – cakelover Jul 24 '19 at 10:57
  • $\begingroup$ @cakelover Good point! See the edit. $\endgroup$ – Nephente Jul 24 '19 at 11:08
  • $\begingroup$ @Nephente My point is that there is a relation between the two but that they do not determine each other. A good skier has his skis parallel but the skis do not determine each other's direction. The directions are determined by the skier. In EM, E and B are determined by the four potential, or by the charge-current. $\endgroup$ – my2cts Jul 24 '19 at 14:33
  • $\begingroup$ @my2cts Of course, I never disagreed about that. But OPs question is clearly about vacuum solutions and for those, the fields determine each other mutually, because there is no other source except through induction. $\endgroup$ – Nephente Jul 24 '19 at 16:07
  • $\begingroup$ @Nephente The fields do not determine each other. In this special case there is a simple relation but this is an exception not a rule. This exercise may well give students a wrong idea, which is the opposite of education. $\endgroup$ – my2cts Jul 24 '19 at 19:24

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