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Good day, I am a student of Physics at the university of Padova, I must solve this problem for my exam of electromagnetic fields, but I have got different problems. The text is the follower:

The electric field of an electromagnetic wave is: $$E = E_0\sin \omega t (\sin \omega z, \cos \omega z, 0 ). $$ I have to find the magnetic field $B$, then I have to verify the Maxwell equation for $E$ and $B$, and finally I have to find the 4-potential $A^\mu$ in the Lorenz gauge.

First of all, I have considered $n$ the direction of propagation of the wave:

$n=(0,0,1)$

So that I thought B is the cross product of $n$ and E, I have obtained:

$B= n \times E = E_0\sin \omega t (-cos \omega z, sin \omega z, 0 )$

And this result seemed me reasonable, because the scalar product between E and B is null.

In empty space, I expect that the divergence of E and B is null, and in this case is verified. The other two Maxwell's equation establish:

$\nabla \times E + \frac{\partial B}{\partial t} =0 $

$E_0 \omega \sin \omega t (sin \omega z, cos \omega z, 0 ) + E_0 \omega \cos \omega t (-cos \omega z, sin \omega z, 0 ) = E_0 \omega (-cos(\omega t + \omega z), sin(\omega t + \omega z),0) $

But in this way the sine and the cosine can not vanish simultaneously, they have got the same argument !

$\nabla \times B - \frac{\partial E}{\partial t} =0 $

$E_0 \omega \sin \omega t (-cos \omega z, sin \omega z, 0 ) - E_0 \omega \cos \omega t (sin \omega z, cos \omega z, 0 ) = E_0 \omega (-sin(\omega t + \omega z), -cos(\omega t + \omega z), 0)$

There is the same problem as before.

Then I have tried to obtain $A^\mu$.

$A = \frac{B}{\omega}$

$\nabla \times A = \nabla \times \frac{B}{\omega} = \frac{1}{\omega} \nabla \times B = B$

Because I noticed

$\nabla \times B = \omega B$

Then I wanted to obtain $A^0$

$E = \frac{\partial A}{\partial t} - \nabla A^0$

But I stopped there because in my opinion there are too many errors in my reasoning. The Maxwell equation are not verified.

If someone has the solution, I will be to him infinitely grateful

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  • $\begingroup$ It seems strange to have $\omega z$ in the expression for the electric field. Shouldn't there be the wave vector, $k$? Your guess, $B=n\times E$ seems unjustified and also runs into problems with dimensions. Other than this making B perpendicular to E it is not clear how you come up with this. $\endgroup$ – user1583209 Jan 30 '17 at 22:51
  • $\begingroup$ Pinging @user1583209 , I included the derivation in my answer, if you are interested in it. You can avoid the $c$ in non-SI units where $\vec B$ and $\vec E$ have the same units; in SI units it's $\vec B = c^{-1} \hat k \times \vec E$ where $\hat k$ is the unit vector in the direction of $\vec k.$ $\endgroup$ – CR Drost Jan 30 '17 at 23:31
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There are at least two ways to solve this; one was given by @Ismasou above.

Another is the method that you seem to be using, the method of plane waves. The wave equation in vacuum $\nabla^2 E = c^2 \ddot E$ is being solved by terms which look like $f(\vec k \cdot \vec r - \omega t)$ for $\omega / |\vec k| = c.$ We can choose our coordinates so that the electric field is in the $\hat x$ direction and the propagation is in the $\hat z$ direction thus for some phase angle we have $\vec E = E_0 ~\hat x~ \cos(kz - \omega t + \varphi)$ or so. We do what Ismasou is suggesting with this simpler wave and we find that $\partial_t {\vec B} = -\hat y(\partial_z E_x - \partial_x E_z)$ which is straightforwardly $+\hat y ~E_0~k~\sin(kz - \omega t + \varphi)$. Integrate with respect to $t$ and you find $\vec B =\vec B_0 + \hat y ~E_0 ~(k/\omega)~\cos(\omega t - k z),$ so, ignoring the constant, $\vec B = c^{-1} \hat z \times \vec E.$Since the wave vector in this case is $\vec k = k \hat z$ we can make this fully general and say that for any such plane wave, if $\hat n = \vec k / |\vec k|$, then$$\vec B = \frac1c \hat n \times \vec E.$$ Note that this only holds for plane waves.

OK, so now you come in with this more complicated expression for a standing wave, $$\vec E = E_0 \sin(\omega t) \begin{bmatrix}\sin(kz)\\\cos(kz)\\0\end{bmatrix}.$$This is not a plane wave, so what do we do? We decompose it into plane waves. In general standing waves are sums of plane waves. Let's see how to do this.

We start with the angle sum rules, $$\cos(\omega t \pm k z) = \cos(\omega t) \cos(k z) \mp \sin(\omega t) \sin(k z),\\ \sin(\omega t \pm k z) = \sin(\omega t) \cos(k z) \pm \cos(\omega t) \sin(k z).$$Now we try to "reverse" these using both signs. In your case, $$\sin(\omega t)\sin(k z) = \frac 12 \big [\cos(\omega t - k z) - \cos(\omega t + k z) \big],\\ \sin(\omega t)\cos(k z) = \frac 12 \big [\sin(\omega t - k z) + \sin(\omega t + k z) \big].$$In other words, $$\vec E = \frac{E_0}2 \begin{bmatrix}\cos(\omega t - k z)\\\sin(\omega t - k z) \\ 0\end{bmatrix} + \frac{E_0}2 \begin{bmatrix}-\cos(\omega t + k z)\\+\sin(\omega t + k z) \\ 0\end{bmatrix}.$$ The first of these is propagating in the $\hat n = +\hat z$ direction as you can see from the opposite signs of the argument $f(\omega t - k z).$ (If you've never practiced this, meditate on the sentence "to see what it is at this $z$ at some time $t + dt$, look at what it is at this $t$ but at some place $z - c~dt$" until you see that this means that the wave is moving towards $+z$.) But the second of these is propagating in the $\hat n = -\hat z$ direction as you can see from the same signs of the argument $f(\omega t + k z)$.

So by superposition, the $\vec B$ field for the sum is the $\vec B$ field for each individual one, and you can find it with $\frac 1c \hat n \times \vec E$ for each of those individually, then sum them together. Then you should be able to use those angle-sum rules again in the "forwards" direction to get the correct standing-wave expression, and what was wrong with your approach was that you treated the standing wave as if it were a forwards-travelling plane wave, when in fact it is this superposition of forwards and backwards waves.

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  • $\begingroup$ You are right ! I should have I should considered a superposition of plane waves, not a plane wave, I really had a bad start ... anyway thanks a lot! I am grateful and satisfied $\endgroup$ – Pico Jan 31 '17 at 10:35
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I would rather use faraday law to get the magnetic field: $$\nabla \times E = -\frac{\partial B}{\partial t}= E_0 \omega\sin\omega t (\sin\omega z, \cos\omega z, 0) $$ $$B=E_0 \cos\omega t (\sin\omega z, \cos\omega z, 0) $$

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