A plane electromagnetic wave has the shape:
\$\vec{E}(\vec{r},t)=E_0\cdot cos(\vec{k}\vec{r}-\omega t)\cdot \vec{e}_y\$
\$\vec{B}(\vec{r},t)=(B_1\cdot cos(\vec{k}\vec{r}-\omega t)+B_2\cdot sin(\vec{k}\vec{r}-\omega t))\cdot \vec{e}_z=B_0\cdot cos(\vec{k}\vec{r}-\omega t+\phi)\cdot \vec{e}_z\$
In what direction is \$\vec{k}\$ facing?
Determine the amplitude \$B_0\$ and the phase change \$\phi\$ between \$\vec{E}\$ and \$\vec{B}\$.
I asked this question in a physics forum but they said that the actual exercise is more mathematical because of trigonometric identities for finding \$B_1\$,\$B_2\$ and \$B_0\$ as well as the phase angle, so I'm sorry if this is also the wrong place.
We got this one in our lecture of experimental physics and it kind of bugs me because I can't find the right approach to this one.
Since \$\vec{E}\$ only got a component in y-direction and \$\vec{B}\$ got only a direction in z-component that the k is either in -x or x-direction.
About the second part: In general \$E\$ and \$B\$ are in phase, right? Since that's not the case here, can I assume that it results from a reflection on a surface? But how would I get to \$B_0\$ and the phase change without explicit values for \$\omega\$ and such?
Best Answer
True. Imagine a simple wave
$$A(t)=\sin\left( \vec{k}\vec{x}-\omega t \right)$$
with \$\vec{k}=\begin{pmatrix}\pi/2\\0\end{pmatrix}\$ at \$t=0\$.
Draw a grid with x, y in the range [-4;+4] and calculate the field at each position. (25 position, but you'll see, you don't have to calculate 25 values).
You will notice wave fronts parallel to the y-axis, because the scalar product \$\vec{k}\vec{x}\$ only depends on the x-coordinate. This means, the wave travels parallel to the x-axis, so parallel to \$\vec{k}\$.
What happens a little bit later, say for \$\omega t=\pi/2\$ ? The wave will be shifted to the right, i.e. it travels to the right. And what if you invert \$\vec{k}\$ ?
You'll see, the direction of \$\vec{k}\$ is the direction where the wave travels.
No, not in general but for a single free wave
No, it's not a reflection, because a reflected wave travels into the opposite direction. While the incident wave is \$\sin(kx-\omega t)\$, the reflected is \$\sin(-kx-\omega t)\$ (see above). Because the arguments of your sine and cosine function are equal, there is no reflection.
Also, if you think there's a reflection in B, why isn't there an evidence for a reflection in E?
The only thing one can see is that B is written with a (not yet defined) phase shift relative to E. This can either be expressed as sum of sine and cosine function with different factors, or by a sine function with that phase shift build-in. By the trigonometric identity
$$\sin \theta + \sin \varphi = 2 \sin\left( \frac{\theta + \varphi}{2} \right) \cos\left( \frac{\theta - \varphi}{2} \right)$$
you can change between both representations. You also do not need to know \$\omega\$ for this.
I'm not yet sure what else you should do, may be because you didn't post the full question. In general, giving the E-field, you can calculate the B-field via Maxwell's equations, which will lead to the dependency between \$E_0\$ and \$B_0\$ was well as \$\phi=0\$ . It will also give you the helicity (if the wave moves away from you and the maximum of E points upwards, does the maximum of B point to the right or left?).
By the way: There is no phase shift for a single wave. It is interesting, that for a standing wave (sum of wave and reflected wave of same amplitude), E and B are phase shifted by 90°.