What is the electromagnetic wave propagation direction and significance of $\vec{k}$-vector?

Question

Consider the electric field of electromagnetic wave to be of form $\vec{E} = E_°\cos(ax+bz) \hat{{i}}$ at $t= 0$ seconds. How to see in which direction wave is propagating? Does the value of $\vec{k}$ have a special significance with respect to this? How to find that if it is?

Answer 1

The equation of a plane sinusoidal wave propagating in the direction given by the unit vector $\mathbf{\hat k}$ is $$\mathbf E = \mathbf E_0 \cos\left(\frac{2\pi}{\lambda}\mathbf{\hat k} \cdot \mathbf r -\frac{2\pi}{T} t+\phi_0\right)$$ Here, $\mathbf{\hat k}$ is the unit vector at right angles to the wavefront, whose perpendicular distance from the origin – as you'll see if you draw a diagram – is $\mathbf{\hat k} \cdot \mathbf r$. Writing, for convenience, $\mathbf k=\frac{2\pi}{\lambda} \mathbf{\hat k}$ and $\omega=\frac{2\pi}{T}$, $$\mathbf E = \mathbf E_0 \cos(\mathbf k \cdot \mathbf r -\omega t+\phi_0)$$ Expanding the dot product: $$\mathbf E = \mathbf E_0 \cos(k_xx+k_yy+k_zz -\omega t+\phi_0)$$ Comparing with the wave you are considering, we see that

$$k_x=a, k_y=0, k_z=b$$ So the components of $\mathbf{\hat k}$ are $$\frac {a\lambda}{2\pi}, 0, \frac {b\lambda}{2\pi}$$ The propagation direction therefore lies in the x-z plane, at an angle $\arctan{\tfrac ba}$ to the x axis.

We note that, since $\mathbf{\hat k}$ is a unit vector, $$\left(\frac {a\lambda}{2\pi}\right)^2+ 0^2+\left( \frac {b\lambda}{2\pi}\right)^2=1.$$

Answer 2

You can't know in which direction the wave is propagating (or if it is even propagating) just from its snapshot at one instant. You need to know how the wave changes over time.

Let's write the electric field vector at the position $(x,y,z)$ at time $t$ as $\vec{E}(x,y,z,t)$. Let us rewrite your equation as \begin{equation} \vec{E}(x,y,z,0) = E_0 \cos(a x + b z) \hat{i}. \end{equation} Suppose you measure the electric field vector as a function of the position $(x,y,z)$ at other instants and found a relation, \begin{equation} \vec{E}(x,y,z,t) = E_0 \cos(a x + b z +\phi(t)) \hat{i}. \end{equation} That is, the phase in the cosine function is shifted by $\phi(t)$, a function of time, which satisfies $\phi(0)=0$. The wave form won't change suddenly in the next instant of $t=0$ (by physical intuition), and hence your measurement will indicate $\phi(\epsilon) = -\omega \epsilon$ at time $t=\epsilon$ for sufficiently small $|\epsilon|$. Here $\omega$ is a constant real number. [The negative sign in front of $\omega$ is put there just to arrive at simpler expression at the end. The symbol $\omega$ was chosen just to follow a convention.]

Now let us try to relate $\vec{E}(x,y,z,\epsilon)$ to $\vec{E}(x,y,z,0)$. By the above consideration, \begin{equation} \vec{E}(x,y,z,\epsilon) = E_0 \cos(a x + b z -\omega\epsilon) \hat{i}. \end{equation} We want to rewrite this as \begin{equation} \vec{E}(x,y,z,\epsilon) = E_0 \cos(a [x-v_x \epsilon] + b [z-v_z \epsilon]) \hat{i}. \end{equation} In fact, this is possible by identifying $v_x = a\omega/(a^2+b^2)$ and $v_z=b\omega/(a^2+b^2)$. Then, we can say that \begin{equation} \vec{E}(x+v_x \epsilon,y,z+v_z \epsilon,\epsilon) = E_0 \cos(a x + b z) \hat{i} = \vec{E}(x,y,z,0) . \end{equation} We can view this relation as the vector $\vec{E}$ we saw at $(x,y,z)$ at $t=0$ has moved to $(x+v_x \epsilon, y, z+v_x \epsilon)$ over time $\epsilon$. That is, the wave is moving by the velocity $(v_x,0,v_z)$ around $t=0$.

The vector $\vec{k} = (k_x, k_y, k_z) = (a,0,b)$ is called the wave vector. We have seen above that this vector is related to the velocity of the wave propagation as $v_x = k_x\omega/k^2$, $v_z = k_z\omega/k^2$. That is \begin{equation} \vec{v} = (v_x, 0, v_z) = \frac{\omega}{k} \hat{k}, \end{equation} where $\hat{k} = (1/k) \vec{k}$ is the unit vector in the direction of the wave vector. We see that $v = \omega/k$. [I expressed the magnitude of a vector $\vec{g}$ by $g$.]

It is possible that your measurement of the electric field over time yields \begin{equation} \vec{E}(x,y,z,t) = F_0(t) \cos(a x + bz) \hat{i}, \end{equation} where, for example, \begin{equation} F_0(t) = E_0 \cos(\Omega t) \end{equation} for some real number $\Omega$. This wave (called standing wave) looks as if it does not propagate but stays at the same position while the overall amplitude changes in time. Note that this $\vec{E}(x,y,z,t)$ gives the same $\vec{E}(x,y,z,0)$ as in your question. Therefore, you need to measure the wave form at different instants in order to judge whether the wave is propagating or not.