Understanding the mathematical representation of a travelling plane wave

Question

A travelling wave in 3D can be represented as the following: $\vec{\Psi}(\vec{r},t) = \vec{A}e^{i(\vec{k} \cdot \vec{r}-\omega t)}$. I’m not sure I fully understand this expression.

$\vec{A}$ to me means the amplitude vector, which gives the magnitude of the wave in different components of space eg $A_x,A_y,A_z$, in space described by cartesian coordinates. The exponential term I am assuming to be the phase term which tells us how the phase of the wave evolves with time.

My issue is that when writing out the x component of the above, I would’ve thought $\Psi_x = A_xe^{i(k_x r_x -wt)}$, whereas in reality, $\Psi_x = A_xe^{i(\vec{k} \cdot \vec{r}-wt)}$.

Can someone explain why we include the full wave vector in the exponent, rather than its component when describing any one component of the plane wave?

Answer 1

The exponential factor is a scalar, with $\vec k$ pointing in the direction of propagation, but otherwise $e^{i(\vec k\cdot \vec r-\omega t)}$ is a scalar, like any $f(x,y,z)$, and you would not think of taking the $x$-part only of $f(x,y,z)$. There is no a priori reason to suggest this contains any information about the components of the amplitude of the wave.

In the same way the $\hat y$ component of a vector can depend on $x$ or $z$ and there’s no reason to think this component depends only on $\hat y$: for instance the magnetic field about an infinitely long wire is in the $\hat\phi$ direction but the magnitude depends on the radial distance only.

In the well-known case of an E&M, the amplitude vector $\vec A$ is in fact orthogonal to the direction of propagation: $\vec A\cdot \vec k$, so that for a wave travelling along $\hat z$ we have $A_z=0$.

Answer 2

You have to disentangle the meaning of two different vectors, $\vec A$ and $\vec k$. Probably, an analysis which does not introduce components at all is easier to visualize and to understand.

1. The amplitude, $\vec A$, of the field is a vector whose direction tells us what is the direction of the instantaneous field vector. For a pure plane wave it is a constant and uniform vector providing the uniform direction of the field eveywhere.
2. The wave-vector $\vec k$ provides information about the direction of the wave. For a pure plane wave, it is the direction orthogonal to the constant phase planes. Such a direction in general has no relation at all with the direction of the field. Only for some specific kind of waves there could be some relation between direction of the amplitude and of the wave-number. For example, in the case of purely transverse waves ($\vec A \cdot \vec k = 0$), or purely longitudinal waves ($\vec A \times \vec k = 0$). For a generic wave, there no special relation.

Notice that the third vector in this expression, $\vec r$, is the vector specifying the point where the wave is sampled and its direction does not play any role

Answer 3

The equation describes a vectorial field, where for any ($t,x,y,z$) there is a vector. Each component of the vector is a complex number.

The meaning of $\mathbf k$ in the exponent is the direction of the maximum change of each component of the field, for a given time $t_0$. So, if we take the gradient of one of the components:

$\nabla \psi_x = i\mathbf k\psi_x$

On the other hand, planes normal to $\mathbf k$ for a given $\mathbf r_0$, all $\mathbf r$ such that $\mathbf k \cdot(\mathbf r - \mathbf r_0) = 0$ have a constant value for the vector component in a given time. It is a level surface.

The field can be understood as a travelling wave because an observer travelling with a velocity $v = \frac{\omega}{|\mathbf k|}$ will see a static field.