What does $\vec{\nabla}^2 \vec{E} = \vec{\nabla}^2 \left[ f(\vec{k} \cdot \vec{r} - \omega t) \vec{E}_0 \right]$ mean?

Question

$\vec{E} = f(\vec{k} \cdot \vec{r} - \omega t) \vec{E}_0$ with the constant vector field $\vec{E}_0$

I only know the case if I apply the Laplacian operator on a scalar field, in this case it is a vector field $\vec{\nabla}^2 \vec{E} = \vec{\nabla}^2 \left[ f(\vec{k} \cdot \vec{r} - \omega t) \vec{E}_0 \right]=\text{div} \left[\text{grad}( f(\vec{k} \cdot \vec{r} - \omega t) \vec{E}_0)\right]$ makes no sense because $\text{grad}( f(\vec{k} \cdot \vec{r} - \omega t) \vec{E}_0)$ is not defined.

What does $\vec{\nabla}^2 \vec{E} = \vec{\nabla}^2 \left[ f(\vec{k} \cdot \vec{r} - \omega t) \vec{E}_0 \right]$ mean?

https://de.wikipedia.org/wiki/Laplace-Operator#Definition

Answer 1

The vector operator $\vec\nabla^2\{\cdot\}$ can be expressed (and is defined in general coordinates) in $\mathbb{R^3}$ as

$$\begin{align} \vec\nabla^2\{\cdot\}&=\text{Grad}(\text{Div} \{\cdot\})-\text{Curl}\left(\text{Curl}\{\cdot\}\right)\\\\ &=\nabla \left(\nabla\cdot\{\cdot\}\right)-\nabla\times\left(\nabla\times\{\cdot\}\right) \end{align}$$

In Cartesian coordinates in $\mathbb{R}^N$, the vector Laplacian is given by

$$\begin{align} \vec\nabla^2\{\cdot\}&=\sum_{i=1}^N\frac{\partial^2 \{\cdot\}}{\partial x_i^2}\\\\ \end{align}$$

So, we have for a vector field $\vec A$, in $\mathbb{R^N}$

$$\begin{align} \nabla^2 \vec A(\vec r)&=\sum_{i=1}^N\sum_{j=1}^N\hat x_j \frac{\partial^2A_j(\vec r)}{\partial x_i^2} \\\\ &=\sum_{j=1}^N \hat x_j \nabla^2 A_j(\vec r) \end{align}$$

For the specific problem, $\vec A(\vec r)=\vec E(\vec r)$, and $\vec E(\vec r)=\vec E_0 f(\vec k\cdot \vec r-\omega t)$, with $\vec r\in \mathbb{R}^3$. Therefore, the vector Laplacian applied to $\vec E(\vec r)$ simplifies to

$$\begin{align} \vec \nabla^2\vec E(\vec r)&=\nabla \left(\nabla\cdot\vec E(\vec r)\right)-\nabla\times\left(\nabla\times\vec E(\vec r)\right)\\\\ &=\vec k(\vec k\cdot \vec E_0)f''(\vec k\cdot\vec r-\omega t)-\vec k\times(\vec k\times \vec E_0)f''(\vec k\cdot\vec r-\omega t)\\\\ &=|\vec k|^2f''(\vec k\cdot\vec r-\omega t)\vec E_0 \end{align}$$

Answer 2

Without going into the generalities of tensors and covariant derivatives, it just means the vector-valued function obtained by taking the Laplacian of each component.

For any function $F:U\subset\Bbb{R}^n\to\Bbb{R}^n$, we can define the vector Laplacian component wise: \begin{align} \nabla^2F&:=(\nabla^2(F_1),\dots, \nabla^2(F_n)) \end{align} i.e $\nabla^2F$ is the vector-valued function whose $i^{th}$ component is obtained by calculating the Laplacian of the real-valued function $F_i:U\to\Bbb{R}$, $\nabla^2(F_i)=\text{div}(\text{grad}(F_i))$.

In your particular case, it boils down to taking the Laplacian of the real-valued function $\vec{r}\mapsto f(\vec{k}\cdot\vec{r}-\omega t)$ (which again gives you a real-valued function) and then multiplying it by the constant vector $\vec{E_0}$ (i.e the constant vector comes out).