Show that solutions to the wave equation are invariant under Poincaré-Transforms In my abstract algebra lecture I have to solve this exercise that I have problems with because I‘m a bit rusty on vector analysis:
A Poincaré-Transform is a map: $\mathbb{R}^4 \rightarrow \mathbb{R}^4$, such that $x \mapsto Ax+b, \quad A \in O(1,3), b \in \mathbb{R}^4$.
Where $O(1,3)$ is the group of all linear operators on $\mathbb{R}^4$ that preserve the Minkowski Metrik. That is to say $(Ax,Ay)=(x,y)=x_0y_0-x_1y_1-x_2y_2-x_3y_3$.
I now have to show that $u(x)$ is a solution to the wave equation:
$\partial^2_{x_0}u-\sum_{i=1}^3 \partial_{x_1}^2u=0$ if and only if $u(Ax+b)$ is one.
I‘d be very happy if someone could tell me how exactly I‘m supposed to differentiate here, my first instinct would be:
$\partial_{x_i}u(Ax+b)=\nabla u(Ax+b)\cdot A_i$ and $\partial^2_{x_i}u(Ax+b)=\Delta u(Ax+b)|A_i|$
Where $\Delta$ is the laplacian (or its 4-dim equivalent if the laplacian is only defined for functions of 3 variables), $\cdot$ the standard dot product, $||$ the norm associated with the dot product, and $A_i$ the i th column in the matrix representation of $A$.
 A: Let $x_i' = A_{ij}x_j + b_i$, and let the Lorentz metric tensor $\eta$ be defined such that $(x,y) = \eta_{ij}x_iy_j$. Then, the operators $A\in O(1,3)$ satisfy
\begin{align}
(x, y) &= (Ax,Ay) \\
\implies \sum_{k,l=0}^3\eta_{kl}x_kx_l &= \sum_{i,j,k,l=0}^3\eta_{ij}A_{ik}A_{jl}x_kx_l \\
\implies \eta_{kl} &= \sum_{i,j=0}^3\eta_{ij}A_{ik}A_{jl} \\
&= \sum_{i,j=0}^3\eta_{ij}A_{kl}A_{lj}
\end{align}
where the last equality can be checked by explicit computation.
Also, the wave equation becomes, taking $\partial_{x_i} = \partial_i$
\begin{equation}
\left(\partial^2_{x_0} - \sum_{i=1}^3\partial^2_{x_i}\right)u(x) = \sum_{i,j=0}^3\eta_{ij}\partial_{x_i}\partial_{x_j} u(x)
\end{equation}
Using this, the partial derivatives transform under the Poincare transform as
\begin{align}
\partial_{x_i}u(x) &= \sum_{k=0}^3\frac{\partial x_k'}{\partial x_i}\partial_{x_k'}u(x') \\
&= \sum_{k=0}^3A_{ki}\partial_{x_k'}u(x') \\
\implies \sum_{i,j=0}^3\eta_{ij}\partial_{x_i}\partial_{x_j}u(x) &= \sum_{i,j,k,l=0}^3\eta_{ij}A_{ki}A_{lj}\partial_{x_k'}\partial_{x_l'}u(x') \\
&= \sum_{k,l=0}^3\eta_{kl}\partial_{x_k'}\partial_{x_l'}u(x')
\end{align}
Thus, $u(x)$ is a solution of the wave equation iff $u(Ax+b)$ is.
