What is the relationship between the Laplace equation and the Wave equation?

Question

What is the relationship between the Laplace equation: $$ (\delta^2_x + \delta^2_y)\phi = 0 $$ and the Wave equation: $$ (\delta^2_x - \delta^2_y)\phi = 0 $$ What is the relationship between the Laplace equation: $$ (\delta^2_{x0} + \delta^2_{x1} + \delta^2_{x2} + \delta^2_{x3})\phi = 0 $$ and the Wave equation: $$ (\delta^2_{x0} - \delta^2_{x1} - \delta^2_{x2} - \delta^2_{x3})\phi = 0 $$

Answer 1

Making the substitutions: $ x:= x $; $ y:= iy $ transforms the 2-dimensional Laplace equation into the 2-dimensional Wave equation. Making the substitutions: $ x_0:= x_0 $; $ x_1:= ix_1 $; $ x_2:= jx_2 $; $ x_3:= kx_3 $ transforms the 4-dimensionsl Laplace equation into the 4-dimensional Wave equation. These transformations move the Laplace equation from Euklidean space into reciprocal space. The Wave equation is thus just Laplace's equation in reciprocal space!

Answer 2

Another take is $$ \Delta=\partial_{xx}+\partial_{yy} $$ is the Laplacace operator with respect to the Euclidean metric $$ ds^2=(dx)^2+(dy)^2 $$ and $$ \square=\partial_{xx}-\partial_{yy} $$ is the Laplacace operator with respect to the pseudo-Riemannian metric $$ ds^2=(dx)^2-(dy)^2. $$ In both case we use the definition $$ \Delta=\nabla^* d u $$ or $$ \square=\nabla^* d u $$ where $\nabla^*$ denotes the adjoint with respect the inner product induced Euclidean or pseudo-Riemannian metric and $d$ is the differential.

Answer 3

Some interesting physical relationships can be obtained by naively playing around with Maxwell's equations;

$$\tag{1}\space\space\space\nabla \times \vec{B}=\mu_{0}\varepsilon_0\partial_t\vec{E}+\mu_0\vec{J}$$ $$\tag{2}\space\space\space\nabla\times\vec{E}=-\partial_t\vec{B}$$

Letting $\vec{J}=0$, we can use the Levi-Civita symbol, $\epsilon_{ijk}$, to write $(1)$ as;

$$\mu_{0}\varepsilon_0\partial_tE^i=\epsilon_{ijk}\partial_jB^k$$

Now, using $(2)$ we can find $B^k$ in terms of the components of $\vec{E}$;

$$-\partial_tB^k=\epsilon_{kmn}\partial_mE^n\implies B^k=-\epsilon_{kmn}\int\partial_jE^ndt$$

So, we will have;

$$\mu_{0}\varepsilon_0\partial_tE^i=-\epsilon_{ijk}\epsilon_{kmn}\int\partial_{mj}E^ndt$$

$$\therefore\mu_{0}\varepsilon_0\partial^2_tE^i=-\epsilon_{ijk}\epsilon_{kmn}\partial_{mj}E^n$$

Utilizing the identity $\epsilon_{ijk}\epsilon_{kmn}=\delta_{im}\delta_{jn}-\delta_{in}\delta_{jm}$ yields;

$$\mu_{0}\varepsilon_0\partial^2_tE^i=-\big[\partial_{ij}E^j-\partial^2_{j}E^i\big]$$

$$\therefore\underbrace{\big[\mu_{0}\varepsilon_0\partial^2_t-\partial^2_j\big]}_{:=\space\square}E^i=-\partial_{ij}E^j$$

In total, we then have;

$$\square\vec{E}=-\nabla(\nabla\cdot\vec{E})$$

In vacuum, we will have $\nabla\cdot\vec{E}=0$ which will yield the wave equation for the electric field;

$$\square\vec{E}=0$$

However, in the presence of stationary charge we obtain;

$$\square\vec{E}=-\nabla(\nabla\cdot\vec{E})=\Delta\vec{E}-\nabla\times\nabla\times\vec{E}$$

Assuming a constant $\vec{B}$ field causes $\nabla\times\vec{E}=0$ which then gives us the very clean identity; $$\square\vec{E}=\Delta\vec{E}$$

Additionally, we can re-write $\small\square\vec{E}=-\nabla(\nabla\cdot\vec{E})$ as;

$$\square(E^ie_i)=-\partial_ie_i\otimes(\partial_je_j\cdot E^me_m)=-\underbrace{(\nabla\otimes\nabla)}_{:=\space\textbf{H}}\cdot E^me_m$$

$$\therefore \square\vec{E}=-\textbf{H}\cdot \vec{E}$$

Where $\textbf{H}$ is the Hessian operator defined as;

$$\mathbf{H}=\begin{bmatrix} \partial^2_x & \partial_x\partial_y & \partial_x\partial_z\\ \partial_y\partial_x & \partial^2_y & \partial_y\partial_z\\ \partial_z\partial_x & \partial_z\partial_y & \partial^2_z \end{bmatrix}$$

Answer 4

Comment: All formula of electrodynamics and Special Relativity follow out of the 4-dimensional Wave equation, which is the 4-dimensional Laplace equation in reciprocal (hyperbolic) space. A conformal mapping of the unit vectors $ x_0:= x_0; x_1:=ix_1; x_2:= jx_2; x_3:=-kx_3 $ carries the 4-dimensional Wave equation from the reciprocal space of Special Relativity over into the Euclidean space of the 4-dimensional Laplace equation. This latter can be generally solved by factoring the operator in Quaternion space, using D'Alembert's method of characteristics. The general solution is a combination of forward- and backward-rotating functions, which may interfere with each other and also give rise to self-interference (stationary solutions). Mapping back by $ x_0:= x_0; x_1:=-ix_1; x_2:= -jx_2; x_3:= -kx_3 $ gives then the results in the hyperbolic space of Special Relativity.

Answer 5

A more philosophical consideration about measurement: Physical measurements are always made against a standard reference: either counting events per standard reference time unit, or dividing a quantity by a standard reference quantity. Physical measurements thus always involve a division. The measurement result is a number. The "space" of the measured event and the "space" of the reference quantity are therefore in a reciprocal relationship: "space" and "reciprocal space" I did not find any litterature on this, but it seems obvious to me that the "energy", i.e. the quantity of relativistic movement (which is maintained in a closed physical system: see P.A.M. Diracs relativistic energy-momentum-relationship) is described in an Euclidean space, where you can sum up, and where no upper limit exists, whereas the measurement scale is described in a hyperbolic (i.e. reciprocal) space, where asymptotes exist for the maximum speed, because measurement implies a division. It seems also obvious to me from the same facts, that the 3 known division algebras (complex numbers, quaternions, and octonions) represent particularly favorable frameworks to describe physical measurements, because they can handle division in multidimensional spaces. (The 3 known division algebras have their roots in 3 algebraic number identities, which define their multiplicaion rules: the complex numbers in the 2-squares-identity found by Diophantes; the quaternions in the 4-squares-identity found by Leonhard Euler; and the octonions in the 8-squares-identity found by Ferdinand Degen. As A. Hurwitz has shown in 1898, the mentioned 3 algebraic number identities are mathematical exceptions, in that there is no other such identity, except these three. Which means that there is no other division algebra either except the complex numbers, the quaternions and the octonions, and theirs homomorph mathematical constructions.)