5

Consider a complex-valued function of the form $$ \newcommand{\bfx}{\mathbf{x}} \newcommand{\bfk}{\mathbf{k}} f(t,\bfx)=\int d^Nk\ g(\bfk)\exp\big(-i\omega(\bfk)t-i\bfk\cdot\bfx\big) \tag{1} $$ where boldface denotes a list of $N$ real variables, the dot-product is defined as usual, and $$ \omega(\bfk)\equiv \sqrt{\strut{}1+\bfk\cdot\bfk}. \tag{2} $$ (I'm calling this a "wave," but notice the constant term under the square root.) Suppose that $f(0,\bfx)$ is nonzero at least for some $\bfx$. Can we choose $g(\bfk)$ so that $f(t,\bfx)$ and $df(t,\bfx)/dt$ both have compact support in $\bfx$ at $t=0$?

The answer must be no, because otherwise I could use the Paley-Wiener theorem to construct a contradiction to the Reeh-Schlieder theorem. But that's a very indirect argument that uses relativistic quantum field theory, which surely isn't necessary for the simple question I'm asking here! How can we prove more directly that no such $g(\bfk)$ exists?

Chiral Anomaly
  • 840
  • I am trying to understand the same issue but for more general functions, since recently I found the paper Finite Time Differential Equations I am trying to figure out if their findings can be extended to more than one dimensions (unsuccessfully so far). Maybe it could help you. – Joako Feb 21 '22 at 02:52
  • I am not fully sure about this, but I think it is possible to adjust the function shown in this answer to a question I made to be a counter-example to your question. – Joako Mar 27 '24 at 02:11
  • @Joako Thank you for the comment! I don't know how to adjust that example to construct a counter-example, though. If we omit the "$1$" term under the square root in my equation (2), then any function of the form $f(kx-ct)$ with compact support at $t=0$ would be a counter-example with $N=1$. It's that pesky "$1$" term under the square root that makes things difficult. – Chiral Anomaly Mar 28 '24 at 23:29
  • @Joako By the way, the function $E(x,y,z,t)=f(k_xx-ct, k_yy-ct, k_zz-ct)$ shown in that example doesn't satisfy the usual wave equation. Not sure what the author of that answer meant. – Chiral Anomaly Mar 28 '24 at 23:29
  • If I remember I do checked on Wolfram-Alpha at least it do works in two dimensions (only variables $x$ and $t$ considered), but knowing myself I probably checked it also on 2D $(x,\ y,\ t)$... also the traveling way $f(kx-ct)$ is a classic solution for the wave equation in 1D so it should be right: maybe the value of $k$ is not well calculated? (I remember having problem with it - as you could see on the comments) – Joako Mar 28 '24 at 23:58
  • with unitary constants I do check it in 1 spatial dimension here, but you are right, it looks it fails for more dimensions, I don't know why – Joako Mar 29 '24 at 01:23
  • @Joako In case you're interested, here's a relatively easy way to prove that the version with $\geq 2$ dimensions doesn't satisfy the wave equation. In $N$-dimensional space, the usual wave equation with $c=1$ is $$ \partial_t^2E(t,x_1,...,x_N) =(\partial_1^2+\cdots+\partial_N^2)E(t,x_1,...,x_N) $$ with $\partial_n\equiv \partial/\partial x_n$. Continued... – Chiral Anomaly Mar 29 '24 at 02:29
  • 1
    @Joako Now let $f(x_1,...,x_N)$ be any smooth function, and suppose $$ E(t,x_1,...,x_N)=f(x_1-t,x_2-t,...,x_N-t). $$ Any such function $E(t,x_1,...,x_N)$ satisfies $$ \partial_t^2 E = (\partial_1+\partial_2+\cdots+\partial_N)^2 E. $$ If $E$ also satisfies the wave equation, then this immediately implies $$ \sum_{j\neq k} \partial_j\partial_k E = 0. $$ By inspection, the example described in the other answer doesn't satisfy this when $N\geq 2$, not even when $t=0$. – Chiral Anomaly Mar 29 '24 at 02:29
  • @ChiralAnonaly I think your last hypothesis is not right (if I understood properly: here is an example of a function in $R^{2+1}$ that fulfill the wave equation, (...) – Joako Mar 29 '24 at 04:09
  • (...) but the sum of their crossed-derivatives is not zero – Joako Mar 29 '24 at 04:09
  • @Joako Good example! The other answer didn't specify the values of the coefficients $k_x,k_y,k_z$, so for simplicity I set them all equal to $1$. That choice was essential in my proof. Your example has the form $E(t,x,y)=f(k_xx- t,k_yy- t)$ with $k_x=k_y\neq 1$, and it shows that my proof really does rely on the simplification $k_x=k_y=1$. Your example doesn't have compact support along the $x-y$ direction, because it's independent of that combination of coordinates, but it at least exposes an important limitation of my proof. – Chiral Anomaly Mar 29 '24 at 13:16
  • it is just a simple plane wave as example for the cross-derivatives sum, not for the compact-support function. The classic plane wave is $\exp(i\vec{k}\vec{x}-ict)$ such as $|\vec{k}|=1$, so the components of vector $\vec{k}$ only will be unitary on one dimension – Joako Mar 29 '24 at 13:37

0 Answers0