In thinking about the superposition principle I ran into the following question: Say we take two known waves:

\begin{align}f(x,t) &= \sin(x-B\cdot t)\\ g(x,t) &= A\cdot x^2 + C\cdot t^2\end{align}

Then, by the superposition principle, their sum should be a solution. But directly applying the wave equation gives a propagation velocity, $v = \sqrt{\frac{2A-\sin(x-Bt)}{2C-B^2 \sin(x-Bt)}}$

But this cannot exist for all x and t so the sum cannot be a wave for any $A,B,C$.

Does superpositon actually fail or have I made some other mistake?

  • $\begingroup$ How do you intend $g$ to exist even by itself? $\endgroup$ – John Dvorak Dec 16 '16 at 4:52
  • 2
    $\begingroup$ How is $g(x,t)$ a wave? The superposition principle is the consequence of a mathematical principle: the solutions of certain types fo DE add up to form a full solution. $\endgroup$ – Gert Dec 16 '16 at 5:12
  • $\begingroup$ $g$ doesn't satisfy the wave equation. $\endgroup$ – DanielSank Dec 16 '16 at 5:14
  • $\begingroup$ Read up on: en.wikipedia.org/wiki/Superposition_principle $\endgroup$ – Gert Dec 16 '16 at 5:15
  • $\begingroup$ In your case, $f$ and $g$ cannot be particular solutions to the same DE (the wave equation), thus you can't simply add them up. $\endgroup$ – Gert Dec 16 '16 at 5:17

If $g$ is to satisfy the equation $v^2\dfrac{\partial^2 g}{\partial x^2}= \dfrac{\partial^2 g}{\partial t^2}$, then $v = \pm\sqrt{C/A}$. If $f$ is to satisfy the same equation, then $v = \pm B$. So your equations are (taking one choice for the signs) \begin{equation*} f(x,t) = \sin(x-vt)\\ g(x,t) = A(x^2+v^2t^2) = \frac{A}{2}\left[(x+vt)^2+(x-vt)^2\right], \end{equation*} where the last form for $g$ just shows how to write it as a superposition of functions moving to the left with speed $v$ and to the right with speed $v$. If you add these together and substitute them into the wave equation you should find that the sum does in fact satisfy the equation.

(I think your expression for $v$ is upside-down; but if you turn it over and substitute $C = Av^2$ and $B = v$, I think you get $v = v$.)

  • $\begingroup$ Thanks, but why is your substitution valid? Also, plugging them in gives v =abs(v), implying v can only be positive. So the wave can only move to the right. $\endgroup$ – John Doe Dec 16 '16 at 15:07
  • $\begingroup$ Your premise is that g is a wave. But that's only true if $C = Av^2$, as I showed above. Similarly, $f$ is only a wave if $B = \pm v$. (As for abs(v), your expression should strictly speaking have abs(v) on the LHS.) As an alternative argument, for your expression to come out to be a constant for $v$ (once you've turned it up the right way), you have to have $C = AB^2$, and then abs(v) comes out to be $B$. $\endgroup$ – pwf Dec 16 '16 at 17:00

The wave function is not arbitrary, it must describe a particular system, i.e. satisfy the Schrodinger equation $$ i\hbar \frac{\partial \psi}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2}+V\psi $$ Since there is an $i$ on the LHS and no $i$ on the RHS, the wave function cannot be real unless it describes a zero-energy state. Since your time-dependent functions $f(x,t)$ and $g(x,t)$ are real, they can't be solutions to some Schrodinger equations, not to mention to the same equation.

  • $\begingroup$ The OP is talking about solutions to the wave equation, not solutions to the Schrodinger equation. $\endgroup$ – march Dec 16 '16 at 18:27
  • $\begingroup$ @march Oh, I made a silly mistake... $\endgroup$ – Xavier Dec 17 '16 at 1:32

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