The solution of the wave equation, $$\frac<\partial^2><\partial x^2>\psi(x,t)=\frac\frac<\partial^2><\partial t^2>\psi(x,t)$$ by separation of variables is $$\psi(x,t)=Ae^
$\begingroup$ Look at it this way, your function has two independent variables $x$ and $t$. So your function $\phi(x,t)$ has a 2D basis. The 2D basis results in 4 co-ordinate quadrants. each of the terms $\pm kx \pm \omega t$ occupies one of the four quadrants $\endgroup$
Commented Jan 10, 2017 at 5:25$\begingroup$ @ubuntu_noob Quite right. Would you be willing to expand that thought into an answer? $\endgroup$
Commented Jan 10, 2017 at 6:22$\begingroup$ @DanielSank Yeah yeah doing it below. Thanks for making me less lazy btw ;) $\endgroup$
Commented Jan 10, 2017 at 19:25It depends of the initial and boundary conditions. The solution,
is the general solution of the wave equation and thus carries no particular direction. It is possible to have a wave function that travels both to the left and to the right such as:
Which can create a stationary wave in some cases (e.g. if $A=B$). If one specifies initial conditions such that
$$\psi(x,t)= A\sin(kx-\omega t)$$
then one has a wave that moves forward expicitly.
answered Jan 11, 2017 at 3:06 962 5 5 silver badges 15 15 bronze badges $\begingroup$One reason why the directionality is not obvious to you is because you are solving a 1-dimensional case. For instance, if you were to write your answer in 3D, you would get:
The initial choice of the coordinate system with respect to the geometry of the problem defines the sign of the scalar product $\pmb k \cdot \pmb r = |\pmb k| |\pmb r| \cos \phi_$ , where in case with co-aligned $\pmb k$ and $\pmb r$ the cosine will end up being $\pm1$ defining the relative propagation direction of the wave.
One other directional phase component is $\omega t$ , where angular frequency $\omega$ , by convention, is always positive. Hence, by choosing $\pm t$ you define the direction of time flow.