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The relation

$$\psi=Ce^{i/\hbar(Et-\mathbf{p}\cdot\mathbf{x})}\tag{1}$$

satisfies the Klein Gordon equation on the mass shell, i.e. for $E^2=p^2+m^2$.

Now let's think in the reverse direction. Relation (1) should satisfy the PDE:

$$\frac{\partial^2 \psi}{\partial E^2}-\frac{\partial^2 \psi}{\partial p_x^2}-\frac{\partial^2 \psi}{\partial p_y^2}-\frac{\partial^2 \psi}{\partial p_z^2}=0\tag{2}$$

on the "coordinate shell": $$ t^2-x^2-y^2-z^2=0 $$ Could relation (2) be related to a gravitational wave?

[You may assume that observation is being made from a Local Inertial Frame(your lab). Alternatively you may think of upgraded forms of the given equations in the curved spacetime context]

Allied issue: Going "off the mass shell" has produced interesting results in particle physics?What about the prospects of going "off the coordinate shell"?

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Just a note: your "coordinate shell" is the light cone of the origin. – David Zaslavsky♦ Sep 7 '12 at 4:44
Also, did you mean to write something different for equation (2)? That doesn't look like the analogue of the KG equation in momentum space, I would expect a differential operator applied to $\psi$. – David Zaslavsky♦ Sep 7 '12 at 4:46
I have used the term analogue/counterpart in the sense that the variables get interchanged,for example E<-->t;p(x)<---->x etc. We get the wave equation[speed=c=1] with the variables interchanged. – Anamitra Palit Sep 7 '12 at 4:49
As the $\psi$ wave passes through a point the values of energy and momentum(determining the spacetime curvature) can change in a specific manner as given by PDE(2) if we are to stay on the "coordinate shell" – Anamitra Palit Sep 7 '12 at 5:30
$\frac{\partial E}{\partial t}=\frac{\partial E}{\partial \psi}\frac{\partial \psi}{\partial t}$Or,$\frac{\partial^2 E}{\partial t^2}=\frac{\partial^2 E}{\partial \psi^2}(\frac{\partial \psi}{\partial t})^2+\frac{\partial E}{\partial \psi}\frac{\partial^2 \psi}{\partial t^2}$Similarly:$\frac{\partial^2 E}{\partial x^2}=\frac{\partial^2 E}{\partial \psi^2}(\frac{\partial \psi}{\partial x})^2+\frac{\partial E}{\partial \psi}\frac{\partial^2 \psi}{\partial x^2}$ We finally have$\frac{\partial^2 E}{\partial t^2}-\frac{\partial^2 E}{\partial x^2}=0$ for points where mass=0. – Anamitra Palit Sep 7 '12 at 8:30

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