@Ross7
The perpendicular/angle/Pythagoras thingy is tricky. To do it properly, you have to solve for the geometry of the waveguide, knowing the trajectories of the photons.
There is one limiting case where it's obvious that the geometry isn't important. In this case, the photon emitter is a flat surface in the Y axis, towards the large end (left) of the waveguide. It blasts photons out to the left and right, along the X axis. Most of the ones travelling to the right hit the sloping wall of the waveguide. In this case, there is no Y component to the photon momentum, all the momentum goes along the X axis, and there's no net left/right force.
Anything else is much more complicated. To solve these cases for a conical waveguide, you would define an annular ring in the Y axis, where the edge of the ring has height dy, calculate the force along the X axis against this dy ring, and integrate along the length of the cone. The fundamental insight here is that we use the same dy anywhere along the length of the cone.
Since dy is the same for all our rings, dx must vary depending on the precise geometry of the cone at that point on the X axis. In other words, we have a little triangle, with photons entering along the hypotenuse, and we need to calculate the force normal to the dy side (along X). You're right that Pythagoras says that only a proportion of the force goes to the right, but the insight is that the hypotenuse is longer than dy, so more photons than 'necessary' enter the triangle, and it all balances out when you take the X component of these extra photons. This isn't obvious, but consider the simple case where dx and dy are the same (the wall is 45 degrees at this point), and the photons enter perpendicular to the hypotenuse. In this simple case, all the root 2's cancel, and it's exactly the same as "viewing either end in 2D" as you put it.
Ok, a general proof is impossible because it depends on the location and geometry of the photon source, the geometry of the waveguide, extra reflections, and so on. In any event, this is the wrong way to do it - the right way is to solve for the fields inside the waveguide, and find the forces at the boundaries.
Shawyer hasn't done any of this. He didn't attempt any particle-like force calculations along the sloping walls, and he didn't attempt to solve for the fields, so his force calculations are incorrect. In any event, it's unnecessary to solve anything if you accept that the waveguide is a closed system. All the actions and reactions inside balance out, so there's no net movement. He glosses over this bit by saying that Special Relativity "implies" that the system isn't closed, without stating in what way it isn't closed. He has to demonstrate that there's a matter or energy transfer in or out of the waveguide to make it an open system, and he completely omits to say what this transfer is. And, of course, he only invokes special relativity *after* his derivation of the net force imbalance; he can hardly use it to justify his result.
> I still struggle to understand how photons that are reflected and retain the same
> energy content can produce any thrust - you're creating energy. However, that's
> probably why nobody pays me to do physics :)
There is a general conception that photons only transfer energy in an absorption process (and not in reflection or transmission). I don't understand where this comes from - maybe a pre-quantum view? When a photon strikes a surface and transfers momentum via radiation pressure, momentum is conserved, but this doesn't mean that there's no energy transfer. If the surface it strikes is free to move (a solar sail, for example) it takes on twice the momentum of the photon, and must therefore take some of its energy. The photon reflects with a longer wavelength to take account of the lost energy. In this case, the waveguide doesn't move, but some energy will be transferred to the waveguide surface, and the photon will again lose energy.
