At 07:14 AM 11/8/2009, Michel Jullian wrote:
2009/11/8 Abd ul-Rahman Lomax <[email protected]>:
...
> Suppose the spinning of the assembly was caused by the counter-rotation of
> the motor shaft and what is attached to it. What would happen when the pump
> was turned off? The rotation would stop as the pump rotor slowed down and
> stopped. Conservation of angular momentum. If part of you spins one way,
> suspended in space you are, then part of you must spin the other way, so
> that the sum of angular momenta remains constant.
A very good point Abd, counter-rotation of the device balancing the
internal rotations of the shaft assembly and of the cooling fluid in
the cooling circuit seems a perfectly good explanation for the
phenomenon.
It could be. The point, really, is that the demonstration is far from
convincing, and combine that with the doubletalk in the theoretical
explanations, combined with effusive predictions of imminent major
success, one gets a pretty ordinary and standard picture.
Look, consider that cavity. Instead of making this difficult
contraption, pouring lots of power into it, why not just put in less
power. The pressure should be linear with the energy in the cavity,
there is no issue of the microwaves going faster with more power
(which is why his invocation of relativistic effects is so totally
bogus, he's dealing with EM radiation, which is beyond relativistic!)
It would be fairly simple, I'd think, to measure the radiation
pressure at various points in the cavity, and, in particular, along
the sides. I'm certainly not a microwave engineer....
But, frankly, there is no reason at all to suspect that anything but
standard physics is operating in that cavity. It's just a waveguide,
and how to analyze waveguides, I'm pretty sure, is well-known, if he
has the waveguide at resonance, what's going on at every point in it
should be accurately known. Where would standard calculations break
down and how? He hasn't shown that.
The technique described in this paper uses
radiation pressure, at microwave frequencies,
in an engine which provides direct conversion
from microwave energy to thrust, without the
need for propellant.
Okay, this is his claim, though you can produce thrust without the
need for propellant. If that was just a microwave horn, open at one
end, and the microwaves are emitted preferentially in one direction,
there would be thrust. Just not much!
The concept of the microwave engine is
illustrated in fig 1. Microwave energy is fed
from a magnetron, via a tuned feed to a closed,
tapered waveguide, whose overall electrical
length gives resonance at the operating
frequency of the magnetron.
Okay, seems pretty simple. Isn't done very often, to be sure, so if
there was some effect, one could imagine that it would have been
overlooked. However, this is where he goes off the deep end:
The group velocity of the electromagnetic
wave at the end plate of the larger section is
higher than the group velocity at the end plate
of the smaller section.
"Group velocity." What group velocity? Group velocity refers to the
apparent velocity of an interference pattern, or modulation. That
pattern can "move" at any speed, including faster than light. It
isn't anything but a pattern. It does not carry energy, as such, so
it will not exert a pressure.
Thus the radiation
pressure at the larger end plate is higher that
that at the smaller end plate.
He says so. That's certainly not my intuition of the situation; if
anything, the pressure at the larger end would be lower. If you have
photons bouncing back and forth in that cavity, i.e., if we look at
it from a particle nature of light POV, the particles would be
striking both ends at the same rate, and if they had the same energy,
they would create lower pressure at the larger end.
If this were a liquid in a channel that was shaped like a
cross-section of the waveguide, gravity waves would bounce back and
forth, and the wave height would be lower at the large end than at
the small end.
Remember, this is his explanation. It doesn't get any clearer.
The resulting
force difference (Fg1 -Fg2) is multiplied by the
Q of the resonant assembly.
Why? Q represents (inversely) the rate at which the oscillation in
the cavity will die away. At any given moment, the most that high Q
could do would be to sustain oscillation, not amplify the force
exerted by it. High Q would mean that the power being pumped into the
cavity would accumulate, and, yes, that would make the forces
stronger, but that simply means that the values of Fg1 and Fg2 would
be higher. They would be the same as with high power in the first place.
This force difference is supported by
inspection of the classical Lorentz force
equation (reference 1).
F = q(E + vB) (1)
If v is replaced with the group velocity vg of
the electromagnetic wave, then equation 1
illustrates that if vg1 is greater than vg2, then Fg1
should be expected to be greater than Fg2.
I.e., he's using an equation applicable to a charged particle as if
it applied to an electromagnetic wave phenomenon. q is the electric
charge of the particle in that equation, and the electric charge of
the wave is zero. He's expecting a pressure from a group velocity phenomenon.
However as the velocities at each end of the
waveguide are significant fractions of the
speed of light, a derivation of the force
difference equation invokes the difference in
velocities and therefore must take account of
the special theory of relativity.
The velocities of what? Apparently, we are talking about "group
velocity," and it might be appropriate to review the current
Wikipedia article, http://en.wikipedia.org/wiki/Group_velocity, it
seems to be a reasonable explanation.
Relativity theory implies that the
electromagnetic wave and the waveguide
assembly form an open system. Thus the force
difference results in a thrust which acts on the
waveguide assembly.
Quite simply, he states his desired conclusion, grammatically
connected with the premise, which is here utterly unestablished (how
does "relativity theory" imply that the wave and the waveguide
assembly for an "open system." He's stated it's a closed waveguide.
In what way is it open?
He wants to say it is open because then he can site "open" as meaning
that the conservation of momentum in a closed system doesn't apply.
He simply defines it away. It's not closed because it's closed, it's
open because of relativity theory.
Ah! I got it. This is how the drive works. He didn't mention it, but
there is a window in the end of the cavity, and he throws relativity
theory out the window. As theory has no mass, but it is neverheless
heavy, he has a reactionless drive due to the reaction to the theory.
