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#2600 by deltaMass on 06 May, 2015 03:49
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The power input P dictates the available thrust. If the steady state thrust exceeds P/c, then you're violating conservation law.
But if I were you I'd disavow myself of the photon rocket model entirely, for the reasons I've given. It's a dead end. -
#2601 by ppnl on 06 May, 2015 03:59
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As far as I can tell, I am not the only one to have thought to a general relativity effect here. This paper uses a modified version of Einstein equations to explain what is going on in a conical resonant cavity without violating momentum conservation. The point is that the author uses a weak perturbation approximation and I do not know if this is fully justified.
Yeah...
No.
The blasted thing goes faster and that is an increase in momentum. Unless you show where the opposite momentum is then it violates conservation of momentum. It can't get any simpler than that.
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#2602 by ThereIWas3 on 06 May, 2015 04:02
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Rather than using stiff coax to bring the microwaves onto the balance beam and to the chamber, it might be easier to actually mount the RF generator on the balance beam and then only bring line power to it, through a lightweight piece of stranded, unshielded, zip-cord (no thicker than 12 or 14 gauge). That will be a lot more flexible than any kind of shielded cable.
You might also beam the power to the apparatus, through open air, at right angles to the pivot. A feedhorn at the fulcrum picks up the power and connects to the waveguide leading to the frustrum. No wires at all. The air gap could be just a few millimeters. The trick is in avoiding reflections from the impedance discontinuity and cooking your magnetron.
One of my most enjoyable college courses, back when the world was young, was the 3rd section of Electromagnetic Field Theory, taught by Dr John Kraus himself. He made it not only interesting, but fun, using a Socratic teaching method. It was really cool having one of the leading authorities on antenna design teaching a course on antenna and waveguide design to us undergraduates. -
#2603 by deltaMass on 06 May, 2015 04:09
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As far as my experience with exotic propulsion devices goes (20 years), the best way of all to construct a debunk-free apparatus to unambiguously demonstrate thrust is to make the whole thing completely self-contained.
That means battery power and the whole kit and kaboodle sealed in a box.
Accept no substitutes. -
#2604 by sanman on 06 May, 2015 04:10
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The blasted thing goes faster and that is an increase in momentum. Unless you show where the opposite momentum is then it violates conservation of momentum. It can't get any simpler than that.
Well, if EMdrive is a photon rocket, then the opposite momentum is in the photons that are leaving. Except that photons with lower momentum are just photons with lower energy. So if you want to blame something for violating Conservation of Momentum, then why not blame the lousy photons for becoming fainter instead of behaving more intuitively like particles of matter do?
Just quoting from EMdrive's site:
http://www.emdrive.com/QuoteSatellite Propulsion Research Ltd (SPR Ltd) a small UK based company, has demonstrated a remarkable new space propulsion technology. The company has successfully tested both an experimental thruster and a demonstrator engine which use patented microwave technology to convert electrical energy directly into thrust. No propellant is used in the conversion process. Thrust is produced by the amplification of the radiation pressure of an electromagnetic wave propagated through a resonant waveguide assembly.
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#2605 by PushHigher on 06 May, 2015 04:11
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Warp Drive is a purely FTL thing and doesn't claim to have anything to do with conventional displacement or momentum as we know it. And neither Mach Effect nor EMdrive claim to be able to go FTL, even though they are claiming a novel mechanism for displacement.
Regarding FTL - can I beg to differ?
Everything in the universe warps space-time. A rocket at rest warps space-time. An accelerating rocket warps it more.
IMHO - warping space-time to infinity still brings us to the speed of light because the speed of gravity (the warp) is limited to c. The EM Drive has nothing to do with FTL and I personally don't think FTL exists; in my mind the speed of light is the speed of reality.
http://en.wikipedia.org/wiki/Speed_of_gravity -
#2606 by ppnl on 06 May, 2015 04:13
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I hear you, but the main salient thing for me is that the EMdrive is supposed to be a photon rocket - the resonant cavity is not a closed chamber but is actually a chamber with a hole in it (the waveguide end), just as the combustion chamber of a conventional rocket has a hole in it.
So photons are shooting out one end of the EMdrive, just as photons would shoot out one end of a photon rocket, just as propellant shoots out the end of a conventional chemical rocket. What differentiates EMdrive from a traditional photon rocket, is that the photons have been drained of a lot more momentum due to bouncing around in the waveguide first before exiting. This is analogous to the way an expansion nozzle and bell allow the propellant exhaust particles to bounce around more and transfer more of their momentum to the rocket before exiting.
So maybe it's best to start with the picture of a photon rocket in your head, since a photon rocket doesn't violate Conservation of Momentum.
The thing is a photon rocket would require a huge number of photons that would probably burn down your lab. The power requirements would be monstrous. The only way to make a "practical" photon rocket is if you had a supply of antimatter. And it would be entirely useless for tooling around the solar system. It would only be used for interstellar travel.
Mix a great deal more matter in to reduce the energy of the exhaust and you could do interplanetary. -
#2607 by sanman on 06 May, 2015 04:15
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Warp Drive is a purely FTL thing and doesn't claim to have anything to do with conventional displacement or momentum as we know it. And neither Mach Effect nor EMdrive claim to be able to go FTL, even though they are claiming a novel mechanism for displacement.
Regarding FTL - can I beg to differ?
Everything in the universe warps space-time. A rocket at rest warps space-time. An accelerating rocket warps it more.
IMHO - warping space-time to infinity still brings us to the speed of light because the speed of gravity (the warp) is limited to c. The EM Drive has nothing to do with FTL and I personally don't think FTL exists; in my mind the speed of light is the speed of reality.
http://en.wikipedia.org/wiki/Speed_of_gravity
But what I said was that EMdrive has nothing to do with Alcubierre Warp Drive or its proposed FTL. That was the point I was trying to make. On that, we seem to agree. -
#2608 by sanman on 06 May, 2015 04:27
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So the crux of the issue for me, is that photons and matter particles behave differently, and are Apples & Oranges which cannot be judged by the same standards:
When you fully drain the momentum of a particle of propellant/matter (wrt your reference frame), then its velocity drops all the way down to stationary - but when you fully drain a photon of its momentum (wrt your reference frame), then it drops out of existence (ie. fades to nothing), because it loses energy rather than losing velocity.
So:
Photon Propellant != Matter Propellant
Matter is persistent, and you can see that exhaust hanging around after it's left the rocket.
Photons are not persistent, so they can disappear and leave you scratching your head about who/what was pushing your rocket to begin with.
Maybe the experimenters need to vary the angles of that waveguide, so that you have a series of cases where you tap greater and greater amounts of energy from the photons. The steeper and longer the waveguide, the more energy you're draining (transferring to your rocket/apparatus as momentum). The shallower the waveguide, the less energy you're draining (transferring to your rocket/apparatus as momentum).
Shouldn't the experimenters be trying to measure the energy of the photons that leave the resonant cavity via the waveguide exit? Wouldn't establishing some sort of correlation there be useful for proof of principle?? -
#2609 by warpmebro on 06 May, 2015 04:34
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Seems to me like too many people are worried about "laws" being "violated" instead of figuring out the new science behind this. Half the discussion here is about "this can't happen because of this "law" we came up with." Quit trying to fit within the "current model" of physics. Please, let me explain.
Throughout history, these "laws" have constantly changed. The earth is square, the earth is the center of the universe, (not the greatest examples but you get the idea) whatever the hell other dumb idea we can think of. On and on. And in each "era" they thought they knew it all. There's a few people here who know that we DON'T know it all, and these "laws" aren't really "laws" at all, it's ideas we've created to that fit our CURRENT UNDERSTANDING of the universe. Well guess what? We really don't know shit about the universe, honestly, and you should expect everything you know to change. That's the way it's been throughout history. It's a simple concept. Expect the unexpected, right?
We have an opportunity before us that nobody in history as been privileged with...worldwide collaboration via the internet, by the greatest minds of our age. Let's take advantage of it.
Maybe I'm just spitting out nonsense, and maybe I'm out of line, but I know one thing for sure...putting my ideas into writing isn't my strong-point, so take it for what it's worth.
The only physics I've studied is caveman physics, so this "idea" of mine is all I can "contribute", but I hope it can spark an idea or two, or a different way of thinking about all of this. -
#2610 by PushHigher on 06 May, 2015 04:41
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So the crux of the issue for me, is that photons and matter particles behave differently, and are Apples & Oranges which cannot be judged by the same standards:
When you fully drain the momentum of a particle of propellant/matter (wrt your reference frame), then its velocity drops all the way down to stationary - but when you fully drain a photon of its momentum (wrt your reference frame), then it drops out of existence (ie. fades to nothing), because it loses energy rather than losing velocity.
So:
Photon Propellant != Matter Propellant
Matter is persistent, and you can see that exhaust hanging around after it's left the rocket.
Photons are not persistent, so they can disappear and leave you scratching your head about who/what was pushing your rocket to begin with.
Agreed. If EM is just vibrations of space-time then EM is the (very small) waves of the ocean and mass is the boat. The enclosed frustum with concave/convex ends bottles this up.
The reason I use the gravity gradient analogy so much is because if you imagine a wave pool that has a narrow and wide end - the waves would be stronger toward the narrow end and weaker toward the wide end. -
#2611 by WarpTech on 06 May, 2015 05:12
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....
Todd,
It is exciting to read your explanation about exponentially decaying waves that have been squeezed beyond their cut-off diameter in the waveguide. You write very clearly.
It's great that you found your way into this thread
Please see the following reference (https://www.osapublishing.org/oe/fulltext.cfm?uri=oe-17-1-34&id=175583, click "Get PDF" to download the paper for free):
It is shown that all modes run continuously from travelling waves through a transition to an evanescent (exponentially decaying) wave region and the value of the attenuation increases as they approach the cone vertex.
A strict distinction between pure travelling waves and pure evanescent (exponentially decaying) waves cannot be achieved for conical waveguide.
One mode after the other reaches cutoff in the tapered hollow metallic waveguide as they approach the cone vertex.
Unfortunately, this analysis is for an open waveguide, not for a closed cavity, but the fact that a strict distinction between pure travelling waves and pure evanescent waves cannot be achieved for a conical waveguide, also has implications for modes approaching cutoff in the truncated cone cavity.
ADDENDUM:
One thing that has not been explored is whether these truncated cones are being prematurely ended towards the cone vertex. The tested designs are almost cylindrical.
Roger Shawyer has progressively (but very slowly with time) increased the cone angle of his truncated cones, culminating in the superconducting design he unveiled last October 2014. NASA Eagleworks and Yang in China have truncated cone designs that look like earlier Shawyer designs, with smaller cone angles.
For reference. the tangent of the cone's half angle thetaw and the cone's half angle thetaw, in ascending order, for the following cases are:
(Notice how Shawyer progressively increased the cone's half-angle, with time, in his experimental designs, by a factor of 7 in the tangent of the half-angle)
Example (and geometry) { Tan[thetaw],thetaw (degrees) }
Shawyer Experimental {0.104019, 5.93851}
Shawyer Fligth Thruster {0.19086, 10.8055}
Shawyer Demo {0.219054, 12.3557}
NASA Eagleworks frustum {0.263889, 14.7827}
Egan's example {0.36397 , 20}
Prof. Juan Yang (2014) {0.4538, 24.4 }
Shawyer Superconducting 2014 {0.7002, 35}
The people looking at running experiments here are (understandably) also looking at earlier designs with small cone angles and prematurely terminated before reaching the vertex. The group from a university was looking at running experiments with perfectly cylindrical geometry.
Given the latest write-up by Shawyer (concerning his choice of R1 being too large) it appears that what you are bringing up concerning modes near cutoff has not yet been appreciated or explored, as the researchers are not focusing on general wave solutions in the complex plane containing simultaneously both real and imaginary components.
Best regards,
PS: @aero has valiantly attempted to run a full analysis, of general waves, containing simultaneously both real and imaginary components, using MEEP. Unfortunately MEEP is finite difference code and hence he has only been able to run 2-D simulations (due to computer time limitations). It is known that this problem (truncated cone) is 3-D, as a 2-D analysis cannot simulate very important features of the geometry. The MEEP solutions have also been very difficult to interpret, because of the lack of suitable post-processing software to explore the solutions.
@Mulletron envisioned a completely conical EM Drive, terminating at the vertex (at the beginning of the thread when we were all trying to explore all possibilities with an open mind). A perfect cone may not be the best solution because in a perfect cone ALL modes are cut-off and hence there will not be any resonance, but in reality it is impossible to have a perfect cone, since the vertex will always terminate with a finite dimension (not a point). I explored some time ago some geometries, and it looks like there is plenty of room to explore truncated cones that terminate at different distances from the cone vertex.
Thank you Dr. Rodal for a very informative post!
I've been studying the reference you provided, to Zeng and Fan. You cannot imagine how coincidental it is, but sometimes the universe works that way. Their equations 8 thru 11, are simple enough to understand without too much difficulty. These are "effectively" the same equations that govern gravity in the Engineering model of GR I work with, which is based on the PV Model. Gravity, as a refractive index, appears as the Damping function that governs the attenuation of the wave functions, and the ZPF acts as the Driving function that keeps it all afloat at "our" relative vacuum energy level. The two are in equilibrium, in what QED calls the fluctuation-dissipation relationship, and gravity is the asymmetry between the two that occurs wherever you have matter that filters the modes. It's pretty simple and intuitive to understand, but nobody seems to get it.
You asked about the truncated cones. From an engineering perspective, if it is not truncated it will have a difficult time resonating at any mode. The convex-concave end plates would seem to be necessary to maximize energy storage as spherical harmonics. So then, what modes do we want to attenuate? That would depend on what modes we can inject that will sustain resonance. It won't resonate when the angle is increased too much, but if we have attenuation factor equations from this paper, then I believe it can be modeled.
Getting back to the paper, based on their graphs for attenuation, it would seem a small angle is preferred. A large angle approximates a flat plate. Anything greater than pi/6 is not much better than bouncing photons off of a flat plate. However, for theta = pi/24, the attenuation is very high at much shorter wavelengths, and very high at longer wavelengths. It needs to strike a balance between energy storage and thrust at the modes available to us.
Again, gravity acts on the wave functions through the metric, transforming the (E,p) 4-vector. The metric is a refractive index. The effect on the wave function is equivalent to a Damping factor, in the damped harmonic oscillator equation. I see the attenuation factor in their plots as "similar" to that effect, acting on the microwaves in the cavity near the cut-offs. As the waves are attenuated, their momentum is absorbed as wave velocity goes to zero, just like light falling into a black hole. The result is propulsion. The bonus is that in such a space-time where the speed of light is variable, momentum conservation is dependent on the group velocity. It's not Newtonian anymore, because velocity is not a constant.
The light is being squeezed by the slowing of the group velocity, and since Energy is conserved, momentum must increase to compensate for reduced wave velocity. Another way to look at it is, photons in the waveguide "gain" an "effective mass".
I see a lot of people arguing over photon rockets, despite the evidence that the thrust is orders of magnitude larger. No rocket nozzle is going to change that as long as the speed of light is considered to be constant, even if it captured all the energy from all the reflections produced. In order to get the thrust values they are seeing you must consider the reduction in wave velocity inside the waveguide, and that attenuation is asymmetrical, just like it is in a gravitational field.
Thanks again, for some very interesting new information.
Best Regards,
Todd D.
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#2612 by ppnl on 06 May, 2015 05:28
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So the crux of the issue for me, is that photons and matter particles behave differently, and are Apples & Oranges which cannot be judged by the same standards:
When you fully drain the momentum of a particle of propellant/matter (wrt your reference frame), then its velocity drops all the way down to stationary - but when you fully drain a photon of its momentum (wrt your reference frame), then it drops out of existence (ie. fades to nothing), because it loses energy rather than losing velocity.
So:
Photon Propellant != Matter Propellant
Matter is persistent, and you can see that exhaust hanging around after it's left the rocket.
Photons are not persistent, so they can disappear and leave you scratching your head about who/what was pushing your rocket to begin with.
Maybe the experimenters need to vary the angles of that waveguide, so that you have a series of cases where you tap greater and greater amounts of energy from the photons. The steeper and longer the waveguide, the more energy you're draining (transferring to your rocket/apparatus as momentum). The shallower the waveguide, the less energy you're draining (transferring to your rocket/apparatus as momentum).
Shouldn't the experimenters be trying to measure the energy of the photons that leave the resonant cavity via the waveguide exit? Wouldn't establishing some sort of correlation there be useful for proof of principle??
What you are describing is just red shift. I think. It is very confused.
In an ordinary rocket the propellent has no momentum with respect to the rockets reference frame. You burn it and give it a huge momentum with respect to the rockets reference frame. In reaction the rocket gets a huge momentum in the other direction. In the original reference frame they both have huge momentum gain in opposit directions. That is how conservation of momentum works.
A photon rocket is not much different. A photon has some amount of momentum despite being massless. As the photon rocket accelerated away you would see the photons slowly red shifting. But you could not be "scratching your head about who/what was pushing your rocket" until it reached the speed of light which would be never. Instead you would be scratching your head wondering what vaporized the asteroid behind you. Hint: the photons didn't just disappear. A photon rocket has an exhaust that is hotter and more destructive than any other kind of rocket giving the same thrust. -
#2613 by ipcalit on 06 May, 2015 05:37
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Someone posted the theory of a warp bubble happening around the device as a means for propulsion. I recall seeing some interferometry setup with the half of a laser beam going through the device...
Did anybody test splitting the beam and sending each half very close (0.1mm) to the front and back of the device while keeping the device stationary in hard vacuum? This should be easy to do with the existing setup, just move some mirrors around. -
#2614 by sanman on 06 May, 2015 06:08
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What you are describing is just red shift. I think. It is very confused.
In an ordinary rocket the propellent has no momentum with respect to the rockets reference frame. You burn it and give it a huge momentum with respect to the rockets reference frame. In reaction the rocket gets a huge momentum in the other direction. In the original reference frame they both have huge momentum gain in opposit directions. That is how conservation of momentum works.
A photon rocket is not much different. A photon has some amount of momentum despite being massless. As the photon rocket accelerated away you would see the photons slowly red shifting. But you could not be "scratching your head about who/what was pushing your rocket" until it reached the speed of light which would be never. Instead you would be scratching your head wondering what vaporized the asteroid behind you. Hint: the photons didn't just disappear. A photon rocket has an exhaust that is hotter and more destructive than any other kind of rocket giving the same thrust.
But I'm not talking about light that the rocket-rider would see, I'm talking about light that the guy back on the launchpad would see. The launchpad guy would see the photon exhaust as red-shifted for a different reason, which is that the photons would have transferred some of their energy (momentum) to the rocket - ie. those photons are coming out of the rocket with lower energy than what they had when they were originally created inside the cavity.
That's not really a red-shift.
ie. energy transfer to a 3rd party (Mr Rocket) is not the same as a Relativistic Doppler effect.
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#2615 by PushHigher on 06 May, 2015 06:23
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I tried to search for this but I think the thread search is broken because I was only seeing one result incorrectly.
Has anyone given any thought to using a a tapered electromagnet against a cylinder versus using a frustum? Would this allow for similar attenuation while having the added benefit of being variable and tunable? -
#2616 by sanman on 06 May, 2015 06:23
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If I pretend that the EmDrive is a photon rocket, for which the thrust F = P/c, then the value of k in N/W = 1/c = 3.3 10-9 N/W. That is at least 3 orders smaller than the experimentally determined values of k that we've seen. And yet you wish to model the EmDrive as a diluted form (in some sense I do not understand) of a photon rocket? That would seem to imply a k value lower than 1/c, but in fact we see a higher one.
I can't bring myself to agree.
And about this "exhaust hole" of yours: this implies that the placement of the hole (or the waveguide power feed, you assert) must align with the resultant thrust vector. But experiment shows that this is not the case.
Again, I cannot agree with you.
But the useful angle for the waveguide is the inverse of what would be the useful angle on a bell.
Light hitting the waveguide is producing momentum transfer in the same (parallel) direction as the photon emissions.
Exhaust from matter propellant is producing momentum transfer in the opposite (anti-parallel) direction as the exhaust stream.
It's because the mechanism of momentum transfer is different for a waveguide vs a bell.
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#2617 by ppnl on 06 May, 2015 06:25
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What you are describing is just red shift. I think. It is very confused.
In an ordinary rocket the propellent has no momentum with respect to the rockets reference frame. You burn it and give it a huge momentum with respect to the rockets reference frame. In reaction the rocket gets a huge momentum in the other direction. In the original reference frame they both have huge momentum gain in opposit directions. That is how conservation of momentum works.
A photon rocket is not much different. A photon has some amount of momentum despite being massless. As the photon rocket accelerated away you would see the photons slowly red shifting. But you could not be "scratching your head about who/what was pushing your rocket" until it reached the speed of light which would be never. Instead you would be scratching your head wondering what vaporized the asteroid behind you. Hint: the photons didn't just disappear. A photon rocket has an exhaust that is hotter and more destructive than any other kind of rocket giving the same thrust.
But I'm not talking about light that the rocket-rider would see, I'm talking about light that the guy back on the launchpad would see. The launchpad guy would see the photon exhaust as red-shifted for a different reason, which is that the photons would have transferred some of their energy (momentum) to the rocket - ie. those photons are coming out of the rocket with lower energy than what they had when they were originally created inside the cavity. That's not really a red-shift.
That's what I'm talking about. The guy on the launch pad would see ever more redshifted light as the rocket accelerated away. And the loss of the momentum of the redshifted photons would exactly match the gain in momentum of the photon rocket. If you did not see those photons then you are violating conservation of momentum. The redshifted photons are as necessary for the photon rocket as the drifting hot gasses of the regular rocket is to it. They are what conserves momentum.
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#2618 by sanman on 06 May, 2015 07:11
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Hmm, so let me maybe replace the photon rocket idea with a solar sail.
So you've got a solar sail that is moving because photons are hitting it. But what if your photon source isn't some external sun/star, but is actually sitting onboard the ship attached to the solar sail? Then it sounds like you can't generate any net thrust, since whatever momentum your photons transfer to the sail is offset by the momentum that was lost when the photons popped out of your onboard emitter.
But suppose your sail could experience multiple collisions with each photon that came out of the emitter? So that's the waveguide that's allowing this to happen. And each of those collisions is transferring some momentum. So the sum total of all momentum that a photon can transfer to the waveguide can't exceed the original momentum of the photon when you first produced/emitted it. So net thrust is zero because of the anti-thrust from producing the original photon.
Hmm, so now it doesn't work again...
I better go ask GoatGuy.
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#2619 by ppnl on 06 May, 2015 07:39
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Hmm, so let me maybe replace the photon rocket idea with a solar sail.
So you've got a solar sail that is moving because photons are hitting it. But what if your photon source isn't some external sun/star, but is actually sitting onboard the ship attached to the solar sail? Then it sounds like you can't generate any net thrust, since whatever momentum your photons transfer to the sail is offset by the momentum that was lost when the photons popped out of your onboard emitter.
But suppose your sail could experience multiple collisions with each photon that came out of the emitter? So that's the waveguide that's allowing this to happen. And each of those collisions is transferring some momentum. So the sum total of all momentum that a photon can transfer to the waveguide can't exceed the original momentum of the photon when you first produced/emitted it. So net thrust is zero because of the anti-thrust from producing the original photon.
Hmm, so now it doesn't work again...
I better go ask GoatGuy.
You get thrust when the photon is produced. With a solar sail or mirror you can change the direction of that thrust. Nothing will cancel out as long as you let the photon go into deep space.
Multiple reflections of a photon help nothing. You get as much thrust from that photon as you can ever hope to get simply by letting it go on its merry way. If you screw with it you are likely to get less thrust or none at all.
Let the light shine through.
