Author Topic: EM Drive Developments - related to space flight applications - Thread 2  (Read 3321828 times)

Offline Mulletron

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I had the lightweight spreader bar part of the balance flying overnight on the Spiderwire reviewed below. I was surprised that the bar did find stable equilibrium on the very thin .011" diameter line. I figured it would sit there and oscillate all over the place.

I got started on the laser screen/camera setup.

Spent the day attaching hardware and stress testing the balance to make sure it will be safe to fly the frustum. The all up weight of the frustum +balance and everything else required for it to run is ~7lbs.

I have all this stuff weighed down to the gram btw but I haven't posted the details of it yet. I'm writing the weights of all the items on each item with a sharpie in grams with the precision of .1 gram.

I was shooting for a consistent safety factor of at least 80% of breaking strength to ensure no accidents. I used Spiderwire .011" 10lb test monofilament. This stuff broke at >10lbs all up weight two times, which is good. It does stretch quite a bit. I expected that and it really isn't a problem.

But I had two breaks at ~8 pounds. One break was with a Improved Clinch Knot, the other with the Palomar knot. It can certainly hold up the experiment, but that is too close for comfort for me, so it is out. That inconsistency in the break has me worried to use it. Honestly now that I know how much weight I'm trying to fly, I need to up the tensile strength anyway.

So I spent the day hanging and then breaking my balance.

I don't want to ride the razor's edge of possibly having a dented frustum, so I'm going overkill on tensile strength while maintaining the same diameter.

I had one slip (too many) with the Improved Clinch Knot, so I'm going to go with the double Palomar, which is a must for this new line I'm getting, and it's good that this line has it's own dedicated tying method:
http://www.netknots.com/fishing_knots/nanofil-knot

I have some Berkley Nanofil .01" 17lb test on the way.
http://www.amazon.com/Berkley-Nanofil-Uni-Filament-010-Inch-Diameter/dp/B009UCPF2C
This is about as good as it gets without going to a thicker line or a braid. I don't want to resort to a braid because I don't think a braided line is suitable for a torsion experiment.

I'm trying to use very thin line for now until I get proof that there is a force being generated or not. I only have 2 watts of power to play with. What happens next will inform whether a thicker line or a metallic wire is warranted.

Well anyway, this is a minor setback. I'm still going to fly the thing using some very strong braided Nylon line (yellow stuff in the photo coming from the spreader bar) in place of the mono torsion line, for now. If it moves (which it seems crazy that it would) there won't be any reliable force measurements.

I hooked up all the electronics and the amp to the frustum and let it run overnight and nothing failed or burst into flames, so that's good. Sadly it didn't hover or even wiggle.

https://drive.google.com/folderview?id=0B4PCfHCM1KYoTl90eDBuMklOeTg&usp=sharing&tid=0B4PCfHCM1KYoTXhSUTd5ZDN2WnM
« Last Edit: 04/03/2015 10:00 pm by Mulletron »
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Offline Rodal

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....
I hooked up all the electronics and the amp to the frustum and let it run overnight and nothing failed or burst into flames, so that's good. Sadly it didn't hover or even wiggle.

https://drive.google.com/folderview?id=0B4PCfHCM1KYoTl90eDBuMklOeTg&usp=sharing&tid=0B4PCfHCM1KYoTXhSUTd5ZDN2WnM
All the above refers to the truncated cone without the HDPE polymer dielectric segment inside it, correct?

Offline Mulletron

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....
I hooked up all the electronics and the amp to the frustum and let it run overnight and nothing failed or burst into flames, so that's good. Sadly it didn't hover or even wiggle.

https://drive.google.com/folderview?id=0B4PCfHCM1KYoTl90eDBuMklOeTg&usp=sharing&tid=0B4PCfHCM1KYoTXhSUTd5ZDN2WnM
All the above refers to the truncated cone without the HDPE polymer dielectric segment inside it, correct?

Yep. Empty. It wasn't on the balance. Just sitting on the floor.
« Last Edit: 04/04/2015 12:02 am by Mulletron »
And I can feel the change in the wind right now - Rod Stewart

Offline RotoSequence

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....
I hooked up all the electronics and the amp to the frustum and let it run overnight and nothing failed or burst into flames, so that's good. Sadly it didn't hover or even wiggle.

https://drive.google.com/folderview?id=0B4PCfHCM1KYoTl90eDBuMklOeTg&usp=sharing&tid=0B4PCfHCM1KYoTXhSUTd5ZDN2WnM
All the above refers to the truncated cone without the HDPE polymer dielectric segment inside it, correct?

Yep. Empty. It wasn't on the balance. Just sitting on the floor.

The entire frustum was on the floor, or just the HDPE disk? Sorry, I'm a little confused about what test you conducted.  :-[
« Last Edit: 04/04/2015 12:33 am by RotoSequence »

Offline Mulletron

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....
I hooked up all the electronics and the amp to the frustum and let it run overnight and nothing failed or burst into flames, so that's good. Sadly it didn't hover or even wiggle.

https://drive.google.com/folderview?id=0B4PCfHCM1KYoTl90eDBuMklOeTg&usp=sharing&tid=0B4PCfHCM1KYoTXhSUTd5ZDN2WnM
All the above refers to the truncated cone without the HDPE polymer dielectric segment inside it, correct?

Yep. Empty. It wasn't on the balance. Just sitting on the floor.

The entire frustum was on the floor, or just the HDPE disk? Sorry, I'm a little confused about what test you conducted.  :-[

The HDPE disc wasn't part of this. All the test was just hooking everything up to make sure it didn't overheat or fail. Just turned it all on for the first time.
And I can feel the change in the wind right now - Rod Stewart

Offline mlindner

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It appears my previous post got deleted.

So I'll ask again. Has this psudoscience nonsense been found to be false yet? Have any proper scientists tested this contraption so that they can show that it doesn't work?
LEO is the ocean, not an island (let alone a continent). We create cruise liners to ride the oceans, not artificial islands in the middle of them. We need a physical place, which has physical resources, to make our future out there.

Offline Notsosureofit

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It appears my previous post got deleted.

So I'll ask again. Has this psudoscience nonsense been found to be false yet? Have any proper scientists tested this contraption so that they can show that it doesn't work?

There has as yet been no nullifying experiment.  We keep at it though !

Offline QuantumG

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Have any proper scientists tested this contraption so that they can show that it doesn't work?

The real scientists are too busy doing grade school experiments to test something as unimportant as this.  :P
Human spaceflight is basically just LARPing now.

Offline jmossman

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I found the following paper  http://www-ssc.igpp.ucla.edu/personnel/russell/papers/skip_ed/node4.html, ( http://onlinelibrary.wiley.com/doi/10.1029/98JA02101/pdf  ) for example, that confirms the simple result I discussed above, for standing waves:

Quote
First, we may obtain some information from the simultaneous Poynting vectors as shown in Figures 5 and 6. If we consider a transverse wave causing field line oscillations, the Poynting vectors behave very differently depending on whether the wave is traveling or standing. Figure 9 is a diagram of the Poynting vectors for the two different schemes. Even though the wave amplitudes for both conditions are set to be the same and the magnitude of Poynting vector oscillations is consequently the same, the traveling wave propagates energy, while the standing wave produces no net energy flux. The Poynting vectors in Figures 5 and 6 more resemble the traveling wave pattern. Thus for the Pc3-4 wave activities in our observations the traveling wave component is stronger. We may also estimate the resonant condition by examining the phase difference between dE and dB [e.g., Singer et al., 1982]. If the phase difference is 90, the wave is standing and a resonant condition is reached.


   TRAVELLING WAVE                                                   STANDING WAVE (EM Drive)

  Poynting Vector time average is (+1/2)                      Poynting Vector time average is zero

  (Cos[ ω t])^2  =( 1+Cos[ 2 ω t]) /2                          Sin[ ω t] Cos[ ω t]  = Sin[ 2 ω t] /2



So it is as simply as this: to transfer energy or momentum from virtual particles in the Quantum Vacuum, as proposed by Dr. White, a traveling wave would be needed, but then one would have no resonance, and no Q.

If one has a cavity EM Drive then resonance can take place, and hence a high Q, but that precludes the possibility of transferring energy or momentum, according to Maxwell's equations.

Dr. Rodal, I am not an RF engineer and haven't performed math with Maxwell's equations many years.  My following thoughts may have completely overlooked a fundamental issue, so please take them with the appropriate degree of skeptism.  :)

Doesn't the above description (regarding the mathematical basis for the standing wave, and resulting zero time-average Poynting vector) require an ideal resonator with no losses?

http://www.physicsclassroom.com/class/waves/Lesson-4/Traveling-Waves-vs-Standing-Waves
Quote from: web link=http://www.physicsclassroom.com/class/waves/Lesson-4/Traveling-Waves-vs-Standing-Waves
When the proper frequency is used, the interference of the incident wave and the reflected wave occur in such a manner that there are specific points along the medium that appear to be standing still. Because the observed wave pattern is characterized by points that appear to be standing still, the pattern is often called a standing wave pattern...  These points vibrate back and forth from a positive displacement to a negative displacement; the vibrations occur at regular time intervals such that the motion of the medium is regular and repeating.

In the excited modes demonstrated by the IR thermal imagery, the frustum doesn't appear to be behaving like an ideal resonator cavity.  I would suspect that the time average reflected power from the "hot" end of the frustum is lower than the average power of the incident wave.  Doesn't that imply that the average energy of the Dr. White's QV collisions from the reflected wave will be lower than the average the energy of the QV collisions from the incident wave?  While the delta in power between incident and reflected might be close enough for many to consider EM drive as a "standing wave" within the frustum resonator, wouldn't a non-zero average be more accurate?  (i.e. a very low power traveling wave?)

Phrased a little differently, doesn't the frustum "hot" end look more like a "resistive" [1] load, thereby invalidating the simplistic ideal resonator and allowing for a non-zero time average Poynting vector?

[1]  a better word than "resistive" escapes me;  the induced currents are "draining" energy from the cavity and acting as a load

Best regards,
James
« Last Edit: 04/04/2015 01:49 am by jmossman »

Online aero

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I found the following paper  http://www-ssc.igpp.ucla.edu/personnel/russell/papers/skip_ed/node4.html, ( http://onlinelibrary.wiley.com/doi/10.1029/98JA02101/pdf  ) for example, that confirms the simple result I discussed above, for standing waves:

Quote
First, we may obtain some information from the simultaneous Poynting vectors as shown in Figures 5 and 6. If we consider a transverse wave causing field line oscillations, the Poynting vectors behave very differently depending on whether the wave is traveling or standing. Figure 9 is a diagram of the Poynting vectors for the two different schemes. Even though the wave amplitudes for both conditions are set to be the same and the magnitude of Poynting vector oscillations is consequently the same, the traveling wave propagates energy, while the standing wave produces no net energy flux. The Poynting vectors in Figures 5 and 6 more resemble the traveling wave pattern. Thus for the Pc3-4 wave activities in our observations the traveling wave component is stronger. We may also estimate the resonant condition by examining the phase difference between dE and dB [e.g., Singer et al., 1982]. If the phase difference is 90, the wave is standing and a resonant condition is reached.


   TRAVELLING WAVE                                                   STANDING WAVE (EM Drive)

  Poynting Vector time average is (+1/2)                      Poynting Vector time average is zero

  (Cos[ ω t])^2  =( 1+Cos[ 2 ω t]) /2                          Sin[ ω t] Cos[ ω t]  = Sin[ 2 ω t] /2



So it is as simply as this: to transfer energy or momentum from virtual particles in the Quantum Vacuum, as proposed by Dr. White, a traveling wave would be needed, but then one would have no resonance, and no Q.

If one has a cavity EM Drive then resonance can take place, and hence a high Q, but that precludes the possibility of transferring energy or momentum, according to Maxwell's equations.

I'm not sure that I am in 100% agreement with your result but my computational results do show that any force created without a QV model is far to small to be considered the answer to the cause of the measured experimental thrusts.

My current thoughts still hinge on the evanescent waves through the gaps in the structure as being key to the operation of this device. That is, evanescent waves are unidirectional, they do not average to zero but do collapse to zero at some distance from the cavity. This distance is considerably greater than one third or even one full wavelength.

My current thought is that perhaps, as the evanescent waves pass through the gaps, they apply force to the virtual particles in one direction only and then collapse as we know they do. However, the virtual particles have already dissipated back into the QV so only the momentum of the cavity remains.

I hope to run some cases testing this in the near future but a Mathematica model would be much more desirable, and telling, than a Meep test result. Fortunately, Meep does provide a QV model in 2-D - it was prepared in order to measure Casimir forces but should be useful for our situation, too. Unfortunately, only 1-D and 2-D models were implemented. Once I figure out how to run the model, and what it means, I will be ready to attempt some results.

I would be very happy to see you and Mathematica get the jump on me and provide some estimates before I can do so.
Retired, working interesting problems

Offline ThinkerX

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So...then, how is this device generating thrust if the laws of thermodynamics rule out the electromagnetic approach?

Plus, how is it that the predictions from the now invalid approach are...at least in the same ballpark as the reported experimental results? 

Aero's suggestion might be the only way to salvage this. 

Offline Mulletron

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It appears my previous post got deleted.

So I'll ask again. Has this psudoscience nonsense been found to be false yet? Have any proper scientists tested this contraption so that they can show that it doesn't work?

I'm sure some "proper scientists" will come round for a look eventually. In the meantime, history has a lot to teach us about these sorts of things:

http://amasci.com/weird/vindac.html
http://www.lifehack.org/articles/lifestyle/6-world-changing-ideas-that-were-originally-rejected.html
http://www.cracked.com/article_18822_5-famous-scientists-dismissed-as-morons-in-their-time.html

Quote
...so that they can show that it doesn't work

Just curious, what makes you so certain that it doesn't work? I mean, I have no idea if it works or not. The only certain thing I know here is that I don't know for certain if it works or doesn't.

There is a growing body of evidence which suggest that it does work. It hasn't been proven by anybody that it doesn't work.


« Last Edit: 04/04/2015 02:39 am by Mulletron »
And I can feel the change in the wind right now - Rod Stewart

Offline Rodal

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Dr. Rodal, I am not an RF engineer and haven't performed math with Maxwell's equations many years.  My following thoughts may have completely overlooked a fundamental issue, so please take them with the appropriate degree of skeptism.  :)

Doesn't the above description (regarding the mathematical basis for the standing wave, and resulting zero time-average Poynting vector) require an ideal resonator with no losses?

...
Hi James,

No, there is no requirement to have an ideal resonator with no losses, meaning an infinite Q.

The Finite Element Analysis (FEA) using COMSOL performed by NASA took into account tan delta losses to compute a finite Q.  Still, COMSOL FEA solves the standard Maxwell equations, and it contains no esoteric physics whatsoever.

One can also consider a non-ideal resonator using an exact solution.  One must use complex variables.  The losses responsible for a finite Q are due to the imaginary part, which is responsible for the material property "loss tangent" or "tan delta". 

See for example:  http://web.mit.edu/22.09/ClassHandouts/Charged%20Particle%20Accel/CHAP12.PDF

Maxwell's equation mandating that the Curl of the electric field E must equal the negative of the time derivative of the magnetic field B still must be satisfied.  See Equation 12.34 in the above link.



The electric permittivity describes the interaction of a material with an electric field E and is a complex quantity.

The real part of permittivity is a measure of how much energy from an external electric field is stored in a material. The imaginary part of permittivity is called the loss factor and is a measure of how dissipative or lossy a material is to an external electric field. The imaginary part of permittivity is always greater than zero and is usually much smaller than the real part. The loss factor includes the effects of both dielectric loss and conductivity.

Similarly, real materials have a magnetic susceptibility which is a complex quantity.   The complex permeability consists of a real part (µ') that represents the energy storage term and an imaginary part (µ'') that represents the energy loss term.

An analysis taking into account the complex (real and imaginary parts) of these physical properties does not change Maxwell's equations.
« Last Edit: 04/04/2015 03:57 am by Rodal »

Offline jmossman

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Dr. Rodal, I am not an RF engineer and haven't performed math with Maxwell's equations many years.  My following thoughts may have completely overlooked a fundamental issue, so please take them with the appropriate degree of skeptism.  :)

Doesn't the above description (regarding the mathematical basis for the standing wave, and resulting zero time-average Poynting vector) require an ideal resonator with no losses?

...
Hi James,

No, there is no requirement to have an ideal resonator with no losses, meaning an infinite Q.

The Finite Element Analysis (FEA) using COMSOL performed by NASA took into account tan delta losses to compute a finite Q.  Still, COMSOL FEA solves the standard Maxwell equations, and it contains no esoteric physics whatsoever.

One can also consider a non-ideal resonator using an exact solution.  One must use complex variables.  The losses responsible for a finite Q are due to the imaginary part, which is responsible for the material property "loss tangent" or "tan delta". 

See for example:  http://web.mit.edu/22.09/ClassHandouts/Charged%20Particle%20Accel/CHAP12.PDF

Maxwell's equation mandating that the Curl of the electric field E must equal the negative of the time derivative of the magnetic field B still must be satisfied.  See Equation 12.34 in the above link.



The electric permittivity describes the interaction of a material with an electric field E and is a complex quantity.

The real part of permittivity is a measure of how much energy from an external electric field is stored in a material. The imaginary part of permittivity is called the loss factor and is a measure of how dissipative or lossy a material is to an external electric field. The imaginary part of permittivity is always greater than zero and is usually much smaller than the real part. The loss factor includes the effects of both dielectric loss and conductivity.

Similarly, real materials have a magnetic susceptibility which is a complex quantity. 

An analysis taking into account the complex (real and imaginary parts) of these physical properties does not change Maxwell's equations.

Which part of Maxwell's equations account for the observed thermal losses due to the induced currents in the frustum's large base?  Thermal losses are coming from the field within the cavity, so I'm not sure how the idealized concept of a standing wave (i.e. perfect reflection at the cavity boundary) can be satisfied and still account for the thermal energy loss.

I'm not trying to challenge Maxwell's equations, but I don't see the entire system being modeled.  Once the losses in the frustum are accounted for, my (perhaps overly simplistic) mental math suggests that the energy of the reflected wave won't be exactly equal to the energy of the incident wave.  Doesn't every reflection reduce the total energy, with the maximum energy reduction occurring on the face with the largest induced currents (which also generate thermal heat losses)?  Over time wouldn't this incident-vs-reflected energy delta accumulate and erode the ideal assumption behind the perfect standing wave?  (and therefore provide a mathematical explanation for a small non-zero time average Poynting vector?)

Regards,
James

Offline Rodal

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Dr. Rodal, I am not an RF engineer and haven't performed math with Maxwell's equations many years.  My following thoughts may have completely overlooked a fundamental issue, so please take them with the appropriate degree of skeptism.  :)

Doesn't the above description (regarding the mathematical basis for the standing wave, and resulting zero time-average Poynting vector) require an ideal resonator with no losses?

...
Hi James,

No, there is no requirement to have an ideal resonator with no losses, meaning an infinite Q.

The Finite Element Analysis (FEA) using COMSOL performed by NASA took into account tan delta losses to compute a finite Q.  Still, COMSOL FEA solves the standard Maxwell equations, and it contains no esoteric physics whatsoever.

One can also consider a non-ideal resonator using an exact solution.  One must use complex variables.  The losses responsible for a finite Q are due to the imaginary part, which is responsible for the material property "loss tangent" or "tan delta". 

See for example:  http://web.mit.edu/22.09/ClassHandouts/Charged%20Particle%20Accel/CHAP12.PDF

Maxwell's equation mandating that the Curl of the electric field E must equal the negative of the time derivative of the magnetic field B still must be satisfied.  See Equation 12.34 in the above link.



The electric permittivity describes the interaction of a material with an electric field E and is a complex quantity.

The real part of permittivity is a measure of how much energy from an external electric field is stored in a material. The imaginary part of permittivity is called the loss factor and is a measure of how dissipative or lossy a material is to an external electric field. The imaginary part of permittivity is always greater than zero and is usually much smaller than the real part. The loss factor includes the effects of both dielectric loss and conductivity.

Similarly, real materials have a magnetic susceptibility which is a complex quantity. 

An analysis taking into account the complex (real and imaginary parts) of these physical properties does not change Maxwell's equations.

Which part of Maxwell's equations account for the observed thermal losses due to the induced currents in the frustum's large base?
In order to solve differential equations one must provide boundary conditions.  The thermal losses due to the induced eddy-currects in the copper are a result of the imaginary part of the material properties appearing in the solution of the boundary conditions to solve Maxwell's differential equations.

That is how COMSOL's Finite Element Analysis provided the solution for the thermal losses and the predicted temperature for NASA's EM Drive truncated cone.

The frequencies obtained by COMSOL's FEA were obtained by solving an eigenvalue problem.

Damping is responsible for the finite amplitude of the response, but the presence of damping does not preclude standing waves, waves which have fixed nodes and anti-nodes.  The boundary conditions due to the end plates do not disappear due to heat production.  For example, one of the boundary conditions is that the tangent electric field must be continuous at the material interface.  This boundary condition (and therefore the node produced in that component of the electric field) does not cease to exist due to heat generation.  That only affects the amplitude.  The only way to remove that boundary condition would be to remove the end plate, and if you would do that, the EM Drive would no longer be a completely enclosed cavity.
« Last Edit: 04/04/2015 04:36 am by Rodal »

Offline Mulletron

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BIG CORRECTION

It has bothered me that, if the Poynting vector (ExB) would be a quadratic function of the harmonic function so that it never changes sign and therefore does not change orientation with time, even for an AC field, as for example discussed in http://en.wikipedia.org/wiki/Poynting_vector#Time-averaged_Poynting_vector, then the operation of the EM Drive would not be a subject of so much controversy.

For the Poynting vector to vary like (Cos[ ω t + phaseAngle])^2 one needs that the E and B fields to be in phase with each other, as shown in the following image for example:



But then it dawned on me, that the E and B fields cannot be in-phase with each other (therefore the above image is only true for a travelling wave and it is inappropriate for an EM Drive cavity which instead has standing waves), because Maxwell's equation states that they must 90 degrees out of phase with each other:


One of Maxwell's equations (http://en.wikipedia.org/wiki/Faraday%27s_law_of_induction#Maxwell.E2.80.93Faraday_equation) states that:

Curl E = - d B / dt



so, for example if the magnetic field B varies as Cos[ ω t], then the electric field must vary as its time derivative:
- d(Cos[ ω t])/dt =  ω Sin[ ω t] , and therefore the Poynting vector ExB should vary as
Sin[ ω t] Cos[ ω t]  = Sin[ 2 ω t] /2, which oscillates at twice the frequency of the electromagnetic fields and has a time average value of zero.

Since the Poynting vector has a time average of zero, there cannot be any net energy flow out of the EM Drive.

This is due to the fact that the waves inside the EM Drive are standing waves.  Therefore the Poynting vector is just describing how energy is transferred between the electric and magnetic fields.

Also this means that there cannot be momentum outflow either, due to the Poynting vector, if the electromagnetic fields are harmonic functions of time.

Imagine, for discussion's sake, that it could indeed be possible that virtual electron-positron pairs would materialize out of the Quantum Vacuum, and that when such a pair materializes the Poynting vector is pointing towards the big base of the truncated cone EM Drive.  Then the electron-positron pair would be transported by the Poynting vector field towards the big base of the truncated cone, and shortly during that transport the electron-positron would cease to exist, returning back to the vacuum.  Then (as shown by Einstein himself in a though-model he proposed a long time ago concerning light particles being transported within a friction-less railroad car) the truncated cone would experience a recoil -simultaneous with the transport of the electron-positron pair-, which would result in a net force towards the small base of the truncated cone.  If the Poynting vector would always be pointing towards the big base, this would function as proposed by Dr. White.

Unfortunately, the standing waves within an EM Drive cavity are such that the E and B  fields must be 90 degrees out of phase with each other (due to Maxwell's equations), and this dictates that the Poynting vector is changing direction at a frequency twice as high as the frequency of the electromagnetic fields.  Therefore, if electron-positron pairs would materialize such as in the thought-model discussed above resulting in a recoil of the EM Drive towards the small base, it would occur just as often that electron-positron pairs would be transported in the completely opposite direction and the EM Drive would experience a force in the opposite direction.  Therefore what would be expected (out of the Quantum Vacuum model) is to have forces in the EM Drive pointing towards the small base just as often as having forces pointed in the opposite direction towards the big base, and this would result in no net transport of the EM Drive over a period of such oscillations.

I will need to correct some of my previous postings concerning the Poynting vector for the EM Drive: for a cavity like the EM Drive, the Poynting vector oscillates with time as  Sin[ 2 ω t] /2: therefore the time average of the Poynting vector must be zero.   

This article in Wikipedia does not apply to a resonating cavity like the EM Drive:

http://en.wikipedia.org/wiki/Poynting_vector#Time-averaged_Poynting_vector

because it does not obey Maxwell's equation  Curl E = - d B / dt which must be obeyed for a resonating cavity.

So now that you're officially abandoning reliance on the Poynting vector, what do you think of those other ideas like spectral non-reciprocity due to broken symmetry? That's the reason why I asked your opinion of how you thought the magnetic fields might play into this here:
http://forum.nasaspaceflight.com/index.php?topic=36313.msg1352873#msg1352873
http://forum.nasaspaceflight.com/index.php?topic=36313.msg1352987#msg1352987

It's been shown over and over again that the electric and magnetic fields time average to zero, and now the Poynting vector too (thanks :) ), so it is clear that we're witnessing "other than usual" symmetry conditions at work.


« Last Edit: 04/04/2015 04:48 am by Mulletron »
And I can feel the change in the wind right now - Rod Stewart

Offline Rodal

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...

My current thoughts still hinge on the evanescent waves through the gaps in the structure as being key to the operation of this device. That is, evanescent waves are unidirectional, they do not average to zero but do collapse to zero at some distance from the cavity. This distance is considerably greater than one third or even one full wavelength.

My current thought is that perhaps, as the evanescent waves pass through the gaps, they apply force to the virtual particles in one direction only and then collapse as we know they do. However, the virtual particles have already dissipated back into the QV so only the momentum of the cavity remains.

I hope to run some cases testing this in the near future but a Mathematica model would be much more desirable, and telling, than a Meep test result. Fortunately, Meep does provide a QV model in 2-D - it was prepared in order to measure Casimir forces but should be useful for our situation, too. Unfortunately, only 1-D and 2-D models were implemented. Once I figure out how to run the model, and what it means, I will be ready to attempt some results.



To have confidence on numerical results one should start by comparing the results with known solutions for (at least) simpler cases.  (Particularly for a numerical method, like the Finite Difference Method used in MEEP, that as we have discussed before one can not be assured to converge to a solution? )

 For example, I compared my exact solutions, and they were within 1% or so of COMSOL FEA results for NASA's truncated cone and their experimental results.

From what I recall, your MEEP calculations were very far away from the exact solutions for the cylindrical cavity case and very far away from the COMSOL FEA results for the truncated cone.  Also, the plots for the electromagnetic fields you posted did not look right. But that was some time ago when you were using a 2-D model, and I don't recall seeing recent postings of your MEEP solutions.

Have you now obtained MEEP solutions for the truncated cone and the cylindrical cavity that are in good agreement with known 3D solutions that give you such strong confidence on MEEP's calculations for much more complicated cases like the ones you are discussing above?

« Last Edit: 04/04/2015 05:13 am by Rodal »

Offline jmossman

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Dr. Rodal, I am not an RF engineer and haven't performed math with Maxwell's equations many years.  My following thoughts may have completely overlooked a fundamental issue, so please take them with the appropriate degree of skeptism.  :)

Doesn't the above description (regarding the mathematical basis for the standing wave, and resulting zero time-average Poynting vector) require an ideal resonator with no losses?

...
Hi James,

No, there is no requirement to have an ideal resonator with no losses, meaning an infinite Q.

The Finite Element Analysis (FEA) using COMSOL performed by NASA took into account tan delta losses to compute a finite Q.  Still, COMSOL FEA solves the standard Maxwell equations, and it contains no esoteric physics whatsoever.

One can also consider a non-ideal resonator using an exact solution.  One must use complex variables.  The losses responsible for a finite Q are due to the imaginary part, which is responsible for the material property "loss tangent" or "tan delta". 

See for example:  http://web.mit.edu/22.09/ClassHandouts/Charged%20Particle%20Accel/CHAP12.PDF

Maxwell's equation mandating that the Curl of the electric field E must equal the negative of the time derivative of the magnetic field B still must be satisfied.  See Equation 12.34 in the above link.



The electric permittivity describes the interaction of a material with an electric field E and is a complex quantity.

The real part of permittivity is a measure of how much energy from an external electric field is stored in a material. The imaginary part of permittivity is called the loss factor and is a measure of how dissipative or lossy a material is to an external electric field. The imaginary part of permittivity is always greater than zero and is usually much smaller than the real part. The loss factor includes the effects of both dielectric loss and conductivity.

Similarly, real materials have a magnetic susceptibility which is a complex quantity. 

An analysis taking into account the complex (real and imaginary parts) of these physical properties does not change Maxwell's equations.

Which part of Maxwell's equations account for the observed thermal losses due to the induced currents in the frustum's large base?
In order to solve differential equations one must provide boundary conditions.  The thermal losses due to the induced eddy-currects in the copper are a result of the imaginary part of the material properties appearing in the solution of the boundary conditions to solve Maxwell's differential equations.

That is how COMSOL's Finite Element Analysis provided the solution for the thermal losses and the predicted temperature for NASA's EM Drive truncated cone.

The frequencies obtained by COMSOL's FEA were obtained by solving an eigenvalue problem.

Damping is responsible for the finite amplitude of the response, but the presence of damping does not preclude standing waves, waves which have fixed nodes and anti-nodes.  The boundary conditions due to the end plates do not disappear due to heat production.  For example, one of the boundary conditions is that the tangent electric field must be continuous at the material interface.  This boundary condition (and therefore the node produced in that component of the electric field) does not cease to exist due to heat generation.  That only affects the amplitude.  The only way to remove that boundary condition would be to remove the end plate, and if you would do that, the EM Drive would no longer be a completely enclosed cavity.

You accurately and completely answered the questions I asked.  Unfortunately, I didn't ask my questions very well.  :)

http://forum.nasaspaceflight.com/index.php?topic=36313.msg1354235#msg1354235

A constant-amplitude standing wave does indeed result in a zero time-average Poynting vector.  However, I am questioning your conclusion that a constant-amplitude standing wave accurately represents a real resonator cavity such as the as-tested EM drive frustum.  Instead, I would expect a decaying amplitude standing wave to be a more accurate model/plot (as would be derived from a full solution to Maxwell's equations with proper boundary conditions such as non-zero resistance, etc).

Once a time-decaying standing wave is used for computation of a time-average Poynting vector, I'm having trouble seeing how the incident and reflected energy can perfectly cancel and become zero.  I'll readily admit I may be oversimplifying and/or missing a fundamental concept;  it's been a long time since I actually computed time constants for resonant cavities using Maxwell's equations and non-zero resistances.

Phrased a bit differently, I believe only excited modes with current/thermal losses in the base plates will significantly weight the direction of the time-average Poynting vector.  Each pair of incident/reflected waves would have a larger energy loss at the base plate with the excited E field (and therefore excited currents) than the energy loss at the opposing base plate.  For modes with near-zero E fields at the base plate boundaries, each incident/reflected wave pair would have a near equal energy delta regardless of which base plate they came in contact with;  the resulting time-averaged direction would be random and magnitude limited by the energy lost in the very first reflection (randomly either the large or small base, with a magnitude very close to zero).

I view this Poynting vector discussion to be completely independent of whether Dr. White's QV interactions, or some other classical physics can explain the EM drive anomalous thrust.  Just wanted to chime in on a what appeared to be the use of a constant-amplitude standing wave to describe a real-world system.  Your earlier observation of a non-zero time averaged Poynting vector seemed like a reasonable statement given that only excited modes with current/thermal losses in the base plates would quickly diverge from the simplified constant-amplitude standing wave model. 

Regards,
James
« Last Edit: 04/04/2015 06:13 pm by jmossman »

Offline mlindner

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It appears my previous post got deleted.

So I'll ask again. Has this psudoscience nonsense been found to be false yet? Have any proper scientists tested this contraption so that they can show that it doesn't work?

I'm sure some "proper scientists" will come round for a look eventually. In the meantime, history has a lot to teach us about these sorts of things:

http://amasci.com/weird/vindac.html
http://www.lifehack.org/articles/lifestyle/6-world-changing-ideas-that-were-originally-rejected.html
http://www.cracked.com/article_18822_5-famous-scientists-dismissed-as-morons-in-their-time.html

Every previous world changing idea didn't try to violate a founding principal of all physics, namely CoE and CoM.

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...so that they can show that it doesn't work

Just curious, what makes you so certain that it doesn't work? I mean, I have no idea if it works or not. The only certain thing I know here is that I don't know for certain if it works or doesn't.

There is a growing body of evidence which suggest that it does work. It hasn't been proven by anybody that it doesn't work.

I'm still waiting for an actual test of the operation of this craft. There haven't been any non-faulty experiments done yet that actually show it producing any thrust. Namely it must be tested in a vacuum. I don't try to merge philosophy and science and currently this "EM Drive" is purely in the realm of philosophy with no actual data yet. Thus I dismiss it just like the people claiming they made an anti-gravity drive in their garage.

It's rather insulting that this forum topic even exists here.
« Last Edit: 04/04/2015 07:44 am by mlindner »
LEO is the ocean, not an island (let alone a continent). We create cruise liners to ride the oceans, not artificial islands in the middle of them. We need a physical place, which has physical resources, to make our future out there.

Offline Star One

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The people who do decide if a thread/topic exists or not on here seem happy for it too. I suggest we leave it at that.

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