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#80 by flux_capacitor on 08 Aug, 2015 20:04
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I have a real concern with TheTraveller's Excel spreadsheet. The values I get from the first basic dimensions are inconsistent. I'm talking of the file EMDriveCalc20150617b.xls available from emdrive.wiki as well as TT's Gdrive.
Let's take know values, for example Eagleworks' frustum:
Db = 0.2794 m
Ds = 0.15875 m
Frustum length = 0.2286 m
cone half-angle = 14.78°
Input the first three values, and the spreadsheet returns a cone half-angle of 24.5°
Calculate the hypotenuse or draw the plan in a CAD software with the know values, you will easily get the frustum side length at 0.2364256 m. But the spreadsheet returns 0.2584848 m!
The formula for the cone half-angle (cell D8) in the spreadsheet is :
=DEGREES(ATAN((D3÷2)÷((D5×(D4÷2))+((D3÷2)−(D4÷2))+D5)))
Whereas it could use arccosine, frustum centre length (diameter center to diameter center) and frustum side length:
=DEGREES(ACOS(D5/D9)
Talking about the frustum side length (cell D9), its formula is wrong:
= SQRT(D5^2+(D3−D4)^2)
The correct formula should use end radii squared instead of end diameters squared:
= SQRT(D5^2+((D3−D4)÷2)^2)
How is the rest right or wrong? I can't even get Df right with the available spreadsheet.When inputing the Baby EmDrive data for example, Df becomes negative which is impossible (it should be comprised between 0 and 1)EDIT: my mistake, 24 GHz instead of 2.4 GHz resolved this issue.
Whatever, I don't get the same Df as TheTraveller for the same untouched spreadsheet and same input values. See fourth attachement below. Those differences are quite small, but everything else following in the spreadsheet gets very different values from those discrepancies.
@TheTraveller: can you please double-check those basic values in the spreadsheet, and upload a corrected version to the emdrive.wiki? This would be much appreciated by the EmDrive community
Below, I show two hypothesis for TT's EmDrive Mark 2, according to how the "Frustum centre length" is defined in the spreadsheet.
- The first with Frustum centre length = 208.71 mm has a cone half-angle (corrected formula) of 30°
- The second with Frustum centre length = 240.7 mm has a cone half-angle (corrected formula) of 26.6° (instead of 27.7° with the wrong angle formula).
What is important to note is that "Frustum centre length" as defined in the spreadsheet is the length between the centers of the two end diameters, and not the length defined by TheTraveller in his drawing where it is the apex r2-r1 length. All the misunderstanding comes from the difference in that drawing (attached in third position below).
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#81 by Star One on 08 Aug, 2015 20:06
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By the way those of you supporting the EM drive want an example of a theory that started out on the fringes but has gradually moved more centre wise then they only have to look at holographic theory for the universe.
http://www.sciencedaily.com/releases/2015/04/150427101633.htm -
#82 by Rodal on 08 Aug, 2015 20:15
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...
In a cylinder, the longitudinal component of momentum is the wave propagating down the waveguide. The momentum is in the z direction, not radial toward the walls. In Z&F, they are depicting a waveguide, not a cavity. The momentum is propagating down the waveguide. Changing the angle > 0, changes the momentum in the z direction for a tapered waveguide, relative to a cylindrical waveguide. In the case of the wave moving toward the apex, what Z&F show is that the waves are attenuated and all of the momentum in the z direction is phase shifted and absorbed.
Essentially, attenuation of the E field in the perpendicular directions, "IS" attenuation of momentum in the r direction, because Sr = E x H perpendicular to S. The wave traveling toward the apex is losing momentum to the waveguide, in the z direction. The wave traveling toward the big end is gaining momentum in the z direction. In a cylinder, neither is true because the wave does not change momentum in either direction. So the k-vector reflecting off the wall is not (theta, phi) in Z&F, it is in the r direction. So it is r*Cos(theta) to get z direction thrust.
Todd
A cylinder is a geometrical concept. It is perfectly OK for me to speak of a cylinder as something that has open ends, it carries no implications of a being a cavity.
As to Zeng and Fan, as I said, I would have to re-derive their equations and see whether they satisfy the boundary conditions (Ricvl said that they don't) or if there is something else amiss.
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#83 by X_RaY on 08 Aug, 2015 20:18
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Yeah right it's negligible for huge values of r, sometime the difference of r1 and r2 reach Plank length and there is no longer a difference in a physical sense....
I had that discussion with Todd in a previous thread, where he wrote the same thing you wrote above.
You can't form a cylinder this way, r is the distance from the apex
that's never parallel. For the boundary conditions i have to think about, but your comment 2) make sense to me
That's wrong.
I submitted a formal proof that a cylinder is the limit for r2 ->Infinity with (r2 - r1) kept constant and theta -> 0. Although I derived my proof independently you can also find in Euclidean geometry books, it emanates from the last of Euclid's postulate (parallell lines never meet) that is supplanted in Lobatchesky and Riemannian geometries.
For me this point is done!
Thanks Doc -
#84 by Rodal on 08 Aug, 2015 20:39
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... Anything that can be done to minimize the "z" component of this reflection, will add to the thrust.
Since it is unlikely that I will have the time to re-derive Zeng and Fan's equations any time soon, please allow me to pursue this (it is much less time consuming
Todd )
1) What, specifically, can be done to minimize the "z component of the reflection at the Big End, to add to thrust?"
2) Do I understand you correctly that you think that the net force is pointing from the Big End towards the Small End? (in the opposite direction to Shawyer's thrust which he posits to be pointed from the Small End to the Big End?) -
#85 by X_RaY on 08 Aug, 2015 20:43
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need it 3D
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#86 by WarpTech on 08 Aug, 2015 21:03
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...
In a cylinder, the longitudinal component of momentum is the wave propagating down the waveguide. The momentum is in the z direction, not radial toward the walls. In Z&F, they are depicting a waveguide, not a cavity. The momentum is propagating down the waveguide. Changing the angle > 0, changes the momentum in the z direction for a tapered waveguide, relative to a cylindrical waveguide. In the case of the wave moving toward the apex, what Z&F show is that the waves are attenuated and all of the momentum in the z direction is phase shifted and absorbed.
Essentially, attenuation of the E field in the perpendicular directions, "IS" attenuation of momentum in the r direction, because Sr = E x H perpendicular to S. The wave traveling toward the apex is losing momentum to the waveguide, in the z direction. The wave traveling toward the big end is gaining momentum in the z direction. In a cylinder, neither is true because the wave does not change momentum in either direction. So the k-vector reflecting off the wall is not (theta, phi) in Z&F, it is in the r direction. So it is r*Cos(theta) to get z direction thrust.
Todd
A cylinder is a geometrical concept. It is perfectly OK for me to speak of a cylinder as something that has open ends, it carries no implications of a being a cavity.
As to Zeng and Fan, as I said, I would have to re-derive their equations and see whether they satisfy the boundary conditions (Ricvl said that they don't) or if there is something else amiss.
I think that you would not need to confirm anything more than equations 8 and 9. Forget the Hankel functions. Z&F say this equation for E is validQuoteFor all electric field components of the spherical TE and TM modes mentioned previously, one can expressed them as... [equation 8]
You have the exact solution for E and H in a frustum cavity. Simply plug in the E vector into equation 9,
kr = j*(1/E)*dE/dr
where E is the electric field vector, all 3 components, which you already have in Mathematica. I think it should be just a few lines of code to take the derivative of a vector you already have, and multiply by the inverse. Once you have kr, take the derivative of that wrt time to get the force equation, and multiply by Cos(theta) to get the force in the z direction. No?
Todd
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#87 by WarpTech on 08 Aug, 2015 21:46
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... Anything that can be done to minimize the "z" component of this reflection, will add to the thrust.
Since it is unlikely that I will have the time to re-derive Zeng and Fan's equations any time soon, please allow me to pursue this (it is much less time consuming
Todd )
1) What, specifically, can be done to minimize the "z component of the reflection at the Big End, to add to thrust?"
2) Do I understand you correctly that you think that the net force is pointing from the Big End towards the Small End? (in the opposite direction to Shawyer's thrust which he posits to be pointed from the Small End to the Big End?)
1) At one time I suggested cutting slots in it to eliminate the DC component of current density. Settling for a lower Q and more leakage out the big end. Or a partially reflecting surface, like a Laser uses to allow a beam to escape. I am also thinking that the angle of incidence is important too, after looking at the Meep gifs. The energy does not appear to reflect perpendicular to the big end.
2a) I support the idea that dp/dt of the "frustum" is always forward, meaning small end leading.
2b) The dp/dt of the EM waves inside the frustum are such that dp/dt is positive when the wave is moving backwards, and negative when the wave is moving forward. So in one direction it is like "Shawyer's notion of thrust" and in the other direction it is like a solar wind behind a sail. Waves moving in BOTH directions, exert a force Forward on the frustum. It is only when the wave reflects from the big end that a force is being exerted that opposes that force.
I have calculated the DC B-field inside a frustum cavity and proven to myself that Maxwell's equations yield 0 net force on the system, provided I do not include any kind of nonlinear GR or PV affects on the permeability function. However, if the current is interrupted through the big end, i.e., if there is leakage flux escaping at the big end, then the force is asymmetrical.
Todd -
#88 by Ricvil on 08 Aug, 2015 21:48
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...
In a cylinder, the longitudinal component of momentum is the wave propagating down the waveguide. The momentum is in the z direction, not radial toward the walls. In Z&F, they are depicting a waveguide, not a cavity. The momentum is propagating down the waveguide. Changing the angle > 0, changes the momentum in the z direction for a tapered waveguide, relative to a cylindrical waveguide. In the case of the wave moving toward the apex, what Z&F show is that the waves are attenuated and all of the momentum in the z direction is phase shifted and absorbed.
Essentially, attenuation of the E field in the perpendicular directions, "IS" attenuation of momentum in the r direction, because Sr = E x H perpendicular to S. The wave traveling toward the apex is losing momentum to the waveguide, in the z direction. The wave traveling toward the big end is gaining momentum in the z direction. In a cylinder, neither is true because the wave does not change momentum in either direction. So the k-vector reflecting off the wall is not (theta, phi) in Z&F, it is in the r direction. So it is r*Cos(theta) to get z direction thrust.
Todd
A cylinder is a geometrical concept. It is perfectly OK for me to speak of a cylinder as something that has open ends, it carries no implications of a being a cavity.
As to Zeng and Fan, as I said, I would have to re-derive their equations and see whether they satisfy the boundary conditions (Ricvl said that they don't) or if there is something else amiss.
Yes Dr Rodal, to me Zeng and Fan don't write the correct expressions for the problem of propagation on a tapered conical waveguide.
One way is solve the helmoltz equations respecting directly the physical boundary conditions under the geometry of problem.
This will implies that standard solutions of wave equation in spherical coordinates ( in general, bessel and legendre functions of integer order), must be subtituted by specific solutions with fractionary order legendre/ bessel functions, to respect the angular (theta) PEC boundary condition.
Other way, more complicated, is find solutions using the called "coupled-mode theory in instantaneous eigenmode (quasimode) basis ", where one can use a superposition cilindrical modes, evolving a coupled-mode equation along the longitudinal axis of propagation, where each cilindrical mode has a instantaneous dependence on the radii of tapered section of conical waveguide.
Of course, there are many others forms to solve this problem.
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#89 by WarpTech on 08 Aug, 2015 21:58
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...
In a cylinder, the longitudinal component of momentum is the wave propagating down the waveguide. The momentum is in the z direction, not radial toward the walls. In Z&F, they are depicting a waveguide, not a cavity. The momentum is propagating down the waveguide. Changing the angle > 0, changes the momentum in the z direction for a tapered waveguide, relative to a cylindrical waveguide. In the case of the wave moving toward the apex, what Z&F show is that the waves are attenuated and all of the momentum in the z direction is phase shifted and absorbed.
Essentially, attenuation of the E field in the perpendicular directions, "IS" attenuation of momentum in the r direction, because Sr = E x H perpendicular to S. The wave traveling toward the apex is losing momentum to the waveguide, in the z direction. The wave traveling toward the big end is gaining momentum in the z direction. In a cylinder, neither is true because the wave does not change momentum in either direction. So the k-vector reflecting off the wall is not (theta, phi) in Z&F, it is in the r direction. So it is r*Cos(theta) to get z direction thrust.
Todd
A cylinder is a geometrical concept. It is perfectly OK for me to speak of a cylinder as something that has open ends, it carries no implications of a being a cavity.
As to Zeng and Fan, as I said, I would have to re-derive their equations and see whether they satisfy the boundary conditions (Ricvl said that they don't) or if there is something else amiss.
Yes Dr Rodal, to me Zeng and Fan don't write the correct expressions for the problem of propagation on a tapered conical waveguide.
One way is solve the helmoltz equations respecting directly the physical boundary conditions under the geometry of problem.
This will implies that standard solutions of wave equation in spherical coordinates ( in general, bessel and legendre functions of integer order), must be subtituted by specific solutions with fractionary order legendre/ bessel functions.
Other way, more complicated, is find solutions using the called "coupled-mode theory in instantaneous eigenmode (quasimode) basis ", where one can use a superposition cilindrical modes, evolving a coupled-mode equation along the longitudinal axis of propagation, where each cilindrical mode has a instantaneous dependence on the radii of tapered section of conical waveguide.
Of course, there are many others forms to solve this problem.
I set to bold because, to my understanding, this is precisely what they did. Look at equations 2 & 3, and the associated text that follows it, through to equation 8. This part of their work is not that difficult to follow, so I don't see precisely where they make any error or assumptions that would cause it to be wrong. Their answer includes everything you just said, and the derivatives are Hankel functions of "fractional order". Where is the error?
Todd
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#90 by SeeShells on 08 Aug, 2015 21:59
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I have a real concern with TheTraveller's Excel spreadsheet. The values I get from the first basic dimensions are inconsistent. I'm talking of the file EMDriveCalc20150617b.xls available from emdrive.wiki as well as TT's Gdrive.
Let's take know values, for example Eagleworks' frustum:
Db = 0.2794 m
Ds = 0.15875 m
Frustum length = 0.2286 m
cone half-angle = 14.78°
Input the first three values, and the spreadsheet returns a cone half-angle of 24.5°
Calculate the hypotenuse or draw the plan in a CAD software with the know values, you will easily get the frustum side length at 0.2364256 m. But the spreadsheet returns 0.2584848 m!
The formula for the cone half-angle (cell D8) in the spreadsheet is :
=DEGREES(ATAN((D3÷2)÷((D5×(D4÷2))+((D3÷2)−(D4÷2))+D5)))
Whereas it could use arccosine, frustum centre length (diameter center to diameter center) and frustum side length:
=DEGREES(ACOS(D5/D9)
Talking about the frustum side length (cell D9), its formula is wrong:
= SQRT(D5^2+(D3−D4)^2)
The correct formula should use end radii squared instead of end diameters squared:
= SQRT(D5^2+((D3−D4)÷2)^2)
How is the rest right or wrong? I can't even get Df right with the available spreadsheet. When inputing the Baby EmDrive data for example, Df becomes negative which is impossible (it should be comprised between 0 and 1)
@TheTraveller: can you please double-check those basic values in the spreadsheet, and upload a corrected version to the emdrive.wiki? This would be much appreciated by the EmDrive community
Below, I show two hypothesis for TT's EmDrive Mark 2, according to how the "Frustum centre length" is defined in the spreadsheet.
- The first with Frustum centre length = 208.71 mm has a cone half-angle (corrected formula) of 30°
- The second with Frustum centre length = 240.7 mm has a cone half-angle (corrected formula) of 26.6° (instead of 27.7° with the wrong angle formula).
What is important to note is that "Frustum centre length" as defined in the spreadsheet is the length between the centers of the two end diameters, and not the length defined by TheTraveller in his drawing where it is the apex r2-r1 length. All the misunderstanding comes from the difference in that drawing (attached in third position below).
Didn't the traveler say this came from Shawyer and SPR and the software they used?
Shell
P: Good catch //flux_capacitor -
#91 by SeeShells on 08 Aug, 2015 22:03
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By the way those of you supporting the EM drive want an example of a theory that started out on the fringes but has gradually moved more centre wise then they only have to look at holographic theory for the universe.
I honestly love this theory, simply it fits so much (but what do I know?). I've been following it for the last few months. Good post Star One!
http://www.sciencedaily.com/releases/2015/04/150427101633.htm
I've been kicked out of the shop while other work goes on so I've been reading way too much tech.
Shell -
#92 by SeeShells on 08 Aug, 2015 22:05
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need it 3D
Needs a Heat Shield...... -
#93 by SeeShells on 08 Aug, 2015 22:14
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Something has been bothering me since last night and I couldn't help but watching the modes change and flip in meep.
How can you calculate seriously any Q in a cavity that simply jumps around from one T mode decaying and building into another in such a short time? And they all do it, every simulation with varying speeds.
Just something to mull over on a day away from the shop.
Shell
Added: Back to lurk mode and I'll be quiet. -
#94 by deltaMass on 08 Aug, 2015 23:04
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If EmDrive "thrust" is an artifact of the experimental measurement process, as most people believe it is, then what is your favoured candidate for such an artifact?
I vote for thermal-related. -
#95 by Ricvil on 08 Aug, 2015 23:08
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...
In a cylinder, the longitudinal component of momentum is the wave propagating down the waveguide. The momentum is in the z direction, not radial toward the walls. In Z&F, they are depicting a waveguide, not a cavity. The momentum is propagating down the waveguide. Changing the angle > 0, changes the momentum in the z direction for a tapered waveguide, relative to a cylindrical waveguide. In the case of the wave moving toward the apex, what Z&F show is that the waves are attenuated and all of the momentum in the z direction is phase shifted and absorbed.
Essentially, attenuation of the E field in the perpendicular directions, "IS" attenuation of momentum in the r direction, because Sr = E x H perpendicular to S. The wave traveling toward the apex is losing momentum to the waveguide, in the z direction. The wave traveling toward the big end is gaining momentum in the z direction. In a cylinder, neither is true because the wave does not change momentum in either direction. So the k-vector reflecting off the wall is not (theta, phi) in Z&F, it is in the r direction. So it is r*Cos(theta) to get z direction thrust.
Todd
A cylinder is a geometrical concept. It is perfectly OK for me to speak of a cylinder as something that has open ends, it carries no implications of a being a cavity.
As to Zeng and Fan, as I said, I would have to re-derive their equations and see whether they satisfy the boundary conditions (Ricvl said that they don't) or if there is something else amiss.
Yes Dr Rodal, to me Zeng and Fan don't write the correct expressions for the problem of propagation on a tapered conical waveguide.
One way is solve the helmoltz equations respecting directly the physical boundary conditions under the geometry of problem.
This will implies that standard solutions of wave equation in spherical coordinates ( in general, bessel and legendre functions of integer order), must be subtituted by specific solutions with fractionary order legendre/ bessel functions.
Other way, more complicated, is find solutions using the called "coupled-mode theory in instantaneous eigenmode (quasimode) basis ", where one can use a superposition cilindrical modes, evolving a coupled-mode equation along the longitudinal axis of propagation, where each cilindrical mode has a instantaneous dependence on the radii of tapered section of conical waveguide.
Of course, there are many others forms to solve this problem.
I set to bold because, to my understanding, this is precisely what they did. Look at equations 2 & 3, and the associated text that follows it, through to equation 8. This part of their work is not that difficult to follow, so I don't see precisely where they make any error or assumptions that would cause it to be wrong. Their answer includes everything you just said, and the derivatives are Hankel functions of "fractional order". Where is the error?
Todd
Not only half interger order Todd, but a real order associated legendre and bessel functions.
I need go to a party now
, but latter I will try explain better, but basicaly, boundary conditions under a adapted coordinate system for the problem, are satisfied only with the use of the free parameters of the general solution, and must have no dependence with the independent variables of the differential problem, or will not sove bc conditions and the differential equation simultaneously.
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#96 by Rodal on 08 Aug, 2015 23:26
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Something has been bothering me since last night and I couldn't help but watching the modes change and flip in meep.
How can you calculate seriously any Q in a cavity that simply jumps around from one T mode decaying and building into another in such a short time? And they all do it, every simulation with varying speeds.
Just something to mull over on a day away from the shop.
Shell
Added: Back to lurk mode and I'll be quiet.
Excellent question. I hope that you don't shy away from asking such questions because it is only this way that one can understand what is being output.
My understanding (aero to confirm) is that the quality factor is calculated by aero using the routine Harminv.
Please notice that Meep has this disclaimer for using Hamrinv to calculate the quality factor Q:
http://ab-initio.mit.edu/wiki/index.php/Meep_Reference#HarminvQuoteImportant: normally, you should only use harminv to analyze data after the sources are off. Wrapping it in (after-sources (harminv ...)) is sufficient.
Thus, the Quality factor and the frequency (also obtained by Harminv) should properly be obtained after the sources are OFF, not when they are on.
This takes us back to the whole discussion about Q quality factor in experiments. The problem is not only how to best experimentally measure and report Q, but it seems not ideal to me to discuss and report a Q (as first done by Shawyer and then imitated by all other EM Drive experimenters) with the source ON.
It seems to me that proper measurement of Q also in experiments should be done upon turning the source off and examining the decay (as posted by Frobnicat in a separate post).
With the sources OFF, the definition of Q (inverse to damping) is well-posed.
With the sources ON, the meaning of Q is tricky. It seems to me that when people are measuring Q with the sources on they are assuming a well-posed problem with "nice" properties (symmetry, etc.) that may not be fulfilled. This is particularly contradictory with TheTraveller: who rejects Finite Element solutions (using COMSOL or ANSYS) and exact solutions of the problem saying that only a solution that calculates a force can properly calculate the frequency and at the same time Shawyer "measures" Q with the RF feed ON which assumes a well posed nice solution amenable to presentation of Q as if it would be the same Q as the one calculated with the RF feed off.
The way that the Q is being measured experimentally by Shawyer and others using S11 and the 3db width is similar to a common method in structural vibration analysis. Of course, this presupposes a steady-state response. In reality with the RF feed ON, the measurement may be a transient instead.
_______________________________________________________
QUESTION TO aero: have you been calculating the quality factor Q with Harminv, and if so, have you been doing so with the sources ON or OFF. Have you been wrapping with (after-sources (harminv ...)) ? -
#97 by WarpTech on 08 Aug, 2015 23:33
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I set to bold because, to my understanding, this is precisely what they did. Look at equations 2 & 3, and the associated text that follows it, through to equation 8. This part of their work is not that difficult to follow, so I don't see precisely where they make any error or assumptions that would cause it to be wrong. Their answer includes everything you just said, and the derivatives are Hankel functions of "fractional order". Where is the error?
Todd
Not only half interger order Todd, but a real order associated legendre and bessel functions.
I need go to a party now , but latter I will try explain better, but basicaly, boundary conditions under a adapted coordinate system for the problem, are satisfied only with the use of the free parameters of the general solution, and must have no dependence with the independent variables of the differential problem, or will not sove bc conditions and the differential equation simultaneously.
Just tell me if equations 8 and 9 are correct or not please. It is a simple enough wave function and its first derivative. Is it representative of the electric field for TE and TM modes as they say, or isn't it? I don't need anything past equation 9 from Z&F to calculate the momentum and forces from the wave vector. @Rodal has an exact solution for the fields in his Mathematica program, and we have numbers from Meep. Therefore, regardless if Zeng and Fan derived those fields correctly, if equation 9 is correct, (which I think it is) then what's the problem? Just plug in the E vector from Mathematica or Meep, and its inverse, and take a couple of derivatives... and we have a force equation in both directions of propagation.
Todd
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#98 by rfmwguy on 08 Aug, 2015 23:42
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I opted out of a summer end party tonight mainly because Monday is a milestone birthday. Think I will celebrate with the sense that the next generation will question everything and challenge the status quo. Here's to nonconformity......
In a cylinder, the longitudinal component of momentum is the wave propagating down the waveguide. The momentum is in the z direction, not radial toward the walls. In Z&F, they are depicting a waveguide, not a cavity. The momentum is propagating down the waveguide. Changing the angle > 0, changes the momentum in the z direction for a tapered waveguide, relative to a cylindrical waveguide. In the case of the wave moving toward the apex, what Z&F show is that the waves are attenuated and all of the momentum in the z direction is phase shifted and absorbed.
Ui
Essentially, attenuation of the E field in the perpendicular directions, "IS" attenuation of momentum in the r direction, because Sr = E x H perpendicular to S. The wave traveling toward the apex is losing momentum to the waveguide, in the z direction. The wave traveling toward the big end is gaining momentum in the z direction. In a cylinder, neither is true because the wave does not change momentum in either direction. So the k-vector reflecting off the wall is not (theta, phi) in Z&F, it is in the r direction. So it is r*Cos(theta) to get z direction thrust.
Todd
A cylinder is a geometrical concept. It is perfectly OK for me to speak of a cylinder as something that has open ends, it carries no implications of a being a cavity.
As to Zeng and Fan, as I said, I would have to re-derive their equations and see whether they satisfy the boundary conditions (Ricvl said that they don't) or if there is something else amiss.
Yes Dr Rodal, to me Zeng and Fan don't write the correct expressions for the problem of propagation on a tapered conical waveguide.
One way is solve the helmoltz equations respecting directly the physical boundary conditions under the geometry of problem.
This will implies that standard solutions of wave equation in spherical coordinates ( in general, bessel and legendre functions of integer order), must be subtituted by specific solutions with fractionary order legendre/ bessel functions.
Other way, more complicated, is find solutions using the called "coupled-mode theory in instantaneous eigenmode (quasimode) basis ", where one can use a superposition cilindrical modes, evolving a coupled-mode equation along the longitudinal axis of propagation, where each cilindrical mode has a instantaneous dependence on the radii of tapered section of conical waveguide.
Of course, there are many others forms to solve this problem.
I set to bold because, to my understanding, this is precisely what they did. Look at equations 2 & 3, and the associated text that follows it, through to equation 8. This part of their work is not that difficult to follow, so I don't see precisely where they make any error or assumptions that would cause it to be wrong. Their answer includes everything you just said, and the derivatives are Hankel functions of "fractional order". Where is the error?
Todd
Not only half interger order Todd, but a real order associated legendre and bessel functions.
I need go to a party now , but latter I will try explain better, but basicaly, boundary conditions under a adapted coordinate system for the problem, are satisfied only with the use of the free parameters of the general solution, and must have no dependence with the independent variables of the differential problem, or will not sove bc conditions and the differential equation simultaneously. -
#99 by garrybinder on 08 Aug, 2015 23:43
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I love this thread and am totally addicted, can't put it down. Thanks for all the work and information! Discussions about Q, oxidation, broad / narrow spectrum and the shape of the ends have got me thinking about an alternate configuration. It has many of the same characteristics but might be slightly easier to build / find parts / work with. It is an ordinary cylinder with a cone inside. Maybe something to consider once folks are enjoying consistent thrust (lifting small objects and pets, etc... ;)
Keep up the great work!
that's never parallel. For the boundary conditions i have to think about, but your comment 2) make sense to me
, but latter I will try explain better, but basicaly, boundary conditions under a adapted coordinate system for the problem, are satisfied only with the use of the free parameters of the general solution, and must have no dependence with the independent variables of the differential problem, or will not sove bc conditions and the differential equation simultaneously.