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#1760 by meberbs on 15 Jan, 2017 16:32
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Sorry, but statements from one person do not override demonstrated data, even if said person had any credibility left. Also why should a 2D solution be trusted over the full and validated 3D solution? Especially when that person also makes statements like "guide wavelength terms are strongly non-linear" when Maxwell's equations are linear, and guide wavelength is not properly defined for a closed resonant cavity.The resonant frequency has been checked how?
Did post this before:QuoteAs you will discover the solutions to the guide wavelength terms are very non-linear, which makes the accurate design of a tapered cavity particularly difficult. At SPR Ltd we have developed our own software to give a 2D high resolution numerical solution to the problem. This proprietary design software has been validated for a number of different cavities by ourselves, as well as different research groups operating under commercial agreements with SPR Ltd.
The KISS Thruster spherical dimensions will resonant in TE013 mode at 2.45GHz, despite what other analytical methods suggest.
This also doesn't actually answer the question. Did you send the dimensions to Shawyer, and he confirmed the results of your spreadsheet? In that case it is reasonable to assume he just used the spreadsheet himself, and your answer as shown multiple times will be close (better than I would have initially guessed), but not as good as the other available methods. -
#1761 by WarpTech on 15 Jan, 2017 16:33
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KISS Thruster update:
Latest dimension as attached.
Will probably be using either the Yonlit 10W or 25W Rf amp http://www.yonlit.com/Products.asp?ClassID=7 with the KISS Thruster as it combines the:
1) pre amp
2) attenuator
3) power amp
4) forward and reflected power monitor
into 1 package.
Result are 3 electronic modules: Freq Gen, Rf amp & Freq Tracker instead of 6. KISS = fewer components.
Waiting on pricing quote to decide on which amp to run with.
Another added benefit is the experimenter will be able to choose a Yonlit 5W to 120W Rf amp package and have the same DB15 interface to the freq tracker module.
Please note that in spherical end plate thrusters, the effective resonant length is the frustum height (axial length) and NOT the constant separation between the spherical end plates. Which means the FEKO calculated resonance is not accurate and other means are needed to calculate the resonance freq. The resonant freq of the attached dimensions has been checked.
Are these internal dimensions, or external dimensions?
What is the metal thickness?
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#1762 by M.LeBel on 15 Jan, 2017 16:39
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Steady state means that the total energy in the cavity isn't increasing or decreasing, it doesn't mean that energy isn't flowing in and out. And it doesn't mean that the fields aren't periodically fluctuating. It just means past the fill time for the cavity, which should be very short.
The B field for a TE mode is given by Egan's equation of B = (1/ω) curl E = (1/ω) [(1/r) Rα(r) Q1(θ) er + Rα1(r) Q'(θ) eθ ] (see his page for proper formatting).
Thanks! Can you also tell us what is the function Br(r,θ,φ), for a TE013 mode, at all points in the frustum, including the antenna feed, so it can be integrated? THAT is the part I don't know, not the integrals. Eagan's equations do not include a source antenna term, and then he hand-waves the answer. As such, he just refers to the solutions as being periodic and therefore, they cancel out over 1 full cycle. That's not very useful unless the equation includes the source terms from the antenna. I don't have that either.
I have done the integral for a simple TM field (not a mode) in a cone, and shown that for a perfect conductor, the field pressure on the inside of the cone SUMs to zero. No problem! But I don't have a source term, or any losses in that integral, and it's not a real mode shape. If I "assume" the surface resistance losses for the applied B field are also uniform, then that integral will SUM to zero as well. But what about when it's not uniform, such as the side walls are made of thick copper, the big end is made of very thin copper on FR4 board which is much hotter, and the small end has a dielectric that keeps it cool? Then what is the Br(r,θ,φ) field to integrate and the surface resistance on all surfaces? I have no idea!
You can neglect the antenna, because it shouldn't have any effect in steady state if it is ideally designed and positioned, and if the antenna does affect the fields, its effect should be on the order of 1/Q compared to the rest of the fields.
I am not sure what handwaving you are referring to from Egan, but it sounds like you are referring to his general proof of no force, which does not involve any handwaving or have any relevance.
I don't understand why you are asking the questions about the heating not being uniform. Egan shows that the heating isn't uniform, and the parts about if the ends are cooled, or have different heat capacities is for you to answer with your theory. (And your theory being unclear on these things is related to my main issue with it.)
The highlighted statement above shows that you do not yet understand my theory. There is no "steady state", steady state means no thrust. The location of the antenna, the location of the losses, the power that is radiated and what is reflected back or absorbed and dissipated on each cycle has everything to do with it. I can't neglect those things and claim this is a representation of my theory because it isn't. The stored and dissipated energy must be increasing or decreasing. Anything "except" steady state.
Warptech: Without a steady-state, the system is “out of control”. In this state the system is going through wide fluctuations maybe “scanning” at time the specific condition that creates thrust. This widely fluctuating state will provide a higher probability of thrust, but appearing only as a minimal side effect. IMO you are looking for a steady-state on a specific asymmetric condition the detail of which is still unknown. ...
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#1763 by TheTraveller on 15 Jan, 2017 17:01
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KISS Thruster update:
Latest dimension as attached.
Will probably be using either the Yonlit 10W or 25W Rf amp http://www.yonlit.com/Products.asp?ClassID=7 with the KISS Thruster as it combines the:
1) pre amp
2) attenuator
3) power amp
4) forward and reflected power monitor
into 1 package.
Result are 3 electronic modules: Freq Gen, Rf amp & Freq Tracker instead of 6. KISS = fewer components.
Waiting on pricing quote to decide on which amp to run with.
Another added benefit is the experimenter will be able to choose a Yonlit 5W to 120W Rf amp package and have the same DB15 interface to the freq tracker module.
Please note that in spherical end plate thrusters, the effective resonant length is the frustum height (axial length) and NOT the constant separation between the spherical end plates. Which means the FEKO calculated resonance is not accurate and other means are needed to calculate the resonance freq. The resonant freq of the attached dimensions has been checked.
Are these internal dimensions, or external dimensions?
What is the metal thickness?
Internal dimensions.
Final thickness to be determined. -
#1764 by TheTraveller on 15 Jan, 2017 17:07
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KISS Thruster update:
Latest dimension as attached.
Please note that in spherical end plate thrusters, the effective resonant length is the frustum height (axial length) and NOT the constant separation between the spherical end plates. Which means the FEKO calculated resonance is not accurate and other means are needed to calculate the resonance freq. The resonant freq of the attached dimensions has been checked.
If I use the dimensions on your image, I do not get TE013 at 2.45Ghz.
FEKO does not prefer frustum height over constant separation between spherical endplates. I'm not sure where you get that from as it was in close agreement with EW's measured TM212 resonance.
You changed a number of the formulas I entered to create the spherical end-plate geometry. This is probably where you are having difficulty. Those formulas, and some of the variables, were derived in another application using very specific geometric parameters. Unless you know how those were set up, randomly changing them is likely to yield unusual results.
I didn't randomly change the numbers and yes I know how the FEKO numbers relate to the thruster dimensions.
FEKO doesn't solve for resonance based on travelling wave equations, which is why it fails to properly show cutoff and resonance.
The dimensions will resonate as indicated. It is my money paying for the tooling.
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#1765 by TheTraveller on 15 Jan, 2017 17:13
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a two-dimensional model is better than a three-dimensional model for a real 3D truncated cone?
Time will tell. If the KISS Thruster resonates in TE013 at 2.45GHz, then yes the 2D travelling wave model is superior to the 3D Maxwell model. -
#1766 by WarpTech on 15 Jan, 2017 17:17
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...
Steady state means that the total energy in the cavity isn't increasing or decreasing, it doesn't mean that energy isn't flowing in and out. And it doesn't mean that the fields aren't periodically fluctuating. It just means past the fill time for the cavity, which should be very short.
Warptech: Without a steady-state, the system is “out of control”. In this state the system is going through wide fluctuations maybe “scanning” at time the specific condition that creates thrust. This widely fluctuating state will provide a higher probability of thrust, but appearing only as a minimal side effect. IMO you are looking for a steady-state on a specific asymmetric condition the detail of which is still unknown. ...
I understand what is meant by steady state. Currently, my "theory" is not complete. I can find both types of solutions, but which one can provide thrust "greater than a photon rocket", I don't know. I admit, while I think the equations I have do represent physical reality, I do not fully understand the equations of motion, or the behaviors predicted by them. I have questions regarding non-intuitive things, where I do not confidently know the answer.
For example: If we have a closed box and inside the box we have a uniform gravito-electric field vector. Will the box move in that direction or are they in equilibrium?
It is easy for me to show now, that the CM will move if we do as Woodward says, "push when it's heavy and pull back when it's light", and the capacitor goes from heavy to light by discharging as heat. The same is true with energy stored in an asymmetrical cavity. Unfortunately, this does not provide thrust greater than a photon rocket, as far as I can tell. My best "guess" is based on the fact that most of the data for the EmDrive show the force is less than that of a photon rocket with power Pin*Q, where Q is the quality factor. If this is the case, then in a pulsed mode, where the cavity energy is either charging or discharging, It can give the illusion of providing more thrust than a photon rocket of power Pin, but it can't do it in steady state!
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#1767 by meberbs on 15 Jan, 2017 17:17
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KISS Thruster update:
Latest dimension as attached.
Please note that in spherical end plate thrusters, the effective resonant length is the frustum height (axial length) and NOT the constant separation between the spherical end plates. Which means the FEKO calculated resonance is not accurate and other means are needed to calculate the resonance freq. The resonant freq of the attached dimensions has been checked.
If I use the dimensions on your image, I do not get TE013 at 2.45Ghz.
FEKO does not prefer frustum height over constant separation between spherical endplates. I'm not sure where you get that from as it was in close agreement with EW's measured TM212 resonance.
You changed a number of the formulas I entered to create the spherical end-plate geometry. This is probably where you are having difficulty. Those formulas, and some of the variables, were derived in another application using very specific geometric parameters. Unless you know how those were set up, randomly changing them is likely to yield unusual results.
I didn't randomly change the numbers and yes I know how the FEKO numbers relate to the thruster dimensions.
FEKO doesn't solve for resonance based on travelling wave equations, which is why it fails to properly show cutoff and resonance.
The dimensions will resonate as indicated. It is my money paying for the tooling.
"FEKO doesn't solve for resonance based on travelling wave equations" is false.
FEKO uses Maxwell's equations which inherently include travelling wave equations. Showing off your ignorance, and ignoring the advice of others is not a wise move. -
#1768 by sanman on 15 Jan, 2017 17:28
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Has anybody ever heard of this:
https://en.wikipedia.org/wiki/Aharonov%E2%80%93Bohm_effect
When Paul March made an analogy of the device pushing on the Virtual Particles of the Vacuum, some critics said it's not possible because the Virtual Particles are all randomly and transiently fluctuating in and out of existence, so there's nothing to really push off of.
But the effect described in the link above says it's possible to interact with something even if its wavefunction is zero (fluctuated out of existence), by coupling with the complex component of its wavefunction:QuoteThe Aharonov–Bohm effect, sometimes called the Ehrenberg–Siday–Aharonov–Bohm effect, is a quantum mechanical phenomenon in which an electrically charged particle is affected by an electromagnetic potential (V, A), despite being confined to a region in which both the magnetic field B and electric field E are zero.[1] The underlying mechanism is the coupling of the electromagnetic potential with the complex phase of a charged particle's wave function, and the Aharonov–Bohm effect is accordingly illustrated by interference experiments.
Is it possible that something similar is happening inside the frustrum?
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#1769 by TheTraveller on 15 Jan, 2017 17:48
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FEKO uses Maxwell's equations which inherently include travelling wave equations. Showing off your ignorance, and ignoring the advice of others is not a wise move.
FEKO calcs the KISS Thruster TE013 resonance incorrectly, so apparently Maxwell fails.
Please do and post whatever analysis you desire and show or not resonance as claimed. Then when the KISS Thrusters arrive, I'll post the real world resonance and we'll see if Roger or Maxwell are correct in how to calc resonance in a tapered waveguide with resonant spherical end caps. -
#1770 by X_RaY on 15 Jan, 2017 18:18
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FEKO uses Maxwell's equations which inherently include travelling wave equations. Showing off your ignorance, and ignoring the advice of others is not a wise move.
FEKO calcs the KISS Thruster TE013 resonance incorrectly, so apparently Maxwell fails.
Please do and post whatever analysis you desire and show or not resonance as claimed. Then when the KISS Thrusters arrive, I'll post the real world resonance and we'll see if Roger or Maxwell are correct in how to calc resonance in a tapered waveguide with resonant spherical end caps.
My last few cavity based sensors with more complicated shapes are working perfect (within K-Band) as calculated via FEA.
For the CNC machine model I have increased the length by 100µm (for tuning reasons), what do you think how much I had to decrease the length of the delivered part to hit the focussed resonant frequency?
Yes allmost exactly 100µm!
Maybe there is somthing wrong with your spreadsheet calculations(or better approximation)?
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#1771 by WarpTech on 15 Jan, 2017 18:19
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Has anybody ever heard of this:
https://en.wikipedia.org/wiki/Aharonov%E2%80%93Bohm_effect
When Paul March made an analogy of the device pushing on the Virtual Particles of the Vacuum, some critics said it's not possible because the Virtual Particles are all randomly and transiently fluctuating in and out of existence, so there's nothing to really push off of.
But the effect described in the link above says it's possible to interact with something even if its wavefunction is zero (fluctuated out of existence), by coupling with the complex component of its wavefunction:QuoteThe Aharonov–Bohm effect, sometimes called the Ehrenberg–Siday–Aharonov–Bohm effect, is a quantum mechanical phenomenon in which an electrically charged particle is affected by an electromagnetic potential (V, A), despite being confined to a region in which both the magnetic field B and electric field E are zero.[1] The underlying mechanism is the coupling of the electromagnetic potential with the complex phase of a charged particle's wave function, and the Aharonov–Bohm effect is accordingly illustrated by interference experiments.
Is it possible that something similar is happening inside the frustrum?
The Aharonov-Bohm effect is the quantum mechanical version of the Lorentz force. See Felsager - Geometry, Particles and Fields for a well written chapter on this subject. In QED, it is the gauge potential Au that is quantized. Where, AuAu is the probability density of photons, and Au is A(w), a function of frequency. The fields, E and B are the derivatives of the quantized field, A(w).
Edit: it is a good example where the derivatives of the field don't exist, but the field potential itself, demonstrably does exist. It is also worth noting that Au, can be "gauge" transformed by the gradient in a scalar function without altering the Lorentz force. That function has units of magnetic flux, as I've put in my equations. In this regard, in QED predicts that there are an infinite number of possible solutions for the "inertial" vacuum field, that are all equivalent "except" for a scaling factor proportional to the photon density. See for starters, Milonni - The Quantum Vacuum, chapter 2 and Appendix B. -
#1772 by TheTraveller on 15 Jan, 2017 18:24
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FEKO uses Maxwell's equations which inherently include travelling wave equations. Showing off your ignorance, and ignoring the advice of others is not a wise move.
FEKO calcs the KISS Thruster TE013 resonance incorrectly, so apparently Maxwell fails.
Please do and post whatever analysis you desire and show or not resonance as claimed. Then when the KISS Thrusters arrive, I'll post the real world resonance and we'll see if Roger or Maxwell are correct in how to calc resonance in a tapered waveguide with resonant spherical end caps.
My last few cavity based sensors with more complicated shapes work perfect (K-Band).
For the CNC machine model I have increased the length by 100µm (for tuning reasons), what do you think how much I had to decrease the length of the delivered part to hit the focussed resonance frequency?
Yes allmost exact 100µm!
Maybe there is somthing wrong with your spreadsheet calculations?
You built frustum shaped resonator cavities with spherical end plates using FEKO? If so please share the dimensions, mode and resonant freq so I can check. -
#1773 by X_RaY on 15 Jan, 2017 18:33
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You don't really believe that I would share any dimensions or whatever was developed at my daily work?!FEKO uses Maxwell's equations which inherently include travelling wave equations. Showing off your ignorance, and ignoring the advice of others is not a wise move.
FEKO calcs the KISS Thruster TE013 resonance incorrectly, so apparently Maxwell fails.
Please do and post whatever analysis you desire and show or not resonance as claimed. Then when the KISS Thrusters arrive, I'll post the real world resonance and we'll see if Roger or Maxwell are correct in how to calc resonance in a tapered waveguide with resonant spherical end caps.
My last few cavity based sensors with more complicated shapes work perfect (K-Band).
For the CNC machine model I have increased the length by 100µm (for tuning reasons), what do you think how much I had to decrease the length of the delivered part to hit the focussed resonance frequency?
Yes allmost exact 100µm!
Maybe there is somthing wrong with your spreadsheet calculations?
You built frustum shaped resonator cavities with spherical end plates using FEKO? If so please share the dimensions, mode and resonant freq so I can check.
My point is that many microwave engineers all over the world use this FEA programs for one reason:
It works very well.
Therefore please show the world in detail where Maxwell fails, instead of coming up with such statements. -
#1774 by meberbs on 15 Jan, 2017 19:01
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What I forgot to mention in my previous posts was that at the top of Egan's page, he has the equation B(t) = B cos(ωt) . This time dependence is simple and independent of spatial location, so the rest of the page just deals with the magnitude. you can add the time dependence back in at the end....
Steady state means that the total energy in the cavity isn't increasing or decreasing, it doesn't mean that energy isn't flowing in and out. And it doesn't mean that the fields aren't periodically fluctuating. It just means past the fill time for the cavity, which should be very short.
Warptech: Without a steady-state, the system is “out of control”. In this state the system is going through wide fluctuations maybe “scanning” at time the specific condition that creates thrust. This widely fluctuating state will provide a higher probability of thrust, but appearing only as a minimal side effect. IMO you are looking for a steady-state on a specific asymmetric condition the detail of which is still unknown. ...
I understand what is meant by steady state. Currently, my "theory" is not complete. I can find both types of solutions, but which one can provide thrust "greater than a photon rocket", I don't know. I admit, while I think the equations I have do represent physical reality, I do not fully understand the equations of motion, or the behaviors predicted by them. I have questions regarding non-intuitive things, where I do not confidently know the answer.
For example: If we have a closed box and inside the box we have a uniform gravito-electric field vector. Will the box move in that direction or are they in equilibrium?
It is easy for me to show now, that the CM will move if we do as Woodward says, "push when it's heavy and pull back when it's light", and the capacitor goes from heavy to light by discharging as heat. The same is true with energy stored in an asymmetrical cavity. Unfortunately, this does not provide thrust greater than a photon rocket, as far as I can tell. My best "guess" is based on the fact that most of the data for the EmDrive show the force is less than that of a photon rocket with power Pin*Q, where Q is the quality factor. If this is the case, then in a pulsed mode, where the cavity energy is either charging or discharging, It can give the illusion of providing more thrust than a photon rocket of power Pin, but it can't do it in steady state!
The energy density equations (that feed into the dissipation) Egan shows the time dependence explicitly, because E and B are out of phase.
The E and B fields here should enable you to plug back in to your equations that you called the Lorentz gauge and show whether using the magnetic flux as the parameter to adjust the gauge actually results in a gauge that meets the Lorentz condition. -
#1775 by Monomorphic on 15 Jan, 2017 19:05
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I didn't randomly change the numbers and yes I know how the FEKO numbers relate to the thruster dimensions.
The dimensions will resonate as indicated. It is my money paying for the tooling.
Your dimensions appear dissimilar to Shawyer's conical frustums. It's too long and the cone angle is different. In the below image I have superimposed your 2.45Ghz TE013 frustum over Shawyer's 3.8Ghz TE013 frustum. Notice how Shawyer's small end is completely different from your design? Perhaps that is an indication that you have misinterpreted the "bread crumbs" Shawyer has given you. You have proposed longer frustums for some time now. I don't understand why since this diverges from Shawyer's known thrusters.
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#1776 by WarpTech on 15 Jan, 2017 20:01
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...
What I forgot to mention in my previous posts was that at the top of Egan's page, he has the equation B(t) = B cos(ωt) . This time dependence is simple and independent of spatial location, so the rest of the page just deals with the magnitude. you can add the time dependence back in at the end.
The energy density equations (that feed into the dissipation) Egan shows the time dependence explicitly, because E and B are out of phase.
The E and B fields here should enable you to plug back in to your equations that you called the Lorentz gauge and show whether using the magnetic flux as the parameter to adjust the gauge actually results in a gauge that meets the Lorentz condition.
That's actually quite easy, since Eagan doesn't have the scalar function term in his equations. His equation for "E" is below. Using the identity that,
E = -dA/dt,
Then, since these are harmonic fields, I use the same trick Eagan used,
A = -E/ω ~ curl(B)
Then, div(A) = 0 since div(curl) = 0 is a vector identity.
Since Eagan's solution has no E fields perpendicular to the walls, there is no charge accumulation and no voltage potential. So dφ/dt = 0,
Hence "his" solution meets the Lorentz gauge condition. Therefore, adding a gauge transformation that is the 4-gradient of a scalar function, with units of magnetic flux, cannot change this, because the derivatives of the gauge transformation terms will cancel out, per the definition of a gauge transformation. NOW, I can say that my solution meets this requirement.
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#1777 by WarpTech on 15 Jan, 2017 20:14
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I didn't randomly change the numbers and yes I know how the FEKO numbers relate to the thruster dimensions.
The dimensions will resonate as indicated. It is my money paying for the tooling.
Your dimensions appear dissimilar to Shawyer's conical frustums. It's too long and the cone angle is different. In the below image I have superimposed your 2.45Ghz TE013 frustum over Shawyer's 3.8Ghz TE013 frustum. Notice how Shawyer's small end is completely different from your design? Perhaps that is an indication that you have misinterpreted the "bread crumbs" Shawyer has given you. You have proposed longer frustums for some time now. I don't understand why since this diverges from Shawyer's known thrusters.
It is also counter to Shawyer's DF, which increases with a shorter, larger Big diameter frustum.
It is also counter to @Notsosureofit's theory, and my own which requires a shorter length and a larger cone angle.
TT's design is closer to McCulloch's theory, by having a longer length and smaller cone angle.
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#1778 by meberbs on 15 Jan, 2017 20:31
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After the gauge transformation div(A) is no longer 0, and V is not 0, you have to do the math, your attempt at a shortcut fails....
What I forgot to mention in my previous posts was that at the top of Egan's page, he has the equation B(t) = B cos(ωt) . This time dependence is simple and independent of spatial location, so the rest of the page just deals with the magnitude. you can add the time dependence back in at the end.
The energy density equations (that feed into the dissipation) Egan shows the time dependence explicitly, because E and B are out of phase.
The E and B fields here should enable you to plug back in to your equations that you called the Lorentz gauge and show whether using the magnetic flux as the parameter to adjust the gauge actually results in a gauge that meets the Lorentz condition.
That's actually quite easy, since Eagan doesn't have the scalar function term in his equations. His equation for "E" is below. Using the identity that,
E = -dA/dt,
Then, since these are harmonic fields, I use the same trick Eagan used,
A = -E/ω ~ curl(B)
Then, div(A) = 0 since div(curl) = 0 is a vector identity.
Since Eagan's solution has no E fields perpendicular to the walls, there is no charge accumulation and no voltage potential. So dφ/dt = 0,
Hence "his" solution meets the Lorentz gauge condition. Therefore, adding a gauge transformation that is the 4-gradient of a scalar function, with units of magnetic flux, cannot change this, because the derivatives of the gauge transformation terms will cancel out, per the definition of a gauge transformation. NOW, I can say that my solution meets this requirement.
You are misunderstanding something greatly if you think that Egan's results meet the Lorentz gauge condition. Egan does not use or define any potentials, so it is nonsensical to say what gauge his solution is in. By the definition of a gauge transformation, performing a gauge transformation means that you are no longer necessarily in the same gauge. This is why it is called a gauge transformation. -
#1779 by WarpTech on 15 Jan, 2017 22:46
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After the gauge transformation div(A) is no longer 0, and V is not 0, you have to do the math, your attempt at a shortcut fails....
What I forgot to mention in my previous posts was that at the top of Egan's page, he has the equation B(t) = B cos(ωt) . This time dependence is simple and independent of spatial location, so the rest of the page just deals with the magnitude. you can add the time dependence back in at the end.
The energy density equations (that feed into the dissipation) Egan shows the time dependence explicitly, because E and B are out of phase.
The E and B fields here should enable you to plug back in to your equations that you called the Lorentz gauge and show whether using the magnetic flux as the parameter to adjust the gauge actually results in a gauge that meets the Lorentz condition.
That's actually quite easy, since Eagan doesn't have the scalar function term in his equations. His equation for "E" is below. Using the identity that,
E = -dA/dt,
Then, since these are harmonic fields, I use the same trick Eagan used,
A = -E/ω ~ curl(B)
Then, div(A) = 0 since div(curl) = 0 is a vector identity.
Since Eagan's solution has no E fields perpendicular to the walls, there is no charge accumulation and no voltage potential. So dφ/dt = 0,
Hence "his" solution meets the Lorentz gauge condition. Therefore, adding a gauge transformation that is the 4-gradient of a scalar function, with units of magnetic flux, cannot change this, because the derivatives of the gauge transformation terms will cancel out, per the definition of a gauge transformation. NOW, I can say that my solution meets this requirement.
You are misunderstanding something greatly if you think that Egan's results meet the Lorentz gauge condition. Egan does not use or define any potentials, so it is nonsensical to say what gauge his solution is in. By the definition of a gauge transformation, performing a gauge transformation means that you are no longer necessarily in the same gauge. This is why it is called a gauge transformation.
YOU keep trying to distinguish between a field of magnetic flux and the scalar function used in the gauge transformation, and are asking me to prove that distinction. There is none! There is no physical test that can reveal the flux, "other than" my hypothesis that gradients in this field are gravitational potentials. I can't prove that, because Maxwell's equations, E & B are unaffected by this transformation, just as they are unaffected by gravitational fields.
See page 9, Felsager - Geometry, Particles and Fields, below. This is the mathematics that says Eagan's solution is unaffected by a gauge transformation.
Edit: The highlighted statement above is an incorrect assumption. The gauge is an "arbitrary choice", which has no effect what so ever on the solutions E and B. So, I can choose whatever gauge I want. In this case, I simply took his harmonic function for "E(ω)" and integrated it to get "A(ω)", in the same way he integrated curl(B) to get "E(ω)". I showed that this vector field indeed meets the Lorentz gauge requirement. We are always free to make a gauge transformation of this field without affecting the physics!
