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#80 by X_RaY on 12 Apr, 2016 19:49
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I have not forgotten this calculation, but have to much other things to do at the moment.If I have the time I could check this using EMPro within the next week's. Using the eigen resonance solver the resonant frequency and the Q will be displayed directly for each mode.
It would be great if you check this with EMPro. It seems to me that to maximize Q one wants to maximize
For now I only have the results of the spreadsheet with the approximation formulas.
∫ElectromagneticEnergy dV/ ∫ ElectromagneticEnergy dA
this means minimizing ∫ ElectromagneticEnergy dA
while maximizing
∫ElectromagneticEnergy dV
this means a mode shape that would have most of the high electromagnetic energy in the interior volume of the cavity instead of the exterior of the cavity near the metal
I will check it if I have time while future calculations at our local university. -
#81 by keithpickering on 22 Apr, 2016 22:31
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Longtime lurker, seldom poster.
I've been quite interested in the fact that the NASA experiments, which show more thrust than any theory yet predicts, also has a cavity only partially filled with dielectric material. Not fully, but partially, at the small end only. That leads to the following thoughts, based on the McCullogh theory.
One thing about a dielectric is that it will have an index of refraction n substantially above 1, which implies that the photons within the dielectric move slower than c by some amount. McCulloghs's theory implies that photons have mass (or perhaps virtual mass may be a better term) that is modulated by Unruh radiation, and that during resonance, the photons gain mass while moving within the big end, and lose mass while moving within the small end. The presence of a partial dielectric indicates that in the NASA experiment, the photons would necessarily be moving faster in the big end (while more massive) and slower in the small end (while less massive). If in fact photons do have mass, this is exactly what we would expect from the relativistic change in v caused by the dielectric. In other words, the "dielectric effect" would enhance the effect from the shape of the frustum alone. So far, so good for McCullogh.
If this is true, this also suggests that perhaps the optimum dielectric would be a layered one, starting with a layer of very high n at the small endplate, followed by another of slightly lower n, and another of slightly lower n, until the entire cavity is filled (with perhaps the final layer being air or vacuum, next to the big endplate). Thus the photons would be in states of continuously differential v during the entire traverse of the cavity.
One potential issue with that approach is that any photons not perfectly axial (or parallel to the axis) would refract in odd ways at the interfaces, possibly increasing the number of wall collisions and hence degrading Q.
But this could be avoided if the sides of the frustum were not flat, but rather flaring, like the bell of a trumpet. By computing the angle of refraction at every interface, (which depends on the differing values of n) it would be possible to build a frustum with a layered dielectric and a flared bell, so that a photon could travel along the wall of the cavity, not touching, as it passed through successive refractions in the dielectric. If the endplates were also spherical instead of flat, perfect reflection would allow the photons to retrace their paths exactly.
Just a thought. -
#82 by Rodal on 30 Apr, 2016 21:48
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The relationship between radiation PRESSURE, ENERGY DENSITY and LAGRANGIAN DENSITY
In several EM Drive anomalous force discussions there is confusion as the relationship of the radiation pressure to the electromagnetic fields and a lack of appreciation of the dependence of the radiation pressure to the energy density. We will address this:
1) ENERGY DENSITY
The electromagnetic energy density is defined as:
where
E is the electric field;
D is the electric displacement field;
B is the magnetic flux density;
H is the magnetic field
For linear, nondispersive and isotropic(for simplicity) materials, the constitutive relations can be written as
,
where
ε is the electric permittivity of the material;
μ is the magnetic permeability of the material
therefore
u = (ε/2)E·E + (1/(2μ))B·B
we will write, as short nomenclature for the dot product of a vector with itself, and where the E and B fields are out of phase by 90 degrees, and have monochromatic (single frequency) dependence on angular frequency ω :
E2 = E·E
=(Er2+Eθ2+Ez2)((Sin(ωt))2 (cylindrical coordinates)
=(Eθ2+Eφ2+Er2)((Sin(ωt))2 (spherical coordinates)
B2 = B·B
=(Br2+Bθ2+Bz2)((Cos(ωt))2 (cylindrical coordinates)
=(Bθ2+Bφ2+Br2)((Cos(ωt))2 (spherical coordinates)
where
((Sin(ωt))2 =½(1- Cos(2ωt))
((Cos(ωt))2=½(1+ Cos(2ωt))
u = ½ (εE2 + (1/μ)B2)
Therefore the energy density varies with time around its cyclic average with a frequency 2ω that is twice the frequency ω of the electromagnetic fields E and B.
The "microscopic" version of Maxwell's equations admits only the fundamental fields E and B, without a built-in model of material media. Only the vacuum permittivity and permeability are used, and there is no D or H
recalling the identity between the electric permittivty, magnetic permeability and the speed of light:
(all terms in this identitity: c, εo and μo are Lorentz invariants: they have the same value for observers in different moving reference frames)
we obtain:
u = (εo/2) (E2 + c2B2)
2) UNITS of PRESSURE and of ENERGY DENSITY
Notice that the units of Energy Density are energy per unit volume Energy/Volume or (Force*Length)/Volume, while the units of pressure are force per unit area Force/Area. Since Volume = Area * Length, it is obvious therefore that pressure and energy density have exactly the same units. This is not a coincidence, as we will show.
3) MAXWELL STRESS TENSOR
Maxwell's stress tensor components, for arbitrary orientation of the unit cube in an ortogonal coordinate system, is defined as:
or, equivalently, in terms of the energy density:
σij = εo EiEj + (1/μo) BiBj - u δij
the pressures are the diagonal Maxwell tensor components, given by
σii = εo (Ei)2+ (1/μo) (Bi)2 - u
(this is an important equation which we will be using in further derivations below)
σii =- u + εo ( (Ei)2+ c2 (Bi)2 )
PRINCIPAL VALUES of stress
At every point in a stressed body there are at least three planes, called principal planes, with normal vectors called principal directions, where there are zero shear stresses (the stress matrix off-diagonal components are zero). The three stresses normal to these principal planes are called principal stresses (they are the only non-zero components in the matrix: the diagonal components).
Every second rank tensor (such as the stress tensor) has three independent quantities, which are invariant under rotation, associated with it. One set of such rotational invariants are the principal stresses of the stress tensor, which are just the eigenvalues of the stress tensor. Their direction vectors are the principal directions or eigenvectors.
The eigenvalues are the roots of the Cayley–Hamilton theorem. The principal stresses are unique for a given stress tensor. A coordinate system with axes oriented to the principal directions implies that the normal stresses are the principal stresses and the stress tensor is represented by a diagonal matrix:
The three eigenvalues (principal values) of Maxwell stress tensor are:
{σ1, σ2,σ3} =
The first term under brackets, and the two terms inside the radical are invariant under rotation(*). The first rotationally invariant term is the energy density: the addition of E⋅E and c2B⋅B. The energy density is invariant under rotations but it is not invariant, in general, under Lorentz transformations (it can change from one frame of reference to another frame of reference in special relativity, and therefore Sommerfeld states that the energy density has no meaning independent of a specific frame of reference). The second invariant, the difference between E⋅E and c2B⋅B, is called the "Lagrange density" by Arnold Sommerfeld in his masterwork "Electrodynamics" for good reason: since it can be shown that this expression is the Lagrangian density of the electromagnetic fields in vacuum (see: http://bado-shanai.net/Map%20of%20Physics/mopEMLagrangianDensity.htm ). It is a scalar that is invariant under a Lorentz transformation (this scalar does not change from one frame of reference to another frame of reference in special relativity) as well as invariant under rotations. The final rotationally invariant term, which is also a Lorentz invariant, is the dot product of the electric vector field with the magnetic vector field: E⋅B, divided by the square of the impedance of the vacuum:
To bring the stress tensor to diagonal form, one must rotate the reference axes to a reference system in which the vectors E and B (at a given point in space and at a given time) are parallel to each other or where one of them is equal to zero. Such a transformation is always possible except when both conditions occur: a) E and B are mutually perpendicular (E⋅B=0) and b) when the Lagrange density is zero, such that E⋅E = c2B⋅B. Both these quantities are Lorentz invariants, so when the electromagnetic fields are mutually perpendicular and equal to each other, they are so in any coordinate system under any Lorentz transformation, and we see that two of the three eigenvalues are zero in that case because the expression under the radical sign is zero.
Comparing the first principal stress with the previous expression for energy density, one readily identifies that the first principal stress is the negative of the energy density (an invariant under rotations but not under Lorentz transformations):
σ1 = - u
another invariant, a Lorentz invariant and rotation invariant: the "Lagrangian Density", the first term inside the square root of the double-valued terms can be simplified by noticing that
- u + εo E2 = (εo E2- (1/μo) B2)/2
therefore:
σ2 =(+u - εo E2)√(1+(((εo/μo)E·B)/(- u + εo E2))2)
σ3 =(- u + εo E2)√(1+(((εo/μo)E·B)/(- u + εo E2))2)
When the dot product (also called scalar product) of the electric and vector fields is zero, that is when
E·B = 0
then one gets the simple relations between the principal stresses and the energy density:
σ1 = - u
σ2 = +u - εo E2
σ3 = - u + εo E2
Notice that, in general, only two of the three principal stresses (σ2 and σ3) are Lorentz invariants, while the first principal stress (σ1), is not, in general, a Lorentz invariant.
Although, in general, the value of u is different for observers in different moving reference frames, the value of
+/- (u - εo E2) is the same for observers in different moving reference frames.
The "volumetric or hydrostatic stress" is the summation of all three principal stresses acting on the unit of volume:
σ1 + σ2 + σ3 = - u
and it is therefore equal to the negative of the electromagnetic energy density.
Also notice that, since the charge density is εo E the second term in the last two expressions is the Coulomb pressure "p", the charge density times the electric field
p=εo E2,
(the Coulomb tensile force per unit area), so we can write:
σ1 = - u
σ2 = + u - p
σ3 = - u + p
In words: when the dot product of the electric and vector fields is zero, the principal stresses are just a function of the energy density and the Coulomb pressure.
The Coulomb pressure p=εo E2 is twice the so-called electrostatic pressure ½ εo E2, which arises from
σ3 = - u + p
= -(εo/2) (E2 + c2B2)+ εo E2
= (εo/2) (E2 - c2B2)
The quantity in brackets: E2 - c2B2 is a "Lorentz" invariant of Maxwell's stress tensor (see p. 378 Chapter 21 of Classical Electricity and Magnetism by Panofsky and Phillips)
Which for zero magnetic field B = 0 gives the electrostatic pressure as:
σ3 = (εo/2) E2
For zero magnetic field B = 0, and a unidirectional electric field E (with no transverse components) the principal axes are oriented such that σ3 is parallel to the direction of E while the axes of σ1 and σ2 are perpendicular to E. The electric field transmits a tension σ3 = (εo/2) E2 in the direction of the E field and transverse pressure of equal absolute magnitude σ1 =σ2 =- (εo/2) E2, in the directions perpendicular to the E field.
In the general case for non-zero E and B, one of the principal stresses is the negative of the energy density, while the other two stresses are the difference between the energy density and the Coulomb pressure, the first one with a plus sign and the second one with a minus sign.
Also notice that in general, under all situations, two of the principal stresses are compressive, while the other principal stress is tensile. The cross-section experiencing a tensile principal stress and an equal compressive principal stress in the perpendicular direction experiences a state of pure shear: whereby a square will deform into a rectangle and a circle will deform into an ellipse:
If the energy density is greater than the Coulomb pressure, then the tensile stress is the second principal stress.
If the Coulomb pressure is greater than the energy density, then the tensile stress is the third principal stress.
When the Coulomb pressure is negligibly small when compared to the magnetic field times the square of the speed of light c , the principal stresses are very simply due to the internal energy: two of the principal stresses are compressive: equal to the negative of the energy density, and the third principal stress is tensile: equal to the energy density.
This explains, for example, when Dr. H. White discusses his theory of the Quantum Vacuum as an explanation for the EM Drive he discusses the importance of the energy density, and the fact that the stress is tensile in one principal direction and compressive in the other two orthogonal directions.
4) TRANSVERSE ELECTRIC (TE) RESONANCE OF A TRUNCATED CONE WITH SPHERICAL ENDS
In order to solve for the stresses in an EM Drive with spherical ends, we’ll need to work in a coordinate system in which the cavity walls have a simple description. I will use the same coordinate system as used by Greg Egan (credit to Egan for this image showing the coordinate system) and several textbooks to solve electromagnetic resonance problems in conical cavities.
Resonant Modes of a Conical Cavity
by Greg Egan
http://gregegan.customer.netspace.net.au/SCIENCE/Cavity/Cavity.html
The truncated spherical cone is positioned so that its longitudinal axis of axi-symmetry lies on the z-axis of our spherical coordinate system, its would-be apex lies at the origin, and in spherical coordinates (r, θ, φ) — where r is the distance from the origin, θ is the polar angle from the z-axis, and φ is the azimuthal angle from the x–z plane — the side walls of the cone will be defined by θ =+/- θw, its narrow end will be defined by r = r1, and its wide end will be defined by r = r2.
BOUNDARY CONDITIONS AT SPHERICAL ENDS
The boundary conditions for the electromagnetic fields at the small end (defined by r = r1), and at the big end (defined by r = r2) are:
Er = 0 because the longitudinal electric field is zero in a TE mode
Eθ = 0 because electric fields parallel to a conductive surface must be zero
Eφ = 0 because electric fields parallel to a conductive surface must be zero
Therefore the electric vector field at both spherical ends, for a TE mode is identically zero:
E = 0
Br = 0 because magnetic fields perpendicular to a conductive surface must be zero
Bφ = 0 because the azimuthal (transverse) magnetic field is zero in a TE mode
Therefore the only electromagnetic field that is non-zero at both spherical ends, for a TE mode, is the magnetic field in the polar direction Bθ. This is a general result, that holds for any and all TE mode shapes.
BOUNDARY CONDITIONS AT CONICAL WALLS
The boundary conditions for the electromagnetic fields at the conical side walls (defined by θ = +/-θw) are:
Er = 0 because the longitudinal electric field is zero in a TE mode
Eθ = 0 because the polar electric field is zero for TEmnp mode shapes with m=0
Eφ = 0 because electric fields parallel to a conductive surface must be zero
Therefore the electric vector field at the conical side walls, for a TEmnp mode with m=0, is
E = 0
Bθ = 0 because magnetic fields perpendicular to a conductive surface must be zero
Bφ = 0 because the azimuthal (transverse) magnetic field is zero in a TE mode
Therefore the only electromagnetic field that is non-zero at the conical side walls, for a TEmnp mode with m=0, is the magnetic field in the radial direction Br.
STRESS AT SPHERICAL ENDS
Since as previously shown, the boundary conditions at the spherical ends are such that there is only one non-zero electromagnetic field component, then:
*the shear stress components are zero, and hence the stress field is a principal stress at the ends
Also the dot product of the electric and magnetic fields is zero: E·B=0
Also, since the electric vector field is zero at the spherical ends:
E=0
it immediately follows that, for all (TE) Transverse Electric mode shapes:
* Coulomb's pressure and the electrostatic pressure are zero at the spherical ends (p=0)
* the stress at the spherical ends is compressive and entirely due to the energy density
* the stress is entirely due to the magnetic component in the polar direction, parallel to the end plate
σ3 = - u
σ3 =- (εo/2) ( c2B2)
σrr = - (1/(2μo)) (Bθ Cos(ωt))2
= - (1/(4μo)) (Bθ)2(1+ Cos(2ωt))
Thus the stress varies from a minimum value of zero to its maximum compressive value, never reversing sign, at a frequency which is twice as high as the frequency of the electromagnetic fields.
STRESS AT CONICAL WALLS
Since as previously shown, the boundary conditions at the conical side walls are such that there is only one non-zero electromagnetic field component, then:
*the shear stress components are zero, and hence the stress field is a principal stress at the conical side walls
Also the dot product of the electric and magnetic fields is zero: E·B=0
Also, since the electric vector field is zero at the conical side walls, for TEmnp mode shapes with m=0:
E=0
it follows that, for TEmnp mode shapes with m=0:
* Coulomb's pressure and the electrostatic pressure are zero at the conical side walls (p=0)
* the stress at the conical walls is compressive and entirely due to the energy density
* the stress is entirely due to the magnetic component in the longitudinal direction, parallel to the conical side walls
σ3 = - u
σ3 =- (εo/2) ( c2B2)
σθθ = - (1/(2μo)) (Br Cos(ωt))2
= - (1/(4μo)) (Br)2(1+ Cos(2ωt))
Thus the stress varies from a minimum value of zero to its maximum compressive value, never reversing sign, at a frequency which is twice as high as the frequency of the electromagnetic fields.
STRESS AT ALL INTERNAL SURFACES FOR TE0np MODES
* all shear stress components are zero, and hence the stress field is a principal stress
* the stress is compressive
* the stress varies from zero to its maximum compressive value, never reversing sign, at a frequency which is twice as high as the frequency of the electromagnetic fields.
* the electric vector field is zero at all internal surfaces
* Coulomb's pressure and the electrostatic pressure are zero
* the stress is entirely due to the energy density
* the stress is entirely due to the magnetic component parallel to the surface
The above applies to the mode shapes used in the EM Drive experiments that have claimed the highest force/InputPower.
Prof. Yang has used TE012 mode shapes in her experimental claims.
Shawyer has used mode shape TE012 in his Demonstrator experimental claim and reportedly used mode shape TE013 (according to NSF user TheTraveller) in his Boeing Flight Demonstrator experimental claim.
NASA's reported experiment with the highest force/InputPower has involved mode shape TE012.
Shawyer's hypothesis that there is no pressure on the side walls is entirely falsified. Shawyer is wrong: there is pressure, and hence a force component on the side walls of a truncated conical cavity with spherical ends: this pressure is entirely due to the magnetic field component parallel to the wall. This pressure has nothing to do with the electric field components.
5) TRANSVERSE MAGNETIC (TM) RESONANCE OF A TRUNCATED CONE WITH SPHERICAL ENDS
BOUNDARY CONDITIONS AT SPHERICAL ENDS
The boundary conditions for the electromagnetic fields at the small end (defined by r = r1), and at the big end (defined by r = r2) are:
Eθ = 0 because electric fields parallel to a conductive surface must be zero
Eφ = 0 because electric fields parallel to a conductive surface must be zero and because TM mode
Br = 0 because magnetic fields perpendicular to a conductive surface must be zero and because TM mode
Bθ = 0 because the polar magnetic field is zero for TMmnp mode shapes with m=0
Therefore the only electromagnetic fields that are non-zero at both spherical ends, for TMmnp mode shapes with m=0, are
* the electric field in the longitudinal direction Er perpendicular to the surface
* the magnetic field in the azimuthal ("transverse") direction Bφ parallel to the surface
BOUNDARY CONDITIONS AT CONICAL WALLS
The boundary conditions for the electromagnetic fields at the conical side walls (defined by θ =+/- θw) are:
Er = 0 because electric fields parallel to a conductive surface must be zero
Eφ = 0 because the azimuthal (transverse) electric field is zero in a TM mode
Br = 0 because the longitudinal magnetic field is zero in a TM mode
Bθ = 0 because magnetic fields perpendicular to a conductive surface must be zero
Therefore the only electromagnetic fields that are non-zero at the conical side walls are:
* the electric field in the polar direction Eθ perpendicular to the surface
* the magnetic field in the azimuthal ("transverse") direction Bφ parallel to the surface
STRESS AT SPHERICAL ENDS
Since as previously shown, the boundary conditions at the spherical ends are such that there is only one non-zero electric field component and only one non-zero magnetic field component, then:
*the shear stress components are zero, and hence the stress field is a principal stress at the ends
Also the dot product of the electric and magnetic fields is zero: E·B=0 because the non-zero electric field component has a zero magnetic field component in the same direction, and vice-versa.
Therefore the state of stress is a principal stress composed simply of the energy density and Coulomb's pressure:
σ3 = - u + p
= -(εo/2) (E2 + c2B2)+ εo E2
using the aforementioned boundary conditions and assuming a standing wave solution such that the electromagnetic fields E and B are 90 degrees out of phase, one obtains:
σrr = (εo/2) ((Er Sin(ωt))2 - c2(Bφ Cos(ωt))2)
= (εo/4) ( ((Er)2 - c2(Bφ )2) - ((Er)2 + c2(Bφ )2)Cos(2ωt)) )
The cyclic-average stress is compressive if the magnetic field in the azimuthal (transverse) direction, times the speed of light, is greater than the electric field perpendicular to the surface, and tensile otherwise:
compressive for c Bφ > Er
zero for c Bφ = Er (zero cyclic average)
tensile for c Bφ < Er
The stress is zero at the very center of the spherical ends, corresponding to the intersection with the axis of axi-symmetry.
Calculations show that tension occurs only in regions that are near the center (closer to the axis of axi-symmetry), while compression occurs further away from the center: further away from the axis of axi-symmetry. The occurrence of tension is also dependent on the truncated cone geometry.
The higher "p" is in mode shapes TM0np, the smaller the amplitude of tension and the larger the region over which there is compression, and the larger the amplitude of compression.
The absolute value of the compressive stress maximum amplitude is always larger than the one for the tensile stress, even for the lowest order mode shape. In other words, the integral of the stress distribution: the force, is compressive, because compression takes place over a larger area, and it has greater absolute amplitude.
STRESS AT CONICAL WALLS
Since as previously shown, the boundary conditions at the conical side walls are such that there is only one non-zero electric field component and only one non-zero magnetic field component, then:
*the shear stress components are zero, and hence the stress field is a principal stress at the conical side walls
Also the dot product of the electric and magnetic fields is zero: E·B=0 because the non-zero electric field component has a zero magnetic field component in the same direction, and vice-versa.
Therefore the state of stress is a principal stress composed simply of the energy density and Coulomb's pressure:
σ3 = - u + p
= -(εo/2) (E2 + c2B2)+ εo E2
using the aforementioned boundary conditions and assuming a standing wave solution such that the electromagnetic fields E and B are 90 degrees out of phase, one obtains:
σθθ = (εo/2) ((Eθ Sin(ωt))2 - c2(Bφ Cos(ωt))2)
= (εo/4) ( ((Eθ)2 - c2(Bφ )2) - ((Eθ)2 + c2(Bφ )2)Cos(2ωt)) )
The cyclic-average stress is compressive if the magnetic field in the azimuthal (transverse) direction, times the speed of light, is greater than the electric field perpendicular to the surface, and tensile otherwise:
compressive for c Bφ > Eθ
zero for c Bφ = Eθ (zero cyclic average)
tensile for c Bφ < Eθ
The higher "p" is in mode shapes TM0np the larger the area over which the stress is tensile, and the higher the amplitude of the tensile stress. However, the amplitude of the compressive stress is always larger than the amplitude of the tensile stress. The integral of the stress distribution: the force, is compressive, because compression takes place over a larger area, and it has greater absolute amplitude.
Following is an image from Greg Egan (cited above) showing the stress integrated over the circumference, and thus showing the force per unit length distribution along the length of the conical side walls. The red curve corresponds to TM011, the green curve corresponds to TM012 and the blue curve corresponds to TM013:
STRESS AT ALL INTERNAL SURFACES FOR TM0np MODES
* all shear stress components are zero, and hence the stress field is a principal stress
* the stress is due to two components: a compressive component due to the magnetic energy density and a tensile component due to the electrostatic pressure
* the compressive stress is entirely due to the magnetic component parallel to the surface, in the transverse (azimuthal) direction
* the tensile stress component is entirely due to the electric field component perpendicular to the surface
* the stress integrated over the surface, the force, is compressive , in other words, the magnetic energy density effect predominates over the electrostatic pressure. The stress can have small tensile regions, where the electrostatic pressure exceeds the magnetic energy density.
Again, Shawyer's hypothesis that there is no pressure on the side walls is entirely falsified. Shawyer is wrong: there is pressure, and hence a force component on the side walls of a truncated conical cavity with spherical ends.
___________________________________
NOTES:
(*) Given the two vector fields E and B, the only way to form rotational invariants is to form dot products, which gives E⋅E, B⋅B, and E⋅B . One can do any arithmetic operations with them that one likes and still get a rotational invariant, although it's not guaranteed to be a Lorentz invariant. It is not guaranteed to be a Lorentz invariant because a magnetic field in one moving frame may be seen as an electric field in a different moving frame, and vice-versa (since E and B are not Lorentz invariant quantities). One observer’s E field is another’s B field (or a mixture of the two), as viewed from different moving reference frames. To form Lorentz invariants one has to be able to express them as a linear combination of the inner products of the field strength tensor or its dual (https://en.wikipedia.org/wiki/Electromagnetic_tensor) with themselves, or between themselves:
Also see: https://en.wikipedia.org/wiki/Classification_of_electromagnetic_fields#Invariants , https://en.wikipedia.org/wiki/Electromagnetic_tensor#Properties and https://en.wikipedia.org/wiki/Lorentz_scalar .
(**) Mode shape nomenclature is adopted as per the cylindrical cavity (with constant circular cross section) designation, because there is no standardized way to number truncated cone mode shapes. I am aware that there is no mode shape for a truncated cone with electromagnetic fields constant in the longitudinal direction, unlike cylindrical cavities which have TM mode shapes with "p=0". Still, because the truncated cone geometries used up to now have shapes that are not too far from a cylinder with constant cross section (because small cone angles are used and the cones are truncated far from the cone vertex) it is possible to use a cylindrical cavity mode shape designation and select m,n,p accordingly. -
#83 by Rodal on 06 May, 2016 12:19
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POYNTING VECTOR
The Poynting vector, in the macroscopic version of Maxwell's equations of Abraham is defined as the cross product of the electric field vector E with the magnetic field vector H :
SA= E x H
while in the Mynkowski version of Maxwell's equations it is defined as:
SM= c2 D x B
In the microscopic version of Maxwell's equations one can use the constitutive equation relating H and B, or relating D and E , to obtain the same expression for both the Abraham and the Mynkowski versions of Maxwell's equations:
where μ0 is the magnetic permeability of the vacuum.
The Poynting vector is a useful expression, because the electromagnetic momentum density in the vacuum is given by the Poynting vector divided by the square of the speed of light in vacuum, c2:
Or, recalling that
pem = εo ExB
Unfortunately, the Poynting vector S is not, in general. a Lorentz invariant. Recall that the energy density u is not, in general, a Lorentz invariant either. However, if one subtracts the dot product of the Poynting vector with itself from the square of the energy density , one obtains the following scalar, which is a Lorentz invariant:
u2 - S · S/c2
Although the value of u and S may be different for observers in different moving reference frames, the value of
u2 - S · S/c2 is the same for observers in different moving reference frames.
(this fact is not as commonly known. I have not found it in Wikipedia, but it can be found here: p.82 of Introduction to Tensor Calculus, Relativity and Cosmology, 3rd ed. Edition, D. F. Lawden, or one can prove it by inspection)
TRAVELLING WAVES
An electromagnetic plane wave is an idealization that, although not practical, is important as the simplest solution to Maxwells' equations, and therefore often used in the literature to explore the fundamental physics of a problem.
Such a "monochromatic" (=single-frequency) electromagnetic plane wave is a one-dimensional wave traveling in the z direction with a phase velocity (the rate at which the phase of the wave propagates in space) equal to the speed of light in vacuum: vp=λ/T=ω/k=c (since the index of refraction of vacuum is n=1, vp=c/n=c). Such a plane electromagnetic wave represents a purely transverse vibration without any longitudinal electromagnetic component along the z axis of propagation. In this restricted sense, it is like the transverse vibration of a stretched string.
It has the electric field E and the magnetic field B oriented perpendicular to each other, in the transverse direction to the direction of propagation. In a single electromagnetic plane wave (rather than a standing wave which can be described as two such waves travelling in opposite directions), E and H are exactly in phase.
Recall that the boundary conditions at the metallic spherical ends of the EM Drive require that the electric fields parallel to a conductive surface be zero. Since the electric field in a plane wave is perpendicular to the direction of travel of the plane wave, this means that the electric field of the plane wave is parallel to the surface and therefore that it must be zero at the metal surface. Recall that plane wave is a travelling wave, so this means that the ideal plane wave cannot really satisfy the boundary conditions at the surface of a conductor, because, for a travelling wave, the amplitude of the fields must vary sinusoidally with time at any specified location, therefore the plane travelling wave cannot satisfy the boundary conditions of a conductive wall:
The fact that a plane electromagnetic wave travelling normal to a conductive surface cannot satisfy the boundary conditions at the conductive wall is not just a problem for this idealized plane wave. It is a problem for any travelling wave that has an electric field perpendicular to the direction of propagation. Therefore, this problem also exists for travelling waves in waveguides: they are only valid solutions for waveguides that do not have conductive walls perpendicular to the direction of propagation. The following image from Wikipedia shows the electric field component perpendicular to the direction of propagagation for mode shape TE31 inside a hollow metal waveguide with rectangular cross-section (observe how the electric field at the end of the waveguide oscillates like a sinusoid and therefore one cannot impose the condition that the electric field should be zero, constant with time, at a conductive surface):
STANDING WAVES
A travelling wave with electric field perpendicular to the direction of propagation cannot satisfy the boundary condition that the electric field parallel to the surface of a conductive material must be zero. To satisfy the boundary condition one needs a standing wave: a standing wave results from the constructive and destructive interference of two counter propagating travelling waves. In the following image (from acs.psu.edu) observe how two counter-propagating travelling waves can create a node at which the electromagnetic field is zero:
Therefore, a standing wave makes it possible to insert a conductor (a conductive wall perpendicular to the standing wave) at any of the nodes where the tangential electric field is zero without changing the structure of the electric field!
Therefore we arrive at the conclusion that Shawyer statement "there are travelling waves and standing waves in the EM Drive" and therefore that self-acceleration of the EM Drive can be explained somehow by single travelling waves in an open waveguide is incorrect, because only standing waves can meet the boundary conditions at the metallic surfaces: resulting from the interference of two travelling waves propagating in opposite directions. This applies to sinusoidal standing waves (as in a cavity with constant cylindrical cross-section) as well as to spherical Bessel function standing waves (as in a truncated cone with spherical ends), or any other kind of standing waves. Single travelling waves cannot meet the boundary conditions. In order to meet the boundary conditions for a problem with conducting walls, standing waves are necessary.
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#84 by Rodal on 10 May, 2016 17:40
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The relationship between radiation PRESSURE and the POYNTING VECTOR
A common misconception is that the radiation pressure can always be obtained from the Poynting vector simply by the cyclic average of the Poynting vector divided by the speed of light:
In a "monochromatic"(=single frequency) plane electromagnetic travelling wave, the Poynting vector points in the direction of propagation while oscillating in magnitude at twice the frequency of the travelling wave. The time-averaged magnitude of the Poynting vector is:
<S> = ½(1/μo)Re([Em] x [Bm*])
where Em is the peak-amplitude of the (complex) electric field Em eiωt, and Bm* is the complex-conjugate peak-amplitude of the magnetic field Bm eiωt, which, in a plane wave are exactly in phase with each other. If the electromagnetic fields are described in terms of their root mean square (rms) values then the factor of ½ should be replaced by a factor of 1.
The problem with using the cyclic average of the Poynting vector divided by the speed of light to obtain the radiation pressure is that the electric field parallel to a conductive surface must be zero, and that the electric field in a plane electromagnetic wave propagating normal to a conductive surface is parallel to the surface, and hence it should be zero. Hence at the surface of the conductor, E=0, therefore Em = 0 and therefore the Poynting vector must be zero to satisfy the boundary conditions at the conductive surface: S=0 and <S>=0. There cannot be real power transmitted by a planar electromagnetic wave into a conductive surface.
One arrives at the conclusion that either
1) the electric field component parallel to the wall must be zero, hence the electromagnetic plane wave has zero amplitude electric field everywhere. Therefore its Poynting vector is zero and the radiation pressure is zero at the wall
or
2) the electromagnetic wave satisfying the boundary condition cannot be an electromagnetic plane wave, as somehow the electromagnetic wave's electric field has to decay to zero to match the boundary conditions at the conductive wall.
Now, you may say: this is similar to the mechanical momentum of a ball hitting a perfectly reflecting surface, it makes sense that the momentum at the wall should be zero, because the velocity of the ball at the surface is zero. What happens at the wall is the interference of two momenta: the momentum of the ball hitting the wall interfering with the opposite momentum of the ball bouncing from the wall. Similarly, in the electromagnetic case of a wave hitting a conductive surface one gets a standing wave, that results from the interference of a travelling wave propagating against the wall and bouncing back, (counter-propagating) from the wall. You may say OK, I agree with that, but why can't I take the momentum of the planar electromagnetic field far away from the wall, in the far-field, its Poynting vector far away, where the electric field has a well-defined sinusoidal value and let's don't worry too much about the details of what really happens to my travelling plane wave as it hits the conductive surface and has to satisfy the boundary condition. I think that this is what many introductory texts may be assuming (actually if one looks at many introductory texts that discuss using the time-average of the Poynting vector as a measure of radiation pressure they seldom discuss a conductor: instead they usually just say: an absorbing surface or a perfectly-reflecting surface).
A single travelling plane electromagnetic wave cannot exist inside a resonant cavity: it cannot meet the boundary conditions at the metal surfaces. The solution for a conical waveguide comprises spherical propagating waves. The solution for a truncated conical cavity comprises spherical standing waves.
Even considering a cavity with constant cross-section, the standing wave solution responsible for transverse electric (TE) or transverse magnetic (TM) modes varies sinusoidally in the lengthwise direction. In a truncated cone the standing wave solution varies like a spherical Bessel function in the lengthwise direction of the cavity, such that the wavelength becomes longer as one approaches the small end of the cone:
The relationship between radiation PRESSURE and the ENERGY DENSITY
There is a more general approach to calculate the radiation pressure, which instead of considering the Poynting vector, considers the pressure as being due to the energy density, as for example in the following discussion by Richard Fitzpatrick, Professor of Physics at The University of Texas at Austin: http://farside.ph.utexas.edu/teaching/em/lectures/node90.html
In this more general approach, the radiation pressure is obtained from the cyclic time-average of the energy density:
Pradiation = <u>
Such an analysis is consistent with the derivation in https://forum.nasaspaceflight.com/index.php?topic=39214.msg1526577#msg1526577 that the radiation pressure, for the case in which the Coulomb pressure is zero, is really due to the energy density.
As we have shown, one cannot simply calculate the radiation pressure in a resonant cavity using the cyclic average of the Poynting vector divided by the speed of light:
since 1) the Poynting vector is zero at the metal walls for TE modes and 2) the Poynting vector does not have a constant magnitude in the longitudinal direction for standing waves in a cavity, for any mode shape. On the other hand, as we have shown, one can certainly calculate the radiation pressure at the walls of resonant cavity based on the energy density value at the wall for TE modes, or in general, based on the energy density and the Coulomb pressure at the wall for any mode shape, hence obtaining the radiation pressure from the energy density is a more general approach than obtaining it from the Poynting vector field. -
#85 by X_RaY on 18 May, 2016 19:06
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http://forum.nasaspaceflight.com/index.php?topic=39214.msg1508105#msg1508105
I have not forgotten this calculation, but have to much other things to do at the moment.If I have the time I could check this using EMPro within the next week's. Using the eigen resonance solver the resonant frequency and the Q will be displayed directly for each mode.
It would be great if you check this with EMPro. It seems to me that to maximize Q one wants to maximize
For now I only have the results of the spreadsheet with the approximation formulas.
∫ElectromagneticEnergy dV/ ∫ ElectromagneticEnergy dA
this means minimizing ∫ ElectromagneticEnergy dA
while maximizing
∫ElectromagneticEnergy dV
this means a mode shape that would have most of the high electromagnetic energy in the interior volume of the cavity instead of the exterior of the cavity near the metal
I will check it if I have time while future calculations at our local university.
http://forum.nasaspaceflight.com/index.php?topic=39214.msg1508376#msg1508376
I got something to share on the question.
Remember, the Basic idea was:
Due to the linearity of the Maxwell equations the mode number of a given mode shape impacts on the resulting Q. In fact, the higher the mode number the higher the Q as long as the cavity dimensions remains constant. In this case we want to focus on the "p" value (of TXmnp*). The mode of discussion is TE01p.
We did calculations using a spreadsheet/exact solution to calculate the Q and the resonant frequency for this mode family for low modenumbers.
Now this calculations has been proved/verified using Keysight-EMPro.
I could follow the TE01 mode for the Brady-cone dimensions (Comsol calculation by Frank Davies, NASA) up to TE016.
I found the Q increase linear with the index "p" for this mode (of course the higher the modenumber the higher the resonant frequency).
For higher values (TE017,TE018,...) the modes was either not present due to degeneration or I simply couldn´t find it. The higher the Resonant frequency the lower the frequency distance to other modes and maybe its unlikely to excite this higher "p" value modes because in relation to the wavelength at higher frequencies the diameters are to big. Its more easy to satisfy lowest energy conditions of higher order modes like TE02p or TE03p for diameters much larger than half the wavelength.
There was a TE0 mode near the calculated TE017 frequency (shown below). Again I couldn't find higher than TE016 due to whatever.
Look at the TE015 mode and its location of the maximum energy density
*Using mode notation for a cylindrical cavity because the cone half angle θ is small and there is no convention for the notation of modes inside of a truncated conical cavity resonator.
Notice the datafiles below the pics.
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#86 by Monomorphic on 18 May, 2016 19:43
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I found the Q increase linear with the index "p" for this mode (of course the higher the modenumber the higher the resonant frequency).
Looks like about a 10.4% increase in Q for each higher mode number. Nice!
Increasing the size of the frustum lowers the resonant frequency. Just build a bigger frustum...
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#87 by X_RaY on 18 May, 2016 20:50
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Yes this was proved before(in this thread). Now we know** the Q also increase with the mode index.I found the Q increase linear with the index "p" for this mode (of course the higher the modenumber the higher the resonant frequency).
Looks like about a 10.4% increase in Q for each higher mode number. Nice!
Increasing the size of the frustum lowers the resonant frequency. Just build a bigger frustum...
That was the thought and I liked to satisfy since a quarter year or so, using professional software without the restrictions of feko lite.
**At least the prediction holds using the Keysight-EMPro FEM algorithm. -
#88 by X_RaY on 19 May, 2016 18:30
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Just picked the TE015 frequency and tried to find it using FEKO Lite and got it during the first run. The Result (even for the coarse mesh in feko lite) is almost as good as the EMPro calculations.
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#89 by X_RaY on 26 May, 2016 20:46
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The relationship between radiation PRESSURE and the POYNTING VECTOR
Is it possible that thrust generation is based on the magnetic field vectors (rotation vector of the standing wave{curl_H}) acting at the sidewall(s) instead of radiation pressure at the end pates??
A common misconception is that the radiation pressure can always be obtained from the Poynting vector simply by the cyclic average of the Poynting vector divided by the speed of light:
In a "monochromatic"(=single frequency) plane electromagnetic travelling wave, the Poynting vector points in the direction of propagation while oscillating in magnitude at twice the frequency of the travelling wave. The time-averaged magnitude of the Poynting vector is:
<S> = ½(1/μo)Re([Em] x [Bm*])
where Em is the peak-amplitude of the (complex) electric field Em eiωt, and Bm* is the complex-conjugate peak-amplitude of the magnetic field Bm eiωt, which, in a plane wave are exactly in phase with each other. If the electromagnetic fields are described in terms of their root mean square (rms) values then the factor of ½ should be replaced by a factor of 1.
The problem with using the cyclic average of the Poynting vector divided by the speed of light to obtain the radiation pressure is that the electric field parallel to a conductive surface must be zero, and that the electric field in a plane electromagnetic wave propagating normal to a conductive surface is parallel to the surface, and hence it should be zero. Hence at the surface of the conductor, E=0, therefore Em = 0 and therefore the Poynting vector must be zero to satisfy the boundary conditions at the conductive surface: S=0 and <S>=0. There cannot be real power transmitted by a planar electromagnetic wave into a conductive surface.
One arrives at the conclusion that either
1) the electric field component parallel to the wall must be zero, hence the electromagnetic plane wave has zero amplitude electric field everywhere. Therefore its Poynting vector is zero and the radiation pressure is zero at the wall
or
2) the electromagnetic wave satisfying the boundary condition cannot be an electromagnetic plane wave, as somehow the electromagnetic wave's electric field has to decay to zero to match the boundary conditions at the conductive wall.
Now, you may say: this is similar to the mechanical momentum of a ball hitting a perfectly reflecting surface, it makes sense that the momentum at the wall should be zero, because the velocity of the ball at the surface is zero. What happens at the wall is the interference of two momenta: the momentum of the ball hitting the wall interfering with the opposite momentum of the ball bouncing from the wall. Similarly, in the electromagnetic case of a wave hitting a conductive surface one gets a standing wave, that results from the interference of a travelling wave propagating against the wall and bouncing back, (counter-propagating) from the wall. You may say OK, I agree with that, but why can't I take the momentum of the planar electromagnetic field far away from the wall, in the far-field, its Poynting vector far away, where the electric field has a well-defined sinusoidal value and let's don't worry too much about the details of what really happens to my travelling plane wave as it hits the conductive surface and has to satisfy the boundary condition. I think that this is what many introductory texts may be assuming (actually if one looks at many introductory texts that discuss using the time-average of the Poynting vector as a measure of radiation pressure they seldom discuss a conductor: instead they usually just say: an absorbing surface or a perfectly-reflecting surface).
A single travelling plane electromagnetic wave cannot exist inside a resonant cavity: it cannot meet the boundary conditions at the metal surfaces. The solution for a conical waveguide comprises spherical propagating waves. The solution for a truncated conical cavity comprises spherical standing waves.
Even considering a cavity with constant cross-section, the standing wave solution responsible for transverse electric (TE) or transverse magnetic (TM) modes varies sinusoidally in the lengthwise direction. In a truncated cone the standing wave solution varies like a spherical Bessel function in the lengthwise direction of the cavity, such that the wavelength becomes longer as one approaches the small end of the cone:
The relationship between radiation PRESSURE and the ENERGY DENSITY
There is a more general approach to calculate the radiation pressure, which instead of considering the Poynting vector, considers the pressure as being due to the energy density, as for example in the following discussion by Richard Fitzpatrick, Professor of Physics at The University of Texas at Austin: http://farside.ph.utexas.edu/teaching/em/lectures/node90.html
In this better approach, the radiation pressure is obtained from the cyclic time-average of the energy density:
Pradiation = <u>
Such an analysis is consistent with the derivation in https://forum.nasaspaceflight.com/index.php?topic=39214.msg1526577#msg1526577 that the radiation pressure, for the case in which the Coulomb pressure is zero, is really due to the energy density.
As we have shown, one cannot simply calculate the radiation pressure in a resonant cavity using the cyclic average of the Poynting vector divided by the speed of light:
since 1) the Poynting vector is zero at the metal walls for TE modes and 2) the Poynting vector does not have a constant magnitude in the longitudinal direction for standing waves in a cavity, for any mode shape. On the other hand, as we have shown, one can certainly calculate the radiation pressure at the walls of resonant cavity based on the energy density value at the wall for TE modes, or in general, based on the energy density and the Coulomb pressure at the wall for any mode shape, hence obtaining the radiation pressure from the energy density is a more general approach than obtaining it from the Poynting vector field.
I mean unbalanced force of eddy currents in the two opposite directions and therefore different values of reactive forces along these directions. -
#90 by Rodal on 26 May, 2016 20:56
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Is it possible that thrust generation is based on the magnetic field vectors (rotation vector of the standing wave{curl_H}) acting at the sidewall(s) instead of radiation pressure at the end pates?
In here https://forum.nasaspaceflight.com/index.php?topic=39214.msg1526577#msg1526577:
I show that for TE modes the stress at the conical walls is compressive and entirely due to the energy density, due to the magnetic component in the longitudinal direction, parallel to the conical side walls. Therefore if thrust is real for mode shapes TE012 and TE013 used by Shawyer, it must be due to the magnetic field vector parallel to either the conical walls, and/or the magnetic field vector parallel to the end plates. -
#91 by VAXHeadroom on 29 May, 2016 01:19
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Is it possible that thrust generation is based on the magnetic field vectors (rotation vector of the standing wave{curl_H}) acting at the sidewall(s) instead of radiation pressure at the end pates?
In here https://forum.nasaspaceflight.com/index.php?topic=39214.msg1526577#msg1526577:
I show that for TE modes the stress at the conical walls is compressive and entirely due to the energy density, due to the magnetic component in the longitudinal direction, parallel to the conical side walls. Therefore if thrust is real for mode shapes TE012 and TE013 used by Shawyer, it must be due to the magnetic field vector parallel to either the conical walls, and/or the magnetic field vector parallel to the end plates.
In the meep simulation which I have subsequently animated (December 2015), the magnetic fields are most definitely NOT parallel to the small end or the side sloped walls. They ARE parallel to the big end plate.
Original Post: http://forum.nasaspaceflight.com/index.php?topic=39004.msg1467619#msg1467619
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#92 by Rodal on 29 May, 2016 01:57
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Is it possible that thrust generation is based on the magnetic field vectors (rotation vector of the standing wave{curl_H}) acting at the sidewall(s) instead of radiation pressure at the end pates?
In here https://forum.nasaspaceflight.com/index.php?topic=39214.msg1526577#msg1526577:
I show that for TE modes the stress at the conical walls is compressive and entirely due to the energy density, due to the magnetic component in the longitudinal direction, parallel to the conical side walls. Therefore if thrust is real for mode shapes TE012 and TE013 used by Shawyer, it must be due to the magnetic field vector parallel to either the conical walls, and/or the magnetic field vector parallel to the end plates.
In the meep simulation which I have subsequently animated (December 2015), the magnetic fields are most definitely NOT parallel to the small end or the side sloped walls. They ARE parallel to the big end plate.
Original Post: http://forum.nasaspaceflight.com/index.php?topic=39004.msg1467619#msg1467619
The reason for that is because you are solving Meep using Cartesian coordinates and you have the end plates parallel to one of the Cartesian axes while you have the conical walls not oriented parallel to any of the Cartesian axes. This is a bad choice of coordinates and a bad choice for display of the results. In the exact solution of the problem one uses instead spherical coordinates such that the conical walls are aligned along the spherical radius, while the polar angle coordinate is perpendicular to it. This leads to a clear satisfaction of the boundary conditions. The situation for Meep is made worse by the fact that Meep uses a Finite Difference discretization approach and it is not based on a variational principle. I have told this to aero and Shell a long time ago. The situation was so bad that it was almost impossible for aero to tell what mode shape was being excited in Meep (due to the poor choice in what was being displayed). The choice of images is very poor and has lead to misinterpretation of the fields by NSF Meep users. NSF Meep users would be well advised to at least do what FEKO, COMSOL and other Finite Element packages do standard out of the box: instead of plotting the fields in Cartesian axes components, they plot the norm (https://en.wikipedia.org/wiki/Norm_(mathematics)) of the vector field (which is invariant under rotations). This leads to a much better physical understanding of what is going on.
You should be plotting the vector fields with their vector magnitude and vector orientation, in order to get a physical feeling for what is going on.
This is a good example of what a good plot looks like (FEKO) (this is showing the electric vector field: observe how FEKO shows it to be perpendicular to the conical walls, correctly satisfying the boundary conditions):
Also X-Ray has been showing good simulations using FEKO. And previously and concurrently, X_Ray has been also showing great simulation with EM Pro. I was also showing simulations using the exact solution, with the fields in spherical coordinates, and also plotting vector fields. All of these much better than the Meep displays of fields in Cartesian coordinates which lead to rampant confusion. (The human brain is not very good at vector decomposition particularly when the electromagnetic fields are displayed as colored contours in Cartesian axes that are not oriented along the walls).
Another huge problem with NSF Meep displays has been that the electromagnetic fields have been displayed in colors without any numerical scheme to ascertain the numerical significance of the fields. This is bad practice, FEKO and COMSOL always give you the numerical values corresponding to the contour fields being plotted.
Meep was intended to be used NOT as a black box like FEKO or COMSOL but it was intended as an open source code where students could write their own code to solve particular problems.
In any case, the boundary conditions are what they are, they can be found in standard introductory textbooks and not at all a subject of controversy. The boundary condition dictates that the magnetic field perpendicular to the conductive wall should be zero and the only non-zero component of the magnetic field should be the component parallel to the conductive wall. When you have a wall that is oriented at angle to the Cartesian axes, then the boundary conditions act on both Cartesian components. To satisfy the Boundary Condition at a wall at an angle to the Cartesian axes you naturally have to have non-zero component along both Cartesian directions, in general (vector decomposition). You should be plotting the vector fields with their vector magnitude and vector orientation, in order to get a physical feeling for what is going on. -
#93 by Rodal on 29 May, 2016 14:54
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Further to this issue about Boundary Conditions, Jackson has an excellent discussion in his masterpiece Classical Electrodynamics, 3rd Edition, pages 352 to 356 (ISBN-10: 047130932X; ISBN-13: 978-0471309321). Pay special attention to the graph on page 355, Fig. 8.2 "Fields near the surface of a good, but not perfect conductor". For a good but not perfect conductor, for example copper as used in the EM Drive, and as modeled in Meep with the Drude equation model, a very tiny magnitude electric field parallel to the surface will be present, as well as a very tiny magnetic field perpendicular to the surface. The electric field parallel to the surface is inversely proportional to the square root of the conductivity:
This solution exhibits the expected rapid exponential decay (skin depth), and Pi/4 phase difference. For a good conductor, the fields inside the metal conductor are parallel to the surface and propagate normal to it, with magnitudes that depend only on the tangential magnetic field parallel to the surface, that exists just outside the surface of the metal conductor.
As shown in the graphs in Jackson's book, and as one can readily calculate these fields are practically zero, insignificant for copper, due to its very high conductivity. Therefore it is not a surprise that Meep does not show a significant difference in the fields calculated using the perfect conductor model vs. the Drude model. The main influence of the finite conductivity Drude model is to allow the calculation of a finite Q (instead of an infinite Q). But again, the boundary condition is such that the electric field parallel to the surface and the magnetic field perpendicular to the surface must be practically zero at the surface, due to the very high conductivity of copper. This is particularly so for the experiments of EM Drive where experimenters seek a high Q, which is tantamount to practically zero fields for these variables. It is completely inconsistent for EM Drive experimenters to advocate a high Q (Shawyer even claiming to research superconducting EM Drives) and not realize that these boundary conditions are such that these fields must be practically zero at the surface of the good conductor.
NOTE: for those not having ready access to Jackson's monograph, the following discussion by a professor at Duke University is also good: https://www.phy.duke.edu/~rgb/Class/phy319/phy319/node59.html -
#94 by X_RaY on 29 May, 2016 15:42
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My biggest problem with the meep pics was that the single pics only represent single vector components (most of what we saw was autoscaled), so this has generated a lot of confusion.
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#95 by VAXHeadroom on 29 May, 2016 17:50
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My biggest problem with the meep pics was that the single pics only represent single vector components (most of what we saw was autoscaled), so this has generated a lot of confusion.
Just to clear up any possible confusion as to what the animation in that post shows:
http://forum.nasaspaceflight.com/index.php?topic=39004.msg1467619#msg1467619
The animation in the linked post is 3D vectors plotted with origin, direction, and amplitude, of the H field for two slices across the X plane (x=0, x=10). There is a min(always 0) and max (for scale) in numbers along with the frame number on the top left. All values are in 'meep units' which aero tells me should be multiplied by 3.33 to get to 'engineering units'.
The meep simulation is 291x390x327 elements (~37M elements) (on the order of 1/100 of the wave size), and ~3.2 degrees of phase per frame (112 frames within one waveform). The E fields are indeed parallel to the surfaces. The H fields are not. -
#96 by Rodal on 29 May, 2016 18:00
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Sorry to be blunt about this, but the following is my professional opinion from decades of experience with authoring and running Finite Difference and Finite Element methods. I am being blunt in order to be helpful and clarify the issues.My biggest problem with the meep pics was that the single pics only represent single vector components (most of what we saw was autoscaled), so this has generated a lot of confusion.
Just to clear up any possible confusion as to what the animation in that post shows:
http://forum.nasaspaceflight.com/index.php?topic=39004.msg1467619#msg1467619
The animation in the linked post is 3D vectors plotted with origin, direction, and amplitude, of the H field for two slices across the X plane (x=0, x=10). There is a min(always 0) and max (for scale) in numbers along with the frame number on the top left. All values are in 'meep units' which aero tells me should be multiplied by 3.33 to get to 'engineering units'.
The meep simulation is 291x390x327 elements (~37M elements) (on the order of 1/100 of the wave size), and ~3.2 degrees of phase per frame (112 frames within one waveform). The E fields are indeed parallel to the surfaces. The H fields are not.
That output is all wrong: it indicates something very bad going on in the modeling input and/or the postprocessing calculations leading to those bad results. The E fields cannot be parallel to the conductive surface as shown in any basic course on Electromagnetism. The H fields, on the other hand should BE parallel to the surface. You write that the model shows the opposite. Obviously there is something very wrong with the way that Meep calculations have been run at NSF EM Drive thread and / or with the calculations implemented to show those electromagnetic fields.
I suggested a long time ago to aero that he should run a comparison between his Meep simulations and an exact solution. I never saw any successful comparison of these NSF EM Drive Meep runs with any exact solution. On the contrary, the comparisons were far off the mark: he could not even match the natural frequency for simple problems. Instead of working on learning what has to be done to get a correct comparison with an exact solution, the emphasis appears to have been to model DIY builds, without prior validation that the modeling is being done correctly.
If so, this is not Meep's fault: rather it may be a modeling problem due to lack of previous experience and training with running Finite Difference computer models (so many things can go wrong in the modeling process): computer models need to be validated prior to proceeding to attempt to model actual experiments, as the computer paradigm is "Garbage In = Garbage Out". The first step in anybody attempting to learn how to run Finite Difference and Finite Element models should be first to run a comparison between a computer model and an exact solution, for example the exact solution of resonance in a cylindrical cavity. Then proceed to a comparison with Greg Egan's exact solution for a spherical truncated cone. Then proceed to a comparison with COMSOL's FE analysis for several mode shapes by Frank Davis at NASA. Otherwise one ends up with what you are discussing above: a model with a mesh having 291x390x327elements(~37Melements) (do you mean nodes ? there are no elements in Meep, since it is a Finite Difference program, not a Finite Element program) (*), and images where the NSF modeler cannot tell what mode shape has been excited and where the boundary conditions appear to not have been correctly satisfied. How is one going to get the correct solution to a system of partial differential equations if the boundary conditions are not correctly satisfied?
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(*) It is also important that computer modelers understand the software they are using: a Finite Difference code like Meep does not have elements with interpolating basis functions (usually polynomials) as in Finite Element analysis. Instead a Finite Difference code like Meep relies on finite differences at each node, to model the partial differential equations.
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#97 by X_RaY on 29 May, 2016 19:05
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I am sure you (and aero) put many work into these calculations as well as in the visualisation. I am sorry but Dr.Rodal is absolutely right about the related physical basics. The E-field vector close to a conductive wall can´t be parallel to that wall and so on.. Boundary conditions of electromagnetic fields are non-negotiableMy biggest problem with the meep pics was that the single pics only represent single vector components (most of what we saw was autoscaled), so this has generated a lot of confusion.
Just to clear up any possible confusion as to what the animation in that post shows:
http://forum.nasaspaceflight.com/index.php?topic=39004.msg1467619#msg1467619
The animation in the linked post is 3D vectors plotted with origin, direction, and amplitude, of the H field for two slices across the X plane (x=0, x=10). There is a min(always 0) and max (for scale) in numbers along with the frame number on the top left. All values are in 'meep units' which aero tells me should be multiplied by 3.33 to get to 'engineering units'.
The meep simulation is 291x390x327 elements (~37M elements) (on the order of 1/100 of the wave size), and ~3.2 degrees of phase per frame (112 frames within one waveform). The E fields are indeed parallel to the surfaces. The H fields are not.
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#98 by VAXHeadroom on 30 May, 2016 00:20
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I am sure you (and aero) put many work into these calculations as well as in the visualisation. I am sorry but Dr.Rodal is absolutely right about the related physical basics. The E-field vector close to a conductive wall can´t be parallel to that wall and so on.. Boundary conditions of electromagnetic fields are non-negotiableMy biggest problem with the meep pics was that the single pics only represent single vector components (most of what we saw was autoscaled), so this has generated a lot of confusion.
Just to clear up any possible confusion as to what the animation in that post shows:
http://forum.nasaspaceflight.com/index.php?topic=39004.msg1467619#msg1467619
The animation in the linked post is 3D vectors plotted with origin, direction, and amplitude, of the H field for two slices across the X plane (x=0, x=10). There is a min(always 0) and max (for scale) in numbers along with the frame number on the top left. All values are in 'meep units' which aero tells me should be multiplied by 3.33 to get to 'engineering units'.
The meep simulation is 291x390x327 elements (~37M elements) (on the order of 1/100 of the wave size), and ~3.2 degrees of phase per frame (112 frames within one waveform). The E fields are indeed parallel to the surfaces. The H fields are not.
Thank you both, back to the books
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#99 by Rodal on 30 May, 2016 15:01
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Professor Melcher at MIT (and previously Prof. Chu and others at MIT's Radiation Laboratory) was already showing decades ago in MIT classes that the Einstein-Laub formulation of electrodynamics is invalid since it yields a stress-energy-momentum tensor that is not frame invariant. (It is an interesting historical vignette that Einstein did not realize this at the time he wrote the paper with Laub)
The momentum of light in media remains one of the most controversial topics in physics [1–6]. The debate has continued for more than a century since Minkowski and Abraham formulated 4 × 4 energy-momentum tensors in the early 1900s [7–9].
https://www.researchgate.net/publication/292342272_Kinetic-energy-momentum_tensor_in_electrodynamics
They do credit Prof. Chu with the correct invariance relations. But that is one of the reasons why Prof. Chu at MIT developed his formulation.
This was known by students that listened to the lectures of Prof. Chu and Melcher at MIT
Reference: Prof. Melcher's masterpiece "Continuum Electromechanics" which he wrote in 1972-1973 while he was in sabatical at Cambridge University working with Sir Taylor and G. Batchelor
Correct, although The Einstein-Laub formulation of electrodynamics does work in a real world lab. 1973, Ashkin and Dziedzic performed a experiment in which they focused a green laser beam on the surface of water and saw what they called "the toothpaste tube effect" where a bulge appeared in the surface of the water. Using a Lorentz formula the existence of expansive and compressing forces effectively cancel out, negating a possible bump forming on the water surface. The real world lab test showed where pure theoretical extraction detailing out why there are thrusts from a asymmetrical cavity are lacking. . .
This is why lab data is king right now.
Shell
Back to work... will be on later. Have company helping me move my new frame for the lab.
Ashkin and Dziedzic used an argon-ion laser source to investigate the pressure on solid dielectric spheres immersed in liquid, and the pressure on a liquid-air interface owing to the passage of a beam of radiation.
Ashkin and Dziedzic's experiment cannot distinguish between the various stress tensors (Abraham, Minkowski, Einstein-Laub, etc.), since it gives no information about the direction of the local surface force.
Ashkin and Dziedzic observed the liquid interface to bend outwards in the liquid both when the light enters and leaves the liquid. This is contrary to the conservation law he derives from a Lagrangian formalism, and which predicts inward bending.
Loudon (see: Loudon, R., Radiation pressure and momentum in dielectrics. Fortschr. Phys., 2004. 52(11-12): p. 1134-1140) concluded that:Quote from: LoudonThe Ashkin and Dziedzic experiment observed an outward bulge on an illuminated water surface, apparently consistent with the sign of the Poynting surface force and the value of the Minkowski momentum. In agreement with Gordon, it is shown here that the effect is governed by a radial force and it provides no information on the longitudinal force associated with the linear momentum of light.
(bold added for emphasis)
One of the reasons why this Abraham Minkowski controversy persists is because of problems with the experiments:
1) Most experiments in the Abraham-Minkowski controversy (including Ashkin and Dziedzic) ignore electrostriction and magnetostriction. A term should be added for the Helmholtz electrostriction term to Abraham’s and Minkowski’s tensor whenever necessary. Both Abraham and Minkowski’s tensors do not describe electrostriction or magnetostriction. Their tensors, therefore, are unable to give a complete description of the local electromagnetic state in the medium. It is unfortunate that electrostriction and magnetostriction are ignored, as one of the things that Prof. Woodward has realized with his Woodward/Mach Effect experiments is the importance of electrostriction or magnetostriction. It can be misleading to perform an experiment and extract conclusions without assessing the significance of electrostriction or magnetostriction.
2) The problem with Abraham-Minkowski experiments performed to date (and this goes also for EM Drive experiments) is that they are usually trying to measure only one term of the equation of motion or are trying to measure a boundary of a fixed block of dielectric material – so it is very difficult to consider all forces at the boundary. Instead of resolving the problem, contradictory experimental results have often clouded the picture further, because of incomplete analysis of very small forces, where not all the forces are properly taken into account.
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At MIT, Professors P. Penfield Jr., H.A.Haus, Electrodynamics of Moving Media. 1967, Cambridge,Mass.: MIT Press. 241-243 provided a very detailed discussion showing that the force densities each comprise about 20 terms, most of the terms correspond to very small forces.
In Abraham-Minkowsk experiments, the majority of these forces are being ignored by the authors !
Authors in China, more than 50 years after MIT's work, are re-discovering this:
http://arxiv.org/pdf/1504.06437
Analytic derivation of electrostrictive tensors and their application to optical force density calculations
Wujiong Sun1,2, S. B. Wang2, Jack Ng3, Lei Zhou1, and C. T. Chan2*
1 State Key Laboratory of Surface Physics, Key Laboratory of Micro and Nano Photonic Structures (Ministry of Education), and Collaborative Innovation Center of Advanced Microstructures, Fudan University, Shanghai 200433, China
2 Department of Physics and Institute for Advanced Study, The Hong Kong University of Science and Technology, Clear Water Bay, Hong Kong
3 Department of Physics and Institute of Computational and Theoretical Studies, Hong Kong Baptist University, Hong Kong
