On permittivity and permeability
Timo Nieminen
permeability of free space are zero, and that the concepts of
permittivity and permeability are somehow intrinsically related to matter
or ether.
So, what ARE permittivity and permeability all about?
Consider that the Maxwell equations:
curl H - dD/dt = J
curl E + dB/dt = 0
div D = rho
div B = 0
contain 4 fields: E, D, B, and H, which, being 3D vector fields means that
there are 12 unknown quantities. Since the 2 vector equations + 2 scalar
equations only provides a total of 8 equations in vector components,
something more is needed. Enter the consitutive relations:
D = eE
B = mH
where e and m are the permittivity and permeability respectively. (We
could also add J = sE, where s is the conductivity.) Finally, we have the
Lorentz force law:
F = rho E + JxB
This can be read as a definition of E and B in terms of forces on charges
and currents.
From the Lorentz force law, we see that E must have dimensions of
force/charge, while from div D = rho, D must have dimensions of
charge/area. The permittivity of free space is essentially a conversion
between these two quantities, so e has dimensions of
charge^2/(force*area). One can of course choose units such that e0 has a
numerical value of one, allowing it, perhaps unwisely, to be omitted from
equations.
Exercise: In the conventional SI choice of units, with E in V/m, show that
e can be written in F/m. (That's farads/m; recall that Q=CV.)
What if we have some matter around? The fields can result in an induced
electrical polarisation in the matter. Note that dipole moment per unit
volume has dimensions of charge/area, the same as D. So, the usual
D = e0 E + P
where P is the induced polarisation per unit volume, makes sense. If we
write the polarisation in terms of a dielectric susceptibility X,
we have
D = ( e0 + X ) E = e E
In the absence of matter, X = 0, and we have e = e0. Thus, while e is
related to the induced electric polarisation in matter, it is more
fundamentally a conversion factor between the dimensionally-different E
and D, which are two different forms of writing essentially the same
thing: the electric field.
Exercise: repeat for B and H.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
FrediFizzx
news:Pine.LNX.4.50.0510131623280.28876-100000@localhost...
Hi Timo,
There is no system of units that I know of that sets e0 = 1. In
gaussian cgs units and natural units of hbar = c = 1, e0 = 1/4pi. There
is a big mistake in Jackson's appendix on units Table 2 where he shows
e0 = 1. At least it is this way in the third ed. I think that should
read k1 and k2 at the top of the table and not eps0 and mu0.
Of course it is my belief that quantum "vacuum" charge = +,-
sqrt(hbar*c) easily explains all of this. ;-)
FrediFizzx
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf
or postscript
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps
Androcles
"Timo Nieminen" <ti...@physics.uq.edu.au> wrote in message
news:Pine.LNX.4.50.0510131623280.28876-100000@localhost...
| permeability of free space are zero, and that the concepts of
| permittivity and permeability are somehow intrinsically related to
matter
| or ether.
HEY! Give credit. I, ANDROCLES, made that claim, not some vague
recent posts.
|
| So, what ARE permittivity and permeability all about?
|
| Consider that the Maxwell equations:
|
| curl H - dD/dt = J
| curl E + dB/dt = 0
| div D = rho
| div B = 0
Where is mu0, epsilon0 in those equations?
| contain 4 fields: E, D, B, and H,
But there are only 3 fields, gravitism, electrism and magnecity,
so Einstein was correct one hundred years ago,
"It is known that Maxwell's electrodynamics--as usually understood at
the present time--when applied to moving bodies, leads to asymmetries
which do not appear to be inherent in the phenomena."
[snip potty mouthing]
Androcles.
John C. Polasek
It's easier to show that D = Ee0:
D (coul/m^2) = E (volt/m)*e0(coul/volt*meter)
(since farad = coul/volt) = Ee0(coul/m^2)
and solve for e0.
>What if we have some matter around? The fields can result in an induced
>electrical polarisation in the matter. Note that dipole moment per unit
>volume has dimensions of charge/area, the same as D. So, the usual
>
>D = e0 E + P
>
>where P is the induced polarisation per unit volume, makes sense. If we
>write the polarisation in terms of a dielectric susceptibility X,
>we have
>
>D = ( e0 + X ) E = e E
>
>In the absence of matter, X = 0, and we have e = e0. Thus, while e is
>related to the induced electric polarisation in matter, it is more
>fundamentally a conversion factor between the dimensionally-different E
>and D, which are two different forms of writing essentially the same
>thing: the electric field.
No it's much more specific than that if we're talking real material
and real physics: E is the stress, and D is resulting strain. It is a
mistake to use one for the other, Similarly, H is the stress and B is
the strain. Gaussian cgs theory essentially denies these real
properties and then arranges things so everything works out with
hybrid units etc.
Maxwell's divergence equation requires that there be charge density
for real D:
Div D = rho
dD/dx = rho
Int dD = Int rho dx = D = rho delta
where delta is the deflection of an individual charge or charge pair.
This requirement for charge density presents a real problem for a true
vacuum which nevertheless possesses u0 and m0.
The rho is composed of electron positron pairs in quantum vacuum or
what I call Espace in my theory. They neutralize each other before
polarization.
See the permittivity paper on my website. The required density is
enormous to fit e0. It is a dual space.
>>
>Exercise: repeat for B and H.
John Polasek
http://www.dualspace.net
Androcles
"John C. Polasek" <jpol...@cfl.rr.com> wrote in message
news:4f0tk19u0bu9kd1j9...@4ax.com...
[snip idiot Nieminen]
| No it's much more specific than that if we're talking real material
| and real physics: E is the stress, and D is resulting strain. It is a
| mistake to use one for the other, Similarly, H is the stress and B is
| the strain. Gaussian cgs theory essentially denies these real
| properties and then arranges things so everything works out with
| hybrid units etc.
| Maxwell's divergence equation requires that there be charge density
| for real D:
| Div D = rho
| dD/dx = rho
| Int dD = Int rho dx = D = rho delta
| where delta is the deflection of an individual charge or charge pair.
Ok so far.
| This requirement for charge density presents a real problem for a true
| vacuum which nevertheless possesses u0 and m0.
Why do you think that a vacuum possesses properties?
| The rho is composed of electron positron pairs in quantum vacuum or
| what I call Espace in my theory.
Why do you suppose something filled with mass in the form of
electron positron pairs is called a vacuum?
Androcles.
John C. Polasek
wrote:
>
>"John C. Polasek" <jpol...@cfl.rr.com> wrote in message
>news:4f0tk19u0bu9kd1j9...@4ax.com...
>
>[snip idiot Nieminen]
>
>| No it's much more specific than that if we're talking real material
>| and real physics: E is the stress, and D is resulting strain. It is a
>| mistake to use one for the other, Similarly, H is the stress and B is
>| the strain. Gaussian cgs theory essentially denies these real
>| properties and then arranges things so everything works out with
>| hybrid units etc.
>| Maxwell's divergence equation requires that there be charge density
>| for real D:
>| Div D = rho
>| dD/dx = rho
>| Int dD = Int rho dx = D = rho delta
>| where delta is the deflection of an individual charge or charge pair.
>
>Ok so far.
>
>| This requirement for charge density presents a real problem for a true
>| vacuum which nevertheless possesses u0 and m0.
>
>Why do you think that a vacuum possesses properties?
Look at my Permittivity paper. There you will see a vacuum capacitor
whose dimensions have been calculated use e0*A/g and which in a teacup
size can carry 135 amperes through the vacuum at 5KV. Permittivity is
real.
>
>| The rho is composed of electron positron pairs in quantum vacuum or
>| what I call Espace in my theory.
>
>Why do you suppose something filled with mass in the form of
>electron positron pairs is called a vacuum?
It's why I discovered the space that is dual to our vacuum.
Permittivity does appear to happen in a vacuum, but it can't happen in
a vacuum, so further investigation led to Espace.
Espace contains an image of our universe that originated from
positrons rather than the electrons that are the basis of all mass in
our universe. (?Why arent there as many positrons as electrons?)
See my permittivity paper #1. I have those papers on a website because
it would be too tedious to cover complex material in a language that
is 7 bits wide.
>Androcles.
John Polasek
http://www.dualspace.net
FrediFizzx
news:qEr3f.16124$ey6....@fe1.news.blueyonder.co.uk...
|
| "Timo Nieminen" <ti...@physics.uq.edu.au> wrote in message
| news:Pine.LNX.4.50.0510131623280.28876-100000@localhost...
| | In some recents posts, the claim was made that the permittivity and
| | permeability of free space are zero, and that the concepts of
| | permittivity and permeability are somehow intrinsically related to
| matter
| | or ether.
|
| HEY! Give credit. I, ANDROCLES, made that claim, not some vague
| recent posts.
Too bad that you are wrong about permittivity and permeability of free
space being zero. Quantum "vacuum" charge = +,- sqrt(hbar*c) rules,
baby!
| | So, what ARE permittivity and permeability all about?
| |
| | Consider that the Maxwell equations:
| |
| | curl H - dD/dt = J
| | curl E + dB/dt = 0
| | div D = rho
| | div B = 0
|
| Where is mu0, epsilon0 in those equations?
Too bad that you snipped out the part that shows where they are. Read
Jackson 3rd ed. and Griffiths' "Intro. to Electrodynamics". Please.
| | contain 4 fields: E, D, B, and H,
|
| But there are only 3 fields, gravitism, electrism and magnecity,
LOL!
Timo Nieminen
> "Timo Nieminen" <ti...@physics.uq.edu.au> wrote:
> |
> | So, what ARE permittivity and permeability all about?
> |
> | Consider that the Maxwell equations:
> |
> | curl H - dD/dt = J
> | curl E + dB/dt = 0
> | div D = rho
> | div B = 0
>
> Where is mu0, epsilon0 in those equations?
They're not. They're in the constitutive relations (which you cut).
> | contain 4 fields: E, D, B, and H,
>
> But there are only 3 fields, gravitism, electrism and magnecity,
Classical electrodynamics only deals with the latter 2. Fundamentally, the
last 2 are the same thing; easy enough to write them unified as the
4-potential or the rank-2 field tensor. Which basically followed from
Einstein's clarification of the electrodynamics of moving media:
> so Einstein was correct one hundred years ago,
> "It is known that Maxwell's electrodynamics--as usually understood at
> the present time--when applied to moving bodies, leads to asymmetries
> which do not appear to be inherent in the phenomena."
And since there are only two fields, E and D must be fundamentally the
same thing, ditto for B and H. But they're dimensionally different, so a
conversion factor is needed. Which is what the constitutive relations are
in free space. So e and m can't be zero, can they?
> [snip potty mouthing]
Ah, discussion of physics is bad language fit to be cut to Androcles,
while he considers his usual emissions to be proper!
--
Timo
Androcles
| Look at my Permittivity paper. There you will see a vacuum capacitor
| whose dimensions have been calculated use e0*A/g and which in a teacup
| size can carry 135 amperes through the vacuum at 5KV. Permittivity is
| real.
You are talking about the Meivac capacitor, obviously.
My TV tube is a 26 KV capacitor, 5 times better than yours.
It leaks a few milliamps through the dielectric.
I use that to paint the raster, but I get nothing like 135 amps.
Permittivity your arse!
Androcles.
Androcles
"FrediFizzx" <fredi...@hotmail.com> wrote in message
news:3r7o4pF...@individual.net...
| "Androcles" <Androcles@ MyPlace.org> wrote in message
| news:qEr3f.16124$ey6....@fe1.news.blueyonder.co.uk...
||
|| "Timo Nieminen" <ti...@physics.uq.edu.au> wrote in message
|| news:Pine.LNX.4.50.0510131623280.28876-100000@localhost...
|| | In some recents posts, the claim was made that the permittivity and
|| | permeability of free space are zero, and that the concepts of
|| | permittivity and permeability are somehow intrinsically related to
|| matter
|| | or ether.
||
|| HEY! Give credit. I, ANDROCLES, made that claim, not some vague
|| recent posts.
|
| Too bad that you are wrong about permittivity and permeability of free
| space being zero. Quantum "vacuum" charge = +,- sqrt(hbar*c) rules,
| baby!
Androcles.
Timo Nieminen
> On Thu, 13 Oct 2005 16:58:34 +1000, Timo Nieminen
> <ti...@physics.uq.edu.au> wrote:
>>
>> D = ( e0 + X ) E = e E
>>
>> In the absence of matter, X = 0, and we have e = e0. Thus, while e is
>> related to the induced electric polarisation in matter, it is more
>> fundamentally a conversion factor between the dimensionally-different E
>> and D, which are two different forms of writing essentially the same
>> thing: the electric field.
>
> No it's much more specific than that if we're talking real material
> and real physics: E is the stress, and D is resulting strain. It is a
> mistake to use one for the other,
In matter, there is the complication that one includes the polarisation.
One cannot subsitute one for the other; they're geometrically different
entities (as you note).
> Maxwell's divergence equation requires that there be charge density
> for real D:
[cut]
Somewhere, somewhen, there must be/have been non-zero charge density. div
D = rho is just a statement about sources/sinks of D, and all sorts of
fields manage to exist just fine away from their sources.
Nonetheless, that's a comment made before reading:
> See the permittivity paper on my website. The required density is
> enormous to fit e0. It is a dual space.
[cut]
> http://www.dualspace.net
so perhaps I missed your point. Will look when time permits!
Androcles
"Timo Nieminen" <uqtn...@mailbox.uq.edu.au> wrote in message
news:2005101404...@emu.uq.edu.au...
| On Thu, 13 Oct 2005, Androcles wrote:
|
| > "Timo Nieminen" <ti...@physics.uq.edu.au> wrote:
| > |
| > | So, what ARE permittivity and permeability all about?
| > |
| > | Consider that the Maxwell equations:
| > |
| > | curl H - dD/dt = J
| > | curl E + dB/dt = 0
| > | div D = rho
| > | div B = 0
| >
| > Where is mu0, epsilon0 in those equations?
|
| They're not.
| > | contain 4 fields: E, D, B, and H,
| >
| > But there are only 3 fields, gravitism, electrism and magnecity,
|
| Classical electrodynamics only deals with the latter 2.
So you need two equations, the rest are the excrement of the male
bovine.
| Fundamentally, the
| last 2 are the same thing;
Therefore as superfluous as the luminiferous aether to which they refer.
End of discussion, agreement reached.
Androcles.
John C. Polasek
wrote:
>
There are several things wrong with your capacitor. The capacity can't
be determined, it carries a few nanoamperes, and it has a huge gap
which makes it a very small capacitor. The 26KV has nothing to do with
capacity, it is only the voltage configured by the manufacturer for
100 or so uA of beam current as a CRT. My 135A is true capacitive
current that depends on permittivity.
John Polasek
http://www.dualspace.net
Androcles
| On Thu, 13 Oct 2005, John C. Polasek wrote:
|
| > On Thu, 13 Oct 2005 16:58:34 +1000, Timo Nieminen
| > <ti...@physics.uq.edu.au> wrote:
| >>
| >> D = ( e0 + X ) E = e E
| >>
| >> In the absence of matter, X = 0, and we have e = e0. Thus, while e
is
| >> related to the induced electric polarisation in matter, it is more
| >> fundamentally a conversion factor between the
dimensionally-different E
| >> and D, which are two different forms of writing essentially the
same
| >> thing: the electric field.
| >
| > No it's much more specific than that if we're talking real material
| > and real physics: E is the stress, and D is resulting strain. It is
a
| > mistake to use one for the other,
|
| In matter, there is the complication that one includes the
polarisation.
Electrolytic capacitors are indeed polarized. My TV tube leaks
current through the dielectric, I use it to paint the raster.
Polasek's capacitor leaks no current through the dielectric, although
he claims 135 amps passes.
I can't test his theory, I don't have a 675 kVA supply to my home.
Androcles.
Ken S. Tucker
Some things below...
I think Timo's "P" vector is equivalent to the polarization
tensor, just as E and B are F_0i and F_ij respectively, ok,
but P is fine, (as a density in GR).
Let's look at the "piezoelectric effect", where an applied
mass distorts the relations of a charge latice to produce
a voltage, that is the basis of electronic weigh scales
unless Timo likely understands better, but that's what I
understand.
In the presence of the Sun's "MASS" , light bends
(deflection) and slows down, (Shapiro), indicating
an alteration to "e" and "m" that seems to vary the
refractive index by the presence of matter, from
that PoV.
But this effect might also appear as a polarization,
due to mass, like the "piezo" does, you see Timo, GR
doesn't affix a certain mechanical reason to P, on
the contrary I think P indicates the presence of
matter, _even in a so-called vacuum_!
Anyway, that's why I think Timo's post is cool, and
I hope he will explain "piezoelectrical" voltage.
Regards
Ken S. Tucker
...
Androcles
| On Thu, 13 Oct 2005 19:11:50 GMT, "Androcles" <Androcles@ MyPlace.org>
| wrote:
|
| >
| >"John C. Polasek" <jpol...@cfl.rr.com> wrote in message
| >news:uj6tk15eidge18tue...@4ax.com...
| >
| >| Look at my Permittivity paper. There you will see a vacuum
capacitor
| >| whose dimensions have been calculated use e0*A/g and which in a
teacup
| >| size can carry 135 amperes through the vacuum at 5KV. Permittivity
is
| >| real.
| >
| >You are talking about the Meivac capacitor, obviously.
| >My TV tube is a 26 KV capacitor, 5 times better than yours.
| >It leaks a few milliamps through the dielectric.
| >I use that to paint the raster, but I get nothing like 135 amps.
| >Permittivity your arse!
| >Androcles.
| There are several things wrong with your capacitor. The capacity can't
| be determined, it carries a few nanoamperes, and it has a huge gap
| which makes it a very small capacitor.
Yeah, I said it leaked current, but I can disconnect the cathode
heater.
| The 26KV has nothing to do with
| capacity, it is only the voltage configured by the manufacturer for
| 100 or so uA of beam current as a CRT. My 135A is true capacitive
| current that depends on permittivity.
I'll bet your capacitor can't pass DC like mine. It won't "permit"
it to flow. It has no "permittivity".
Androcles.
Timo Nieminen
> "Timo Nieminen" <ti...@physics.uq.edu.au> wrote:
>
> | The permittivity of free space is essentially a conversion
> | between these two quantities, so e has dimensions of
> | charge^2/(force*area). One can of course choose units such that e0 has
> a
> | numerical value of one, allowing it, perhaps unwisely, to be omitted
> from
> | equations.
>
> There is no system of units that I know of that sets e0 = 1. In
> gaussian cgs units and natural units of hbar = c = 1, e0 = 1/4pi. There
> is a big mistake in Jackson's appendix on units Table 2 where he shows
> e0 = 1. At least it is this way in the third ed. I think that should
> read k1 and k2 at the top of the table and not eps0 and mu0.
Jackson is correct in that table; in Gaussian units, D is defined
differently, namely div D = 4pi * rho, so the 4pi is taken care of there.
The Coulomb constant is set to equal 1 in Gaussian units, and with the 4pi
given to D, then e0 = 1/k = 1.
Heaviside-Lorentz units also have e0 = 1, and sort-of-Gaussian units with
c=1.
FrediFizzx
Now Timo, that just doesn't make sense. If Coulomb's constant, k_e = 1
in gaussian cgs and k_e = 1/(4pi*eps0) in SI, then it is pretty plain to
see that SI's eps0 = 1/4pi in gaussian cgs. And that is also what
Griffiths claims in "Intro. to Electrodynamics" 3rd ed. page 558 and is
confirmed by my MathCad program when going between SI and gaussian cgs
units. It is easier to see if we go to div E = rho/eps0 in SI and div E
= 4pi*rho in gaussian for free space. Clearly, this way it is easy to
see that eps0 = 1/4pi in gaussian. Jackson has to be wrong in his Table
2 about eps0 and mu0. Unless he is changing the definitions of eps0 and
mu0 between the unit systems to be more synonymous with k_e and k_m.
| Heaviside-Lorentz units also have e0 = 1, and sort-of-Gaussian units
with
| c=1.
Now what is going on here with H-L units? The 4pi is absorbed into
charge density it looks like. So it is going to be the same thing.
SI's eps0 = 1/4pi since the difference in the two rho's will be a factor
of 4pi still.
Timo Nieminen
Yes, the definitions of lots of things change. Jackson 1st ed and
Griffiths both give div E = 4pi * rho for free space, and
div D = 4pi * rho. Since D = eE, then e0 = 1 in Gaussian units. In
Gaussian units, D = E + 4pi * P (hmm, is this the same P?). Now, a
statvolt is AFAIK a statcoulomb/cm (which makes the unit of capacitance
the cm, so e0, with dimensions of capacitance/length, is unitless), so E
has units statvolt/cm, and D statcoulomb/cm^2, all of which reminds me why
I don;t like to use Gaussian units :(
Fundamentally, yes, e0 is a different thing in the 2 systems, since D is
defined differently.
> | Heaviside-Lorentz units also have e0 = 1, and sort-of-Gaussian units
> with
> | c=1.
>
> Now what is going on here with H-L units? The 4pi is absorbed into
> charge density it looks like. So it is going to be the same thing.
> SI's eps0 = 1/4pi since the difference in the two rho's will be a factor
> of 4pi still.
I've never had the great joy of actually working with H-L units, so I
can't say off-hand. I think I'll stick to SI units.
FrediFizzx
In free space for gaussian units, D = E not D = eE since P is taken to
be zero. You have hit on John Polasek's big gripe with gaussian units.
;-) Yes, capacitance is length in gaussian so it is easy to see that
SI's farad/meter will be dimensionless in gaussian. But it is not 1.
It is 1/4pi keeping SI's definition of eps0. A lot of people think that
SI's eps0 disappears in gaussian and natural units but it doesn't. It
just becomes a dimensionless factor of 1/4pi. I highly suspect it is
related to quantum spin.
| Fundamentally, yes, e0 is a different thing in the 2 systems, since D
is
| defined differently.
So I think we can agree that *SI's* eps0 is equal to 1/4pi in gaussian
cgs units. I think it would have been more clear if Jackson had just
stuck to his k1 and k2 for Table 2, instead of redefining eps0 and mu0.
Which is what he must be doing but never says it.
| > | Heaviside-Lorentz units also have e0 = 1, and sort-of-Gaussian
units
| > with
| > | c=1.
| >
| > Now what is going on here with H-L units? The 4pi is absorbed into
| > charge density it looks like. So it is going to be the same thing.
| > SI's eps0 = 1/4pi since the difference in the two rho's will be a
factor
| > of 4pi still.
|
| I've never had the great joy of actually working with H-L units, so I
| can't say off-hand. I think I'll stick to SI units.
I have only run across H-L units a couple of times in some old
literature. Basically the value of rho is just 4pi times the value of
rho in gaussian. Heck, I will use either gaussian cgs or SI or natural
units. MathCad makes it easy to convert so no problem.
John C. Polasek
Gaussian has another problem: its capacitance is in cm but it refers
to the selfcapacitance of a sphere, but as far as I know it has
nothing useful to say about the parallel plate capacitor. No question
it gives D as Q/4pir^2 for a sphere, or should very logically, but to
get electric field E you must divide by e0, which has been banished.
JP
>| Fundamentally, yes, e0 is a different thing in the 2 systems, since D
>is
>| defined differently.
>
snips.
>
>
>FrediFizzx
>
>http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf
>or postscript
>http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps
>
>http://www.vacuum-physics.com
John Polasek
http://www.dualspace.net
FrediFizzx
To add to this, there is also a "loose cannon" factor of 2 (or is it
1/2) involved with the magnetic constant that seems to appear and
disappear depending on what you are doing. ;-)
| | > | Heaviside-Lorentz units also have e0 = 1, and sort-of-Gaussian
| units
| | > with
| | > | c=1.
| | >
| | > Now what is going on here with H-L units? The 4pi is absorbed
into
| | > charge density it looks like. So it is going to be the same
thing.
| | > SI's eps0 = 1/4pi since the difference in the two rho's will be a
| factor
| | > of 4pi still.
| |
| | I've never had the great joy of actually working with H-L units, so
I
| | can't say off-hand. I think I'll stick to SI units.
|
| I have only run across H-L units a couple of times in some old
| literature. Basically the value of rho is just 4pi times the value of
| rho in gaussian. Heck, I will use either gaussian cgs or SI or
natural
| units. MathCad makes it easy to convert so no problem.
I take that back about just old literature. Weinberg uses H-L units
with c = 1 in his "Gravitation and Cosmology" book.
Timo Nieminen
> Let's look at the "piezoelectric effect", where an applied
> mass distorts the relations of a charge latice to produce
> a voltage, that is the basis of electronic weigh scales
> unless Timo likely understands better, but that's what I
> understand.
Luckily, most of the stuff that I deal with falls into the
can-ignore-electrostriction etc category. Landau has a brief coverage of
such things, and notes that stresses due to applied fields are
proportional to E^2 in isotropic dielectrics, but proportional to E in
piezoelectrics. As field -> deformation, so deformation -> field, E
proportional to the deformation. I don't know the microscopic details, but
one can go quite some way without them (as per Landau). [That's Landau vol
8, Electrodynamics of continuous media.]
> In the presence of the Sun's "MASS" , light bends
> (deflection) and slows down, (Shapiro), indicating
> an alteration to "e" and "m" that seems to vary the
> refractive index by the presence of matter, from
> that PoV.
[Caveat: I'm rather non-expert in GR, but here goes ... ]
Local Lorentz invariance, with e0 and m0 being 4-scalars, would mean the
"local observer" would disagree on there being a refractive index other
than 1, while the "distant observer" can happily model the effect as a
change in refractive index. I saw many years ago a nice plastic lens that
modelled gravitational lensing.
> But this effect might also appear as a polarization,
> due to mass, like the "piezo" does, you see Timo, GR
> doesn't affix a certain mechanical reason to P, on
> the contrary I think P indicates the presence of
> matter, _even in a so-called vacuum_!
In non-GR electrodynamics, P is only non-zero in the presence of matter.
I'd guess that in GR, P is only non-zero according to the "observer on the
spot" in the presence of what everybody will agree is matter. To that
extent, I think it's pretty clear that nonzero P indicates the presence of
matter. As for what a distant observer sees, I dunno.
Ken S. Tucker
Timo Nieminen wrote:
> On Fri, 13 Oct 2005, Ken S. Tucker wrote:
>
> > Let's look at the "piezoelectric effect", where an applied
> > mass distorts the relations of a charge latice to produce
> > a voltage, that is the basis of electronic weigh scales
> > unless Timo likely understands better, but that's what I
> > understand.
>
> Luckily, most of the stuff that I deal with falls into the
> can-ignore-electrostriction etc category.
Well Timo, that's ambiguous, I think everyone in the
group would agree you have a penetrating mind for
detail, like when you posted about mag-monopoles,
so giving consideration to the physical function of
an electronic weigh scale, ought to be fun, and it
wasn't meant as trick cause they're commonly used.
> Landau has a brief coverage of
> such things, and notes that stresses due to applied fields are
> proportional to E^2 in isotropic dielectrics, but proportional to E in
> piezoelectrics. As field -> deformation, so deformation -> field, E
> proportional to the deformation. I don't know the microscopic details, but
> one can go quite some way without them (as per Landau). [That's Landau vol
> 8, Electrodynamics of continuous media.]
>
> > In the presence of the Sun's "MASS" , light bends
> > (deflection) and slows down, (Shapiro), indicating
> > an alteration to "e" and "m" that seems to vary the
> > refractive index by the presence of matter, from
> > that PoV.
>
> [Caveat: I'm rather non-expert in GR, but here goes ... ]
>
> Local Lorentz invariance, with e0 and m0 being 4-scalars, would mean the
> "local observer" would disagree on there being a refractive index other
> than 1, while the "distant observer" can happily model the effect as a
> change in refractive index. I saw many years ago a nice plastic lens that
> modelled gravitational lensing.
Agreed, the local observer is saturated by his/her local
e0, m0.
> > But this effect might also appear as a polarization,
> > due to mass, like the "piezo" does, you see Timo, GR
> > doesn't affix a certain mechanical reason to P, on
> > the contrary I think P indicates the presence of
> > matter, _even in a so-called vacuum_!
>
> In non-GR electrodynamics, P is only non-zero in the presence of matter.
> I'd guess that in GR, P is only non-zero according to the "observer on the
> spot" in the presence of what everybody will agree is matter. To that
> extent, I think it's pretty clear that nonzero P indicates the presence of
> matter. As for what a distant observer sees, I dunno.
Well based on light deflection data, I take confidence
the e0 and m0 may be regarded as non unity near the sun
on account of the mass of the sun, ok.
((Tucker tries to relate gravitation and electricity)).
So I'll ask a direct question, related to common
electronic weigh scales...
The application of mass (weight) deforms the lattice,
is it reasonable to suggest the permittivity e0 and
the m0 acquire polarized characteristics? and thus
become vectors in 3D apart from scalars?
IOW's the permittivity e0 in the deformed crystal
depends upon direction.
Regards
Ken S. Tucker
Timo Nieminen
> So I'll ask a direct question, related to common
> electronic weigh scales...
>
> The application of mass (weight) deforms the lattice,
> is it reasonable to suggest the permittivity e0 and
> the m0 acquire polarized characteristics? and thus
> become vectors in 3D apart from scalars?
>
> IOW's the permittivity e0 in the deformed crystal
> depends upon direction.
In classical macroscopic electrodynamics, in free space, we have E
parallel to D, always, so we can write D = e0 E, with scalar E. In matter,
we have D = e0 E + P, but often enough, especially in crystals, P is not
parallel to E. So, e, defined by D = e E, is a tensor quantity, but I
wouldn't say that that does anything to e0, which by definition, is
unaffected by matter or things done to matter. If you meant to say e
rather than e0, well, there are lots of materials that are isotropic when
unstressed that become anisotropic (ie tensor e) when stressed, so it's a
perfectly reasonable suggestion. I have a vague memory of piezoelectric
necessarily being anisotropic even when unstressed. Thinking about
symmetry and producing an E field - a directed thing - by squeezing a slab
makes me agree with this memory.
Ken S. Tucker
on this subject, good stuff, I'm a bit crude.
Timo Nieminen wrote:
> On Mon, 16 Oct 2005, Ken S. Tucker wrote:
>
> > So I'll ask a direct question, related to common
> > electronic weigh scales...
> >
> > The application of mass (weight) deforms the lattice,
> > is it reasonable to suggest the permittivity e0 and
> > the m0 acquire polarized characteristics? and thus
> > become vectors in 3D apart from scalars?
> >
> > IOW's the permittivity e0 in the deformed crystal
> > depends upon direction.
>
> In classical macroscopic electrodynamics, in free space, we have E
> parallel to D, always, so we can write D = e0 E, with scalar E. In matter,
> we have D = e0 E + P, but often enough, especially in crystals, P is not
> parallel to E. So, e, defined by D = e E, is a tensor quantity, but I
> wouldn't say that that does anything to e0, which by definition, is
> unaffected by matter or things done to matter. If you meant to say e
> rather than e0, well, there are lots of materials that are isotropic when
> unstressed that become anisotropic (ie tensor e) when stressed, so it's a
> perfectly reasonable suggestion. I have a vague memory of piezoelectric
> necessarily being anisotropic even when unstressed. Thinking about
> symmetry and producing an E field - a directed thing - by squeezing a slab
> makes me agree with this memory.
For notational clarity, allow me to suggest the
speed of light "c" in vacuum changes in a dense
medium to C by the sqrt(em)with "e" being permat-
tivity and m being permeability.
Let's try a gedanken experiment.
I'll take a mass and place it on a PiezoElectrical
(supposedly to compress the z axis) crystal so that
it now measures X=Y=Z by some fella with a ruler
exterior of the cube, ie a perfect cube,
also the cube is transparent.
Here's the question. Will the velocity of light
directed through the crystal, in the directions
X,Y,Z be equal?
That's not a trick question. One can attach voltage
sensors to all 6 sides and find variations across
the crystal in the 3D.
Timo, from your post above, I got the impression
you may think so, and I think is outstanding if
true.
Recall the Michelson-Morley Experiment, (MMX),
well it nulled on perpendicular axes, and that
underwrites SR. However, I think the gedanken
above may produce a positive result!
If so, then the speed of light C is deflected
by the crystal voltage. If someone can firmly
establish that link, hell that's near Nobel
prize stuff!
But more importantly (to me) it would establish, by
experimental means a tensor nature in permittivity
and permeability, and from that define the nature
of the spacetime field better.
Over-Excitedly yours,
Ken S. Tucker
Androcles
"Ken S. Tucker" <dyna...@vianet.on.ca> wrote in message
news:1129674042.7...@g47g2000cwa.googlegroups.com...
Don't get too excited.
Birefringence is an optical property possessed by a material
which has more than one index of refraction. This anisotropy
in the index of refraction is dependent on the crystallographic
projection. (Googled).
Androcles.
Mark Fergerson
No, they won't.
> | That's not a trick question. One can attach voltage
> | sensors to all 6 sides and find variations across
> | the crystal in the 3D.
Yes, but not really relevant; light does not care that a DC bias
is present (superposition of fields). Also piezoelectricity is not
seen in crystals with cubical symmetry.
> | Timo, from your post above, I got the impression
> | you may think so, and I think is outstanding if
> | true.
Calm down; it's a direct consequence of the changed density of
material the light has to traverse in different directions.
> | Recall the Michelson-Morley Experiment, (MMX),
> | well it nulled on perpendicular axes, and that
> | underwrites SR. However, I think the gedanken
> | above may produce a positive result!
But vacuum is not crystalline, nor is it (generally) anisotropic.
> | If so, then the speed of light C is deflected
> | by the crystal voltage. If someone can firmly
> | establish that link, hell that's near Nobel
> | prize stuff!
Yep, been done.
> | But more importantly (to me) it would establish, by
> | experimental means a tensor nature in permittivity
> | and permeability, and from that define the nature
> | of the spacetime field better.
> |
> | Over-Excitedly yours,
> | Ken S. Tucker
>
> Don't get too excited.
> Birefringence is an optical property possessed by a material
> which has more than one index of refraction. This anisotropy
> in the index of refraction is dependent on the crystallographic
> projection. (Googled).
Yes. Birefringence and piezoelectricity are both due to
anisotropy in crystal structure. I mention this only because of the
"perfect cube" gedankenexperimental material above; piezoelectricity
is not seen in crystals with cubic symmetry.
I know, it looks like a pedant point, but it's really not.
Mark L. Fergerson