Anyone interested,

As a kid, I learned that the speed of light is not an independent constant, but a constant derived from two other constants. The formula being:

c=1/sqrt(u0*e0)

Where u0 is the magnetic permeability of free space, and e0 is the electric permittivity of free space.

u0=4*Pi * 10^-7
e0=8.854 * 10^-12

Doesn't this mean that aether does exist since space has these unique electric and magnetic properties?? Wouldn't areas with no space at all have both the magnetic permeability and electric permittivity equal to zero?? Theoretically, in these areas of nothingness, the speed of light would be infinite ( It would travel 1 meter and 1 kilometer in the same time). Isn't the fact the the speed of light in a vacuum is finite, proof, in itself, that aether exists??

Tom

Uh.........Uh........ actually

C = 1/sqrt(u*e) NOT C = 1/sqrt(u0*e0) where

u = ur*u0 and e = er*e0

uo = 4*pi*e-7 in vaccum. ur = 1 in vacuum. ur = 1 in air as well because air is not magnetic.

e0 = 8.85e-12 (in SI unit) er varies with different medium. er is 2.2 for teflon, 1 for air, 9.2 for silicon germanium, and etc....

in vacuum both ur and er = 1

Joeman,

If "empty space" consisted of nothing, wouldn't ur and er both equal 0 for empty space???

Tom

Wait, I don't think what I said is quite right. It just doesn't feel right. Let me ask my collegues.

In vaccum, I always use er = ur = 1 for calculation. Since I am an engineer I don't really know the derivation.

I just talked to collegue. What I said was correct. Er = 1 in vacuum. Er = 1.006 in air. Er is never less than 1. Ur = 1 for both vacuum and air since they are both non-magnetic. Ur can be sometimes slightly less at one. According to my collegue, Er has no physic meanings. It is just a constant out front to make an equation correct.

Joeman,

According to my collegue, Er has no physic meanings. It is just a constant out front to make an equation correct.

Er has a very important physical meaning. As you indicated before, different materials have different Er values. Here is a list of some:

Hard Vacuum 1.0
Pure Teflon® 2.1
Type GY Teflon®-Glass 2.2 - 2.3
Type GX Teflon® Glass 2.55
Cyanate Ester/Glass 3.2 - 3.6
Cyanate Ester-Quartz 2.8 - 3.4
Polyimide-Quartz 3.5 - 3.8
Polyimide-Glass 4.0 - 4.6
Epoxy-Glass (FR-4) 4.4 - 5.2
Non-woven Aramid Epoxy 3.8 - 4.1
Woven Aramid Epoxy 3.8 - 4.1
Ceramic-Filled Teflon® 6.0 - 10.2
Water 70.0


Tom

Adam dons his silly black ninja outfit, and hides in the shadows, awaiting his opportunity...

"HAH!" Adam strikes, sword slicing cleanly through Prosoothus!

Ninja Adam dissolves seemlessly back into the shadows, as a body in two halves flops to the sciforums floor...

prosoothus

Doesn't this mean that aether does exist since space has these unique electric and magnetic properties?? Wouldn't areas with no space at all have both the magnetic permeability and electric permittivity equal to zero?? Theoretically, in these areas of nothingness, the speed of light would be infinite ( It would travel 1 meter and 1 kilometer in the same time).

These properties would only imply that light travels through a vacuum at 299792458 meters/second regardless if there is an aether or not. It is the state of motion of that medium and the source of the light which would determine whether an aether existed with respect to the rest frame of the observer. At this speed, it would be relative to the absolute rest frame of the aether. In any other reference frames, the speed of light would differ depending on the direction of travel. Measurements taken with extremely precise interferometers found essentially no directional variation regardless of the direction of travel.

Isn't the fact the the speed of light in a vacuum is finite, proof, in itself, that aether exists??

One can see this in the aberration of starlight, which can only be explained in an aether theory if it is assumed that an observer on the Earth's surface is not at rest with respect to the local aether.

<i>As a kid, I learned that the speed of light is not an independent constant, but a constant derived from two other constants. The formula being:

c=1/sqrt(u0*e0)</i>

No. Actually things are the other way round.

This formula establishes a set of electromagnetic units. In the SI system, you start with the second, then use c to define what a metre is.

Having c, you then define u0, which sets up some magnetic units for you. That's why u0 in the SI system is <b>exactly</b> 4 pi &times; 10^-7. It isn't measured - it's defined.

Given c and u0, both defined quantities, e0 is set using the above equation.

James R,

Given c and u0, both defined quantities, e0 is set using the above equation.

You would be correct if e0 was a meaningless constant, but it is a very important constant. It is used in Coulomb's Law for calculating the force between two charged objects. It is just as important as u0 is to the magnetic interaction or g is to the gravitational interaction. In other words, e0 is a more fundamental constant than c.

It is very likely that the interaction between the electric and magnetic properties of space (uo and e0) and the magnetic and electric fields of photons, is what limits a photon's speed in a vacuum.

The slowing down of photons in certain materials appears to be caused by the differen ur and er values of the material, not a delay caused by the absorbtion and re-emmitance of photons in that material.

Tom

Q,

Measurements taken with extremely precise interferometers found essentially no directional variation regardless of the direction of travel.

We all agree that the direction of emmited photons is dependent on the speed and direction of the object emitting them. If this is really the case, then not even an interferometer can be used to confirm the existance or non-existance of aether.

Tom

Adam,

Be carefull!!! We don't want you to get a hernia! :)

Tom

Tom,

<i>You would be correct if...</i>

I am correct. If you don't think I'm correct, show me the flaw in my argument.

<i>The slowing down of photons in certain materials appears to be caused by the differen ur and er values of the material, not a delay caused by the absorbtion and re-emmitance of photons in that material.</i>

No, wrong way round again. The different values of ur and er are caused by the absorption and re-emittance of photons in the material.

James,

I am correct. If you don't think I'm correct, show me the flaw in my argument.

I already showed you. I pointed out that e0 is used in Coulomb's Law. Coulomb's law deals with electric fields which are one of the fundamental interactions. Electric fields are MORE fundamental than the electric and magnetic fields of photons. Photons are more complex than simple electric fields. This is why e0 is more fundamental than c.

If you don't believe me, look it up. This is the formula for Coulomb's Law:

F=q1*q2/4*Pi*e0* r^2

It's like the chicken and the egg problem, except everyone knows what came first. The electric field came first and photons come later.

Tom

Non photon objects can never exist at the speed of light in their current states, as the process of a mass gaining velocity is the increased presence of photons absorbed by that mass, now recall what I said how photons are half matter - half antimatter, which says to me that for something to travel at c the mass must be hald matter half antimatter, a perfect, yet opposing balance. Thus the ratio of antimatter to normal matter dictates the velocity/energy of a mass, reaching c when the ratio is equal.

Let us look at a mass of matter, studying its velocity as its internal energy increases. The fact is that as a mass's velocity increases, it's mass increases, thus pertaining to the mass of the photons which are absorbed by the mass, providing the energy for the velocity. Lets consider the structure of the photon, its matter half can be considered the 'dead weight' stopping the antimatter side from reacting with normal matter, but the antimatter half is what provides us with the energestic abilities of photons. But our problem in exeeding c, lies in the fact that the antimatter portion of the photon, our energy, is equally balanced by the dead weight matter. So thus when we apply photons to a normal mass attempting to propell it past c, that mass of 'dead weight' matter portion, will add to the mass of objects matter, thus never attaining a balance with the antimatter, thus never reaching c but always increasing in mass to infinety. Even in the case of an object moving at 0.999999999999 c, the mass of the original object would be utterly insignificant, to the mass of the absorbed photons, but would still push the scales in favour of the matter, preventing the non-photon object from ever reaching c. (Loud exhale)

Please feel free to comment on, (try to) discredit, or add ideas.
Thnx D.P.

Originally posted by Prosoothus
[B]James,



I already showed you. I pointed out that e0 is used in Coulomb's Law.


Actually, James R is exactly right. Check out Classical Electrodynamics by J D Jackson for a development of E+M from the fundamentals.


Coulomb's law deals with electric fields which are one of the fundamental interactions. Electric fields are MORE fundamental than the electric and magnetic fields of photons. Photons are more complex than simple electric fields. This is why e0 is more fundamental than c.


No, the classical electromagnetic interaction has not been considered fundamental for about 5 decades now. QED is more fundamental than Maxwell's equations because Maxwell's equations can be derived from QED in the high-n limit ("n" is the eigenvalue of the number operator for photons, in this case). James R is correct again.


If you don't believe me, look it up. This is the formula for Coulomb's Law:

F=q1*q2/4*Pi*e0* r^2


How does merely quoting the formula prove the fundamentality of Coulomb's law? I can just as easily write the QED Lagrangian and say, "see, it's more fundamental". Now QED is more fundamental classical E+M, but that would not be the reason. The reason is, as I stated above, is that you can derive classical E+M from QED, but the reverse is not true.


It's like the chicken and the egg problem, except everyone knows what came first. The electric field came first and photons come later.


Err....no. The theory of classical electromagnetic fields predates QED for sure, but now that photons have been discovered, it is safe to assume that they always existed.

Tom,

<i>I already showed you. I pointed out that e0 is used in Coulomb's Law...

F=q1*q2/4*Pi*e0* r^2</i>

Shouldn't that be:

F=u0 c^2 q1*q2/4*Pi r^2

?

See? u0 and c^2 are more fundamental! :)

Tom2,

Welcome to sciforums!!

I really don't see how photons can be more fundamental than electric fields, when photons create electric fields.

If e0 is not fundamental in Coulombs Law, what would you replace it with?

Tom

James,

I already showed you. I pointed out that e0 is used in Coulomb's Law...

F=q1*q2/4*Pi*e0* r^2

Shouldn't that be:

F=u0 c^2 q1*q2/4*Pi r^2

?

See? u0 and c^2 are more fundamental!



Why would two charged objects at rest care what the magnetic permeability of space around them is when they are not producing magnetic fields? Shouldn't u0 only be used in situations where magnetic fields are present? :)

Tom

Originally posted by Prosoothus
[B]Tom2,

Welcome to sciforums!!


Thank you!


I really don't see how photons can be more fundamental than electric fields, when photons create electric fields.


You make the point that photons are fundamental yourself when you point out that they "create electric fields". That would be the very definition of fundamentality. However, I would not say that photons create electric fields, rather I would say that photons are the constituents of the electric field.

Also, do you not understand that the fact that Maxwellian electrodynamics can be derived from QED necessarily implies that QED is more fundamental?

If you are unfamiliar with these things, then you should check out a standard text such as Relativistic Quantum Fields by Bjorken and Drell.


If e0 is not fundamental in Coulombs Law, what would you replace it with?


James R already alluded to a possibility in his last post. Also, it is possible to re-define the units so that e0 needn't appear at all. That's why I referred you to Jackson's book on classical E+M. For instance, in so-called "electrostatic units", e0 is identically 1.

But this is just a game of units. Neither e0 nor c can tell you whether photons or classical electric fields are more fundamental. You have to examine the field equations, and when you do, it will become obvious that the photon field is the one which gives rise to the classical field.

Tom

Tom2,

You have to examine the field equations, and when you do, it will become obvious that the photon field is the one which gives rise to the classical field.

Are you saying that electric fields are made of electromagnetic photons??

That's hard for me to accept since the electric fields of particles are stable, while for photons they are oscillating.

You do know that the scientific community doesn't know what makes up electric fields. Assuming that they are made of electromagnetic photons is a stretch, especially since no magnetic fields are measured in a motionless electric field. (Photons have both electric and magnetic fields).

If electric fields couldn't exist without magnetic fields, then I might consider your assumption. However, everyone knows that electric fields can exist without magnetic fields.

Tom

Hi Tom,

"If electric fields couldn't exist without magnetic fields, then I might consider your assumption. However, everyone knows that electric fields can exist without magnetic fields."

Electric and magnetic fields are relative to the observer. Some observers might measure only an electric field (stationary relative to the field generating object), while others might observe both an electric and magnetic field (all observers that move relative to the field generating object). Hence I would say that it is rather exceptional not to measure a magnetic field when an electric field is present :).

Bye!

Crisp

Crisp,

Some observers might measure only an electric field (stationary relative to the field generating object), while others might observe both an electric and magnetic field (all observers that move relative to the field generating object).

You're wrong. Thank you for proving relativity is incorrect. If you study electromagnetics you will find that in order for two charged particles to interact magnetically with each other, BOTH of them must be moving or spinning. If one of them is moving while the other is not, there is no magnetic interaction, only an electric interaction. This means that a charged particle must be moving or spinning relative to an absolute frame of reference in order for a magnetic field to be generated.

However, relativity, as you pointed out, claims that if an observer is moving so that in his frame of reference both particles are moving, there should be a magnetic interaction between the particles. However, it has been found that this is not the case.

Conclusion: Magnetic fields are proof that relativity is incorrect. :)

Tom

<i>relativity, as you pointed out, claims that if an observer is moving so that in his frame of reference both particles are moving, there should be a magnetic interaction between the particles. However, it has been found that this is not the case. </i>

Your last sentence here is false.

James,

"relativity, as you pointed out, claims that if an observer is moving so that in his frame of reference both particles are moving, there should be a magnetic interaction between the particles. However, it has been found that this is not the case. "

Your last sentence here is false.

Since you didn't understand what I was saying, I'll have to rub it in. :)

Example: Lets say you have two large metal spheres, one is right behind the other in your frame of reference. The two spheres are charged, so there is an attractive electric force between them. Next let's say that you as an observer start spinning around so that, to you, both spheres appear to be spinning. Relativity claims that your frame of reference is as real as any frame of reference, so therefore the two spheres should start producing magnetic fields in your frame of reference. This means that the attractive force between the spheres should increase because, where there was only the electric attractive force, now there is an electric and a magnetic attractive force.

Question: Do the attractive forces between the spheres increase as you start spinning around, or not??

Tom

Since you didn't understand my objection, I'll have to rub it in. :)

When you spin around, the electric field produced by your spheres is observed to <b>decrease</b>. The magnetic field <b>increases</b>. The resulting force between the spheres (ignoring relativistic effects) is exactly the same as before.

You can use Maxwell's equations to show that this occurs. There's no need to bring any relativity into it.

James,

When you spin around, the electric field produced by your spheres is observed to decrease. The magnetic field increases. The resulting force between the spheres (ignoring relativistic effects) is exactly the same as before.

Got you!!!

I was expecting you to say that. That's why I put one sphere in front of the other. The electric field between the spheres does not get converted into the magnetic field, the electric fields that the observer sees coming out radially from the spheres get converted.

It's the right angle rule: A magnetic field appears at right angles to the electric field and the motion of the charge. This means that the electric field between the spheres DOES NOT decrease in strength.

Tom

The Michaelson Morley experiment was designed to measure aether drift, by comparing interferometric light speeds at right angles.

it was assumed that because light speed was constant in both directions that aether did not exist.

however this assumtion did not take into account the possibility
that aether has no drift, or that the entire experiment was in motion in the direction of aether drift.

in either case the experiment was non conclusive at best.

ultravioletten,

There has to be a flaw in the Michelson-Morley experiment because I know there is a delay between the two beams of light in the experiment. Somehow, the device doesn't detect it.

When Einstein developed his assumption on the invariance of light (light travels at c in all frames of reference), he used time dilation and length contraction to counteract this delay.

Therefore, if there is no aether, Einstein's formulas for length contraction and time dilation become unneccessary.

Tom

Originally posted by Prosoothus
Are you saying that electric fields are made of electromagnetic photons??


Yes.


That's hard for me to accept since the electric fields of particles are stable, while for photons they are oscillating.


What do you mean by "stable"? The electric field described by Coulomb's law is only a macroscopic average of a field that undergoes quantum fluctuations on a microscopic scale. The classical EM field is no more stable than, say, classical orbits of planets, which are also macroscopic averages of quantum rules.


You do know that the scientific community doesn't know what makes up electric fields. Assuming that they are made of electromagnetic photons is a stretch, especially since no magnetic fields are measured in a motionless electric field. (Photons have both electric and magnetic fields).


This is a complicated issue, and unfortunately with no equation editor I cannot explain it adequately, but here's the gist of it. Virtual photons have a property called multipolarity. An "E1" photon, for instance, corresponds to electric monopoles. It is entirely possible for such photons to be responsible for the electrostatic interaction, and indeed this is part of QED.

Consider something else: You have your electrostatic field, which you don't see how can be made of virtual photons because there is no magnetic field. Now, start moving with respect to that static field. What happens? You see a magnetic field, in addition to the electric field. Now, do you think that, just because you start moving, the field turns from "classical" to "quantum"? If so, then you need to check out one of those books I mentioned.

Originally posted by Prosoothus
You're wrong.


No, Crisp is right. An electric field on one frame will indeed appear as an electric and magnetic field in any other frame, and this is true even in Galilean relativity.


Thank you for proving relativity is incorrect.


Oh, no, not another unbeliever!


If you study electromagnetics you will find that in order for two charged particles to interact magnetically with each other, BOTH of them must be moving or spinning.


It is obvious that a stationary charge distribution in one frame will be seen as a combination electric/magnetic field in any other frame, because a moving charge distribution is actually a current distribution which, if you studied electromagnetics, gives rise to a magnetic field.


However, relativity, as you pointed out, claims that if an observer is moving so that in his frame of reference both particles are moving, there should be a magnetic interaction between the particles. However, it has been found that this is not the case.


No, it has been found that it is the case. Relativistic dynamics of charged particles is well tested and, as I pointed out, this phenomenon doesn't even require relativity to be understood qualitatively. It is a different matter altogether for a quantitative understanding, but right now you are challenging the very existence of the phenomenon, so a qualitative approach will do.


Conclusion: Magnetic fields are proof that relativity is incorrect. :)


Not even close!

ultra

however this assumtion did not take into account the possibility
that aether has no drift, or that the entire experiment was in motion in the direction of aether drift.

You are inferring the aether drifts along with the exact rotation of the Earth and the exact orbit of the Earth around the Sun and the exact movement of the Solar System around the Galactic plane and...

That is highly unlikely.

Tom2,

Did you read my post on this thread regarding the two charged spheres? What is your answer to that example: Does the force between the two spheres increase or stay the same if you start spinning around??

Tom2,

"Are you saying that electric fields are made of electromagnetic photons??"


Yes.

Let's see.....An electric field is made of photons. But a photon has an electric field, so the electric field of the photon is made of even smaller photons. But these photons also have electric fields so their electric fields are made of even smaller photons............

It doesn't seem to work that way. The electric field would have to be primary, while the photon is secondary.

Hi Tom,

"Let's see.....An electric field is made of photons. But a photon has an electric field, so the electric field of the photon is made of even smaller photons."

Who ever said that a photon has an electric field ? It has no charge, why would there be an electric field involved ?

Bye!

Crisp

cause it's called "electro magnetic radiation"?

The example where you spin around near two charged objects is complicated, because you are observing the objects from a non-inertial reference frame as you spin. Therefore, "inertial" forces arise, such as a centripetal acceleration on the objects in this case. That complicates the question of electric and magnetic fields.

James R,

The example where you spin around near two charged objects is complicated, because you are observing the objects from a non-inertial reference frame as you spin.

Actually, I would be in an inertial frame of reference as long as my spin remains constant.

Tom

Crisp,

Who ever said that a photon has an electric field ?

As c'est moi pointed out, it's a common fact that a photon produces oscillating magnetic and electric fields at right angles to each other and to the motion of the photon.

It has no charge, why would there be an electric field involved ?

I have no idea. You're guess is as good as mine. :)

Tom

Originally posted by Prosoothus
Tom2,



Let's see.....An electric field is made of photons. But a photon has an electric field, so the electric field of the photon is made of even smaller photons. But these photons also have electric fields so their electric fields are made of even smaller photons............

It doesn't seem to work that way. The electric field would have to be primary, while the photon is secondary.

No, a photon has neither an electric nor magnetic field. I'll post more on this later, but one thing you should realize is that a photon is not to be thought of as a small segment of a classical EM field. The photon field is something from which that field is built up, and yes, it is more fundamental than the Maxwellian E-field.

Is there a way to post equations here?

<i>Actually, I would be in an inertial frame of reference as long as my spin remains constant.</i>

I suggest you read up on the meaning of "inertial reference frame". A rotating frame is non-inertial.

James R,

You and Crisp have been debating with me for months that the universe is not absolute, it is relative based on the observer.

However, now you claim that relativity doesn't apply in non-inertial frames of reference such as spin.

If the universe was truly relative, wouldn't relativity apply in all frames of reference including non-inertial? Isn't the fact that relativity doesn't apply in non-inertial frames of reference proof that the universe is actually absolute?

There can only be two models: An abslolute universe OR a relative universe, not half-and-half. Obviously, as you pointed out, the correct model can't be the relative model because of the big, fat exception called "non-inertial frames of reference".

Tom

Hi Tom,

Special Relativity is only a "special" case of the theory of General Relativity. I never got into it deeper, but General Relativity can handle accelerating frames (such as spinning frames).

Bye!

Crisp

Tom,

<i>However, now you claim that relativity doesn't apply in non-inertial frames of reference such as spin.</i>

It does apply. All I said was that a spinning frame is non-inertial. It was you who called it inertial, not me.

<i>If the universe was truly relative, wouldn't relativity apply in all frames of reference including non-inertial?</i>

Yes. It does, although the laws of physics appear in modified form in non-inertial reference frames. It complicates things. That's why we like to use inertial frames - they make things much simpler in terms of the required maths.

Now you're talking 'bout inertial and non-inertial frames, I have some questions appearing in my mind

1) isn't a non-inertial frame like the earth who's spinning an inertial frame? i mean, it does have inertia no? From where the idea of making a distinction?

2) I recall the Sagnac experiment. The debate ended when James R said that since the earth is non-inertial, Special relativity does not apply. What has general relativity to tell about it? We can calculate the velocity of the earth, without needing any other frame than the earth itself. That's absolute.

Originally posted by c'est moi

[quote] 1) isn't a non-inertial frame like the earth who's spinning an inertial frame? i mean, it does have inertia no? From where the idea of making a distinction?

It has nothing to do with inertia, oddly. An inertial frame is one with zero or constant velocity. Anything spinning will have acceleration and hence be non-inertial. Same for any frame where Gravity is dominant. Strictly speaking any frame on Earth is non-inertial, gravity is everywhere. But for a sufficiently small region (about Lab sized) we can assume it is inertial. The difference in the maths is so small we can ignore it.

2) I recall the Sagnac experiment. The debate ended when James R said that since the earth is non-inertial, Special relativity does not apply. What has general relativity to tell about it? We can calculate the velocity of the earth, without needing any other frame than the earth itself. That's absolute. [/B]

I have to admit I don't know the ins and outs of the Sagnac affect. You may want to read this page about it though (http://www.mathpages.com/rr/s2-07/2-07.htm). Very informative.