Magnetism as a "relativistic effect"

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alyosha

10-28-06, 01:41 PM

Trilairian

10-28-06, 04:13 PM

Any frame is correct. If you look at it from the frame of the wire what you see is
(vp = 0)....p...p...p...p...p...p...p...p...p...p...p... p...p...p...
(ve = <-- )e...e...e...e...e...e...e...e...e...e...e...e...e ...e...

(vp = 0)....p...p...p...p...p...p...p...p...p...p...p... p...p...p...
(ve = <-- )e...e...e...e...e...e...e...e...e...e...e...e...e ...e...

Since there is one proton for each electron this frame sees no electric field but somehow the electrons try to pull each to the other wire and puls their wire along with them so we must experimentally find that there is a force of attraction and so we call it the magnetic field, but look at it now from the frame of the electrons. What you then see is
(vp = --> )p..p..p..p..p..p..p..p..p..p..p..p..p..p...
(ve = 0)....e....e....e....e....e....e....e....e....e... ..

(vp = --> )p..p..p..p..p..p..p..p..p..p..p..p..p..p...
(ve = 0)....e....e....e....e....e....e....e....e....e... ..
Now the electrons are stationary and there is no force from a magnetic field on a stationary particle. Instead what we see is that the protons are all length contracted together and the electrons therefor see an electric field coming from the other wire attracting them to it. The end result is the same as befor where the electrons are attracted to the other wire and drag their wire along with them, but according to this frame the attraction is electric not magnetic. This means that every electronic device that makes use of a magnetic field is experimental proof of relativity in its prediction of length contraction.

alyosha

10-28-06, 07:16 PM

That is a very fascinating line of reasoning. However, would this then be to say that in order for the currents of electrons to influence eachother they must necessarily be moving with respect to a wire? Because certainly, if we simply "picked up" the entire wire and placed it upon a high speed aircraft we could then say the electrons are "moving" with respect to us, but no longer with respect to the wire.

Pete

10-29-06, 06:44 PM

Pete

10-29-06, 07:15 PM

The case of electrons moving freely without a wire is different again.
In this case, the electron streams feel a very strong electrostatic repulsion. I'm not sure how to calculate this force properly, but a rough calc puts in in the order of 1016 N/m. This makes the magnetic attraction between the streams pretty insignificant - about 1029 times smaller than the electrostatic repulsion for a current of 1 amp.

For incredibly high currents, the magnetic attraction will be significant, reducing then net repulsive force. For example, consider a reference frame in which the electrons are moving at 2.6×108m/s. In this frame, the electrons are packed together at twice the density (by length contraction), and the resulting current is about 8.3×1011 amps.

I don't know the quantitative relationships from here, but qualitatively here are the things that come into consideration:
The high current gives a high magnetic attraction between the streams.
The high electron density (due to length contraction) gives a higher electrostatic repulsion between the streams.
Length contraction also gives a high mass density between the streams, which reduces the effect of the resulting force between the streams.
The effect of the resulting force is also reduced due to the high relativistic momentum of each electron in the streams.
Time dilation might also need to be considered. I'm very fuzzy on the details from here on. Suffice to say that it's not intuitively simple!

CANGAS

10-31-06, 02:43 AM

Take two long thin parallel wires, 1m apart in vaccuum, both carrying 1 amp in the same direction. These wires experience an attractive magnetic force of 2×10-7 N/m

For typical wires, there might be 1022 free electrons per metre in each wire, with an average velocity of 0.625 mm/s.
This gives a relativistic gamma factor of about 1 + 2×10-24. In the average electron rest frame, this means that there is one extra proton for every five metres or so of wire.

This isn't going to make much difference - the resulting electrostatic force between the wires is about 4×10-29 N/m, about 5×1021 times smaller than the original magnetic force. Even worse, the force is in the wrong direction. Both wires have a net positive charge, so the force acts to push them away from each other.

So, in the electron rest frame the force between the wires is not due to length contraction.

It's due to the moving positive charges. In the electron average rest frame, the positive charges are moving in the opposite direction, producing a magnetic field which in turn produces an attractive force between the wires.

Say what?

Pete, you did a good job of looking up the published numbers, initially provided by CANGAS free of charge ( pun unintentional ), but; you say " the positive moving charges". In Special Relativity, which you have already proved that you love and obey, either the free electrons ( actually NEGATIVE moving charges ) or the ionized lattice atoms ( actually the real POSITIVE charges) must be eligible to be called the MOVING charges.

And, in case any smarty pants wants to dig up the matter of the twin paradox in which the ACCELERATED body wins, let me explain that the combined gravitational and electromagnetic interraction between participating parties means that BOTH the free electrons in the electron gas and the bound atoms in the lattice are mutually accelerating, unless some smarty Pete wants to explain to us how Newton's action and reaction is not working.

Pete

10-31-06, 04:25 AM

In the wire's rest frame the electrons are moving.
Int the electron rest frame, the wire is moving.
It's not difficult to grasp, CANGAS... it's basic Galilean relativity.

CANGAS

11-02-06, 02:52 AM

In the wire's rest frame the electrons are moving.
Int the electron rest frame, the wire is moving.
It's not difficult to grasp, CANGAS... it's basic Galilean relativity.

Anything, and I mean ANYTHING that Pete can understand is a genuinely simple concept.

Pete

11-02-06, 05:21 AM

Throwaway insults? Please...

Come on, cagnas, you're not even trying anymore.

CANGAS

11-03-06, 12:36 AM

Throwaway insults? Please...

Come on, cagnas, you're not even trying anymore.

You are right. And I have never been trying. I am glad to see that you understand such a riduculously simple concept.

Pete, you have obviously been off your meds for some time now. It is typical for someone with your kind of condition to honestly believe that you can do without them. I have been closely acquainted with two people during my lifetime who were identical in their behavior. YOU CANNOT REALLY DO WITHOUT YOUR MEDS! Get it?

Trilairian

11-03-06, 09:28 AM

Take two long thin parallel wires, 1m apart in vaccuum, both carrying 1 amp in the same direction. These wires experience an attractive magnetic force of 2×10-7 N/m

For typical wires, there might be 1022 free electrons per metre in each wire, with an average velocity of 0.625 mm/s.
This gives a relativistic gamma factor of about 1 + 2×10-24. In the average electron rest frame, this means that there is one extra proton for every five metres or so of wire.

This isn't going to make much difference - the resulting electrostatic force between the wires is about 4×10-29 N/m, about 5×1021 times smaller than the original magnetic force. Even worse, the force is in the wrong direction. Both wires have a net positive charge, so the force acts to push them away from each other.

So, in the electron rest frame the force between the wires is not due to length contraction.

It's due to the moving positive charges. In the electron average rest frame, the positive charges are moving in the opposite direction, producing a magnetic field which in turn produces an attractive force between the wires.You are guessing, and wrong. The protons of one wire don't experience a force from the other. In their frame they are at rest and so experience no magnetic force, and the opposite wire is neutral and so experience no electric force from the opposite wire. Length contraction on the electrons yielding a magnetic field is the answer has been quantitatively demonstrated as the answer and is the only answer. For more info see wiki's section "The origin of magnetic forces"
http://en.wikipedia.org/wiki/Relativistic_electromagnetism
This also settles the silly question whether length contraction is real.

Pete

11-03-06, 10:33 AM

The protons of one wire don't experience a force from the other.
Sure they do. In all frames, they are repelled from the protons in the other wire, and attracted to the electrons in the other wire.

In their frame they are at rest and so experience no magnetic force
Right - in their rest frame. But in the electron's rest frame, the protons do experience a magnetic force, which is the question under discussion.

Length contraction on the electrons yielding a magnetic field is the answer has been quantitatively demonstrated as the answer and is the only answer.
As I demonstrated, the numbers don't work out. It's short by a factor of 1021.

The discussion in Wikipedia is correct, but different to this discussion - it describes a test charge beside a current-carrying wire, rather than the two parallel wires of our thread.

Trilairian

11-03-06, 07:26 PM

Sure they do. In all frames, they are repelled from the protons in the other wire, and attracted to the electrons in the other wire.
Don't be rediculous, the net is zero.

Right - in their rest frame. But in the electron's rest frame, the protons do experience a magnetic force, which is the question under discussion.
Its not under question. I told you the net was zero. In this frame the net including electric must also be zero.

As I demonstrated, the numbers don't work out. It's short by a factor of 1021.
You demonstrated no such thing. You were wrong.

The discussion in Wikipedia is correct, but different to this discussion - it describes a test charge beside a current-carrying wire, rather than the two parallel wires of our thread.
Its not different. It is an element of this discussion. Integrate it. Wiki proved you wrong. I proved you wrong. Relativity proved you wrong.

Pete

11-03-06, 07:58 PM

Its not under question. I told you the net was zero. In this frame the net including electric must also be zero.
The net electric is not quite zero in both frames - the not quite is a tiny amount due to length contraction.
In both frames, the net magnetic is significant. In the wire's rest frame, th magnetic force comes from the moving electrons, in the electron rest frame it comes from the moving positive charges.

Trilairian

11-03-06, 08:25 PM

The net electric is not quite zero in both frames - the not quite is a tiny amount due to length contraction.
In both frames, the net magnetic is significant. In the wire's rest frame, th magnetic force comes from the moving electrons, in the electron rest frame it comes from the moving positive charges.

No, the math proved you wrong.

Pete

11-03-06, 09:44 PM

Trilairian

11-03-06, 10:05 PM

What maths, Trilarian? The only maths in this thread has come from me, and it showed that your explanation gives a force in the wrong direction, and too low by a factor of a thousand million million million.
No, you didn't do any math and I meant the math in the actual analysis.
Your speculation on that factor is plain wrong.

Pete

11-03-06, 10:24 PM

Perhaps you could share your mathematical expertise with us, Trilarian. All we've heard from you is a rather unconvincing repetition of "you're wrong... no, you're wrong... you're wrong".

Trilairian

11-03-06, 10:28 PM

Perhaps you could share your mathematical expertise with us, Trilarian.
All right, I'll write it up and post it by Sunday night.

vx220

11-04-06, 08:30 AM

Trilairian

11-04-06, 08:59 AM

... or the above paragraph would be incorrect and it can hardly be incorrect.

Sure it can be.

vx220

11-04-06, 09:21 AM

So why would it be incorrect? I'm not looking for a fight or a wisdom contest or whatever.

Trilairian

11-04-06, 10:14 AM

It will be sufficient to prove my point to consider the simple case of identical infinite parallel wires carrying the same amount of current. In the wire frame the number of carriers per volume will be n and the current will be I. Doing a velocity boost the amount of the drift velocity v d a wire will have a "drift electron" number density of n/g and an unpaired proton number density density of ng. This yields a charge density of 
eng - en/g = eng(1 - g-2) = eng(1 - (1 - v2/c2)) = engvd2/c2
The field of an infinite line charge easily obtained from Gauss's law is
E = (l/2pe0)(1/r)
and in our case 
l = Aengvd2/c2
So 
E = [Aengvd2/(2pe0c2)](1/r)
The force per length on one wire's drift electrons due to this electric field from the other wires unpaired charges is
(enAg)E = A2e2n2g2vd2/(2pe0c2)](1/r)
and is attractive. 
To transform this force per length back to the wire frame first consider that a ordinary force perpendicular to the wire on a test charge transforms as 
f ' = g -1 f/(1 - uxv/c2)
Since ux = v = vd for the transformation of the force on a drift electron back to the wire frame this would be
f ' = g f
f = g -1 f '
On the other hand we are actually transforming a force per length not a net force to the wire frame and so we pick up one factor of g in transforming the density due to length contraction.
This yields a force per length of
A2e2n2g2vd2/(2pe0c2)](1/r) ~ A2e2n2vd2/(2pe0c2)](1/r)
Now Anevd we call the current I and we have
I2/(2pe0c2)](1/r)
Using m0e0 = 1/c2
we have 
m0I2/(2pr)
Now remember that. This used NO MAGNETIC FIELDS. It was only length contraction and the electric field that yielded this force per length.
Now that aside calulating the magnetic field using Gausses law from one wire you get
B = m0I/(2pr)
and the force per length due to the field on a current I will be
IB = m0I2/(2pr)
But there it is again.
So I mathematically proved you wrong. When I assert something with certainty regarding relativity or the history of religion always assume that I am right and ask why it is that you are wrong when it doesn't seem so.
In the end this shows that length contraction though relative is real as the existence of the magnetic field proves it so.

vx220

11-04-06, 10:38 AM

Ok. I am wrong, heh :)

How does time dilation enter into the equations?
Would time dilation cause an observer to see a field to be stronger or weaker than some other observer would see?

Trilairian

11-04-06, 03:14 PM

Ok. I am wrong, heh :)

How does time dilation enter into the equations?
Would time dilation cause an observer to see a field to be stronger or weaker than some other observer would see?

Above it yields the factors of g that enter into the transformation of ordinary force. In short, Newton's second law of motion is actually correct for relativity as long as the time derivatives are done with respect to proper time, the time according to the test mass.
In other words, the law of motion for relativity is that four-vector force is mass which is invariant(does not change with speed) times four-vector acceleration.
F = mA
In special relativity the components of the four-force are related to the components of four momentum by
Fm = dpm/dt
and the four-acceleration is the second proper time derivative of position
Am = d2xm/dt2
So Newton's second law for special relativity then becomes
dpm/dt = md2xm/dt2
This looks just like Newtonian dynamics so far. The key is that when you observe the behavior in the lab, you are recording speeds and accelerations using your coordinate time, not the particle's time and so the dynamics looking so much different than Newton would have thought is due in entirety to time dilation. So, every experiment which someone wrongly claims is a verification of mass dilation, is actually a demonstration of time dilation. But, my point was that in the force transformation the factors of g were due to time dilation, so lets look at how the force transforms. The four-force is a four vector which is a tensor and therefor Lorentz transforms in special relativity.
F ' m = (¶x'm/¶xn)Fn
dp'm/dt = (¶x'm/¶xn)dpn/dt
Now enters time dilation
gu'(dp'm/dt') = gu(¶x'm/¶xn)dpn/dt
Then enters a g transformation
gu' = g(1 - uxv/c2)gu
g(1 - uxv/c2)gu(dp'm/dt') = gu(¶x'm/¶xn)dpn/dt

g(1 - uxv/c2)(dp'm/dt') = (¶x'm/¶xn)dpn/dt
Then look at a perpendicular component
g(1 - uxv/c2)(dp'y/dt') = (¶x'y/¶xy)dpy/dt + 0 + 0 + 0
g(1 - uxv/c2)(dp'y/dt') = (1)dpy/dt
But these are the ordinary forces according to special relativity
g(1 - uxv/c2)f'y = fy
and you see the only factors remainging in the transformation were due to time dilation.
How it effects how fields look in general is a bit more complicated. In fact even stating that magnetism is due to length contraction is a tad oversimplified. Really the field is unified into the electromagnetic field which is a rank two tensor and in special relativity Lorentz transforms. Thus length contractions and time dilations and even relative simultaneity all enter into how the components of both the electric and magnetic fields transform. In the end if relativity wasn't right, not of our electronic devices would work at all. They all work the way they do because relativity's effects are real. (even though relative)

Pete

11-05-06, 06:53 AM

Thanks Trilarian. Respect to you!
Is it just me, or is the Symbol font not being displayed for others as well? I've transcribed your post so it displays better. Please check to make sure I haven't stuffed it up.

This is a most interesting case. It was extremely puzzling, because while your calculations looked good and your results checked out (by a quick reality check against the SI ampere definition), I didn't understand why you only analysed the force on the electrons, rather than all unpaired charges in the wire. It seemed to me that you considered the force on a free infinite stream of electrons running parallel to an infinite wire, rather than two parallel wires, yet still got the correct force.

However, on close inspection I discovered a small error that (surprisingly) turned what seems to me to be an incorrect analysis into a correct result.

It will be sufficient to prove my point to consider the simple case of identical infinite parallel wires carrying the same amount of current. In the wire frame the number of carriers per volume will be n and the current will be I. Doing a velocity boost the amount of the drift velocity vd a wire will have a "drift electron" number density of n/γ and an unpaired proton number density density of nγ. This yields a charge density of:
enγ - en/γ = enγ(1 - γ-2) = enγ(1 - (1 - v²/c²)) = enγvd²/c²
I was using the assumption that the wire was charge neutral when no current was flowing, but this way is OK too.
The field of an infinite line charge easily obtained from Gauss's law is:
E = (λ/2πε0)(1/r)
...and in our case:
λ = Aenγvd²/c² [where A is the area of cross-section of the wire - Pete]

So:
E = [Aenγvd²/(2πε0c²)](1/r)
The force per length on one wire's drift electrons due to this electric field from the other wires unpaired charges is:
(enAγ)E = A²e²n²γ²vd²/(2πε0c²)](1/r)
...and is attractive.
This is correct. But what about the force per length on the wire's unpaired protons due to this electric field? It will be:
(enA/γ)E = A²e²n²vd²/(2πε0c²)](1/r)
...and repulsive. So the net force per length on the wire in the electron rest frame due to the other wire's electric field will be:
A²e²n²vd²/(2πε0c²)](1/r) (γ² - 1)
...and attractive. Right? Why only consider the force on the electrons?

To transform this force per length back to the wire frame first consider that a ordinary force perpendicular to the wire on a test charge transforms as:
f ' = γ -1 f/(1 - uxv/c²)
This is where I got lost in my rough calculations. I haven't applied a Lorentz transform to a force like this before. I can't quickly confirm it, and haven't time to derive it (I'm weak on relativistic dynamics) so I'll take the expression on faith... can you outline its derivation, or point me to a text or a link?

Since ux = v = vd for the transformation of the force on a drift electron back to the wire frame this would be:
f ' = γ3 f
f = γ -3 f '
Ah! There's the mistake!
f ' = γ3 f --- is incorrect. It should be:
f ' = γ f
f = f ' / γ
Right?
Carrying this through...
On the other hand we are actually transforming a force per length not a net force to the wire frame and so we pick up one factor of γ in transforming the density due to length contraction. This means that to get the force per length in the wire frame we in the end multiply by γ-2. This yields a force per length of:
A²e²n²vd²/(2πε0c²)](1/r)
...we find that the force per length in the wire frame is the same as the force per length in the electron frame (this is surprising, but like I said I haven't got a good intuitive feel for relativistic dynamics).
Now Anevd we call the current I and we have:
I²/(2πε0c²)](1/r)
Using μ0ε0 = 1/c² we have:
μ0I²/(2πr)
So the attractive force on the electrons is:
A²e²n²γ²vd²/(2πε0c²)](1/r)
= γ²μ0²I²/(2πr)
...and the repulsive force on the unpaired protons is:
A²e²n²vd²/(2πε0c²)](1/r)
= μ0²I²/(2πr)
...giving a net attractive force of:
(γ² - 1)μ0²I²/(2πr)

For γ very close to 1, the electrostatic forces on the wire in the electron drift frame are clearly insufficient to account for the magnetic force on the wire in the wire rest frame.

So the next step is to consider the unpaired protons. In the electron drift frame, the unpaired protons are moving with velocity vd.
These moving charges produce a magnetic field, which in turn produces a force on the moving charges (unpaired protons) in the other wire.
Right?

Trilairian

11-05-06, 08:46 AM

Thanks Trilarian. Respect to you!
Is it just me, or is the Symbol font not being displayed for others as well?
You have to set your preferences to accept HTML
I didn't understand why you only analysed the force on the electrons, rather than all unpaired charges in the wire.
Because I knew apriori that the net force on the protons electric and magnetic due to the other wire is zero in every frame or at least very close. This should be obvious. Just look at the wire frame. The protons aren't moving so no magnetic force can be on them. The wires are uncharged no no electric force can be on them. Zero force transforms to zero force in every frame. It makes no difference what electric or magnetic fields they are in in any other frame as the sum of the forces are zero. The only electric field they feel in the wire frame is due to the electrons to which they are immediately bound. Thats why I only looked at the force on those electrons. They are forced and drag the protons in the wire along with them. There is perhaps a miniscule contribution of the order 1-g2 but for all practical purposes can be ignored for this discussion. It would merely serve to correct the factor of g2 (~1 anyway) in the final answer.
In the wire frame the
However, on close inspection I discovered a small error that (surprisingly) turned what seems to me to be an incorrect analysis into a correct result.
Your thought that there must be an error is whats in error. Face it.
But what about the force per length on the wire's unpaired protons due to this electric field? It will be:...No, the net from both fields from the other wire will be zero in every frame. They only feel the field of the nearby electrons in their frame. The net force from the fields of the other wire is zero on them in every frame.
I haven't applied a Lorentz transform to a force like this before. I can't quickly confirm it, and haven't time to derive it (I'm weak on relativistic dynamics) so I'll take the expression on faith... can you outline its derivation, or point me to a text or a link?
I derived it in this very thread! Fine,
http://www.geocities.com/zcphysicsms/chap3.htm
Look at Problem 3.2.5

Ah! There's the mistake!
f ' = γ3 f --- is incorrect. It should be:
f ' = γ f
f = f ' / γ
Right?
Yeah there was an arithemetic sign error carrying over a factor of g2 but that doesn't change the result for realistic drift velocities so I'm still right on the physics. I don't make mistakes conserning the physics of relativity anymore so as I said, just assume I'm right and ask why your wrong when it seems otherwise when I make a statement with certainty conserning relativity or the history of religion. Got bored, snipped the rest.

CANGAS

11-05-06, 11:01 PM

Protons in the conductor are, except in extremely rare and unusual circumstances, imbedded within the lattice atoms, or, we might more accurately say, within the lattice ions.

Lattice ions are in the vast majority deprived of the amount of movement that free, thermal, electrons have the privelege of enjoying. However, the lattice ions are perfectly free to vibrate within their electromagnetic entrapment within the lattice. So, while their movement is much less notorious than the movement of the free thermal electrons making up the electron gas, the lattice ions are certainly subject to acceleration by the electric field, or, electric wave, as the case may be, so that, in terms of whatever Relativity we choose to invoke, we have no choice but to consider either the free electrons or the entrapped protons to be equally guilty of performing acceleration, and therefore being equally guilty of being the proper, or, if preferred, the preferred reference frame.

Of course, if one wishes to think in terms of the ORIGINAL basis of the Fitzgerald Lorentz contraction, the electrons only can be considered to be moving in terms of absolute motion in absolute space.

Pete

11-06-06, 12:02 AM

You have to set your preferences to accept HTML
Theres no such preference here that I can see. I've found the problem - it was to do with Firefox and the way it handles the html4.01 Doctype. I found a resolution here (http://home.att.net/~numericana/about.htm#symbol).)

I derived it in this very thread! Fine,
http://www.geocities.com/zcphysicsms/chap3.htm
Look at Problem 3.2.5
Thanks. There's no need get all defensive. I didn't decipher your response to vx220 because of the problem with the Symbol font.

I don't make mistakes conserning the physics of relativity anymore so as I said, just assume I'm right and ask why your wrong when it seems otherwise when I make a statement with certainty conserning relativity or the history of religion.
In your dreams! :D Claims like that are worthless, as I'm sure you're aware... I'll judge your posts on their merits, just like everyone else's.
How do you ever learn anything if you don't consider the possibility that you might be wrong?

Got bored, snipped the rest.
Well fuck you very much! :p

The difference in our approaches is clear:
You chose to begin by noting that the electrostatic and magnetic force on the positive charges almost or exactly cancels (depending on the precise charge distribution in the wire rest frame), and used the electrostatic force on the electrons to determine the force on the wire.
I chose to being by noting that the electrostatic forces on the positive and negative charges almost or exactly cancel (again, depending on the precise change distribution in the wire rest frame), and used the magnetic force on the positive charges to determine the force on the wire.

Both are correct, since the magnetic force on the positive charges in the electron drift frame is the same as the electrostatic force on the negative charges (except at relativistic drift velocities).

CANGAS

11-06-06, 12:11 AM

Theres no such preference here that I can see. I've found the problem - it was to do with Firefox and the way it handles the html4.01 Doctype. I found a resolution here (http://home.att.net/~numericana/about.htm#symbol).)

Thanks. There's no need get all defensive. I didn't decipher your response to vx220 because of the problem with the Symbol font.

In your dreams! :D Claims like that are worthless, as I'm sure you're aware... I'll judge your posts on their merits, just like everyone else's.
How do you ever learn anything if you don't consider the possibility that you might be wrong?

Well fuck you very much! :p

The difference in our approaches is clear:
You chose to begin by noting that the electrostatic and magnetic force on the positive charges almost or exactly cancels (depending on the precise charge distribution in the wire rest frame), and used the electrostatic force on the electrons to determine the force on the wire.
I chose to being by noting that the electrostatic forces on the positive and negative charges almost or exactly cancel (again, depending on the precise change distribution in the wire rest frame), and used the magnetic force on the positive charges to determine the force on the wire.

Both are correct, since the magnetic force on the positive charges in the electron drift frame is the same as the electrostatic force on the negative charges (except at relativistic drift velocities).

Please quit saying "fu**ck" in an obvious attempt to be excessively profane and rude.

THANK you very much.

CANGAS

11-08-06, 11:58 PM

There is a faction who claim that Lorentz contraction, embodied in toto in Special Relativity, completely explains magnetism as a relativistically transformed way of observing electric fields. Thus, as it is claimed, two isolated charges moving identically through space have no motion relative to each other, therefore have no relativistically transformed way of of observing each other's electric field, thus have no electromagnetic interraction other than the basic Coulomb electric force.

The Lorentz contraction proposition depends upon the electric force interraction between moving charges of the electric current and stationary lattice ions of the conductor. Generally it is explained that the current charges are like a long train whizzing past lattice ions as if they were mile markers: Lorentz contraction compresses the train so that the moving train has more, shorter, boxcars between mile markers than when sitting still. Two adjacent trains would be observed to each be compressed and have abnormally many , shorter, boxcars, if observed by a weary hobo sitting still on the embankment.

So, the Lorentz faction claim that abnormally many charges in the contracted current exert abnormally greater Coulomb force upon an adjacent conductor's abnormally more current charges and upon its NORMALLY numerous lattice ions.

Our weary hobo observes an Eastbound train to have N % more, shorter, boxcars between mile markers. W. H. also observes a Westbound train, traveling at the same speed, to have the same N % more, shorter boxcars between mile markers.

Replacing our weary hobo with a weary stationary observer, the embankment with his laboratory, the boxcars with current charges, and the mile markers with lattice ions, we now ponder relativistic observation of two parallel wires, first carrying direct current in identical directions, then carrying direct current in opposite directions.

First applying current in identical directions, the S. O. observes each wire to contain N % more electrons and therefore to exert more coulomb force than in the absence of current.

Next applying the same amount of current in opposite directions, the S. O. again observes each conductor to contain N % more electrons. The S. O. realizes that the number of excess electrons is identical as in the first test. And the S. O. realizes that the number of lattice ions is the same as in the first test, and the same as when there is no current.

The S. O. realizes that the two tests result in exactly the same quantity and sign of excess Coulomb force between the two wires.

The S. O. then feels very sad for Ampere and every other physicist who have somehow believed that changing the direction of current could possibly makes two wires alternately attract and then repel; Relativity has proved that the laboratory observer must observe an identical result, not an opposite result. but he is cheered by the science job of having to look for his favorite slide rule so that he can sum the Coulomb forces to discover if the net result is for the forces to cancel, or always attract, or always repel. But he is happy that Relativity has shown that the Coulomb force due to Lorentz contraction will always be observed to be the same in every laboratory observation of the Ampere science experiment.

The S. O. considers that if an observer were riding an current electron, the observation might be different. But he has been warned too many times already, by Relativity experts, to avoid mixing or switching frames. And he has been advised that if something happens in one frame, it happens in every frame ( even though such advice directly contradicts Einstein's Relativity of Simultaneity ). Therefore, he decides that if a stationary laboratory observer observes the identical result in both tests, that is good enough. The weary S. O. just wants to quit work and rest and sleep a little while.

It is exactly like what Merlin told King Arthur ( perhaps slightly paraphrased ): " You must choose. You must burn the Relativity. Or, you must burn the Ampere science experiment.".

The real world description of current in a wire is vastly more complicated than a simile to train boxcars. However, the Lorentz explanation crucially depends upon such a simplified metaphor. The Lorentz explanation was first proposed well over a century ago. The Quantum Physics description of current in a wire is vastly more complicated and provides us with much more information than what Lorentz or Einstein had to work with. By the time Feynman was confronted with having to explain the Ampere experiment, he was much too deeply invested in the standard model to risk disagreeing with the Lorentz explanation. In such cases it is not surprising to observe a famous scientist try to jump backward through a hoop and coincidentally bewilder someone trying to figure out what they mean. What they really mean is that they are smart enough to know when a theory stinks, but their career depends too much upon defending it anyway.

In my HUMBLE opinion.

CANGAS

11-12-06, 01:36 AM

I was reading the feynman lectures and he was going over some of the conceptual subtleties of magnetic fields. One point remained unclear to me, however, despite feynmans attempt to clarify it. In the parallel wire experiment, when we allow the currents to flow in the same direction the wires attract, and if they are in the opposite direction, they repel, etc. Essentially the point is that the direction of the magnetic field depends on the direction of the current. When considering our "stationary" point of view with respect to the currents, this makes sense, but from the point of view of the electrons traveling parallel in the same direction, how do they establish movement at all? In fact, even if from our point of view the electrons are moving in the same direction, if one electron is moving faster, then from its point of view, the other electron may as well be moving in the opposite direction, which would predict an oppositely oriented magnetic field. How do we decide on our frame of reference, and if it doesn't make a difference, how is their apparently relative motion accounted for? Maybe the "correct" frame of reference is that of the wire that the electrons are moving upon, but this still doesn't quite explain the magnetic effects of an electron moving freely without a wire. I suppose the question can be summarized as "Motion? With respect to what?"

Feynman is probably, though your incomplete information leaves it to some amount of guesswork, referring to the Special Relativity dictum that ANY and EVERY observer MUST consider himself to be the stationary observer.

An observer may be standing still and watching a moving thing whizz by. The observer MUST consider himself to be stationary, and, according to Special Relativity, the observed moving object is necessarily observed to be time dilated, length contracted, and mass increased.

If the observer has jumped onboard the moving thing and is therefore sitting on it as if a rider is sitting upon a motorcycle, then the observer MUST again consider himself as being stationary. The motorcycle is stationary. It is the road, the landscape, and everything that is actually moving, going rearward past the stationary rider and motorcycle at 90 miles per hour or whatever. According to Special Relativity.

So, a scientist may be in a laboratory watching two wires which are bolted down and immoveable with respect to the laboratory, and the scientist must consider himself to be stationary while he watches the current electrons whizz through the wires.

Or, the scientist may imagine that he has gone through the Disney Shrinkulator and has been shrunk down to electron size, has jumped upon an electron and strapped himself safely in place with duct tape, and is riding upon a current electron which would be considered to be moving through a conductor. In such a case, according to Special Relativity, the scientist MUST consider himself to be stationary. The scientist MUST consider himself to be the stationary observer and MUST consider his ride electron to be stationary while the conductor, the universe, and everything is observed by the scientist to be whizzing past. The conductor atoms ( actually ions ) must be considered to be time dilated, length contracted, and mass increased. And if there is a parallel wire, it must be considered to be moving rearward. If there are current electrons in the parallel wire, their apparent motion MUST be considered by the electron-riding observer to be all THEIR motion. The observer IS STATIONARY.

Always. Acording to Special Relativity.

alyosha

11-12-06, 11:49 AM

About the driving on the road and considering yourself stationary; wouldn't that lead to some strange complications in explaining how nothing stationary with respect to the earth is experiencing a strong outward "force" due to the "rotation" of the earth? Or is that just a failure of special relativity to explain accelerated reference frames?

Pete

11-13-06, 09:11 PM

Hi aloysha,
In accelerated reference frames, "frame forces" need to be introduced to correctly describe how things behave.

Examples of frame forces include centrifugal forces, coriolis forces, and (in a general relativity context) gravity.

CANGAS

11-15-06, 12:00 AM

"Frame forces" is in the same category as Groucho Marx's "Sanity Clause".

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