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Magnetism seems absolute despite being relativistic effect of electrostatics

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kmarinas86
#19
Feb14-12, 06:32 PM
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Quote Quote by DaleSpam View Post
No, I already covered the non-steady state situation in post 8. None of your suggestions are correct, neither in the transient nor in the steady-state conditions.
Can you describe the nature of the length contraction?

The rest frame of the straight conductor is chosen. If the negative charge of the free electron steady current is seen uniformly spread throughout the wire, then I have no choice but to assume that the free electron charge density is inversely proportional to the factor [itex]\sqrt{1-\left(\frac{v}{c}\right)^2}[/itex], where v is the electron drift velocity of the current. Logically then, the amount free electrons in the same conductor must increase to the same degree as the free electron charge density does. The density and count of other charges is not affected. This would violate the neutral wire assumption. If that is not the case, prepare or find a picture of what you think actually happens.

Quote Quote by kmarinas86 View Post
[....] You don't have just the fundamental particles contracting. In the extreme case, going from 0 current to a very high current would cause the following to occur:

This

+       +       +       +       +
-       -       -       -       -
into this

+       +       +       +       +
              -----
or

+       +       +       +       +
-----
or


+       +       +       +       +
                            -----
et cetera
Quote Quote by DaleSpam View Post
It wouldn't be as simple as that. Remember the relativity of simultaneity.

Suppose that you suddenly turn the current on at time t=0 in the lab frame. So at t=0 electrons begin leaving one end of the wire at a certain rate and entering the other end of the wire at the same rate, so there is no net charge. In the moving frame the beginning of the electrons leaving one end is not the same time as the beginning of the electrons entering the other end, so there is a net charge.
Your "uniformity" assumption combined with a length contraction factor of [itex]\sqrt{1-\left(\frac{v}{c}\right)^2}=1/8[/itex], then we would have something like this:

+       +       +       +       +
---------------------------------
DrGreg
#20
Feb14-12, 06:38 PM
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I think the following comment I made 2 years ago applies here:

Quote Quote by DrGreg

I think it is perhaps worth pointing out that some people have a false impression about what Lorentz contraction is. They may think that "when something accelerates it gets shorter". Or to be a bit more precise, if Alice measures (=x) something at rest (relative to Alice) and then later measures (=y) the same thing in motion, the length contracts. There may then be some debate over whether or not the "things" this applies to are just solid objects, or gaps between objects, or "space itself".

The above description of Lorentz contraction is wrong.

In many circumstances, what I said above is true, but reason it is true is not simply Lorentz contraction alone; it is Lorentz contraction plus some other reason combined.

A more accurate description of Lorentz contraction is that when inertial observer Bob measures the length z between two things both at rest relative to Bob, and another inertial observer Alice in relative motion measures the length y between the same two things at the same time, Alice measures a shorter distance than Bob.

So, the situation I described in the first paragraph will arise if there is a reason why Alice's initial "rest distance" x between the two things beforehand is the same as the Bob's final "rest distance" z. For example if the the two things are the two ends of a rigid object that doesn't break into pieces as a result of the acceleration.

The attached illustration emphasises my point. The transformation of x to y is not Lorentz contraction. The transformation of z to y is Lorentz contraction. If there is a reason why x = z, then the transformation of x to y will be a contraction. But if there's no reason, then contraction need not occur.




The application to this thread is that the electrons are not rigidly linked to each other so there's no reason for the rest-distance between them when moving to equal the rest-distance when not. In fact they will spread out to fill whatever space is available to them.

There is no "centre of contraction" because contraction-due-to-acceleration doesn't occur.
kmarinas86
#21
Feb14-12, 07:12 PM
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Quote Quote by DrGreg View Post
I think the following comment I made 2 years ago applies here:

Quote Quote by DrGreg
I think it is perhaps worth pointing out that some people have a false impression about what Lorentz contraction is. They may think that "when something accelerates it gets shorter". Or to be a bit more precise, if Alice measures (=x) something at rest (relative to Alice) and then later measures (=y) the same thing in motion, the length contracts. There may then be some debate over whether or not the "things" this applies to are just solid objects, or gaps between objects, or "space itself".
There "may be" debate about it? You know this, or is that just a way of saying that others may debate about that if they like?

Quote Quote by DrGreg View Post
Quote Quote by DrGreg
The above description of Lorentz contraction is wrong.

In many circumstances, what I said above is true, but reason it is true is not simply Lorentz contraction alone; it is Lorentz contraction plus some other reason combined.
Hmm. Ok.

Quote Quote by DrGreg View Post
Quote Quote by DrGreg
A more accurate description of Lorentz contraction is that when inertial observer Bob measures the length z between two things both at rest relative to Bob, and another inertial observer Alice in relative motion measures the length y between the same two things at the same time, Alice measures a shorter distance than Bob.

So, the situation I described in the first paragraph will arise if there is a reason why Alice's initial "rest distance" x between the two things beforehand is the same as the Bob's final "rest distance" z. For example if the the two things are the two ends of a rigid object that doesn't break into pieces as a result of the acceleration.
That's simply a Lorentz boost.

Quote Quote by DrGreg View Post
Quote Quote by DrGreg
The attached illustration emphasises my point. The transformation of x to y is not Lorentz contraction. The transformation of z to y is Lorentz contraction. If there is a reason why x = z, then the transformation of x to y will be a contraction. But if there's no reason, then contraction need not occur.
The application to this thread is that the electrons are not rigidly linked to each other so there's no reason for the rest-distance between them when moving to equal the rest-distance when not. In fact they will spread out to fill whatever space is available to them.
Let's assume that the total space (defined as the volume inside the conductor) available does not change. Let's assume that we are only viewing things from BOB's perspective. Forget Alice here. So the volume is defined by BOB.

Quote Quote by DrGreg View Post
There is no "centre of contraction" because contraction-due-to-acceleration doesn't occur.
I can imagine a "center of contraction" without any assumptions about acceleration whatsoever:



In this example, the center of contraction exists (NOT "occurs") halfway through the baseball. No assumption about acceleration or deceleration of the baseball is required.

If I had two baseballs moving inline at the same velocity, I might imagine the center of contraction exists (NOT "occurs") between the two baseballs.

I can compare Lorentz contractions of electron velocities at different moments. Let's compare the free electron density before and after switching on the current. The "contraction" we speak of is a NOUN not a verb. We are not concerned with the contracting, but the contracTION.

We can say that before and after turning the current ON, the electrons move at different speeds. It's therefore logical that BOB sees them as length contracted. Again, we are not talking about Alice.

The electrons in motion can be described as a region between two points. The distance between the two points falls.

A baseball contains electrons, protons, and neutrons. Its "length contracTION" is relative to the observer. I believe you agree with that. Or am I mistaken?
DrGreg
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Feb14-12, 08:27 PM
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kmarinas86, I haven't time to make a detailed response now, and I'll be offline for the next 20 hours or so.

In this scenario there is no "front" or "back" of a "train" of electrons. They are continually being pushed into one end of the wire and pulled out of the other. It is a continuous stream, not a finite-length train. So all you can say is that the length of the stream is the same as the length of the wire, in whatever common frame you make both measurements. The electrons spread out to fill whatever "vessel" they are in. Baseballs don't do that. The electrons behave more like a gas than a solid.

For the purpose of the thought experiment you might as well consider the wire bent to form a continuous circular loop. Now which part of the circumference of a circle is its centre?

Quote Quote by kmarinas86 View Post
There "may be" debate about it? You know this, or is that just a way of saying that others may debate about that if they like?
I've witnessed a number of such debates in this forum. All based on a false notion of what "Lorentz contraction" is. It's not a comparison of before and after, it's a comparison from two different observers at the same time.
kmarinas86
#23
Feb14-12, 08:35 PM
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Quote Quote by DrGreg View Post
kmarinas86, I haven't time to make a detailed response now, and I'll be offline for the next 20 hours or so.

In this scenario there is no "front" or "back" of a "train" of electrons. They are continually being pushed into one end of the wire and pulled out of the other. It is a continuous stream, not a finite-length train.
So let's call it a train with no relevant "back" or "front". Let's say it is like the straight part of a tape between the guide rollers in a cassette tape, while ignoring what length is on the reels.

Quote Quote by DrGreg View Post
So all you can say is that the length of the stream is the same as the length of the wire, in whatever common frame you make both measurements.
So let's say that the train cars contract, and the "visible stream" has more cars when the velocity increases.

Quote Quote by DrGreg View Post
The electrons spread out to fill whatever "vessel" they are in. Baseballs don't do that. The electrons behave more like a gas than a solid.
This should only matter if we are changing the "volume" occupied by the electrons.

BTW: The electrons do not always "spread out" to fill whatever "vessel" they are in. Sometimes they must "be compressed inwards" to match the new "specific volume" encountered.

Quote Quote by DrGreg View Post
For the purpose of the thought experiment you might as well consider the wire bent to form a continuous circular loop. Now which part of the circumference of a circle is its centre?
You're starting to understand my concern.

Quote Quote by DrGreg View Post
I've witnessed a number of such debates in this forum. All based on a false notion of what "Lorentz contraction" is. It's not a comparison of before and after, it's a comparison from two different observers at the same time.
If true, then it is logical to say that magnetism CANNOT be explained via "length contracted" charges. Furthermore, it means that Feynman's explanation for the Lorentz force as being due to "length contraction" is completely false.

The problem is that what you say is likely false, considering why the concept of length contraction was conceived in the first place:

http://en.wikipedia.org/wiki/Length_contraction
Quote Quote by Length contraction
==Experimental verifications==
{{See also|Tests of special relativity}}

Since the occurrence of length contraction depends on the inertial frame chosen, it can only be measured by an observer ''not'' at rest in the same inertial frame, ''i.e.'', it exists only in a non-co-moving frame. This is because the effect vanishes after a Lorentz transformation into the rest frame of the object, where a co-moving observer can judge himself and the object as at rest in the same inertial frame in accordance with the relativity principle (as it was demonstrated by the Trouton-Rankine experiment). In addition, even in a non-co-moving frame, ''direct'' experimental confirmations of Lorentz contraction are hard to achieve, because at the current state of technology, objects of considerable extension cannot be accelerated to relativistic speeds. And the only objects traveling with the speed required are atomic particles, yet whose spatial extensions are too small to allow a direct measurement of contraction.

However, there are ''indirect'' confirmations of this effect in a non-co-moving frame. It was in fact the negative result of a famous experiment, that required the introduction of Lorentz contraction: the Michelson-Morley experiment (and later also the Kennedy–Thorndike experiment). In special relativity its explanation is as follows: In its rest frame the interferometer can be regarded as at rest in accordance with the relativity principle, so the propagation time of light is the same in all directions. However, in a frame in which the interferometer is in motion, the propagation time of the transverse beam is time dilation|time dilated, while in the longitudinal direction the interferometer is also contracted, so that speed of light is constant and the propagation time in both directions is the same in this frame as well.

Other indirect confirmations are: Heavy ions that are spherical when at rest should assume the form of "pancakes" or flat disks when traveling nearly at the speed of light. And in fact, the results obtained from particle collisions can only be explained, when the increased nucleon density due to Lorentz contraction is considered.[http://www.bnl.gov/rhic/physics.asp][http://nuclear.ucdavis.edu/~calderon...Research.html][http://www.gsi.de/forschung/kp/kp2/c.../R3B/EME.html][http://arxiv.org/abs/physics/0105022] Another confirmation is the increased ionization ability of electrically charged particles in motion. According to pre-relativistic physics the ability should decrease at high speed, however, the Lorentz contraction of the Coulomb's law|Coulomb field leads to an increase of the electrical field strength normal to the line of motion, which leads to the actually observed increase of the ionization ability.[http://en.wikipedia.org/wiki/Special...es/3528172363] Lorentz contraction is also necessary to understand the function of free-electron lasers. Relativistic electrons were injected into an undulator, so that synchrotron radiation is generated. In the proper frame of the electrons, the undulator is contracted which leads to an increased radiation frequency. Additionally, to find out the frequency as measured in the laboratory frame, one has to apply the relativistic Doppler effect. So, only with the aid of Lorentz contraction and the rel. Doppler effect, the extremely small wavelength of undulator radiation can be explained.[http://hasylab.desy.de/science/stude...ndex_eng.html][http://flash.desy.de/sites/site_vuvf...refrs_web.pdf] (PDF 7.8MB)] Another example is the observed lifetime of muons in motion and thus their range of action, which is much higher than that of muons at low velocities. In the proper frame of the atmosphere, this is explained by the time dilation of the moving muons. However, in the proper frame of the muons their lifetime is unchanged, but the atmosphere is contracted so that even their small range is sufficient to reach the surface of earth.<ref name=sexl />
It doesn't matter if your dealing with different inertial frames of reference or the before and after states. Lorentz contraction is based on relative velocity. That's more general of an applicability than what you insist to be the case.
kmarinas86
#24
Feb14-12, 09:40 PM
P: 1,011
It seems to me that the Ladder paradox is relevant here:

http://en.wikipedia.org/wiki/Ladder_paradox





Quote Quote by Ladder Paradox


Resolution
Figure 7: A ladder contracting under acceleration to fit into a length contracted garage

In the context of the paradox, when the ladder enters the garage and is contained within it, it must either continue out the back or come to a complete stop. When the ladder comes to a complete stop, it accelerates into the reference frame of the garage. From the reference frame of the garage, all parts of the ladder come to a complete stop simultaneously, and thus all parts must accelerate simultaneously.

From the reference frame of the ladder, it is the garage that is moving, and so in order to be stopped with respect to the garage, the ladder must accelerate into the reference frame of the garage. All parts of the ladder cannot accelerate simultaneously because of relative simultaneity. What happens is that each part of the ladder accelerates sequentially, front to back, until finally the back end of the ladder accelerates when it is within the garage, the result of which is that, from the reference frame of the ladder, the front parts undergo length contraction sequentially until the entire ladder fits into the garage.
This would explain the earlier comment by DaleSpam about "relativity of simultaneity".
DaleSpam
#25
Feb14-12, 09:46 PM
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Quote Quote by kmarinas86 View Post
The rest frame of the straight conductor is chosen. If the negative charge of the free electron steady current is seen uniformly spread throughout the wire, then I have no choice but to assume that the free electron charge density is inversely proportional to the factor [itex]\sqrt{1-\left(\frac{v}{c}\right)^2}[/itex], where v is the electron drift velocity of the current. Logically then, the amount free electrons in the same conductor must increase to the same degree as the free electron charge density does. The density and count of other charges is not affected. This would violate the neutral wire assumption.
What you say is correct except that you have the frames wrong. The wire is neutral and carries a current in the lab frame. Charge density transforms like time and current density transforms like space. So there is "charge density dilation" and "current density contraction" in the test-charge frame. The result is that the wire is charged in the moving frame.
kmarinas86
#26
Feb14-12, 10:07 PM
P: 1,011
Quote Quote by DaleSpam View Post
What you say is correct except that you have the frames wrong.
Do I appear choosing the "wrong" frame (whatever -that- is), or do I appear to be mixing frames?

Quote Quote by DaleSpam View Post
The wire is neutral and carries a current in the lab frame. Charge density transforms like time and current density transforms like space. So there is "charge density dilation" and "current density contraction" in the test-charge frame. The result is that the wire is charged in the moving frame.
So let's say the length contraction of the electrons does explain "magnetism". From this point of view, the electric field of each electron is clearly length contracted. Disregarding any changes in the number of electrons between the ends of the wire, the fact is that component of the electric field normal to velocity increases as a charge moves through space:

http://en.wikipedia.org/wiki/Relativ...g_point_charge

Quote Quote by Relativistic electromagnetism#The field of a moving point charge
== The field of a moving point charge ==
|frame|'''Figure 3''': A point charge at rest, surrounded by an imaginary sphere.

|frame|'''Figure 4''': A view of the electric field of a point charge moving at constant velocity.

A very important application of the electric field transformation equations is to the field of a single point charge moving with constant velocity. In its rest frame, the electric field of a positive point charge has the same strength in all directions and points directly away from the charge. In some other reference frame the field will appear differently.

In applying the transformation equations to a nonuniform electric field, it is important to record not only of the value of the field, but also at what point in space it has this value.

In the rest frame of the particle, the point charge can be imagined to be surrounded by a spherical shell which is also at rest. In ''our'' reference frame, however, both the particle and its sphere are moving. Length contraction therefore states that the sphere is deformed into a spheroid, as shown in cross section (geometry)|cross section in Fig 4.

Consider the value of the electric field at any point on the surface of the sphere. Let x and y be the components of the displacement (vector)|displacement (in the rest frame of the charge), from the charge to a point on the sphere, measured parallel (geometry)|parallel and perpendicular to the direction of motion as shown in the figure. Because the field in the rest frame of the charge points directly away from the charge, its components are in the same ratio as the components of the displacement:

:[itex]{E_y \over E_x} = {y \over x}[/itex]

In our reference frame, where the charge is moving, the displacement x' in the direction of motion is length-contracted:

:[itex]x' = x\sqrt{1 - v^2/c^2} [/itex]

The electric field at any point on the sphere points directly away from the charge. (b) In a reference frame where the charge and the sphere are moving to the right, the sphere is length-contracted but the vertical component of the field is stronger. These two effects combine to make the field again point directly away from the current location of the charge. (While the y component of the displacement is the same in both frames).

However, according to the above results, the y component of the field is enhanced by a similar factor:

:[itex]E'_y = {E_y\over\sqrt{1 - v^2/c^2}} [/itex]

whilst the x component of the field is the same in both frames. The ratio of the field components is therefore

:[itex]{E'_y \over E'_x} = {E_y \over E_x\sqrt{1 - v^2/c^2}} [/itex]

So, the field in the primed frame points directly away from the charge, just as in the unprimed frame.
A view of the electric field of a point charge moving at constant velocity is shown in figure 4. The faster the charge is moving, the more noticeable the enhancement of the perpendicular component of the field becomes. If the speed of the charge is much less than the speed of light, this enhancement is often negligible. But under certain circumstances, it is crucially important even at low velocities.
In any case, it would seem that length contraction of the electrons only (rather than the whole bulk of it) would increase the E-field contribution normal to the wire by a certain amount. This amount is [itex]\gamma[/itex]. If, on the contrary, it were actually like I said, with an increase in charge density on top of that due the current, then this amount would be [itex]\gamma^2[/itex].

But is there another affect on the E-field, due to time dilation?

The following graphic suggests to me there is also an effect due to relativistic aberration:


Is there any good summary of these effects?
universal_101
#27
Feb15-12, 12:05 AM
P: 303
Quote Quote by DaleSpam View Post
Yes, when the current is ON then the electrons and the test charge are moving, all these effects are present, and there is a force.

In the lab frame the force is attributed to the magnetic force on the moving charge due to the current in the neutral wire. In the electron/test-charge frame the force is attributed to the electrostatic force on the stationary charge due to the net charge in the wire.
Thanks, Dalespam

But I think what you are describing here is the magnetic force WHEN THE TEST CHARGE IS ALSO MOVING with the charges in wire.

Whereas, my question is, why don't we see this magnetic force to come in play(or act) when the test charge is STATIONARY w.r.t the wire. Since, even without motion of the test charge there should be the length contraction of the moving charges in the wire, when there is current, and therefore there should be a force even on the stationary test charge.
DaleSpam
#28
Feb15-12, 06:01 AM
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Quote Quote by universal_101 View Post
my question is, why don't we see this magnetic force to come in play(or act) when the test charge is STATIONARY w.r.t the wire.
The magnetic force is [itex]qv \times B = F [/itex]. So if it is stationary then F is 0.
Per Oni
#29
Feb15-12, 07:32 AM
P: 262
Quote Quote by DaleSpam View Post
The magnetic force is [itex]qv \cross B = F [/itex]. So if it is stationary then F is 0.
Hey Dale itís obvious that universal was asking about the altered electrostatic force not about any magnetic force.

Also earlier on you said:
Charge density transforms like time and current density transforms like space. So there is "charge density dilation" and "current density contraction" in the test-charge frame. The result is that the wire is charged in the moving frame.
Can you put formulas to these 2 concepts?

Personally I've also got big troubles believing that magnetism is no more then a Lorentz boosted electrostatic field.
DaleSpam
#30
Feb15-12, 07:53 AM
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Quote Quote by Per Oni View Post
Hey Dale it’s obvious that universal was asking about the altered electrostatic force not about any magnetic force.
No, I don't think he was, I think he was asking about the magnetic force. In any case, I will let him clarify his intentions.

Quote Quote by Per Oni View Post
Can you put formulas to these 2 concepts?
Yes, are you familiar with four-vectors or tensors?

In units where c=1 define the four-current-density:
[itex]J^{\mu} = (\rho,j_x,j_y,j_z)[/itex]
where [itex]\rho[/itex] is the charge density and the j's are the current density in each direction in an inertial reference frame.

Then:
[itex]J^{\mu'}=\Lambda^{\mu'}_{\mu}J^{\mu}[/itex]
where [itex]\Lambda[/itex] is the Lorentz transform.
DaleSpam
#31
Feb15-12, 08:19 AM
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In case I was misunderstanding universal's previous question, let me clarify the situation:

Frame 1 (lab frame):
Wire is neutral and carries a current. Test charge is moving. Electrostatic force on test charge is 0 because wire is neutral. Magnetic force on test charge is non-zero since charge is moving.

Frame 2 (test-charge frame):
Wire is charged and carries a current. Test charge is at rest. Electrostatic force on test charge is non-zero because the wire is charged. Magnetic force on test charge is 0 since charge is not moving.
Per Oni
#32
Feb15-12, 09:24 AM
P: 262
Quote Quote by DaleSpam View Post

Yes, are you familiar with four-vectors or tensors?
No, one day Iíll learn that concept, it looks like itís really useful.

I think I have the same (perhaps faulty) thought process as universal, when he says:
Whereas, my question is, why don't we see this magnetic force to come in play(or act) when the test charge is STATIONARY w.r.t the wire. Since, even without motion of the test charge there should be the length contraction of the moving charges in the wire, when there is current, and therefore there should be a force even on the stationary test charge.
This situation refers to your frame 1 but without the test charge moving. ďIfĒ thereís a Lorentz boost as seen from the test charge, it should feel an extra electrostatic force.

My take on it is (and the reality is) that no such extra electrostatic force is present therefore this whole idea of a magnetic field being a Lorentz boosted electrostatic field is for a lot of us hard to believe.
harrylin
#33
Feb15-12, 10:00 AM
P: 3,184
You refer perhaps to explanations (often accompanied by nice looking calculations) according to which magnetism is claimed to be a kind of illusion due to length contraction.

The most basic and simple case (although very high tech) that I can imagine, as it completely avoids issues with electron source and drain, is that of a closed loop superconductor in which a current is induced.

We thus start with, I think, an insulated wire containing a number of electrons N and an equal number of protons N.

I think that the following situation sketch is correct:

In the wire's rest frame:
- length contraction can play no role at all
- a magnetic field is observed

In any inertial moving frame:
- length contraction plays a role in predicting non-zero electric fields
- a magnetic field is observed that can't be transformed away

Is that correct?
Such a magnetic field looks reasonably "absolute" to me.

Harald
DaleSpam
#34
Feb15-12, 10:05 AM
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Quote Quote by Per Oni View Post
This situation refers to your frame 1 but without the test charge moving. “If” there’s a Lorentz boost as seen from the test charge, it should feel an extra electrostatic force.
Why would there be a Lorentz boost for a stationary test charge?

In this physically different situation there is no force since the wire is uncharged (no electrostatic force) and the test charge is not moving (no magnetic force). In other frames there will be both an electrostatic and a magnetic force, but they will cancel each other for 0 total EM force.

Quote Quote by Per Oni View Post
My take on it is (and the reality is) that no such extra electrostatic force is present
This is completely incorrect. If there is an EM force on a charged particle then it is always possible to transform to a frame where the particle is at rest. In this frame the magnetic force is 0 so the force on the particle is entirely due to the electrostatic force.

Quote Quote by Per Oni View Post
therefore this whole idea of a magnetic field being a Lorentz boosted electrostatic field is for a lot of us hard to believe.
Many things are hard for people to believe, that does not make them wrong.

However, I would also disagree with the idea of a magnetic FIELD being a boosted electrostatic FIELD, the math doesn't support that. The correct statement would be that a magnetic FORCE is a "boosted" electrostatic FORCE, which the math supports.

If you just have the field then you don't have a rest frame. You need a test charge on which there is a force so that you can boost to the test charge's rest frame where there is no magnetic force.
Per Oni
#35
Feb15-12, 11:19 AM
P: 262
Quote Quote by DaleSpam View Post
Why would there be a Lorentz boost for a stationary test charge?
Because of the fact that the test charge sees a current flowing in the conductor.
This is completely incorrect. If there is an EM force on a charged particle then it is always possible to transform to a frame where the particle is at rest. In this frame the magnetic force is 0 so the force on the particle is entirely due to the electrostatic force.
Letís look at the situation I thought I was referring to, which is a test charge stationary wrt a wire. What I ment to say is that in reality there are no different (although I said extra) electrostatic forces on the test charge whether a current flows or not. Can we agree on this?
universal_101
#36
Feb15-12, 12:22 PM
P: 303
Quote Quote by DaleSpam View Post
In case I was misunderstanding universal's previous question, let me clarify the situation:

Frame 1 (lab frame):
Wire is neutral and carries a current. Test charge is moving. Electrostatic force on test charge is 0 because wire is neutral. Magnetic force on test charge is non-zero since charge is moving.

Frame 2 (test-charge frame):
Wire is charged and carries a current. Test charge is at rest. Electrostatic force on test charge is non-zero because the wire is charged. Magnetic force on test charge is 0 since charge is not moving.
First, Yes, you did misunderstand my question,

Second, I believe that whole point of this debate/discussion is that we want to understand magnetic force as a relativistic effect of electrostatics. But, you are explicitly using the two types of forces to explain the experimental observations. Which puts the magnetic force in the categories of absolute forces. Because, if we are unable to explain the motion of charged particles without introducing the magnetic force then it compels us to think of it as an absolute property.


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