Mentor

## Magnetism seems absolute despite being relativistic effect of electrostatics

 Quote by kmarinas86 going from 0 current to a very high current would cause the following to occur: ...
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.

 Quote by DaleSpam 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 $\sqrt{1-\left(\frac{v}{c}\right)^2}$, 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 by kmarinas86 [....] 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 Code: + + + + + - - - - - into this Code: + + + + + ----- or Code: + + + + + ----- or Code: + + + + + ----- et cetera
 Quote by DaleSpam 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 $\sqrt{1-\left(\frac{v}{c}\right)^2}=1/8$, then we would have something like this:

Code:
+       +       +       +       +
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I think the following comment I made 2 years ago applies here:

 Quote by DrGreg link to original thread 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.

Quote by DrGreg
I think the following comment I made 2 years ago applies here:

 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 by DrGreg
 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 by DrGreg
 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 by DrGreg
 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 by DrGreg 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?

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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 by kmarinas86 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.

 Quote by DrGreg 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 by DrGreg 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 by DrGreg 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 by DrGreg 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 by DrGreg 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
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.

This would explain the earlier comment by DaleSpam about "relativity of simultaneity".

Mentor
 Quote by kmarinas86 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 $\sqrt{1-\left(\frac{v}{c}\right)^2}$, 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.

 Quote by DaleSpam 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 by DaleSpam 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 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: :${E_y \over E_x} = {y \over x}$ In our reference frame, where the charge is moving, the displacement x' in the direction of motion is length-contracted: :$x' = x\sqrt{1 - v^2/c^2}$ 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: :$E'_y = {E_y\over\sqrt{1 - v^2/c^2}}$ whilst the x component of the field is the same in both frames. The ratio of the field components is therefore :${E'_y \over E'_x} = {E_y \over E_x\sqrt{1 - v^2/c^2}}$ 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 $\gamma$. 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 $\gamma^2$.

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?

 Quote by DaleSpam 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.

Mentor
 Quote by universal_101 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 $qv \times B = F$. So if it is stationary then F is 0.

 Quote by DaleSpam The magnetic force is $qv \cross B = F$. So if it is stationary then F is 0.
Hey Dale it’s 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.

Mentor
 Quote by Per Oni 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 by Per Oni 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:
$J^{\mu} = (\rho,j_x,j_y,j_z)$
where $\rho$ is the charge density and the j's are the current density in each direction in an inertial reference frame.

Then:
$J^{\mu'}=\Lambda^{\mu'}_{\mu}J^{\mu}$
where $\Lambda$ is the Lorentz transform.

 Mentor 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.

 Quote by DaleSpam Yes, are you familiar with four-vectors or tensors?
No, one day I’ll learn that concept, it looks like it’s 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. “If” there’s 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.

 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

Mentor
 Quote by Per Oni 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 by Per Oni 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 by Per Oni 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.