Magnetism seems absolute despite being relativistic effect of electrostatics

*kmarinas86*

Quote by harrylin

Your argument doesn't apply to more general cases as I illustrated. Magnetic and electric fields are relative in the way that length contraction and simultaneity are relative; it doesn't mean that one of the concepts should be discarded.

Right, length contraction is relative.

The problem is:
* I can have the positive charge have a greater length contraction in the frame of the negative charge.
* I can have the negative charge have a greater length contraction in the frame of the positive charge.

Following the claims of DaleSpam's comments, this would mean that the wire can appear to have net positive charge or a net negative charge, depending on the frame of reference. There is also a frame in which the length contractions of the positive and negative charges match. I suppose that is when the electric-field outside the wire disappears.

Now that I think about it terms of length contraction, the changes of the electric field with respect to the frame given is NOT linear because the equations for length contraction do not have constant derivative with respect to relative velocity with the observer. Therefore, the LT would result in different change "factors" for the electric field of the electrons and the electric field of the protons in the case when there is current.

*kmarinas86*

Quote by jartsa

Are you trying to bully me or what? Collection of free particles may retain its shape when accelerated, without any stresses. When free particles are glued together, we have a rigid body, which must length contract, when accelerated, or else stresses are generated in the body.

I though that force dynamics weren't a part of the Lorentz transformation.

This above sounds to me more like the "deformable electron" concept of Lorentzian-Ether theory.

*kmarinas86*

The idea that the electric field intensity of each charge being variant with respect to the observer isn't strange to me.

What's strange is the idea that steady-state (read: DC) current should somehow be uniform through out the wire when the protons and electrons clearly cannot be subject to the same length contraction.

Quote by kmarinas86

Quote by DaleSpam

Quote by kmarinas86

Let's keep this REALLY simple. Assuming that the wire is neutral (no net charge) and that the wire is 1 meter long and that I have a length contraction of electrons, why should I get from that a uniform charge distribution when the electrons are drifting through wire (current)?

In the steady state the four-current (density) is uniform and constant in the lab frame, therefore it is uniform and constant in the test-charge frame also.

Quote by kmarinas86

I would TOTALLY expect an un-uniform distribution, assuming length contraction applies to the bulk flow of electrons.

Why? Why do you expect a gap of any kind in the steady state?

Quote by kmarinas86

I STILL don't have an answer to my question as to what do the electrons actually length contract towards.

Length contraction occurs, as always, in the direction of motion. The word "towards" doesn't make any sense in this context. The word "towards" implies something changing over time. Length contraction does not change over time in an inertial frame.

Ok, then let me ask it this way: From the lab frame, where is the center of contraction for the bulk of electron flow in a straight wire conductor? The contraction is only "linear", so I assume that this "center" of contraction must be a geometric plane. Where is that located in relation to the observer?

SR says that objects (read: multiple particles) will length contract. So, logically speaking, you can treat the + charges and - charges as two separate "objects" at different speeds. I assume this to mean not only the particles by themselves, but the entire bulks of the particles as a whole. For an object to contract, the distance in-between also has to contract. 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

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.

If the quantity of charge in a length measured by the observer were really to vary depending of the length contraction of each set of charges (the + set vs. the - set) whose length contraction values are different, we would see not only an frame-variant electric field intensity, but also, we would see an frame-variant electric FLUX as well in that length. In reality, if we Lorentz transform a system, we do NOT create positive charge and negative charges out of nowhere. Those additional electrons somehow fitting into the wire must be present with and without the Lorentz transformation. So if the wire was uniformly charged before the length transformation and after it, then some of the - charge that was OUTSIDE the wire without the LT is instead seen as being INSIDE the wire with that LT. The same would go with the + charge.

I guess that the difference of electric flux between different LT frames means that time-retardation effects apply to electric flux as well.

*kmarinas86*

Quote by harrylin

Quote by kmarinas86

Alternatively, if you consider the fact that "ionic current" or "positive charge" current can be just as guilty in producing magnetic fields as the electron current, one would realize that for the case of a neutral wire, different Lorentz transformations do not lead to differences in the magnetic flux. The magnetic field produced by a + charge is equal and opposite of that produced by a - charge if their movements are the same. So the magnetic flux produced by the neutral wire should be frame invariant.[ What changes is the magnetic flux intensity (a.k.a. magnetic flux density) and corresponding area of integration (an area which is itself subject to Lorentz transformations). This is same as with the electric flux; the Lorentz transformation leaves it unaltered (with the electric field intensity (a.k.a. electric flux density) and corresponding integration being subject to exact same transformation as that of their magnetic counterparts).]

For a current-free wire, indeed. That isn't an issue.

I thought that the observed magnetic field was directly proportional to relative velocity [itex]v[/itex]. The electric field's dependence on [itex]\gamma[/itex] should contrast with the magnetic field's dependence [itex]v[/itex].

In that case, I cannot at all see how changes in the E-field can compensate precisely for changes in the B-field. They simply do not match. So it can undershoot or overshoot the requirement for compensating for the difference of the B between different LT frames.

Alternatively, if B varied with the rapidity [itex]\varphi[/itex] (with respect to LT frames, not time or acceleration, mind you), it would not be an exact match either:

Column 1: [itex]v/c[/itex]
Column 2: [itex]\varphi[/itex]
Column 3: [itex]\gamma[/itex]
Column 4: Change in Column 2
Column 5: Change in Column 3
Column 6: Column 2 / Column 3

Code:

0.00	0.00	1.00			
0.10	0.10	1.01	0.10	0.01	19.92
0.20	0.20	1.02	0.10	0.02	6.57
0.30	0.31	1.05	0.11	0.03	3.86
0.40	0.42	1.09	0.11	0.04	2.67
0.50	0.55	1.15	0.13	0.06	1.98
0.60	0.69	1.25	0.14	0.10	1.51
0.70	0.87	1.40	0.17	0.15	1.16
0.80	1.10	1.67	0.23	0.27	0.87
0.90	1.47	2.29	0.37	0.63	0.60

Only one other possibility: The B normal to the wire is proportional to [itex]\gamma_{v\ parallel\ to\ the\ wire}[/itex]. The problem is that I never heard of it.

Meanwhile, in SR, the "relativistic energy" of a particle is relative to LT frames. So the idea that the magnetic field is simply the relativistic component of the electric field appears doomed. SR would have no problem having the change in the E field be more than and/or less than what would be needed to compensate for the magnetic field, for it appears to be required to have the "relativistic energy" of a particle to vary.

By the way, if some E fields and some B fields cannot transform away, then the claim that electric fields and magnetic fields are part of the same "electromagnetic field" seems dubious at best.

Maybe we should move away from the field concepts and stick with the vector potential instead.

DrGreg

Attached to the bottom of this post is a diagram to help explain things. As was mentioned earlier in this thread, one way to approach the problem is to consider it a variant of the ladder paradox, and consider the different definitions of simultaneity.

But my approach here considers length contraction only. And I am going to consider a complete circuit: not just a single wire with a left-to-right electron flow, but also a return wire with a right-to-left flow. Apart from the ends of the wires, we keep the two wires far apart so they have negligible influence on each other. The diagram is a highly idealised simplification, considering just 16 electrons in the circuit. The ends of the wires should be in contact with each other but I've drawn them as separated to keep the diagram simple.

The top left part of the diagram shows the wires with no current flowing, in the rest-frame of the wires. 16 electrons equally spread out along the wire.

The top right part of the diagram again shows the wires with no current flowing, but now in a frame moving at the velocity that electrons would flow in the bottom wire if the current were on. We see length contraction as indicated by the yellow arrows. I'm assuming a Lorentz factor γ=2. So far so good.

The two bottom diagrams now show what happens when the current is flowing.

In the bottom left diagram, as we are told the wires remain electrically neutral, there must still be 16 electrons in the wires. There's no reason for the electrons to bunch together anywhere, they will remain spread out around the whole circuit as shown.

Finally, let's look at the bottom right diagram, which I think some people are having difficulty to imagine. We already know what happens to the red positive ions, their separation contracts just as before. The electrons in the lower wire are now stationary, so their separation must be larger than the bottom left diagram as shown. On the other hand, the electrons in the upper wire are moving faster than in bottom left diagram, so their separation must be less than in bottom left diagram. No electrons have escaped so the total number of electrons in circuit must still be 16. But now there are fewer electrons in the lower wire and more in the upper wire. So the lower wire has a positive charge and the upper wire has a negative charge.

DrGreg

It doesn't seem to have been mentioned in this thread yet. The E and B fields are used to construct a 4×4 matrix[tex]
F^{\mu\nu} = \begin{bmatrix}
0 & -E_x/c & -E_y/c & -E_z/c \\
E_x/c & 0 & -B_z & B_y \\
E_y/c & B_z & 0 & -B_x \\
E_z/c & -B_y & B_x & 0
\end{bmatrix}
[/tex]This is a rank-2 tensor whose components transform as a tensor, i.e. there's a double Lorentz transformation involved.

*kmarinas86*

Quote by DrGreg

Attached to the bottom of this post is a diagram to help explain things. As was mentioned earlier in this thread, one way to approach the problem is to consider it a variant of the ladder paradox, and consider the different definitions of simultaneity.

But my approach here considers length contraction only. And I am going to consider a complete circuit: not just a single wire with a left-to-right electron flow, but also a return wire with a right-to-left flow. Apart from the ends of the wires, we keep the two wires far apart so they have negligible influence on each other. The diagram is a highly idealised simplification, considering just 16 electrons in the circuit. The ends of the wires should be in contact with each other but I've drawn them as separated to keep the diagram simple.

The top left part of the diagram shows the wires with no current flowing, in the rest-frame of the wires. 16 electrons equally spread out along the wire.

The top right part of the diagram again shows the wires with no current flowing, but now in a frame moving at the velocity that electrons would flow in the bottom wire if the current were on. We see length contraction as indicated by the yellow arrows. I'm assuming a Lorentz factor γ=2. So far so good.

The two bottom diagrams now show what happens when the current is flowing.

In the bottom left diagram, as we are told the wires remain electrically neutral, there must still be 16 electrons in the wires. There's no reason for the electrons to bunch together anywhere, they will remain spread out around the whole circuit as shown.

Finally, let's look at the bottom right diagram, which I think some people are having difficulty to imagine. We already know what happens to the red positive ions, their separation contracts just as before. The electrons in the lower wire are now stationary, so their separation must be larger than the bottom left diagram as shown. On the other hand, the electrons in the upper wire are moving faster than in bottom left diagram, so their separation must be less than in bottom left diagram. No electrons have escaped so the total number of electrons in circuit must still be 16. But now there are fewer electrons in the lower wire and more in the upper wire. So the lower wire has a positive charge and the upper wire has a negative charge.

I think that using the return wire to prove a point is cheating. It does nothing for the original scenario without a return wire.

The ladder paradox also has some asymmetries that seem to be missing in your example:

Figure 4: Scenario in the garage frame: a length contracted ladder entering and exiting the garage

Figure 5: Scenario in the ladder frame: a length contracted garage passing over the ladder

The two frames do not see the same number of rungs inside the garage in each case.

If we assumed that the protons were represented as tiles on the garage floor, the garage as the wire, and the ladder as the electron current in and out of the wire, then clearly the charge inside the boundary of the garage is not invariant.

However, considering that the electric field intensity increases by the same amount that the boundary of the garage in the LT frame is length contracted, this would keep the electric flux around that boundary of the garage a constant.

DrGreg

Quote by kmarinas86

I think that using the return wire to prove a point is cheating. It does nothing for the original scenario without a return wire.

Well it has the advantage of charge conservation in a closed system, which doesn't apply to an open-ended wire.

Quote by kmarinas86

The ladder paradox also has some asymmetries that seem to be missing in your example:

The two frames do not see the same number of rungs inside the garage in each case.

If we assumed that the protons were represented as tiles on the garage floor, the garage as the wire, and the ladder as the electron current in and out of the wire, then clearly the charge inside the boundary of the garage is not invariant.

However, considering that the electric field intensity increases by the same amount that the boundary of the garage in the LT frame is length contracted, this would keep the electric flux around that boundary of the garage a constant.

Sorry, somehow the image attachment to my post failed to upload correctly. I have now re-uploaded it and added it to that message.

If you ignore my return wire and concentrated on my lower wire only, it seems to me that my diagram agrees with your ladder diagram, so I haven't grasped what your problem is.

*kmarinas86*

Quote by DrGreg

Well it has the advantage of charge conservation in a closed system, which doesn't apply to an open-ended wire.

Sorry, somehow the image attachment to my post failed to upload correctly. I have now re-uploaded it and added it to that message.

If you ignore my return wire and concentrated on my lower wire only, it seems to me that my diagram agrees with your ladder diagram, so I haven't grasped what your problem is.

The problem is that we are talking about a single current and the fact that charge has to be conserved between frames for that single current.

There can be charge outside the wire ends (say at the ends of a capacitor or what not).

DrGreg

Quote by kmarinas86

The problem is that we are talking about a single current and the fact that charge has to be conserved between frames for that single current.

But both my example (restricted to the highlighted bottom wire) and the ladder example (restricted to the interior of the garage) show that this isn't true.

In my example the number of electrons decreases from 8 to 2. In the ladder example, the number of rungs within the garage decreases from more than 11 to about 7.

(Note: on a technicality "conserved between frames" should really be described as "invariant". "Conservation" refers to lack of change over time within a single frame.)

*kmarinas86*

Quote by DrGreg

But both my example (restricted to the highlighted bottom wire) and the ladder example (restricted to the interior of the garage) show that this isn't true.

In my example the number of electrons decreases from 8 to 2. In the ladder example, the number of rungs within the garage decreases from more than 11 to about 7.

What if we suddenly broke the circuit at two places?

Quote by DrGreg

(Note: on a technicality "conserved between frames" should really be described as "invariant". "Conservation" refers to lack of change over time within a single frame.)

Yes.

DrGreg

Quote by kmarinas86

What if we suddenly broke the circuit at two places?

That would depend on the timing. Simultaneous breaks in one frame would not be simultaneous in another frame.

*kmarinas86*

Quote by DrGreg

That would depend on the timing. Simultaneous breaks in one frame would not be simultaneous in another frame.

Got it. Just like your answer to a similar problem on another thread.

DaleSpam

Quote by kmarinas86

The problem is that we are talking about a single current and the fact that charge has to be conserved between frames for that single current.

This is not true in a couple of ways.

First, there is no such thing as "conserved between frames". Conservation means that something is the same across time. When a quantity is the same in different frames it is called "invariant", not "conserved". The two concepts are completely different.

Second, it is not true that the net charge on the wire is invariant.

I will deal with more of your posts later, but you have really posted a lot of nonsense today.

*kmarinas86*

Quote by DaleSpam

This is not true in a couple of ways.

First, there is no such thing as "conserved between frames". Conservation means that something is the same across time. When a quantity is the same in different frames it is called "invariant", not "conserved". The two concepts are completely different.

I heard the first time, but I made the same mistake accidentally.

Invariant and conserved are different things!
Invariant and conserved are different things!
Invariant and conserved are different things!
....

Quote by DaleSpam

Second, it is not true that the net charge on the wire is invariant.

I have been shown why now.

Quote by DaleSpam

I will deal with more of your posts later, but you have really posted a lot of nonsense today.

I think it has been sufficiently been explained to me at this point. Don't worry about me. I'm done with this topic. I'm satisfied with the answer now.

P.S. I've long used the term "time-invariant" to mean conserved. I must stop doing that.

P.S.S. On another note, I wonder if (http://en.wikipedia.org/wiki/Time-invariant_system) is better termed (time-independent system). (j/k the answer is obvious)

DaleSpam

Quote by kmarinas86

I think it has been sufficiently been explained to me at this point. Don't worry about me. I'm done with this topic. I'm satisfied with the answer now.

Excellent! That is good to hear.

DaleSpam

Quote by universal_101

when there is a current, the charges in the wire start moving in a particular direction, but when there is NO current there is NO motion. Therefore, according to the transformation of one force into other, there should be a force on a stationary charge standing near by, towards the current carrying wire, when there is current.

This is incorrect. In the frame where the test charge is at rest, if the wire is uncharged then there is no force, regardless of the current.

Also, your reasoning doesn't make sense: a current is moving charges, forces transform, therefore there is a force on a stationary test charge. If you could step through your reasoning in a little more detail then I could probably point out where it falls apart, but as it is all I can say is that the premises don't imply the conclusion.

Quote by universal_101

Remembering, that my original post/question is exactly same situation, to which the answer was the transformation of one force into another, to explain the magnetic force.

Sure, relativity can be used to transform a magnetic force in one frame to an electrostatic force in another frame (the rest frame of the particle). It cannot be used to transform no force into some force.

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DrGreg Recognitions: Gold Member Science Advisor	It doesn't seem to have been mentioned in this thread yet. The E and B fields are used to construct a 4×4 matrix[tex] F^{\mu\nu} = \begin{bmatrix} 0 & -E_x/c & -E_y/c & -E_z/c \\ E_x/c & 0 & -B_z & B_y \\ E_y/c & B_z & 0 & -B_x \\ E_z/c & -B_y & B_x & 0 \end{bmatrix} [/tex]This is a rank-2 tensor whose components transform as a tensor, i.e. there's a double Lorentz transformation involved.