# Magnetism seems absolute despite being relativistic effect of electrostatics

P: 1,011
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 $v$. The electric field's dependence on $\gamma$ should contrast with the magnetic field's dependence $v$.

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 $\varphi$ (with respect to LT frames, not time or acceleration, mind you), it would not be an exact match either:

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

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 $\gamma_{v\ parallel\ to\ the\ wire}$. 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.
 Sci Advisor PF Gold P: 1,843 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. Attached Thumbnails
 Sci Advisor PF Gold P: 1,843 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$$F^{\mu\nu} = \begin{bmatrix} 0 & -E_x/c & -E_y/c & -E_z/c \\ E_x/c & 0 & -B_z & B_y \\ E_y/c & B_z & 0 & -B_x \\ E_z/c & -B_y & B_x & 0 \end{bmatrix}$$This is a rank-2 tensor whose components transform as a tensor, i.e. there's a double Lorentz transformation involved.
P: 1,011
 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.

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.
PF Gold
P: 1,843
 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.
P: 1,011
 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).
PF Gold
P: 1,843
 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.)
P: 1,011
 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.
PF Gold
P: 1,843
 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.
P: 1,011
 Quote by DrGreg That would depend on the timing. Simultaneous breaks in one frame would not be simultaneous in another frame.
Mentor
P: 16,967
 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.
P: 1,011
 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)
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P: 16,967
 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.
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P: 16,967
 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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P: 16,967
 Quote by universal_101 Let's consider a simple model of a conducting wire, + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Now, let's suppose there is some current in the wire and the electrons are moving at speed 'v' w.r.t the the wire, secondly, a stationary test charge w.r.t the wire lying around. Naming the above scenario as (1) Now, the test charge starts moving in the direction of electrons with the same speed 'v'. This time in the reference frame of the test charge, electrons are stationary and nucleus(positive charge) is moving at speed 'v'. Naming this scenario as (2) And so the question arise, the two scenario are identical w.r.t principle of relativity. That is, in the first case only negative charges are moving, but there is no force on the charge. But in the second case when positive charges are moving there is a force on the test charge(magnetic force towards wire). Whereas, the two cases are essentially identical w.r.t principle of relativity.
Oops. My apologies universal_101. I did not read your OP closely enough.

Scenario (1) and scenario (2) are NOT identical w.r.t the principle of relativity. They are physically different scenarios. In (1) the test charge is at rest relative to the protons and in (2) the test charge is at rest relative to the electrons. There is no way to Lorentz transform (1) into (2).

If you want the identical scenario then you need to change (2) so that the test charge is moving with the same velocity as the protons. That way the test charge will be at rest wrt the protons in both scenarios.
P: 303
 Quote by DaleSpam Scenario (1) and scenario (2) are NOT identical w.r.t the principle of relativity. They are physically different scenarios. In (1) the test charge is at rest relative to the protons and in (2) the test charge is at rest relative to the electrons. There is no way to Lorentz transform (1) into (2). If you want the identical scenario then you need to change (2) so that the test charge is moving with the same velocity as the protons. That way the test charge will be at rest wrt the protons in both scenarios.
If this is how you see it, then how are you able to explain different scenario with Lorentz transformation. Or, can LT be applied on different scenarios/situations too ?

I'm sure you know this already, but then I can't seem to figure out why are you implying anything like this.
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P: 16,967
 Quote by universal_101 If this is how you see it, then how are you able to explain different scenario with Lorentz transformation. Or, can LT be applied on different scenarios/situations too ? I'm sure you know this already, but then I can't seem to figure out why are you implying anything like this.
Sorry, I don't know if there is a language barrier, but I cannot really parse your post. I will answer what I guess is your question, but if I guess wrong please try to clarify your question carefully.

The LT can be applied to any scenario to generate an infinite number of other scenarios which are, in fact, physically identical to the original scenario. However, two arbitrary scenarios are not necessarily related to each other via a LT. In your case, (1) and (2) are not related by a LT.
P: 303
 Quote by DaleSpam Here is probably the best resource for this question: http://physics.weber.edu/schroeder/mrr/MRRtalk.html Your scenarios are explicitly covered in the section "Magnetism as a Consequence of Length Contraction".
 Quote by DaleSpam Sorry, I don't know if there is a language barrier, but I cannot really parse your post. I will answer what I guess is your question, but if I guess wrong please try to clarify your question carefully. The LT can be applied to any scenario to generate an infinite number of other scenarios which are, in fact, physically identical to the original scenario. However, two arbitrary scenarios are not necessarily related to each other via a LT. In your case, (1) and (2) are not related by a LT.
You described the two scenarios using LT, and now you are implying that the two scenarios are different. But to have a debate, you should stand with only one of the following.

Either the scenarios are different, or, they can be explained by LT.

And if you still think they are different, then please explain, why does the link you provided uses LT to explain different scenarios.

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