# Electromagnetic Field/Force Model

1. Jul 17, 2010

### mysearch

If possible, I would like to confirm or correct my assumptions associated with the model described below and illustrated in the attached diagram. Most descriptions tend to aggregate effects of moving charge as a current flowing in a wire, but the specific purpose of this model is to try to examine the same concept in the isolation of 2 charges.

The model, subject to the Lorentz transforms, is shown in both a stationary and moving reference frame. It consists of a positive charge [Q] moving with velocity [v1] and a negative charge [q] moving parallel to [Q] along the x-axis with velocity [v2]. For the purpose of the model [v1=v2=v], but the subscript is only used to identify the relevance in specific equations cited below.

With reference to the diagram, the model on the left is linked to the stationary frame labelled [O*], while the model on the right is the moving frame labelled [O]. In the stationary [O*] frame there is no B-field and the net force is directly related to the E-field, which points outwards due to the sign of the charge [Q]. In the moving [O] frame, a B-field does exist in the direction shown, i.e. perpendicular in z-axis, and has an associated force that acts inwards, i.e. y-axis, in opposition to the outward E-force [Fe].

[1] $$E= \frac {Q}{4 \pi \epsilon_0 y^2}* \frac {1- \beta^2}{(1- \beta^2 sin^2 \theta)^{3/2}}$$

[2] $$B= \frac {Qv_1}{4 \pi \epsilon_0 c^2 y^2}* \frac {1- \beta^2}{(1- \beta^2 sin^2 \theta)^{3/2}}$$

[3] $$B = \frac {vE}{c^2}$$; where $$sin \theta =1$$

While the form of [1] and [2] is orientated towards the moving frame, the model restricts [r] to the y-axis, which is not subject to length contraction, i.e. y=y*, and therefore the values for the stationary frame can be determined by setting [v=0]. As such, the second term in [1] and [2] disappears and we are left with the standard forms of (E) and (B). However, the corresponding force equations are given below:.

[4] $$F = qE + qv_2Bsin \theta$$

[5] $$F_e =qE; F_b = qv_2B$$

The form of [4] only implies the magnitude of the force, not the direction, which is addressed in the diagram. Again, [5] reduces [4] to the components acting along the y-axis such that $$\theta=90$$. However, it is highlighted that [3] and [5] have cited different velocities [v1] and [v2] in order to clarify an issue.

Conceptually, it seems that the B-field is thought to exist based on [Q] being in motion with velocity [v1]. However, in practice, the existence of a force (Fb) associated with the field can only be measured by a secondary test charge [q] moving through(?) the B-field with velocity [v2]. However, in this case, [Q] and [q] have the same velocity, so is there any need for the velocity of [q] to be relative to the B-field?

The attached diagram assumes that [v1] and [v2] can both be substituted for [v] in [3] and [5] and, as such, the diagram shows the moving frame having a magnetic force (Fb) that opposes (Fe). However, when [v<<c], equation [3]-[5] can be combined and reduced to the following form, again noting the restriction to the y-axis.

[6] $$F_e = \frac {Qq}{4 \pi \epsilon_0 y^2}$$

[7] $$F_b = \frac {Qq}{4 \pi \epsilon_0 y^2} * \frac {v^2}{c^2}$$

Comparing [6] and [7], when (v<<c), suggests that (Fb) may be 16-17 orders of magnitude smaller than (Fe), such that the net force will still repel [q] away from [Q]. Anyway, I would appreciate any correction to the model, where necessary. Thanks.

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2. Jul 17, 2010

### AJ Bentley

As you say, normally one would approach this from the point of view of current and charge densities. I see no reason why your approach should not yield the 'correct' answer, However, it is a complex question and I'm not going to leap in with an immediate response - I'm not sure yet I entirely understand your reasoning.

I will say one thing though. Because there is no preferred frame, the resultant forces in both frames should be identical - however you arrive at them.

3. Jul 17, 2010

### mysearch

I appreciate that it is not always easy to align your understanding with what probably is a non-standard model. However, I was hoping that by reducing a model to the minimum number of components it may have some benefits when trying to understand the fundamental physics at work.

I am not sure I understand the implication of your last statement. Certainly, the E-field is not invariant as the frames are switched and therefore neither can the corresponding electric force, except in the direction of motion. However, the B-field and its associated force will change as a function of velocity and, in this model, will act in opposition to the electric force. Therefore, I am not sure whether these changes cancel each other and imply an unchanged net force?

4. Jul 17, 2010

### AJ Bentley

Imagine a simple apparatus similar to your set-up where you had a pair of unlike charges held apart by a spring. The spring compression being sufficient to hold the two charges at a fixed distance apart.

If you now place that apparatus on a moving platform, relativity stipulates that the spring should show no change of length, otherwise you would be able to work out which was the moving frame.

5. Jul 18, 2010

### mysearch

This seems to be good model to consider and hopefully it will help resolve some of my confusion surrounding the issue of relativistic force. However, first I would like to correct what was a mistake in my original diagram and try to get some basic confirmation, if possible, of the direction of all the vector quantities. In the case of unlike charges, I believe the electric force should have been inwards towards each other. One of the advantages of this type of discrete model is that you have to be explicit about the sign of the charge moving with respect to each other and the resulting direction of the vector quantities. See revised attached diagram to which I have added the opposite case of 2 like charges in-line with example suggested.

Do the vector quantities in this diagram all point in the right direction?

I would also like to try and resolve the issue raised in post #1, which try to highlight that equation [2] and [5] are with respect to 2 different velocities [v1, v2]. In the example, these velocities are equal, which seems to suggest that [q] has no relative velocity to the B-field created by [Q]. So does a magnetic force really exist in this example?
OK, if 2 positive charges were conceptually linked by a spring to hold them together, I am assuming that in the stationary frame the force would be completely defined by a repulsive electrostatic force in the case of 2 positive charges:

[8] $$F_E = \frac {Qq}{4 \pi \epsilon_0 y^2}$$

I have retained [r=y] to indicate that in the orientation shown in the lower attached diagram, (y) would not be subject to length contraction, only (x). In the case of a constant velocity frame, there would be no acceleration of the frame itself plus it is difficult to correlate [8] above to Newton’s 2nd law (F=ma). As such, it would seem that the force linked to [8] is only defined by the separation of the 2 charges [Q,q]. In the stationary frame, this distance is measured as [y], but I would have thought another observer who perceives this frame to be moving with velocity [v] along the x-axis would perceive this distance in terms of:

$$h^2= y^2+ (x \sqrt{1- \beta ^2})^2$$

As such, an observer moving collocated within the frame of the charges would, as a stationary observer, perceive no change in the force, as no relativistic effects would be observed. However, an observer of the moving frame would see a change in the distance, while the charges [Q,q] remain invariant, suggesting a different result to [8]. So while these systems are equivalence under the Lorentz transform, the different observers make different observation (?)

However, I have started to search for other references to either support or reject this argument and I must admit to being slightly confused by some that seem to support the relativistic effect on force and those that don’t. The following link for example seems to suggest invariance, but I haven’t had a chance to work through it in any details. Thanks.
http://arxiv.org/PS_cache/physics/pdf/0507/0507099v1.pdf

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6. Jul 18, 2010

### AJ Bentley

Part of the problem you are having with your model is down to the fact that you are not considering the time-of-travel for the fields. E and B propagate at velocity c.

Here's a reasonably simple explanation of the basic ideas involved.
This is the form that is normally used in explanations which avoid the use of potential energy in favour of force consideration.

http://galileo.phys.virginia.edu/classes/252/rel_el_mag.html" [Broken]

It's reasonably close to the model you want to use, with the exception of a neutralising charge that makes the reasoning simpler.

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7. Jul 19, 2010

### mysearch

That is not the intention. If you could explain where you think propagation delay or time dilation needs to be added to the following description, I would be much obliged.
I have started to look at a number of relativistic models. The one referenced above seems to be a good summary, but, as such, does not seem too specific about each frame of reference under consideration or the overall conclusions. Two other sources that offer more details are:

http://physics-quest.org/Magnetism_from_ElectroStatics_and_SR.pdf
http://physics.weber.edu/schroeder/mrr/MRRnotes.pdf

However, I specifically wanted to try to explain the effects in a discrete charge model using the relativistic equation in post #1, i.e. [1] & [2]. As stated in post #2 “I see no reason why your approach should not yield the 'correct' answer”. If you orientate the model to the y-axis, where $$(sin \theta=1)$$, such that they are comparable to the models in the links above, the equations reduce to the following form:

[10] $$E= \frac {Q}{4 \pi \epsilon_0 y^2}* \frac {1}{(1- \beta^2)^{1/2}}$$

[11] $$B= \frac {Q}{4 \pi \epsilon_0 y^2}* \frac{v}{c^2}* \frac {1}{(1- \beta^2)^{1/2}}$$

[12] $$F_E= qE = \frac {qQ}{4 \pi \epsilon_0 y^2}* \frac {1}{(1- \beta^2)^{1/2}}$$

[13] $$F_B= qvB = \frac {qQ}{4 \pi \epsilon_0 y^2}* \frac{v^2}{c^2}* \frac {1}{(1- \beta^2)^{1/2}}$$

Because [y=y’] and [v] is invariant in all frames, the equations above would appear to be valid for the stationary and moving frames, although (B) and (Fb) will be zero in the stationary frame when [v=0]. If these assumptions are correct, the diagram attached to post #5 also seems correct. Obviously, in the stationary frame, there is only an electric force [Fe]. While a magnetic force [Fb] exists in the moving frame, equations [12] and [13] suggest that the magnetic force is always smaller than the electric force by the factor $$(v/c)^2$$. So given the form of the equation above, I am not sure where the comment “not considering the time-of-travel for the fields” applies. Thanks

Last edited by a moderator: May 4, 2017
8. Jul 19, 2010

### AJ Bentley

It is fairly simple to calculate the E and B fields of a single moving charge.

the fields transform thus:-
$$E' = \gamma(\vec E + \vec v \times \vec B )$$
$$B' = \gamma ( \vec B - \vec v \times \frac{\vec E}{c ^{2}})$$

Since B is zero for the stationary charge, the fields seen in the relatively moving frame are:-

$$E' = \gamma(\vec E )$$
$$B' = \gamma ( - \vec v \times \frac{\vec E}{c ^{2}})$$

However, the other charge is also moving at velocity v so in order to calculate the force it sees, you have to transform back into the original frame. And you simply end up where you began. Not a very interesting result.

Last edited: Jul 19, 2010
9. Jul 22, 2010

### mysearch

I was wondering whether anybody is in a position to comment with any authority on the attached diagram, which is attempting to show the (E) and (B) forces in the stationary and moving frames. The model assumes the (v=0.5c), where (Q) and (q) are both positive charges, which are conceptually tied together to resist the repulsive electrostatic force between them. At relativistic speeds, I am assuming that the inherent propagation speed [c] has to be taken into consideration, such that it takes (t) units of time for the E-field to propagate from (Q) and arrive at (q), where (r=ct).

If the stationary frame is transposed into the moving frame, it must also take a finite time (t’) for the E-field to propagate between (Q) and (q), although this observed time is now subject to time dilation, which increased the path from [r] to [r’=ct’]. Note, the shortest path between (Q) and (P’) is now defined by [r’], which implies that [E’] is reduced at [P’] in the moving frame. However, there is an additional B-field in the moving frame due to the relative velocity [v] of (Q). The direction of the B-field, as shown in the diagram, is defined by the polarity of the charge, its velocity (v) and the effective angle between (P’), (Q) and the axis of motion. The B-field is subject to the same propagation delay as the E-field.

At (t1’), in the moving frame, (q) arrives at (P’). While (Q) will have also moved in position, both the strength and direction of the E and B field vectors are still defined by its position at [t0’]. As far as I can tell, [q] approaches the B-field at [P’] at right-angles, such that the magnetic force on [q] would point vertical downwards, as shown and in accordance with the RH rule. The size of the magnetic force is proportional to the reduced E-force at [P’], but multiplied by the factor (v/c)^2.

The only problem with this model is that I can’t see how the vector addition of E and B forces at [P’] in the moving frame can be reconciled with the force at [P] in the stationary frame. As such, I am assuming the logic is wrong somewhere and would therefore appreciate any insights. Thanks

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