Emitting one or two opposite charges along the x-axis

In summary, the conversation discusses the emission of charged particles and the resulting electric fields in various scenarios. It considers the effects of relativity on the electric fields and makes modifications to the scenarios by placing an observer on the x-axis with different velocities. Ultimately, it raises the question of whether the observer will observe an electric field or not in each scenario.
  • #1
particlezoo
113
4
Let's consider that I have an emitter that can emit both negative and positive electric charges. Here let's consider only scenarios with two particles (one negative and one positive) that start initially at the tip of some electrode, where one or both the charges will separate from at the same time. Let's say the one or two emitted particles are somehow confined to the x-axis. Furthermore, let's say that particles carry a lot of mass with them, so accelerations are negligible, and therefore so is their radiation. Let's make it so that they obtain their final velocities in a very short distance, such that the electric dipole moment that results is vanishingly small. Naturally, we could assume that the velocities therefore are not that large, but we should still consider the effects of relativity on electric fields. Finally, let's consider only scenarios where the relativistic difference in velocities between the particles is some fraction of c, a fraction which is held common between all the scenarios.

I am interested in the x component of the electric field that results from the emission of one or two charged particles with specifications I outlined above. First I will consider two scenarios where the emitter is stationary:

Scenario A) The emitter emits a positive and (an equal and opposite) negative charge at equal and opposite velocities at nearly the same point in spacetime (i.e. the events in spacetime where/when they are emitted from the electrode are separated by a vanishingly small spacetime interval such that they appear to occur at the same time regardless of the frame of observer). The positive charge moves right along the x-axis with velocity +v, while the negative charge moves left along the x-axis with velocity -v. The difference between +v and -v using the composition law of velocities is +u.

Scenario B) The emitter emits only a positive charge right along the x-axis with velocity +u, leaving behind an (equal and opposite) negative charge at the place where it left. The emitter experiences a vanishingly small change in velocity due to its arbitrarily greater mass.

I will now consider the electric fields of the charges in each scenario:

Scenario A's Electric fields) The electric fields of negative charge cancel the electric fields of the positive charge as they are contracted by the same factor.

Scenario B's Electric fields) The electric fields of the negative charge do not cancel the electric fields of the positive charge because the electric fields of the negative charge are not length contracted while the electric fields of the positive charge are length contracted. The electric field components perpendicular to the x-axis point away from the x-axis, while the electric field components parallel to the x-axis point towards the position where the positive charge was emitted. The components change in such a way that electric charge is conserved.

Now I will make some modifications to both scenarios.

In both scenarios, I will position an observer on the x-axis at a coordinate value of x. In both scenarios, I will keep the emitter's velocity constant with time (i.e. its acceleration is vanishingly small).

For each scenario, I will create two sub-scenarios.

In all sub-scenarios, the emitter is exactly the same, the emitter is at rest (or near-rest) relative to the "master observer" (which is wholly independent of the observer that I placed on the x-axis), and in all sub-scenarios in the emitter's frame the observer is placed at the same exact distance away from the emitter along the x-axis.

In each sub-scenario, the observer I positioned on the x-axis will have a velocity equal to either that of the negative charge or that of the positive charge in either Scenario A or Scenario B.

Therefore, there will be four sub-scenarios where the velocities are as follows:

Sub-Scenario A-Left) observer's velocity = -v
Sub-Scenario A+Right) observer's velocity = +v
Sub-Scenario B-Left) observer's velocity = 0
Sub-Scenario B+Right) observer's velocity = +u

Where u = ((+v) - (-v)) / (1 - (+v)(-v)/c^2) = ((+v) + (+v)) / (1 + (+v)(+v)/c^2) per the composition law of velocities.

Therefore, the relative proper velocities between the charges and the observer are the same between Sub-Scenario A-Left and Sub-Scenario B-Left. Similarly, they are the same between Sub-Scenario A+Right and Sub-Scenario B+Right.

In Scenario A, the electric field in the emitter's frame was non-existent due to the electric field of the negative charge cancelling the electric field of the positive charge. There was a magnetic field in Scenario A due to the opposite charges moving at opposite velocities, however, the value of the magnetic field on the x-axis is zero.
Regarding the "placed" observer in Scenario A: The "placed" observer on the x-axis in Sub-Scenario A-Left and Sub-Scenario A+Right will experience neither an electric field nor a magnetic field, as a Lorentz boost along the x-axis does nothing to give rise to an E-field along the axis.

In Scenario B, the electric field in the emitter's frame existed because the positive charge's electric field was stronger perpendicular to the x-axis and weaker parallel to the x-axis, due to its velocity along the x-axis of +u when compared to the electric field of the negative charge whose velocity along the x-axis was 0. So the electric field components perpendicular to the x-axis are generally pointed away from the x-axis, and the electric field components parallel to the x-axis are generally pointed towards the point where the positive charge was emitted.
Regarding the "placed" observer in Scenario B: A Lorentz boost of the observer along the x-axis should not affect the x-component of the electric field which it observes. Since the observer is located on the x-axis, it sees no perpendicular component of the electric field. Therefore, if the observer of Sub-Scenario B-Left were Lorentz boosted such that we get Sub-Scenario B+Right, the electric field it would observe would be the same as in Sub-Scenario B-Left.

Here is the problem in short:

In Sub-Scenario A-Left and Sub-Scenario A+Right, the "placed" observer does not see an electric field.
In Sub-Scenario B-Left and Sub-Scenario B+Right, the "placed" observer does see an electric field.
The relative proper velocities between the charges and the "placed" observer are the same between Sub-Scenario A-Left and Sub-Scenario B-Left.
The relative proper velocities between the charges and the "placed" observer are the same between Sub-Scenario A+Right and Sub-Scenario B+Right.

So we end up with a situation where we can have the same relative proper velocities between the charges and the "placed" observer, but since a Lorentz boost of an observer along the x-axis does not transform the x-component of the electric field that it observes, the x-component of the E-field acting on the our "placed" observer on the axis is the exact same as observed by our "master observer"/emitter frame, on a per scenario basis, yet, the "master observer"/emitter frame does not see an electric field acting on our "placed" observer in scenario A, and yet the "master observer"/emitter frame does see electric field on our "placed" observer in scenario B.

To repeat once more: In all sub-scenarios, the emitter is exactly the same, the emitter is at rest (or near-rest) relative to the "master observer" (which is wholly independent of the observer that I placed on the x-axis), and in all sub-scenarios in the emitter's frame the observer is placed at the same exact distance away from the emitter along the x-axis.
Kevin M.
 
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  • #2
Just use the standard Lienard Weichert fields. Everything will work out correctly.
 
  • #3
Dale said:
Just use the standard Lienard Weichert fields. Everything will work out correctly.

So let me take that and reduce it quite considerably as per the conditions given in the opening post. Per the conditions I brought up, we can disregard the acceleration terms and we can consider the case that beta << 1. Also, we eliminate any electric dipole terms due to v << c such that the obtained charge separation is trivially small over the arbitrary period of time over which the charge is accelerated. So the retarded fields reduce essentially to quasistatic fields. Therefore, the retarded time can be disregarded and we end up with a "quasistatic" field (fair approximation) consisting of superimposed fields due to charge density/current elements in their equilibrium state (constant-with-time velocities and positions whatever frame of reference chosen).

So we have simple electric fields that point radially to/from the source charges, and they may be length contracted or not. There are no transverse electric fields acting on the charges or observer as the charges and observer and their motions are confined to the x-axis of the emitter frame.

In Scenario A, the equal and opposite charges move in opposite velocities in the emitter frame, so they are length contracted by the same amount in that frame. Their electric fields cancel according to the conditions, which include trivial separation distance (i.e. vanishingly small spacetime interval) between the charges.

In Scenario B, the equal and opposite charges do not move in opposite velocities in the emitter frame, but the positive charge moves and the negative charge remains on the emitter. Again, a vanishingly small spacetime interval between the charges applies.

Boosts are considered along the x-axis only, where the particles and observer are confined, thus boosts of our observer do not have an effect on the electric field that it observes. However the electric field observed by the emitter frame to act on this "placed" observer on the x-axis is different between scenario A and B, even though the relative proper velocity between the "placed" observer and charges can be the same in Scenario A and Scenario B (as per the four sub-scenarios I described above).

All velocities below are per the emitter's frame:

Sub-Scenario A-Left) observer's velocity = -v
Sub-Scenario A+Right) observer's velocity = +v
Sub-Scenario B-Left) observer's velocity = 0
Sub-Scenario B+Right) observer's velocity = +u

Where
-v = the velocity of the negative charge in Scenario A
+v = the velocity of the positive charge in Scenario A
0 = the velocity of the negative charge in Scenario B
+u = the velocity of the positive charge in Scenario B
+u = ((+v) - (-v)) / (1 - (+v)(-v)/c^2) = ((+v) + (+v)) / (1 + (+v)(+v)/c^2) per the composition law of velocities.

If we chose Sub-Scenario A-Left, we can see the relative proper velocities are the same as in Sub-Scenario B-Left.

If we chose Sub-Scenario A+Right, we can see the relative proper velocities are the same as in Sub-Scenario B+Right.
 
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  • #4
particlezoo said:
So let me take that and reduce it quite considerably
When you simplify a consistent law and find an inconsistency then you know that you have oversimplified. There is nothing to be done besides avoiding the offending simplification.
 
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  • #5
Dale said:
When you simplify a consistent law and find an inconsistency then you know that you have oversimplified. There is nothing to be done besides avoiding the offending simplification.

I will have to narrow down that then.

The whole point of the simpification of the problem was to start with a less complicated problem. The idea was to create a system were the fields emitted are purely radial to the charges and each charge's field contribution is near-bisymmetrical and exactly cylindrically symmetrical, just as would be for a charge moving at constant velocity (or no velocity). Since charges can be endowed with mass, we can limit their acceleration in principle such that r * a / c << v for the observer at disance r away from the source. Starting with this, would you agree that we can obtain a system of charges confined to the x-axis where these symmetries are respected and where acceleration is not the contibutor to the problem I indicated, but something else?
 
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FAQ: Emitting one or two opposite charges along the x-axis

1. What is meant by "emitting one or two opposite charges along the x-axis"?

"Emitting one or two opposite charges along the x-axis" refers to the process of releasing one or two particles with opposite electrical charges in a specific direction, specifically along the x-axis.

2. What is the purpose of emitting one or two opposite charges along the x-axis?

The purpose of emitting one or two opposite charges along the x-axis is to create an electric field in that direction, which can be used for various applications such as particle acceleration or electrostatic manipulation.

3. How is the strength of the electric field affected by emitting one or two opposite charges along the x-axis?

The strength of the electric field is directly proportional to the magnitude of the charges emitted along the x-axis. Therefore, emitting larger charges will result in a stronger electric field.

4. Can emitting one or two opposite charges along the x-axis be done in any direction?

No, emitting one or two opposite charges along the x-axis specifically refers to releasing the charges in the x-direction. It is possible to emit charges in other directions, but it would not be considered along the x-axis.

5. What is the difference between emitting one charge and emitting two opposite charges along the x-axis?

Emitting one charge along the x-axis will create a unidirectional electric field, while emitting two opposite charges will create a bidirectional electric field. Additionally, the strength of the electric field will be stronger when two opposite charges are emitted compared to one charge.

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