# Propgating Electric Field Lines

1. Jul 7, 2010

### mysearch

Hi,
I am trying to get a better idea of how E-field line diagrams are drawn. The following link appears to be a nice example of field lines associated with a single charge moving linearly with constant velocity [v=0.5c]. By default, after pressing go, the animation displays the field lines for a single charge moving along the x-axis at a constant velocity [v=c/2]. If you stop this animation at any point, the vertical field line is always a line that points straight up. http://www.its.caltech.edu/~phys1/java/phys1/MovingCharge/MovingCharge.html

However, I would like to better understand the rules by which this line is drawn.

Basic Rules:
For each location, the electric field vector is tangent to the direction of the electric field line at any given point? The direction of resultant field vector is the vector sum of all the charges present?

Example:
To describe the problem I am trying to resolve – see attached diagram. This diagram is essentially a freeze frame, which would appear to be equivalent to the animation cited above. There is only 1 charge, so the electric field is assumed to propagate with velocity [c] in all directions, while the charge propagates with a relativistic velocity [v=c/2] from left to right. However, the example freeze frame only focuses on just the vertical field line and attempts to determine the field vector direction at 3 offset points, e.g. p1, p2, p3. At this frozen moment in time, the charge has just arrived a t=0, allowing no time for updates of the electric field to have propagated from this position towards p1, p2, p3. As such, it would seem that the field vector at p1, p2, p3 must reflect the field strength propagated from an earlier position of the charge, which is resolved to a granularity of [dt]. What the diagram suggests is that due to the combination of the charge velocity [v] and the field propagation velocity [c], the field vectors at p1, p2, p3 all point in different directions at the time of the frozen frame, i.e. t=0.

Just such, I don’t understand how the field line can be said to point straight up, if the electric field vector has to be tangent to the direction of the electric field lines at any given point. Therefore, I would appreciate any insights that explain this aspect of field lines. Thanks

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2. Jul 7, 2010

### Bob S

If the charge is moving at a constant velocity, it does not radiate energy, so the vertical field line always remains vertical. If the charge "dragged" the field lines like the shock wave of a bullet in air, then the Poynting vector of the electromagnetic field would point outward, and not directly forward, which would represent radiated energy. The vertical field line remaining vertical is consistent with the Lorentz transform of the symmetric field lines from a stationary charge to a moving frame. See Section 38.1 in

http://pdg.lbl.gov/2009/reviews/rpp2009-rev-kinematics.pdf

If the charge were accelerating, then the Poynting vector would point outward, representing radiated energy. When you suddenly increase the charged particle velocity in the Cal Tech applet, you can see the momentary radiation represented by the outward pointing wavefront (Poynting vector).

Bob S

3. Jul 7, 2010

### mysearch

I agree that the Lorentz transform can convert the picture back to what a stationary or collocated observer would see if moving alongside the moving charge, i.e. an E-field line propagating outwards as a straight line. However, I don’t see how this would be observed within the frame that perceives the charge moving pass with [v=c/2]. In the frame where the charge is moving, the direction of field vector at p1, p2 and p3 at t=0 are all different because they are emitted are earlier times/positions. However, I will think about the transforms some more. Thanks

4. Jul 7, 2010

### Bob S

I have worked a lot with extremely relativistic charged particles inside metallic beam pipes. The Lorentz contraction along the direction of motion reduces the angular distribution of the field lines to a flat pancake perpendicular to the direction of motion. The perpendicular field vector remains perpendicular and terminates on image charges on the beam pipe wall. If there are losses in the beam pipe wall from the image current, the perpendicular field line is dragged a little bit, and the Poynting vector points outward, meaning power is being radiated (dissipated in the beam pipe wall).

Bob S

5. Jul 8, 2010

### mysearch

I don’t have any background experience in this field, so the issues I outlined in post #1 is my present conceptual understanding, which may be proven wrong. However, I would like to try to reach some consistent picture of what is perceived in the stationary and moving frame in this case.

But first, could raise a secondary issues regarding energy/power radiation?

Most standard text seem to suggest that a stationary charge or one moving with constant velocity [v], as in the example cited, does not radiate power, although there would be a magnetic field in the constant velocity frame. However, there does seem to be a general acceptance that any change in the position of the charge must result in a change to the E-field that can be perceived to propagate outwards at [c], in all frames of reference. Given that the strength of the E-field can be related to the potential energy exerted on a test charge, and that any change to the E-field is subject to a propagation delay; there seems to be an implicit suggestion that potential energy is linked to this propagation mechanism. However, this energy is different to that described as EM radiation and linked to acceleration via the Larmor formula.

As such, do the observations in the stationary and moving frames have to be resolved purely in terms of the Lorentz transformation?

There seems to be no issue about the E-field lines propagating along a vertical y-axis within the stationary frame. As such, the E-field vector would conform to the rule that ‘it forms a tangent to the direction of the electric field line at any given point’. Equally, it is accepted that a stationary frame can be converted into a moving frame by applying the Lorentz transforms, but what might be debatable is the scope of the change in the perspective between the 2 frames, if only [c] remains constant

So can the vertical E-field line remain vertical in the moving frame?

The Lorentz factor in the example is $$\sqrt{1- \beta^2}=0.866$$ based on v=c/2 and would result in length contraction along the axis of motion, i.e. x-axis, plus time dilation. Based on limited reading, there also appears to be a strengthening effect of the E-field component in the y-axis. This complementary effect to [x] and [Ey] is said to explain how an E-field line remains perpendicular to the surface of the source charge, as length contraction squashes the circular propagation of the E-field in the stationary frame to an oval shape in the observed moving frame and presumably towards a ‘flat pancake’ at even higher velocities. However, the field line diagrams involving just a single charge are not really making any reference to the E-field strength, just its direction, as such; it would appear that we only need to focus on the central issue:

Can an E-field line propagate vertically at [c], while the frame is seen to be moving with velocity [v]?

At this point, let me reiterate that I am not arguing from a position of certainty, simply based on what I currently understand of the mechanisms involved. As I see it, anything propagating at [c] along the x-axis can be resolved in both frames of reference by the equation: c=x/t=x’/t’, where [x’,t’] reflect the relativistic effects on [x,t]. However, the y-axis seems more problematic as it is not subject to length contraction, such that we would be left to explain the implications of [c=y/t’]. As such, any motion with [c] along the y-axis, while simultaneously propagating with [v] along the x-axis would seem to result in a vector sum that would exceed [c]. Therefore, I don’t see how the Lorentz transforms explain the E-field lines propagating outwards vertically in the moving frame or resolves the directions of the E-field vectors illustrated in post #1. I would welcome any clarification of any mistaken assumptions as this is a learning process to me. Thanks

Last edited: Jul 8, 2010
6. Jul 8, 2010

### mysearch

On further review, I realise that I need to revise a number of the assumptions in post #5. I will post any corrections for clarification when I have worked through the details.

7. Jul 9, 2010

### mysearch

When I first looked at the Caltech animation, see link below, I simply saw the propagation of field lines from a moving charge [Q] without making any assumptions as to the source and all the SINKs [q] of these field lines.
http://www.its.caltech.edu/~phys1/java/phys1/MovingCharge/MovingCharge.html

As such, the field vectors in the diagram attached to post #1 were originally calculated with respect to fixed points in space, e.g. p1, p2, p3. However, I think that the only way the animation makes sense is to assume that both the source [Q] and all the sinks [q1-qn] of the field lines have the same constant velocity [v]. I would welcome any confirmation that this is the case.

As far as I can see, it is not possible to draw field lines of a moving charge [Q] with respect to stationary unit test charges [q] and still maintain the rule that ‘it forms a tangent to the direction of the electric field line at any given point’ while producing a continuous line. Again, if this is not correct, I would like to understand how such field lines are drawn.
As BobS pointed out from the start, if the field lines are associated with ‘consistent’ stationary and moving frames, i.e. Q and q have the same velocity [0 or v], then it should be possible to use the Lorentz transforms to switch between both frames. I have attached a diagram that attempts to illustrate what I think are a consistent model of both frames, but would welcome any corrections. The model on the left is the stationary model in which the field propagates vertically upwards at c=1, such that the arbitrary distance y=10 is covered in 10 units of time. If the moving frame is to be consistent with the stationary frame, the original propagation delay, i.e. t=10, must be equivalent, but subject to time dilation; at v=c/2, this become t’=11.547, which is then converted to the radial distance [r’=ct’]. The logic used to determine [x’] was to assume that it must take the charges [Q,q] the same time to propagate at [v] and so [x’=vt’]. However, I not totally sure about this line of logic, but possible this is an issue for the relativity forum?

Finally, I have started to look at the derivation of an equation of the form:

$$E=\frac{KQ}{r^2}. \frac{1- \beta^2}{(1- \beta^2sin^2 \theta)^{3/2}}$$

If $$[\beta=0]$$ when [v=0], then the equation collapses to [E=F/q]. In the example cited, $$[\beta=0.5]$$ and therefore we might wish to evaluate for [E] along the x-axis, e.g. sin(0)=0, and along the y-axis, e.g. sin(90)=1. If we normalise the [E=F/q] component to 1, the value of E(x)=0.75 would reflect contraction, while E(y)=1.15 would reflect expansion. Again, I would welcome any deeper insights. Thanks.

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8. Jul 10, 2010

### Bob S

I attach an applet thumbnail using the Cal Tech link above. The applet consists of three parts:

1) the particle is stationary (β = 0). All the field lines are radial from the point charge. The electric fields are parallel (longitudinal) to the field lines.

2) The particle is instantly accelerated from β=0 to β=0.335.

3) The particle is moving at β=0.335. The electric fields are still parallel (longitudinal) to the field lines. The field line originally perpendicular to the particle direction of motion is still perpendicular to the velocity.

Back to 2). The zig-zag is due to the acceleration from β=0 to β=0.335. During this instant, the electric fields are perpendicular (transverse) to the field lines and to the wave front propagation away from the particle. Because the magnetic field is azimuthal, there is a non-zero energy (Poynting vector) propagating radially outward from the particle. The energy is being radiated in a sphere away from the moving particle.

Bob S

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9. Jul 11, 2010

### mysearch

Part 1 of 2

With respect to the discussion as a whole, I will split my comments into 2 distinct sections. In the first section, I am assuming the charge has zero or constant velocity [v] such that there is no EM radiation, while in the second section, discussed in my next post, the charge is subject to acceleration. So, as per bullet-1, what we really seem to be describing is a system with 2 charges [Q, q]. Charge [Q] is the central charge that we might consider as the source from which the E-field propagates, while [q] is the sink charge used to measure the resultant E & B fields. When [Q, q] are at rest in the stationary frame, there is no B-field. I use the term ‘propagate’ as implied in post #5.
If we switched to bullet-3, it would seem that this system has to also be described in terms of 2 charges [Q, q], such that both charges are moving with the same velocity [v], which is why this system can be transposed, via the Lorentz transforms, back into the stationary frame as described by bullet-1. My diagram in post #7 attempted to show the basic effects of the transform in which the stationary and moving systems were equivalent and led to my comment:
However, while we are describing equivalent systems, we are still talking about 2 different perspectives, i.e. the stationary and moving frame. Again, I have tried to illustrate what I think the basic perspective ‘looks’ like from the stationary and moving frames in the diagram attached. Even in the stationary frame, it would still take a finite time for the E-field to propagate between Q and q, i.e. dt, but the E-field vector would conform to the rule that it forms a tangent to the direction of the electric field line at any given point.

But what about the moving frame?

With reference to the diagram, the velocity [v] of [Q,q] in the moving frame appears to introduce the effect of a B-field plus the comparative fields at [Q,q] change in both space and time, not just time. As such, an observer of moving frame would ‘see’ [Q] move along a length contracted x-axis, but must still ‘see’ the propagation of the fields from [Q] to [q] proceed at velocity [c], which is resolved via the time dilation, such that [c=r/dt=r’/dt’]. So, as far as I can see, nothing can propagate vertically between [Q] and [q] in the moving frame, because it suggests an infinite velocity. Of course, the description of the moving frame can be transposed back into the stationary frame, where the field propagation and the field lines are allowed to propagate vertically, because there is no change in space, i.e. x.

By way of a tangential issue, I would like to try to clarify the force of the fields on [q] due to the E and B fields:

[1] $$F_E=K_1 \frac {Q}{r^2}$$

This equation is the E-field force implicitly associated with a stationary frame.

[2] $$B=K_2 \frac {Qv_1sin \theta}{r^2}$$

This is the B-field associated with [Q] moving with velocity [v1]

[3] $$F_B=qv_2 \times B$$

This is the B-field force on [q] moving with velocity [v2]. In the examples being discussed [v1=v2=v]. However, I was not sure whether [v2] in equation [3] represented the relative velocity with respect to the B-field being generated by [Q] moving at [v1]? I was also interested in the implications of the following rearrangement of Lorentz force equation:

[4] $$F=q ( E_1 + v_2 \times B)$$

[5] $$F/q= E_1 + v_2 \times B) \equiv E_1 + E_2$$

Given that [E=F/q], the units of each expression in [5] must correspond to an E-field. As such, there is the implication that E1 is the stationary E-field, while [vxB] seems to be an equivalent E-field felt by [q] moving with [v2]?

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10. Jul 11, 2010

### mysearch

Part 2 of 2

The reason I started to look at field lines was due to the derivation of the Larmor formula based on the interpretation of the transverse field – see section 4.2/page 31 of Daniel Schroeder’s pdf document, which can be access via the following link: http://physics.weber.edu/schroeder/mrr/MRRnotes.pdf.

The essence of the derivation seems to centre on the following equations

[1] $$Er = Kq \frac {1}{r^2}$$

[2] $$Et = \frac{Kq}{c^2} \frac {a Sin \theta}{R}$$

Equation [1] is the radial expression that we might associate with a stationary charge linked to the inverse square law, while [2] is the transverse E-field dependent on acceleration [a] and [1/R]. Based on these equations and my initial and limited understanding of field lines I tried to produce s basic animation for a single charge oscillating under the influence of an alternating sine wave voltage. I have tried to attach the gif file animation, but don't know whether this will work via the PF server. Footnote, the server changed the gif file to a jpg file - sorry.

Anyway, the left side of the animation was an initial attempt to show the effect on the electric field lines, while the right was meant to align to an equivalent view based on streams of photons being emitted by the charge. The field lines or photon streams are shown at different angles, e.g. 0, 30 and 60 degrees, from the maximum, which is always perpendicular to the axis. The oscillating red lines on the left reflect the total electric field [E=ER+ET] as a function of distance, such that you see the effects of [ER] reducing by [1/R^2], while [ET] only reduces by [1/R] and so the latter quickly becomes the dominate field as the radius from the charge increases. However, the photon view on the right was thought to better reflect the situation as the charge passes through the centre of oscillation, i.e. maximum velocity, minimum acceleration. Under this condition, i.e. a=0, the Larmor formula would suggest the photon stream should not emit any photons due to zero acceleration. Of course, over an extended period of time, the statistical distribution of photons would still conform to the radiated power predicted by the Larmor formula as represented by the intensity of the yellow shading.

However, due to some of the questions raised in this thread, I began to question my understanding and interpretation of field lines for moving charges. Anyway, appreciated the help extended. Thanks

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11. Jul 11, 2010

### Bob S

Re: Part 2 of 2

Thanks for the link to Schroeder's document. I was unaware of it. It presents a very comprehensive discussion of the radiated field. Schroeder's Fig 4.4 and my Efield3.jpg in my earlier post are the same, except that Schroeder is decelerating the particle, and I am accelerating it.

Your discussion on needing image charges q having the same velocity as the particle charge Q is entirely correct. All field lines need to terminate on charges. With regard to the velocity of the image charges, if they were infinitely far away (implicit in this discussion), they would not have to be moving.

In my view of a charge Q moving inside a perfectly-conducting beam tube, the image charges q are flowing along the beam tube wall at velocity v, equal to the particle velocity. This is equivalent to the TEM mode in a coaxial cable (transmission line) below cutoff. To accelerate the particle, we put an accelerating voltage V on a gap in the beam tube wall. In this case, the zig-zag tom the acceleration has to propagate inward from the beam tube wall to the particle. An alternate view is the voltage V in the gap produces an electric field across the entire interior of the beam tube, which the particle has to cross. Also, if the beam tube wall is resistive, there is a voltage drop and resistive losses in the beam tube. But this voltage drop will not occur until the image charges q reach the resistance in the beam tube. The only way these resistive losses and the resulting voltage drop can reach the particle is via electric field zig-zags propagating inward to Q from the image charges q. So we can have both zig-zags and zag-zigs. So do we accelerate particles with charge Q, or its image charges with charge q? Does Q pull the q's, or vice-versa?

Bob S

12. Jul 12, 2010

### mysearch

I have highlighted the last sentence because I agree that you could put the sink charges at infinity, but I don’t see how it explains the field lines drawn by the following animation at relativistic velocity: http://www.its.caltech.edu/~phys1/java/phys1/MovingCharge/MovingCharge.html

The only way I can see how the field line would propagate vertically towards a sink charge at infinity is that the subtended angle (x with respect to y where y is infinity) is so small that it doesn’t effectively change. However, if this were the case, then the source charge on this scale would not be moving, which is not the case in the default animation above, where v=c/2. For example, I have attached a snapshot from the previous animation, but changed v=0.95c just to accentuate the length contraction. However, for the length contraction to exist, you have to be viewing the field lines from the perspective of an observer that sees the charges moving. As a snapshot, at time [t=0], the field lines from this point have not had time to propagate vertically. As such, the E-field along the vertical y-axis seem to represent the propagated values from earlier times when the ‘moving’ charge was in different positions along the x-axis. As such, I believe the vector direction would also be orientated towards these positions in the moving frame. Again, I believe this is entirely consistent with the Lorentz transforms back into a stationary frame where the field lines would propagate vertically.

I have also attached a second diagram that shows what I believe to be consistent figures for the stationary and moving frames. As such, I still believe that the animation is wrong in its presentation, at least, at relativistic velocities. Of course, when v<<c, the propagation at [c] vertically along the y-axis is so much greater with respect to any horizontal shift along the x-axis due to [v] that the vertical field line is a good approximation. As such, I don’t think this invalidates the diagram presented in post #8, which is also reflected in the Daniel Schroeder document referenced in post #10.

As I only wish to understand the issues, I have emailed the people at Caltech informing them of this discussion. Presumably if they feel that I have misrepresented or misunderstood some aspect of their animation they can post a correction. Thanks.

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