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The electromagnetic field in an accelerated frame

by johne1618
Tags: accelerated, electromagnetic, field, frame
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johne1618
#1
Nov24-12, 04:38 PM
P: 373
Imagine a stationary charge [itex]q[/itex] located on the positive [itex]y[/itex]-axis at a distance [itex]r[/itex] from a stationary observer at the origin.

Let us assume that the distance [itex]r[/itex] is large enough such that the electrostatic field due to the charge is negligible at the origin.

Now let us assume that the charge [itex]q[/itex] is given an acceleration in the positive [itex]x[/itex]-direction.

The electric field at the origin at time [itex]t[/itex] is directed along the [itex]x[/itex]-axis and is given by

[tex]
E_x(t) = \frac{-q}{4 \pi \varepsilon_0 c^2 r} a_x(t-\frac{r}{c}).
[/tex]
This is the standard retarded expression for the electric field due to an accelerated charge measured by a stationary observer at the origin. The electric field occurs at time [itex]t[/itex] after the acceleration of the charge at time [itex]t-r/c[/itex].

Now let us imagine the complementary situation.

We start again with a stationary charge [itex]q[/itex] located on the [itex]y[/itex]-axis at a distance [itex]r[/itex] from a stationary observer at the origin.

Now let us assume that the observer is given an acceleration in the positive [itex]x[/itex]-direction while the charge [itex]q[/itex] remains stationary.

I believe the electric field at the origin at time [itex]t[/itex] in the frame of the accelerated observer is given by

[tex]
E_x(t) = \frac{q}{4 \pi \varepsilon_0 c^2 r} a_x(t+\frac{r}{c}).
[/tex]
This is the advanced expression for the electric field due to a stationary charge as measured by an accelerating observer. From the observer's point of view the charge has an apparent acceleration [itex]-a_x(t+r/c)[/itex]. The electric field occurs at time [itex]t[/itex] before the apparent acceleration of the charge at time [itex]t+r/c[/itex]. We would expect this result as the field should come into existance as soon as the observer accelerates and thus before the apparent acceleration of any charge.

Thus there is an equivalence between charge accelerating/stationary observer and stationary charge/accelerating observer. The former situation is described by a retarded field and the latter by an advanced field.

Is this correct?
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Simon Bridge
#2
Nov24-12, 08:40 PM
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I'm a tad bothered by the assumption that |E| is "negligible" at the origin at t=0 - and then you go on to talk about non-zero fields.

Are you taking time t as from the origin-observer's clock?
r is a function of t as well ... anything happening out at r at time t in the frame of the origin will be seen at the origin no sooner than t+r/c ... because of the light-speed delay. If r is very far away, then r/c will be a very long time.

The first situation has changes in E propagating away from the accelerated charge and the second has the observer moving into and increasing electric field. You'd think that, in the first case, is the charge starts accelerating when the observer's clock reads t=0, then the observer sees the change start at t=r/c wouldn't you? So the equation only applies for t≥r/c.

In the second case, though, the observer starts accelerating at t=0, then the changes should appear right away ... the observer is moving into an electric field that is already there - waiting. Of course, this also means the observer is no longer at the "origin" or in the same frame of reference as before.

So I suspect there is some mix-up of ideas here... but I'm uncertain exactly where.
In relativity you have to be very careful about whose clock and whose rulers etc.

When you want to ask questions like this - it is best you work your reasoning to a conclusion of some kind so we can see if (and where) the important mixups have occurred - otherwise we are left with trying to second guess these things and you get a big list of possible problems.
[edit] upon checking further - you may have provided enough information (see below).

Where were you thinking of going with this?
Simon Bridge
#3
Nov24-12, 09:01 PM
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Checking - I suspect you are thinking in terms of this:
http://www.tapir.caltech.edu/~teviet/Waves/empulse.html
... you wanted r to be large enough that the radiation part of the field equation dominates?
If so you'll see that your equation is not correct ... you have confused the radial and perpendicular components of the field.

Also have some fun with:
http://www.cco.caltech.edu/~phys1/ja...ingCharge.html
... accelerate the charge by selecting user mode and dragging the velocity slider back and forth.

johne1618
#4
Nov25-12, 03:21 PM
P: 373
The electromagnetic field in an accelerated frame

Quote Quote by Simon Bridge View Post
I'm a tad bothered by the assumption that |E| is "negligible" at the origin at t=0 - and then you go on to talk about non-zero fields.
I think this is standard practice when analyzing "far field" effects. The static fields decay as [itex]1/R^2[/itex] and therefore become negligible compared to the acceleration field which decays as [itex]1/R[/itex].

You say that I have confused the radial and perpendicular components of the field.

I don't believe I have. The "static" radial component acts along the z-axis and is taken to be negligible at the origin whereas the "dynamic" perpendicular component [itex]E_x[/itex] acts along the x-axis and is dominant at the origin.

Are you taking time t as from the origin-observer's clock?
Yes. The observer at the origin measures an electric field at time t.

In the first situation the charge must have accelerated at time t-r/c in the observer's past in order that the observer measures a retarded field at time t.

In the second case, though, the observer starts accelerating at t=0, then the changes should appear right away ... the observer is moving into an electric field that is already there - waiting. Of course, this also means the observer is no longer at the "origin" or in the same frame of reference as before.
In the second case the observer accelerates at time t. There is little electric field for him to move into initially as the static field from the charge is negligible at the origin. I believe instead that in his accelerating frame there instantly appears an electric field due to his motion. The rationale for this prediction is that in his frame the charge appears to be accelerating in the opposite direction. An accelerating charge should produce an electric field. Although this is a special situation in which the charge's acceleration is "fictitious" or apparent.

Where were you thinking of going with this?
My hypothesis is that the apparent acceleration of the charge in the second scenario has a real effect: it produces a real electric field in the frame of the accelerating observer. This field is described by an advanced potential rather than the usual retarded potential. I believe this is a novel idea as advanced potentials have not been used in physics before (except for the Wheeler-Feynman theory of 1/2 advanced 1/2 retarded em waves).

So what do you think of this hypothesis? :)
Simon Bridge
#5
Nov25-12, 06:42 PM
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You are right - I misread:
Imagine a stationary charge q located on the positive y-axis at a distance r from a stationary observer at the origin.
... for some reason I was thinking the charge was some distance on the x axis. Excuse. This is why physics is usually explained with diagrams :)

So - in that framework:
The thing to realize is that ##E_r## is not negligible ... it is very small compared with ##E_\perp##. This is important because the whole reason you get acceleration effects when the observer is the one doing the work is that the static field is not negligible. (In other words: "negligible" is a relative term.)

Your formulation looks like you have an instant jump in field strength at the onset of acceleration .... that suggests that an infinitesimal change in speed produces a finite change in the field doesn't it? Does that make sense?

Did you come up with the equation by analogy or by derivation from the physics.
Have you had a look at the case for observers moving at a constant velocity? The E field is changing with time in that case too... so they see a B field as well.
johne1618
#6
Nov26-12, 08:56 AM
P: 373
Quote Quote by Simon Bridge View Post
The thing to realize is that ##E_r## is not negligible ... it is very small compared with ##E_\perp##. This is important because the whole reason you get acceleration effects when the observer is the one doing the work is that the static field is not negligible. (In other words: "negligible" is a relative term.)
I have received an email from Prof. Bahram Mashhoon at the University of Missouri who is an expert in General Relativity and Electrodynamics in accelerated frames of reference. His view supports your position stated above:

I assume that the initial configuration you mentioned is in an inertial frame of reference. According to current ideas (i.e., the standard theory of relativity), the fields as measured by the accelerated observer will still be negligible unless the speed of the observer approaches the speed of light (ultra-relativistic motion). In that case the measured fields (electric and magnetic) will be time-dependent, but they will not be of the radiative type (retarded or advanced). In particular, the electric field in the direction of his motion will always be the same as before, namely, negligible.

The thought experiment you mentioned, namely, that from the viewpoint of the observer, the charge will be accelerating and hence it will radiate, is not an accepted part of the theory of relativity. There is no “relativity” when it comes to acceleration: acceleration is absolute. The charge that is always static in the inertial frame never radiates.

In any case, if a charge is accelerated with respect to an inertial frame, then it radiates and observers will perceive retarded radiation according to current theory, since this is what is dictated by the requirement of causality.
So standard relativity theory says that the accelerated observer will not feel any appreciable electric field.

It still seems to me that the accelerating observer will perceive a changing A-field, due to the changing relative velocity of the charge with respect to himself, and thus an induced electric field along the direction of his motion.

I think it would be good to perform an experiment. One could measure the effective inertia of an electron inside a charged hollow insulating sphere. By my theory the electron should have an extra electromagnetic component to its inertia due to the charge on the sphere.
Simon Bridge
#7
Nov26-12, 10:20 PM
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The self-energy of an electron is problematical enough ... you want to do the experiment with a larger charge. Charge small balls and see if they roll differently down an incline or something (first work out how big your expected effect aught to be.)

That description is tickling the back of my brain though ... I have a feeling it's been done some time ago to find out if charge contributes to inertia in different conditions. I know you'll find discussions in these forums about whether charge adds to mass in some way or another: it is a popular speculation.

I think you still need to address: Your formulation looks like you have an instant jump in field strength at the onset of acceleration .... that suggests that an infinitesimal change in speed produces a finite change in the field doesn't it? Does that make sense?
... i.e. if there is this jump in the field strength as described by the equation, in a vanishingly small time-step, then how does the field know to do this? At t=0 it does not know the observer is about to accelerate (or - how does the observer's field-detector know?)

This is related to the causality comment - the advanced form of the equation implies that the field observed somehow anticipates the acceleration that causes it.

I still think you should work your way through the existing theory and familiarize yourself with the math ... then you can hunt for existing experiments that relate to your ideas if you still maintain them. Note: a moving observer as described will experience a changing scalar potential because ##\phi## depends on ##\vec{r}(t)##.

Good luck - have fun.
johne1618
#8
Nov27-12, 10:14 AM
P: 373
Quote Quote by Simon Bridge View Post
I think you still need to address: Your formulation looks like you have an instant jump in field strength at the onset of acceleration .... that suggests that an infinitesimal change in speed produces a finite change in the field doesn't it? Does that make sense?
... i.e. if there is this jump in the field strength as described by the equation, in a vanishingly small time-step, then how does the field know to do this? At t=0 it does not know the observer is about to accelerate (or - how does the observer's field-detector know?)

This is related to the causality comment - the advanced form of the equation implies that the field observed somehow anticipates the acceleration that causes it.
I think its like

[itex] F = m a[/itex]

[itex] F = m \frac{dv}{dt}[/itex]

[itex] F dt = m dv [/itex]

An infinitesimal change in speed produces an infinitesimal (electromagnetic) inertial impulse.

In answer to Prof Mashhoon's point that acceleration is absolute I replied that:

The Electromagnetic field at a point P due to a distant charge can be described by the Heaviside-Feynman formula. The radiative part of the field at point P depends on the second time derivative of the unit vector pointing from P to the charge. Thus in this formulation there is no dependence on absolute acceleration.

There are two scenarios in which this unit vector can change. In the first scenario the charge can accelerate causing the unit vector at P to change as it tracks the retarded position of the charge. But I believe there is a second way in which the unit vector can change. If the point P itself is accelerated then the unit vector changes immediately as it tracks the advanced position of the charge in P's accelerated frame.
He replied back:

The derivation of such formulas has to do with Maxwell’s equations in an inertial frame of reference. There is no relativity of acceleration. The principle of relativity has to do with global inertial frames of reference and holds only for uniform inertial motion forever.
Simon Bridge
#9
Nov27-12, 09:01 PM
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You are not doing this the right way around - i.e. from understanding the current physical understanding of what is happening. That can be fun, but it is not productive.

Since you are already talking to someone I'll duck out until you need some clarification on what he's saying. Otherwise I'll just confuse you further.


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