The electromagnetic field in an accelerated frameby johne1618 Tags: accelerated, electromagnetic, field, frame 

#1
Nov2412, 04:38 PM

P: 355

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]tr/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? 



#2
Nov2412, 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 nonzero fields.
Are you taking time t as from the originobserver'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 lightspeed 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 mixup 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? 



#3
Nov2412, 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. 



#4
Nov2512, 03:21 PM

P: 355

The electromagnetic field in an accelerated frameYou 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 zaxis and is taken to be negligible at the origin whereas the "dynamic" perpendicular component [itex]E_x[/itex] acts along the xaxis and is dominant at the origin. In the first situation the charge must have accelerated at time tr/c in the observer's past in order that the observer measures a retarded field at time t. So what do you think of this hypothesis? :) 



#5
Nov2512, 06:42 PM

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You are right  I misread:
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. 



#6
Nov2612, 08:56 AM

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It still seems to me that the accelerating observer will perceive a changing Afield, 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. 



#7
Nov2612, 10:20 PM

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The selfenergy 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 timestep, 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 fielddetector 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. 



#8
Nov2712, 10:14 AM

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[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: 



#9
Nov2712, 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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