
#1
Feb1912, 11:41 AM

P: 1,011

Let's say I have a Gaussian sphere 1 light year across with synchronized clocks and sensors all over its surface. All clocks are comoving, not accelerating, and the spatial curvature is negligible. If I have only one charge inside the Gaussian sphere, 1 centimeter from its surface for an entire year, then the integral of the electric field intensity over the surface of that sphere, multiplied by the electric permittivity of free space, should return the value of the single charge. The problem is this: If move the charge out of that sphere and then stop it 1 centimeter outside of it, the electric field at the other side of the sphere does not "update" until nearly 1 year later. I end up with a nonzero integral for electric flux even though the charge is not inside the sphere.
Let's say the sensors record the electric field as a function of time and time stamp it using the synchronized clock data. In about two years, an observer at the place where the electron crossed the sphere will be able to pick up the readings and time stamp information about the measured electric field. That observer would conclude that the readings measured for the electric field on the surface as the charge was displaced from inside to outside the sphere was not a constant. Simultaneity should not be an issue here because all the clocks and sensors share the same inertial frame, and thus are at relative "rest" with respect to one another. The only thing moving here is the charge and the body outside the sphere acting upon it. There is not a whole lot of velocity required, nor a whole lot of time, to make the charge move 2 centimeters. Therefore, no relativistic effects would apply to any appreciable magnitude. 



#2
Feb1912, 12:21 PM

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The fields for an arbitrarily moving point charge are given by the LienardWiechert potentials: http://en.wikipedia.org/wiki/Li%C3%A...hert_potential So either you did not use the correct expression for the fields, or you did the integral wrong. Without seeing your work it is not possible to tell which, but I would expect the former since you didn't mention the LienardWiechert potentials explicitly. 



#3
Feb1912, 12:33 PM

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The hypothetical surface being discussed is 1 light year across. 



#4
Feb1912, 12:44 PM

Sci Advisor
P: 1,562

Why I doubt the generality of Gauss' law: A Gaussian sphere 1 light year across
Most likely you forgot the instant where the charge has to accelerate from 0 velocity to some finite velocity. This will create the disturbance needed to properly balance out the flux through the sphere.
An easier way to do this would be to have the charge moving at constant velocity the entire time, coming in from minus infinity, passing through the sphere, and going off to plus infinity. The instant the charge crosses the boundary of the sphere, you should see the flux jump from 0 to Q, and then jump down again when the charge leaves. 



#5
Feb1912, 12:49 PM

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#6
Feb1912, 12:52 PM

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Even if it did, it's about twice the work. Let's imagine the following. Per OP, the displacement is 2 centimeters. Let's assume non relativistic motion. displacement = (1/2)*a*t^2 2 cm = (1/2)*a*t^2 Let t=1 second 2 cm/s^2 = (1/2)*a 4 cm/s^2 = a a = 0.004 Earth g's Let's calculate number of seconds it takes for a field limited by [itex]c[/itex] to reach the other end of the sphere. 1 year = 31556926 seconds When the time stamp is recorded for the time that the charge left the sphere, less than 0.00001% of the sphere knows that charge even left the sphere. 



#7
Feb1912, 12:53 PM

P: 1,781

If you think about the contributions to the integral you'll see that its quit's plausible that it still works.
The fields far from the source will look nearly identical even after you wait for the 2 years or more. The fields near the source point out of the sphere when it's inside but switch direction and point into the sphere as it crosses out. These terms near the chare are the dominant contribution and don't take but a few nanoseconds to change. The only remaining question is whether the total field just after the charge exit (which consists of a radiation front sweeping over the sphere plus the static field) satisfies Gauss's law exactly or only approximately. I suggest that if it is satisfied at any time then it is always satisfied. You can solve this problem by integrating the fields on an infinite plane for a short time after the charge moves through. Assume the gauss surface is a box instead of a sphere. Then the change hasn't had time to hit the other 5 sides so they remain a constant. You only need find the change to the integral in a small causal sphere on the flat plane. 



#8
Feb1912, 01:00 PM

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#9
Feb1912, 01:06 PM

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P: 1,562

The simple fact is that your charge must accelerate in order to change from 0 velocity to some other velocity. If you look at the LienardWiechart potentials, you will see that there is a term in the E and B fields sourced by the acceleration of the charge. If you have failed to include this term, then you are neglecting an important contribution to the flux integral.
You rightly point out that the E field on the far end of the sphere will not update fast enough. But this is irrelevant. All the necessary changes will happen locally, near the charge. You need to stop responding incredulously to everyone's posts, and actually do a calculation using the full theory with no approximation. 



#10
Feb1912, 01:09 PM

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#11
Feb1912, 01:16 PM

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4chars 



#12
Feb1912, 01:59 PM

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#13
Feb1912, 01:59 PM

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#14
Feb1912, 02:07 PM

P: 1,011

How is it that these changes cancel each other out until the very local moment that the charge crosses the sphere's wall exiting, where all of a sudden, these accelerationinduced fields must some how cancel, not themselves, but field divergence that one might otherwise expect at the surface of the rest of the sphere which is uninformed of this change of side? This is especially inconceivable considering the lack of information present at the crossing concerning how the "updated front" of the absence of the electric charge in the sphere may update the field detected at the overall surface of the sphere at a rate somehow matching in proportion to some function dependent on acceleration of that single charge now outside the sphere. If you guys want me to do the work, I need to know what to model to use. I like exact specifics on how to get to the conclusions through rigorous means. I get to hear these specifics precisely because I have been incredulous. If I don't ask questions, then I don't receive answers as to the context of my question. Without the latter, then how would I even know what to do in order to arrive at the same conclusion? I have encountered a few who have shown the innate tendency of modifying my question before attempting to answer it. That's exactly the kind of thing that I do not respond well to. 



#15
Feb1912, 02:25 PM

P: 1,011

If the charge is at relative rest outside the sphere for a long time, it contributes absolutely nothing to the flux integral on the surface of the sphere. It cancels out. If I have it outside the sphere for x amount of time, there exists a radius around this particle where this potentially cancelling flux may exist. But the initial front leads to net outward field lines from the sphere (or, really, net inward lines towards the negative charge outside), as if it was like having a + charge in the sphere. However, how exactly does that match the amount of integral on the other surfaces of the sphere? If I consider that the charge is now at rest with respect to the sphere, the "positive" contribution from this new field grows as the "negative" contribution from the original field shrinks. If I have a growing positive contribution and shrinking negative contribution, I do not see how those two derivatives can cancel. Instead, both should reinforce each other to gradually decrease the net field through the surface, not instantaneously upon the moment of the charge's crossing. Simultaneity arguments are out the window, I guess, for the Efield sensors in the example are timesynchronized, so there is no problem in determining what the field was at the other side of sphere when the electron crossed the surface. 



#16
Feb1912, 02:38 PM

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You have made a very specific claim: 



#17
Feb1912, 02:48 PM

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#18
Feb1912, 02:56 PM

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