Why I doubt the generality of Gauss' law: A Gaussian sphere 1 light year across

In summary, the electric field at the other side of the sphere does not "update" until nearly 1 year later.
  • #1
kmarinas86
979
1
Let's say I have a Gaussian sphere 1 light year across with synchronized clocks and sensors all over its surface. All clocks are co-moving, 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 non-zero 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.
 
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  • #2
kmarinas86 said:
I end up with a non-zero integral for electric flux even though the charge is not inside the sphere.
Then you made a mistake.

The fields for an arbitrarily moving point charge are given by the Lienard-Wiechert potentials:
http://en.wikipedia.org/wiki/Liénard–Wiechert_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 Lienard-Wiechert potentials explicitly.
 
  • #3
DaleSpam said:
Then you made a mistake.

The fields for an arbitrarily moving point charge are given by the Lienard-Wiechert potentials:
http://en.wikipedia.org/wiki/Liénard–Wiechert_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 Lienard-Wiechert potentials explicitly.

The Lienard-Wiechert potential travels at the speed of light.

The hypothetical surface being discussed is 1 light year across.
 
  • #4
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
kmarinas86 said:
The Lienard-Wiechert potential travels at the speed of light.

The hypothetical surface being discussed is 1 light year across.
Yes. I understood that.
 
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  • #6
Ben Niehoff said:
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 have to do what the OP describes ("this"=what the OP describes)...

Ben Niehoff said:
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.

...which the above does not.

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
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
Antiphon said:
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 closer distance alone wouldn't do it. Most the electric field lines would be within a few millionths of a radian away from tangent with the surface. We're talking about a sphere that is 1/2 a light year in radius and a charge that is placed 1 centimeter from it.
 
  • #9
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 Lienard-Wiechart 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
Ben Niehoff said:
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 Lienard-Wiechart 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.

[...] All the necessary changes will happen locally, near the charge.

If I understand correctly, this would require an effective "positive" charge to appear inside the sphere due to this acceleration. Is this correct?
 
  • #11
kmarinas86 said:
If I understand correctly, this would require an effective "positive" charge to appear inside the sphere due to this acceleration. Is this correct?

No.

4chars
 
  • #12
kmarinas86 said:
The closer distance alone wouldn't do it. Most the electric field lines would be within a few millionths of a radian away from tangent with the surface. We're talking about a sphere that is 1/2 a light year in radius and a charge that is placed 1 centimeter from it.

Wrong. Most field lines are not nearly tangent.
 
  • #13
Ben Niehoff said:
You need to stop responding incredulously to everyone's posts, and actually do a calculation using the full theory with no approximation.
I agree. Since he hasn't shown his work I suspect that he has not used the correct expressions and his mental approximations are leading to incorrect conclusions.
 
  • #14
Ben Niehoff said:
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 Lienard-Wiechart 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.

I also notice that the accelerated charge would certain induce changes to the E-field at the surface of the 1 light year sphere.

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 acceleration-induced 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.
 
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  • #15
Antiphon said:
Wrong. Most field lines are not nearly tangent.

I honestly stand corrected, and you're right about most field lines being normal, but this still doesn't resolve my other points.

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 E-field sensors in the example are time-synchronized, so there is no problem in determining what the field was at the other side of sphere when the electron crossed the surface.
 
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  • #16
kmarinas86 said:
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.
That is why I linked to the Lienard Wiechert potential page in the first response! It contains the exact specific model, expressed in terms of both the potentials and the fields.

You have made a very specific claim:
kmarinas86 said:
I end up with a non-zero integral for electric flux even though the charge is not inside the sphere.
I have challenged you to show your work. Either your work uses the wrong formula (in which case I have pointed you to the correct formulat to use), or you have done the integral incorrectly (in which case I would be glad to help out as much as possible).
 
  • #17
DaleSpam said:
That is why I linked to the Lienard Wiechert potential page in the first response! It contains the exact specific model, expressed in terms of both the potentials and the fields.

You have made a very specific claim:I have challenged you to show your work. Either your work uses the wrong formula (in which case I have pointed you to the correct formulat to use), or you have done the integral incorrectly (in which case I would be glad to help out as much as possible).

By specifics, I am first most concerned about specifics about the phenomenology. I'm not arguing about the math. "What are we computing?" is the first question that comes to mind before I involve myself with mathematical expression of the phenomenology. What about the comment I just made concerning the "growing positive contribution" and the "shrinking negative contribution" and derivative changes to the field measured at the sensors, suggesting a gradual, rather than instantaneous change in the measurements relative to the synchronized time of the sensors?
 
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  • #18
kmarinas86 said:
I'm not arguing about the math.
Yes, you are. You specifically claimed that Gauss' law is wrong because an integral which should be 0 is non-0. That is math.

kmarinas86 said:
"What are we computing" is the first question that comes to mind
If you are looking for violations to Gauss' law then what you should be computing is the electric flux through a closed surface containing a known amount of charge.

kmarinas86 said:
What about the comment I just made concerning the "growing positive contribution" and the "shrinking negative contribution" and derivative changes to the field measured at the sensors, suggesting a gradual, rather than instantaneous change in the measurements relative to the synchronized time of the sensors?
I didn't follow your comments, but the math guarantees that if you use the correct expression for the field then Gauss' law holds.
 
  • #19
DaleSpam said:
Yes, you are. You specifically claimed that Gauss' law is wrong because an integral which should be 0 is non-0. That is math.

I should of have said that example of math.
 
  • #20
kmarinas86 said:
What about the comment I just made concerning the "growing positive contribution" and the "shrinking negative contribution" and derivative changes to the field measured at the sensors, suggesting a gradual, rather than instantaneous change in the measurements relative to the synchronized time of the sensors?

I just want a little recognition and response on this particular point.

EDIT: I guess I couldn't stop asking questions :D
 
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  • #21
The integral form of Gauss' law follows mathematically from the differential form.
So if one form of the law doesn't seem blindingly intuitive, look at the other. (And I suppose it wouldn't hurt to study the proof of how you derive one from the other, as well.)

If both don't seem blindingly intuitive, look at the experimental results supporting them...
 
  • #22
As has been pointed out before, the "field lines" picture will make this bleedingly obvious. Gauss' Law just counts field lines going through a surface. Field lines always end on the charge, no matter how the charge wiggles about. So it becomes a simple topological problem: If the charge is inside the sphere, any field lines going from the charge to infinity must cut through the sphere. Likewise, if the charge is outside, any field lines must either avoid the sphere or cut it twice; once in, once out.

Note: The entire field line picture is dependent on the 1/r^2 nature of the force. 1/r^2 forces in 3 dimensions are very special; they have the property that they can be described by a field line picture. (Mathematically, this has to do with harmonic differential forms.)
 
  • #23
pervect said:
The integral form of Gauss' law follows mathematically from the differential form.
So if one form of the law doesn't seem blindingly intuitive, look at the other. (And I suppose it wouldn't hurt to study the proof of how you derive one from the other, as well.)

If both don't seem blindingly intuitive, look at the experimental results supporting them...

Experimental results supporting Gauss' Law don't normally involve measurements at a frequency of c/d, where d is the distance across volume being measured.

A 10 meter device would require measurements at a rate of nearly 30 million times per second. For, a 1 meter device, it would need to be nearly 300 million times a second. I don't know any specifics, but I suppose it is a safe bet that at those the frequencies, the accuracy (lack thereof) wouldn't be counted on, so actual measurements would not be quick enough to pick up the "time delay" in the update of the electrical field. The "time delay" is the whole point of this argument I'm making.
 
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  • #24
kmarinas86 said:
The "time delay" is the whole point of this argument I'm making.
The time delay is explicitly accounted for in the retarded time of the Lienard Wiechert potential. So the whole point of this argument you're making is already taken into account.
 
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  • #25
kmarinas86 said:
I just want a little recognition and response on this particular point. That is all I am going to ask for the rest of today.
As I said above, the math guarantees that if you use the correct equations for the potential then all of your "positive contributions" and "negative contributions" work out correctly so that Gauss' law holds.
 
  • #26
DaleSpam said:
As I said above, the math guarantees that if you use the correct equations for the potential then all of your "positive contributions" and "negative contributions" work out correctly so that Gauss' law holds.

When passing through the surface of the sphere (0=thickness), is change in the integrated flux instantaneous from one end of the sphere to the other, or is it gradual (per synchronized time)? In other words, could the integral * free-space permittivity be one-half a charge at one point, or does it "jump of a cliff" and go from 1 unit of charge down to 0, while skipping all values in between?

Does the field near the charge, if made stationary after displacement, continue for 1 year to change in a way compensating exactly the field far of it as it updates the far end of the sphere 1 light year away to reflect that the charge had been moved outside the sphere 1 year before?

EDIT: I guess I couldn't stop asking questions :D
 
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  • #27
kmarinas86 said:
When passing through the surface of the sphere (0=thickness), is change in the integrated flux instantaneous from one end of the sphere to the other, or is it gradual (per synchronized time)? In other words, could the integral * free-space permittivity be one-half a charge at one point, or does it "jump of a cliff" and go from 1 unit of charge down to 0, while skipping all values in between?
What does Gauss' law say? (try to answer on your own before reading the spoilers)

kmarinas86 said:
Does the field near the charge, if made stationary after displacement, continue for 1 year to change in a way compensating exactly the field far of it as it updates the far end of the sphere 1 light year away to reflect that the charge had been moved outside the sphere 1 year before?
What does the Lienard Wiechert potential say?

1) For a point charge, yes, the integrated flux is discontinuous. For an extended charge distribution, no.

2) No, the field at a given event depends only on the charge, position, velocity, and acceleration of the charge at the retarded time.
 
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  • #28
DaleSpam said:
kmarinas86 said:
Does the field near the charge, if made stationary after displacement, continue for 1 year to change in a way compensating exactly the field far of it as it updates the far end of the sphere 1 light year away to reflect that the charge had been moved outside the sphere 1 year before?

What does the Lienard Wiechert potential say?

2) No, the field at a given event depends only on the charge, position, velocity, and acceleration of the charge at the retarded time.

Then, by definition the field near the charge, if made stationary after displacement, continues for 1 year to change in a way THAT DOES NOT compensate exactly the field far of it as it updates the far end of the sphere 1 light year away to reflect that the charge had been moved outside the sphere 1 year before. In other words, the measurements at the surface of the sphere would continue to change with respect to synchronized time to reflect the wave of propagation. Also, it makes perfect sense to see that the acceleration of the charge, which occurs for a period of time much shorter than a year, would affect a region of the sphere that intersects a growing "sphere" of some thickness centered on the position of the charge. Such a region would have a shape comparable a rubber band being stretched across from one end of the sphere, to the other end of the sphere, over time.
 
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  • #29
kmarinas86 said:
Then, by definition the field near the charge, if made stationary after displacement, continues for 1 year to change in a way THAT DOES NOT compensate exactly the field far of it as it updates the far end of the sphere 1 light year away to reflect that the charge had been moved outside the sphere 1 year before.
No. It does not continue to change at all. If the charge is made stationary at, say 1 ft away from some location then the field at that location will stop changing 1 ns after the charge becomes stationary.

It does not continue to change for a year simply because you have drawn a Gaussian surface out 1 light year distant. After all, how would it know if you have drawn your surface out 1 year or 2 or 2 million? Furthermore, the fact that it has stopped changing at one part of the surface and continues to change at another part of the surface does not in any way contradict Gauss' law.
 
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  • #30
DaleSpam said:
kmarinas86 said:
Then, by definition the field near the charge, if made stationary after displacement, continues for 1 year to change in a way THAT DOES NOT compensate exactly the field far of it as it updates the far end of the sphere 1 light year away to reflect that the charge had been moved outside the sphere 1 year before.

No. It does not continue to change at all. If the charge is made stationary at, say 1 ft away from some location then the field at that location will stop changing 1 ns after the charge becomes stationary.

I didn't say it would change forever. I said, "Then, by definition the field near the charge, if made stationary after displacement, continues for 1 year to change..." This, means that for 1 year it continues to change.

And the last part, "...to reflect that the charge had been moved outside the sphere 1 year before." is about what happens 1 year later.

DaleSpam said:
It does not continue to change for a year simply because you have drawn a Gaussian surface out 1 light year distant. After all, how would it know if you have drawn your surface out 1 year or 2 or 2 million?

Then it would change the integration. The question is, "How does the value of the integral remain the same after switching the integration?"

DaleSpam said:
Furthermore, the fact that it has stopped changing at one part of the surface and continues to change at another part of the surface does not in any way contradict Gauss' law.

Just how are those changes equal and opposite?

If we go by the "field lines" argument to show how this is done for a negative charge moved out of the sphere, the number of lines into the sphere = the number of those going out, which is easy to imagine, we would have field lines that point away from the "line" overlapping the charge's displacement, poking through a surface immediate to the point through the sphere that the charge crossed. This would be like having an "effective" positive charge inside that part of a sphere and cancels out the "effective" negative charge at the other side of the sphere. If this is what happens, then I can see how one can arrive at Gauss's law. Theory says this is the case, but what about the fact that the Liénard–Wiechert potential is not always valid? Then these field lines may take on a different shape. But why must field lines behave this way?

kmarinas86 said:
Experimental results supporting Gauss' Law don't normally involve measurements at a frequency of c/d, where d is the distance across volume being measured.

A 10 meter device would require measurements at a rate of nearly 30 million times per second. For, a 1 meter device, it would need to be nearly 300 million times a second. I don't know any specifics, but I suppose it is a safe bet that at those the frequencies, the accuracy (lack thereof) wouldn't be counted on, so actual measurements would not be quick enough to pick up the "time delay" in the update of the electrical field. The "time delay" is the whole point of this argument I'm making.

It is not clear to me how one detects "field lines" having the aforementioned quality.
 
  • #31
Hello,

You can try to picture the situation in a more familiar way.

Think about a flat sheet with a stone leaning in some point. The weight of the stone will curve the sheet in such a way that measuring the displacement along a closed line surrounding the stone, with respect the "no-stone" configuration, you can argue the exact value of the mass of the stone (gauss theorem).

When you move the stone to another point a wave is emitted on the sheet surface "updating" the value of the displacement.
The displacement in a given point remains costant except for the time the wave passes onto it; it does not continue to change for the whole duration of the wave (except if the stone continue moving, which is not the case).
At any given time, the displacement on a closed line will tell you whether or not that line surrounds the stone.

The value of the displacemtent ON the wave will compensate for the fact that points not yet reaced by the wave have their "old" value.

P.s. Sorry for my bad english ^^
P.p.s. I hope you didn't need a rigorous treatment. If you are interested in how it can be possible that the filed always satisfy the gauss theorem, the reason is substantially topological.

Ilm
 
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  • #32
kmarinas86 said:
I didn't say it would change forever. I said, "Then, by definition the field near the charge, if made stationary after displacement, continues for 1 year to change..." This, means that for 1 year it continues to change.

And the last part, "...to reflect that the charge had been moved outside the sphere 1 year before." is about what happens 1 year later.
Why would the field 1 ft away from the charge continue to change for 1 year after the charge has stopped moving? What is so special about 1 year?

kmarinas86 said:
Then it would change the integration.
This is what you have claimed multiple times without any proof. Please show your work that shows that the integration is changed. Again, either you are using the wrong equation for the field or you are making a mistake in the integration.

kmarinas86 said:
But why must field lines behave this way?
Because the field lines follow Maxwell's equations and Gauss' law is one of Maxwell's equations. It is essentially tautologically guaranteed. You cannot derive something from equation X and then have it violate equation X without making some math mistake. We cannot spot where you are making the mistake without your posting your work.

EDIT: Btw, perhaps I should ask. Do you understand that the integral and the differential forms of Gauss' law are equivalent?
 
  • #33
kmarinas86 said:
Experimental results supporting Gauss' Law don't normally involve measurements at a frequency of c/d, where d is the distance across volume being measured.

A 10 meter device would require measurements at a rate of nearly 30 million times per second. For, a 1 meter device, it would need to be nearly 300 million times a second. I don't know any specifics, but I suppose it is a safe bet that at those the frequencies, the accuracy (lack thereof) wouldn't be counted on, so actual measurements would not be quick enough to pick up the "time delay" in the update of the electrical field. The "time delay" is the whole point of this argument I'm making.

The traditional form of electromagnetism conserves angular momentum. (Of course you have to include angular momentum in the field).

It's not clear what sort of proposal you're thinking about, but if it's along some von-Flanderen ideas, it's well known that delay in the force will yield a non-conservation of angular momentum.

Anyway, supporting alternate theory development isn't somethign that we really do here, so I'm not going to comment in detail.
 
  • #34
Ben Niehoff said:
As has been pointed out before, the "field lines" picture will make this bleedingly obvious. Gauss' Law just counts field lines going through a surface. Field lines always end on the charge, no matter how the charge wiggles about. So it becomes a simple topological problem: If the charge is inside the sphere, any field lines going from the charge to infinity must cut through the sphere. Likewise, if the charge is outside, any field lines must either avoid the sphere or cut it twice; once in, once out.

Note: The entire field line picture is dependent on the 1/r^2 nature of the force. 1/r^2 forces in 3 dimensions are very special; they have the property that they can be described by a field line picture. (Mathematically, this has to do with harmonic differential forms.)

I think part of the machinery that makes field lines work is that you can write a general two-form as the sum of two wedge products of one-forms in 4 dimensions, and this is peculiair to 4-d. These two wedge products of one forms have a geometric interpretation as electric field lines and magnetic field lines, and their sum gives you the general rank two tensor. Unfortunately I'm going from memory here - I think this was mentioned briefly in MTW, but it's a huge book to search to find the section to refresh my memory about.
 
  • #35
DaleSpam said:
Why would the field 1 ft away from the charge continue to change for 1 year after the charge has stopped moving? What is so special about 1 year?

You did not stick the context of the OP, and it has nothing to do with 1 ft.

kmarinas86 said:
Let's say I have a Gaussian sphere 1 light year across with synchronized clocks and sensors all over its surface. All clocks are co-moving, 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 non-zero 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.
 
<h2>1. Why is Gauss' law doubted?</h2><p>Gauss' law is doubted because it is based on the assumption that the electric field is radial and uniform at all points on a Gaussian sphere. This may not always be the case in real-world scenarios.</p><h2>2. What is a Gaussian sphere?</h2><p>A Gaussian sphere is an imaginary spherical surface used in Gauss' law to calculate the electric field at a point due to a charge distribution.</p><h2>3. Why is the Gaussian sphere chosen to be 1 light year across?</h2><p>The size of the Gaussian sphere is chosen to be 1 light year across in order to encompass a large enough volume to accurately represent the charge distribution in question.</p><h2>4. How does the size of the Gaussian sphere affect the accuracy of Gauss' law?</h2><p>The larger the Gaussian sphere, the more accurate the results of Gauss' law will be. However, if the sphere is too large, it may not accurately represent the local charge distribution and lead to errors in calculations.</p><h2>5. Are there any limitations to Gauss' law?</h2><p>Yes, there are limitations to Gauss' law. It assumes that the electric field is radial and uniform, and does not take into account factors such as non-uniform charge distributions or changing electric fields over time. It is also only applicable in electrostatic scenarios, and does not apply to situations with moving charges or changing magnetic fields.</p>

1. Why is Gauss' law doubted?

Gauss' law is doubted because it is based on the assumption that the electric field is radial and uniform at all points on a Gaussian sphere. This may not always be the case in real-world scenarios.

2. What is a Gaussian sphere?

A Gaussian sphere is an imaginary spherical surface used in Gauss' law to calculate the electric field at a point due to a charge distribution.

3. Why is the Gaussian sphere chosen to be 1 light year across?

The size of the Gaussian sphere is chosen to be 1 light year across in order to encompass a large enough volume to accurately represent the charge distribution in question.

4. How does the size of the Gaussian sphere affect the accuracy of Gauss' law?

The larger the Gaussian sphere, the more accurate the results of Gauss' law will be. However, if the sphere is too large, it may not accurately represent the local charge distribution and lead to errors in calculations.

5. Are there any limitations to Gauss' law?

Yes, there are limitations to Gauss' law. It assumes that the electric field is radial and uniform, and does not take into account factors such as non-uniform charge distributions or changing electric fields over time. It is also only applicable in electrostatic scenarios, and does not apply to situations with moving charges or changing magnetic fields.

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