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

by kmarinas86
Tags: doubt, gauss, gaussian, generality, light, sphere
P: 1,011
 Quote by DaleSpam 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.
P: 1,011
 Quote by kmarinas86 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
 Sci Advisor Emeritus P: 7,204 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...
 Sci Advisor P: 1,516 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.)
P: 1,011
 Quote by pervect 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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P: 15,561
 Quote by kmarinas86 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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P: 15,561
 Quote by kmarinas86 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.
P: 1,011
 Quote by DaleSpam 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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P: 15,561
 Quote by kmarinas86 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)

 Quote by kmarinas86 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?

Spoiler
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.
P: 1,011
Quote by DaleSpam
 Quote by kmarinas86 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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P: 15,561
 Quote by kmarinas86 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.
P: 1,011
Quote by DaleSpam
 Quote by kmarinas86 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.

 Quote by DaleSpam 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?"

 Quote by DaleSpam 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?

 Quote by kmarinas86 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.
 P: 97 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, wich 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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P: 15,561
 Quote by kmarinas86 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?

 Quote by kmarinas86 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.

 Quote by kmarinas86 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?
Emeritus
P: 7,204
 Quote by kmarinas86 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.
Emeritus
P: 7,204
 Quote by Ben Niehoff 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.
P: 1,011
 Quote by DaleSpam 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.

 Quote by kmarinas86 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.
P: 1,011
 Quote by DaleSpam EDIT: Btw, perhaps I should ask. Do you understand that the integral and the differential forms of Gauss' law are equivalent?
I -accept- that they are.

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