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

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.

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.

I didn't follow your comments, but the math guarantees that if you use the correct expression for the field then Gauss' law holds.

kmarinas86

I should of have said that example of math.

kmarinas86

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

EDIT: I guess I couldn't stop asking questions :D

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...

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.)

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.

DaleSpam

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.

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.

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?

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

DaleSpam

What does Gauss' law say? (try to answer on your own before reading the spoilers)

What does the Lienard Wiechert potential say?

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. 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.

DaleSpam

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.

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.

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

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?

It is not clear to me how one detects "field lines" having the aforementioned quality.

Ilmrak

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

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?

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.

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?

pervect

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.

pervect

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.