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

[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, 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

 

[Dale]

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?

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.

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?

 

[pervect]

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.

 

[pervect]

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.

 

[kmarinas86]

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.

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.

 

[kmarinas86]

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.

 

[Dale]

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

I only mention 1 ft because I know that light travels at 1 ft/ns. The speed of light isn't such a nice number in cm.

Do you really think it makes a difference? If so, what is so special about cm and years that EM somehow does things in units of cm and years? I.e. why do you think that the field will take exactly 1 year to update from a distance of 1 cm? If you don't think that it makes a difference then why object?

 

[Dale]

I -accept- that they are.

Then I don't understand your whole question. If it is EM then by definition it satisfies Gauss' law, and it doesn't matter if you express it in the differential or in the integral form.

 

[Khashishi]

If move the charge out of that sphere and then stop it 1 centimeter outside of it

How fast were you planning on moving that charge out of the sphere?

 

[Matterwave]

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.

It's in the chapter on differential forms (chapter 4). But yes, there you have "tubes" (suitably abstracted) of flux, but the number of tubes pierced by a surface must give you the charge enclosed in the hypersurface (with some constants).

 

[kmarinas86]

Do you really think it makes a difference?

Yes. 1 year minus 1 second.

If so, what is so special about cm and years that EM somehow does things in units of cm and years?

Nothing.

I.e. why do you think that the field will take exactly 1 year to update from a distance of 1 cm?

Actually I never said that.

If you don't think that it makes a difference then why object?

You asked me, "Why would the field 1 ft away from the charge continue to change for 1 year after the charge has stopped moving?"

1 foot and 1 year taken together bears no relationship to the speed of light. Instead, it has a physically irrelevant speed of 9.65873546*10^-9 m/s, which has absolutely nothing to do with what is being discussed here.

When you said, "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 was clear to me that:

1) This wasn't a question, so I didn't have to answer it. That's why I didn't even mention 1 ft in my reply.
2) By saying that the field "does not continue to change at all" in conjunction with the claim that "the field at that location will stop changing 1 ns after the charge becomes stationary", I presume that by "does not continue to change at all" that you meant "does not continue to change at all after 1 ns was passed". This was a poor analogy to what I was saying. So I 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.

In other words, the field (on the sphere) continues to change for 1 year of synchronized time. Then the changes (on the sphere) cease as the new stationary field by the charge is "finished" on the sphere.

 

[kmarinas86]

How fast were you planning on moving that charge out of the sphere?

Arbitrarily slow. In the context of the OP, just enough to move it 2 cm within a time span far less than 1 year.

 

[kmarinas86]

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.

A delay in the force will yield a non-conservation of angular momentum if you neglect field momentum. A member of this forum did mention this:

Feynman addresses this (so-called) paradox in the final paragraph of FLP Vol. II Chapter 27, Field Energy and Field Momentum:

"Finally, another example is the situation with the magnet and the charge,
shown in Fig. 27-6. We were unhappy to find that energy was flowing around
in circles, but now, since we know that energy flow and momentum are proportional,
we know also that there is momentum circulating in the space. But
a circulating momentum means that there is angular momentum. So there is
angular momentum in the field. Do you remember the paradox we described in
Section 17-4 about a solenoid and some charges mounted on a disc? It seemed
that when the current turned off, the whole disc should start to turn. The puzzle
was: Where did the angular momentum come from? The answer is that if you
have a magnetic field and some charges, there will be some angular momentum
in the field. It must have been put there when the field was built up. When
the field is turned off, the angular momentum is given back. So the disc in the
paradox would start rotating. This mystic circulating flow of energy, which at
first seemed so ridiculous, is absolutely necessary. There is really a momentum
flow. It is needed to maintain the conservation of angular momentum in the
whole world."

This, however, is merely a qualitative description; sufficient quantitative (mathematical) information about the energy and momentum of the electromagnetic field is given in this chapter that a Caltech sophomore in his second year of the Feynman Lectures course would be expected to be able to find the final angular frequency of the disc as a function of the various parameters it depends on.

Mike Gottlieb
Editor, The Feynman Lectures on Physics, Definitive Edition
---
Physics Department
California Institute of Technology

I don't see the connection between that and the ideas of the not-so-well known "T. Von Flandern".

 

[Ben Niehoff]

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.

This has nothing to do with it. Field lines work because the source-free Maxwell equations imply that the 2-form F is harmonic.

Harmonic forms (in any dimension) have the distinction that they capture purely topological information. If you integrate a harmonic n-form over a closed n-surface, the result is either zero or non-zero, depending on whether the n-surface encloses some topological feature (for example, a 1-surface on a cylinder might wrap around the cylinder...or a 2-surface in R^3 might enclose a charge). Any n-surface that encloses the same set of topological features must give you the same result.

You can think of this as a higher-dimensional analogue of contour integration. In fact, all analytic functions on the complex plane satisfy Laplace's equation, which is why contour integration works.

 

[Dale]

1 foot and 1 year taken together bears no relationship to the speed of light. Instead, it has a physically irrelevant speed of 9.65873546*10^-9 m/s, which has absolutely nothing to do with what is being discussed here.

Similarly with 1 cm and 1 year, so I don't get why you object to replacing the 1 cm with 1 ft. That is all I was doing, because 1 ft is a more convenient small distance than 1 cm.

I said, "Then, by definition the field near the charge, if made stationary after displacement, continues for 1 year to change..." ... In other words, the field (on the sphere) continues to change for 1 year

Sorry about my confusion. I don't know how I could have possibly been so obviously mistaken as to think that "near" referred to the field 1 cm or 1 ft away from the charge when it so obviously should refer to the field 1 ly away

Just to be clear at each point in space the field changes from t_i+d_i/c to t_f+d_f/c where t_i is the time that the charge starts moving, t_f is the time that the charge stops moving, d_i is the initial distance to the charge, and d_f is the final distance to the charge. The field at that point does not change outside of those times.

 

[pervect]

This has nothing to do with it. Field lines work because the source-free Maxwell equations imply that the 2-form F is harmonic.

Harmonic forms (in any dimension) have the distinction that they capture purely topological information. ...

I'd love to read a fuller explanation if you know of any papers or references on the topic of harmonic forms.

 

[Antiphon]

Ok, I have constructed a simple equivalent problem to the original post which has an exact solution and can therefore be tested.

Arrange a dipole oscillating at any frequency you choose at the origin and oriented along the z axis. The complete fields of this dipole including the radiation fields are known exactly.

For the Gaussian surface chose the xy plane enclosed by a hemisphere at infinity. The fields on the hemisphere are radiation fields only and tend to zero.

Therefore you only need to integrate the z component of the dipole electric field over the xy plane.

If Gauss's law holds, this space integral evaluated along time will be +e when the positive charge is above the xy plane and -e when the negative charge pops up. Moreover, the integral should be exactly zero when the dipole charges cross at the origin despite the fact that significant non-zero radiation fields exist on the plane.

Anyone doubting the generality of Gauss's law can prove it using this integral.

I recommend writing the fields in spherical coordinates then performing the integration in cylindrical coordinates over r, theta.

 

[Ilmrak]

I don't get what exactly your claim is, please tell me which of the following:

- Gauss theorem is mathematically wrong, i.e. all mathematicians after Gauss failed in proving that theorem.

- Gauss Theorem is true, but is not applicable to physic, i.e. Maxwell's equations are wrong (or not general).
In this case you obviously can't prove this resolving Maxwell's equations for a oscillating dipole.

As the first of this option isn't actually acceptable to me, I don't see what utility could have to do math in the example you proposed.

If you claim is though that gauss theorem is wrong, I'd suggest to look for an error in the proof (I really can't find it).

Ilm

 

[Matterwave]

This has nothing to do with it. Field lines work because the source-free Maxwell equations imply that the 2-form F is harmonic.

Harmonic forms (in any dimension) have the distinction that they capture purely topological information. If you integrate a harmonic n-form over a closed n-surface, the result is either zero or non-zero, depending on whether the n-surface encloses some topological feature (for example, a 1-surface on a cylinder might wrap around the cylinder...or a 2-surface in R^3 might enclose a charge). Any n-surface that encloses the same set of topological features must give you the same result.

You can think of this as a higher-dimensional analogue of contour integration. In fact, all analytic functions on the complex plane satisfy Laplace's equation, which is why contour integration works.

Just to clarify, by harmonic you mean: (d\delta+\delta d)F=0 where d is the exterior derivative and delta is the co-dfifferential?

Is this true? It seems since dF=0 by definition, then we need to show: d\delta F=-d(*d*F)=4\pi(d**J)=0

Is it true that dJ=0? I can't think of a reason for this...

EDIT: Wait, is that just conservation of charge? It certainly isn't the usual way of expressing it...and I can't seem think think clearly enough at this late hour to figure this out...lol...

 

[Ben Niehoff]

I said source-free Maxwell equations, which are

dF = d * F = 0

 

[Matterwave]

But the source free Maxwell equations can't have a charge inside your surface...? So how does that apply to Gauss's law in what we are talking about here? I guess I don't quite get your argument then.

 

[Ben Niehoff]

The source-free Maxwell equations apply to everywhere where the charge isn't. That is, ALL of space except for the one point where the charge is.

Or in other words, it so happens that J=0 except at one point.

 

[Matterwave]

Certainly one would like Gauss's law to hold even if you are inside a continuous charge distribution though. How does that work then in that case?

 

[Ben Niehoff]

Right. So, with sources, we have

d * F = * J
So now, merely integrate this over some 3-volume V:

\begin{align*} \int_V d * F &= \int_V * J \\ \int_{\partial V} * F &= \int_V * J \end{align*}
which is Gauss' Law.

The reason I focused on the source-free equations, is because it is only when J = 0 that the result of integration doesn't care about the choice of surface. Obviously, if J \neq 0, then different surfaces might contain different amounts of charge, hence giving different results. Gauss' Law still holds, though.

 

[Matterwave]

I guess my question, what does F being harmonic have to do with field lines? It seems to me that Stokes theorem is sufficient in both (empty space and continuous charge) cases to justify the use of field lines.

 

[ARAVIND113122]

i completely agree with kmarinas86. the field SHOULD continue to change for 1 year[approx].so far,none of the members have posted a convincing reason why it shouldn't.after all,information does take time to travel,and nothing happens instantaneously.once the charge is taken out of the sphere,the 'data' from the other side of the sphere would take time[almost 1 year] to reach the side that was initially closer to the charge

 

[Ben Niehoff]

I guess my question, what does F being harmonic have to do with field lines? It seems to me that Stokes theorem is sufficient in both (empty space and continuous charge) cases to justify the use of field lines.

You're right. Harmonic forms are special for other reasons that don't necessarily apply here.

 

[Dale]

i completely agree with kmarinas86. the field SHOULD continue to change for 1 year[approx].so far,none of the members have posted a convincing reason why it shouldn't.after all,information does take time to travel,and nothing happens instantaneously.once the charge is taken out of the sphere,the 'data' from the other side of the sphere would take time[almost 1 year] to reach the side that was initially closer to the charge

The Lienard Wiechert potential is the convincing reason. The LW potential depends only on the motion of the charge at the retarded time. So the field at the near side updates with ~1 ns delay and the field at the far side updates with ~1 year delay. The field does not change outside of the times as indicated above.

If you are not willing to accept standard solutions like the LW potential, then there is little that can or should be done to convince you.