Relativistic corrections to E-fields of charges bounded inside a Gaussian surface

[kmarinas86]

If I have a spread of electrical charges contained inside a Gaussian surface, and if I cause those electrical charges to move at relativistic speeds, the electric fields of those charges should be subject to relativistic contraction. What happens then to electric flux that cuts through that surface? Does it decrease, increase, or remain the same?

 

[Meir Achuz]

The integrated flux of E is still given by the total charge enclosed, which doesn't change.

 

[kmarinas86]

The integrated flux of E is still given by the total charge enclosed, which doesn't change.

We are no longer dealing with an inverse square law force (a basic assumption of Gauss' law) with respect to the charge location, for reasons which are quite obvious.

To show what that looks like in a diagram, here it is using the Liénard–Wiechert Potential:

https://www.physicsforums.com/attachment.php?attachmentid=43069&d=1327380412

Remember that this is a 2D diagram.

A 3D diagram would show a proper account for the area and the corresponding electric field intensities thereof.

The conservation of the integration of the electric flux of a static charge on a closed surface surrounding a fixed quantity of charge should apply even in the most extreme case where v=0.99999c, in which the vast bulk of the weight of the integration is tied to a relatively small band area, in contrast to what would be in the case for a charge stationary with respect to that Gaussian surface.

It remains apparent to me that if the field's force components do not obey the inverse square law, then Gauss' law cannot be applied, and the value of the integration will depend on the surface upon which the integration is done.

 

[Matterwave]

Gauss's law is quite general and does not require the inverse square law potential to hold.

 

[Meir Achuz]

Gauss's law follows by applying the divergence theorem to Maxwell's equation
for div E=4\pi rho, which still holds, no matter how complicated the E field.

 

[kmarinas86]

Gauss's law follows by applying the divergence theorem to Maxwell's equation
for div E=4\pi rho, which still holds, no matter how complicated the E field.

We have a delay in the E-field propagation, so different parts of the surface see the same charge as being in different places within that surface.

So this isn't just a complicated E-field in the sense of its distribution within the surface. The parts of the surface can't even agree on the position of the charge.

So how can you take a statement based on invariance with respect to the distribution of charge and say something to the effect that the same reasoning accounts for invariance of the electric flux integral with respect to a time-retarded electric field? I'm not able to take that leap of faith.

The definition of the Liénard-Wiechert potential of an electric field:

$$\mathbf{E}(\mathbf{r}, t) = \frac{1}{4 \pi \epsilon_0} \left(\frac{q(\mathbf{n} - \boldsymbol{\beta})}{\gamma^2 (1 - \mathbf{n} \cdot \boldsymbol{\beta})^3 |\mathbf{r} - \mathbf{r}_s|^2} + \frac{q \mathbf{n} \times \big((\mathbf{n} - \boldsymbol{\beta}) \times \dot{\boldsymbol{\beta}}\big)}{c(1 - \mathbf{n} \cdot \boldsymbol{\beta})^3 |\mathbf{r} - \mathbf{r}_s|} \right)_{t_r}$$

The divergence of $\mathbf{E}(\mathbf{r}, t)$ is not equal to the RHS of the equation you provided:

$\operatorname{div}\, \mathbf{E}(\mathbf{r}, t) \ne 4\pi \rho$

 

[kmarinas86]

Gauss's law is quite general and does not require the inverse square law potential to hold.

Gauss' law maintains proportionality between charges inside a surface and the flux integral of the field lines of those charges because as radial distance increases, area increases by the square, and the field intensity decreases by the square. So the integral on that surface remains constant regardless of its shape, so as long as that shape is closed (or effectively so as is with the case of two parallel surfaces separated at opposite sides from a charge). However, if we had an inverse cube field, then simply doubling the diameter of a spherical surface surrounding a charge located at its center would cause the integral of the field at that spherical surface to be inverse to its radius.

 

[Bill_K]

Forget the ε₀, I use Gaussian units. Integrate over a sphere centered at r_s. Then E·n = q/((1-n·β)²γ²R²)

∫E·n dS = ∫E·n R² 2π d(cosθ) = 2πq/γ² ∫(1-β cosθ)^-2 d(cosθ) = 2πq/γ² (1/β)[1/(1-β) - 1/(1+β)] = 4πq

 

[Meir Achuz]

I think I'll give that as a homework problem to end this thread.

 

[kmarinas86]

Forget the ε₀, I use Gaussian units. Integrate over a sphere centered at r_s. Then E·n = q/((1-n·β)²γ²R²)

∫E·n dS = ∫E·n R² 2π d(cosθ) = 2πq/γ² ∫(1-β cosθ)^-2 d(cosθ) = 2πq/γ² (1/β)[1/(1-β) - 1/(1+β)] = 4πq

When the charge is not in the center of the sphere, there would be different delay times of the propagation of the charge's electric field as seen from the inertial frame of the sphere itself. This would be particularly important when the charge is accelerating, as β would be time-variant, and thus different points on the sphere would experience fields deriving from different values of β for the charge. However, in the special case where the charge is at the center, this would not seem to matter, and your example would seem to apply.

However, it would seem that the field at ends of the right and left hemispheres of the sphere are separate such that if the electron were to have just reached the center from the end of the left hemisphere at a constant, and still constant, speed, then the end of the left hemisphere would have more a recent field from the charge than the end of the right hemisphere. So when the charge is at the center of the sphere, points on the sphere are receiving the electric field from different points within the sphere, which does not match the formulation in your example.

 

[Matterwave]

We have a delay in the E-field propagation, so different parts of the surface see the same charge as being in different places within that surface.

So this isn't just a complicated E-field in the sense of its distribution within the surface. The parts of the surface can't even agree on the position of the charge.

So how can you take a statement based on invariance with respect to the distribution of charge and say something to the effect that the same reasoning accounts for invariance of the electric flux integral with respect to a time-retarded electric field? I'm not able to take that leap of faith.

The definition of the Liénard-Wiechert potential of an electric field:

$$\mathbf{E}(\mathbf{r}, t) = \frac{1}{4 \pi \epsilon_0} \left(\frac{q(\mathbf{n} - \boldsymbol{\beta})}{\gamma^2 (1 - \mathbf{n} \cdot \boldsymbol{\beta})^3 |\mathbf{r} - \mathbf{r}_s|^2} + \frac{q \mathbf{n} \times \big((\mathbf{n} - \boldsymbol{\beta}) \times \dot{\boldsymbol{\beta}}\big)}{c(1 - \mathbf{n} \cdot \boldsymbol{\beta})^3 |\mathbf{r} - \mathbf{r}_s|} \right)_{t_r}$$

The divergence of $\mathbf{E}(\mathbf{r}, t)$ is not equal to the RHS of the equation you provided:

$\operatorname{div}\, \mathbf{E}(\mathbf{r}, t) \ne 4\pi \rho$

I find this last claim hard to believe considering the Lienard-Wiechert potential was derived explicitly using Maxwell's equations (which includes Gauss's law).

 

[kmarinas86]

Gauss's law follows by applying the divergence theorem to Maxwell's equation
for div E=4\pi rho, which still holds, no matter how complicated the E field.

We have a delay in the E-field propagation, so different parts of the surface see the same charge as being in different places within that surface.

So this isn't just a complicated E-field in the sense of its distribution within the surface. The parts of the surface can't even agree on the position of the charge.

So how can you take a statement based on invariance with respect to the distribution of charge and say something to the effect that the same reasoning accounts for invariance of the electric flux integral with respect to a time-retarded electric field? I'm not able to take that leap of faith.

The definition of the Liénard-Wiechert potential of an electric field:

$$\mathbf{E}(\mathbf{r}, t) = \frac{1}{4 \pi \epsilon_0} \left(\frac{q(\mathbf{n} - \boldsymbol{\beta})}{\gamma^2 (1 - \mathbf{n} \cdot \boldsymbol{\beta})^3 |\mathbf{r} - \mathbf{r}_s|^2} + \frac{q \mathbf{n} \times \big((\mathbf{n} - \boldsymbol{\beta}) \times \dot{\boldsymbol{\beta}}\big)}{c(1 - \mathbf{n} \cdot \boldsymbol{\beta})^3 |\mathbf{r} - \mathbf{r}_s|} \right)_{t_r}$$

The divergence of $\mathbf{E}(\mathbf{r}, t)$ is not equal to the RHS of the equation you provided:

$\operatorname{div}\, \mathbf{E}(\mathbf{r}, t) \ne 4\pi \rho$

I find this last claim hard to believe considering the Lienard-Wiechert potential was derived explicitly using Maxwell's equations (which includes Gauss's law).

The LHS is a function of $\mathbf{r}$ and $t$. What about $\rho$ on the RHS? It is clear that it must be also a function of $\mathbf{r}$ and $t$.

What matters is the integration of $\mathbf{E}(\mathbf{r}, t)$ with respect to the surface. The retarded time $t_r$ of the signal received at $t$ depends on the point in question.

 

[ParticleGrl]

Kmarinas, here is a simple way to think about this. Go to a frame where the charge is at rest- now the (differently shaped) Gaussian surface is moving. Here, it should be obvious that the integral gives the charge enclosed.

Now, we can easily change back since charge is an invariant quantity.

 

[kmarinas86]

Kmarinas, here is a simple way to think about this. Go to a frame where the charge is at rest- now the (differently shaped) Gaussian surface is moving. Here, it should be obvious that the integral gives the charge enclosed.

Now, we can easily change back since charge is an invariant quantity.

Due to time-retardation of the E-field that is limited to speed of light, the points will not agree on the location of the charge.

From the point of view of the stationary charge, the E-field propagates from the source charge, but the relative velocity and angle of a surface with respect to source charge varies. The angle made by each surface with respect to a radial vector from the stationary charge is not the same angle made by the E-field with respect the "effective surface" which is stretched across space and time due to the interception of that moving surface by the propagation of the E-field. Therefore, the E-field intensity must be corrected for each point on the moving surface.

 

[Dale]

If I have a spread of electrical charges contained inside a Gaussian surface, and if I cause those electrical charges to move at relativistic speeds, the electric fields of those charges should be subject to relativistic contraction. What happens then to electric flux that cuts through that surface? Does it decrease, increase, or remain the same?

Gauss' law is one of Maxwell's equations, and Maxwell's equations are inherently relativistic, so how could relativistic motion possibly violate Gauss' law? The idea doesn't make sense.

If a system follows Maxwells equations then it follows both relativity and Gauss' law.

 

[kmarinas86]

Gauss' law is one of Maxwell's equations, and Maxwell's equations are inherently relativistic, so how could relativistic motion possibly violate Gauss' law? The idea doesn't make sense.

One of the strangest things I just now found is that it is claimed that forces are not subject to aberration, and yet light is.

http://en.wikipedia.org/wiki/Field_(physics)#Propagation_of_static_field_effects

http://www.mathpages.com/home/kmath562/kmath562.htm

Since the particle is moving tangentially with speed v, the angle of the incoming light will be affected by aberration, such that the apparent source of the light (from the point of view of the test particle as it crosses the y axis) is at an angle a = arcsin(v/c) ahead of the actual position of the sphere. However, the direction of the electrical force exerted by the sphere on the test particle points directly toward the actual position of the sphere. Thus, the incoming electromagnetic waves from the sphere experience aberration, but the electromagnetic force of attraction to the sphere does not.

The obvious question that follows: How was that concluded?

Isn't a photon a mixture of electric and magnetic fields? Shouldn't they also be subject to aberration?

Aberration, of course, affects the angle light is received. Aberration would of course then affect the surface power density of the received light. The analogy of this for electrical forces would imply that for a given surface, the direction that an E-field of a given intensity intercepts a surface would change, altering the value of electric flux through that surface. For a spherical surface centered on a charge, but moving a great speed, the tendency would be in the reduction of this angle, away from the normal of that surface, making a negative contribution to the effective field integral (electric flux).

 

[Dale]

How is that related to the OP? Do you understand the answer to your question in the OP now?

 

[kmarinas86]

How is that related to the OP?

One cannot desire much to perform the integral of the electric field with a definition that one sees as inadequate or poorly constructed.

So why isn't the electric field subject to aberration?

And why is light-time correction irrelevant to the calculation of the electric field at a surface?

 

[Dale]

And why is light-time correction irrelevant to the calculation of the electric field at a surface?

Because Maxwell's equations are already fully relativistic. There is no need to add any correction terms, they are already correct.

Do you agree or disagree with the following two statements (and if you disagree please explain why):

1) Gauss' law is one of Maxwell's equations, therefore any system which obeys Maxwell's equations necessarily obeys Gauss' law.

2) The form of Maxwell's equations is unchanged under the Lorentz transform, therefore any system which obeys Maxwell's equations automatically obeys SR without further corrections.

If you agree with both of these statements then you should understand that you are guaranteed, from first principles, that that no configuration of relativistically moving charges will ever violate Gauss' law.

 

[kmarinas86]

Because Maxwell's equations are already fully relativistic. There is no need to add any correction terms, they are already correct.

Do you agree or disagree with the following two statements (and if you disagree please explain why):

1) Gauss' law is one of Maxwell's equations, therefore any system which obeys Maxwell's equations necessarily obeys Gauss' law.

2) The form of Maxwell's equations is unchanged under the Lorentz transform, therefore any system which obeys Maxwell's equations automatically obeys SR without further corrections.

If you agree with both of these statements then you should understand that you are guaranteed, from first principles, that that no configuration of relativistically moving charges will ever violate Gauss' law.

I think it is clear at this point that "Special" Relativity implicitly assumes that the electric field is not subject to aberration. Maxwell's equations also assume that the electric field does not suffer aberration. But I do not see how this could not apply to electric fields of charges while it applies to photons, which are both electric and magnetic in nature. I question whether it is really true, for that reason specifically, that "no configuration of [<<no qualifier>>] relativistically moving charges will ever violate Gauss' law", despite my agreement with the two statements you posed.

 

[Dale]

If you agree with the two statements then "no configuration of [<<no qualifier>>] relativistically moving charges will ever violate Gauss' law" is an unavoidable conclusion. If the charges obey Maxwell's equations then they automatically obey relativity too because Maxwell's equations obey relativity.

Look at this as a formal logic proposition:
Let A be the statement "the system obeys Maxwell's equations"
Let B be the statement "the system obeys relativity"
Let C be the statement "the system obeys Gauss' law"
Let D be the statement "the system is a system of charges"
Let E be the statement "the system is a system of relativistically moving charges"

We have the following "axioms"
A->C (since Gauss' law is a subset of Maxwell's equations, see statement 1 above)
A->B (since Maxwells equations are Lorentz invariant, see statement 2 above)
E->D (since relativistically moving charges are a subset of charges)

Proposition
D->A (i.e. Maxwell's equations are correct)
E (i.e. we are dealing with a system of relativistically moving charges)
----------------------
D (E->D is an axiom)
A (D->A by proposition)
C (A->C is an axiom)
B (A->B is an axiom)
Therefore C and B.

So any system of relativistically moving charges follows both Gauss' law and relativity. I don't know how you can maintain any doubt.

 

[kmarinas86]

If you agree with the two statements then "no configuration of [<<no qualifier>>] relativistically moving charges will ever violate Gauss' law" is an unavoidable conclusion. If the charges obey Maxwell's equations then they automatically obey relativity too because Maxwell's equations obey relativity.

Look at this as a formal logic proposition:
Let A be the statement "the system obeys Maxwell's equations"
Let B be the statement "the system obeys relativity"
Let C be the statement "the system obeys Gauss' law"
Let D be the statement "the system is a system of charges"
Let E be the statement "the system is a system of relativistically moving charges"

We have the following "axioms"
A->C (since Gauss' law is a subset of Maxwell's equations, see statement 1 above)
A->B (since Maxwells equations are Lorentz invariant, see statement 2 above)
E->D (since relativistically moving charges are a subset of charges)

Proposition
D->A (i.e. Maxwell's equations are correct)
E (i.e. we are dealing with a system of relativistically moving charges)
----------------------
D (E->D is an axiom)
A (D->A by proposition)
C (A->C is an axiom)
B (A->B is an axiom)
Therefore C and B.

So any system of relativistically moving charges follows both Gauss' law and relativity. I don't know how you can maintain any doubt.

The <<no qualifier>> is key.

Either there is a relativity applying to the whole universe in which observed electric fields are not subject to relativistic aberration or there is a relativity applying to the whole universe in which observed electric fields are subject to relativistic aberration. Those relativities cannot be identical, and they are mutually exclusive if one or the other must apply to the whole universe.

Experimentally, without recourse to theoretical explanations, why doesn't the electric field get subject to aberration? If you can answer that to my satisfaction, then it will address my doubt. Otherwise, I will still doubt.

 

[Dale]

The abberation argument is irrelevant as is the <<no qualifier>> complaint. As you can clearly see, I did not add any qualifiers nor is any consideration of abberation required to reach the conclusion.

Do you dispute the logic? If so, please identify the specific axiom, postulate, or logical inference which is in dispute.

 

[kmarinas86]

The abberation argument is irrelevant as is the <<no qualifier>> complaint. As you can clearly see, I did not add any qualifiers nor is any consideration of abberation required to reach the conclusion.

Do you dispute the logic? If so, please identify the specific axiom, postulate, or logical inference which is in dispute.

It's not the logic that I question. I question the premises behind definitions of two statements in question. Both statement B and statement E permit two possible interpretations; so both the meaning and truth value of A->B would change, while E->D remains as a true axiom due to the obvious grammar involved. Does relativity, as very general concept, not limited to SR, permit aberration of electric fields? If the answer is a definitive yes, then I am not convinced of the validity of the logical inference (A->B) as well as A itself on the basis of incorrect premise (as Maxwell's equations->Gauss' law does not permit aberration of electric fields), but if the answer is a definitive no, then I can agree that is valid (A->B). And if lack of aberration of electric fields is justified in experiment, after taking into account the possible subtle forms that aberration may theoretically take within the substance of matter, then given that I accept A->C as an axiom, I would conclude C as well.

 

[Dale]

It's not the logic that I question. I question the premises behind definitions of two statements in question. Both statement B and statement E permit two possible interpretations; so both the meaning and truth value of A->B would change, while E->D remains as a true axiom due to the obvious grammar involved.

OK, let's focus on B then since, as you mention, E->D remains obviously true regardless.

B is "the system obeys relativity". By that, I meant that the physical laws governing the system do not change their form under the Lorentz transform. Maxwell's equations do not change their form under the Lorentz transform. So if the system obeys Maxwell's equations then the system obeys relativity. A->B.

I hope these comments clarify the meaning I had intended. If you feel that the wording of B was not sufficiently clear to convey my intention then please let me know and I will reword. Perhaps I should have specified "special relativity"?

What are the two possible interpretations of B that you are referring to?

 

[kmarinas86]

OK, let's focus on B then since, as you mention, E->D remains obviously true regardless.

B is "the system obeys relativity". By that, I meant that the physical laws governing the system do not change their form under the Lorentz transform. Maxwell's equations do not change their form under the Lorentz transform. So if the system obeys Maxwell's equations then the system obeys relativity. A->B.

I hope these comments clarify the meaning I had intended. If you feel that the wording of B was not sufficiently clear to convey my intention then please let me know and I will reword. Perhaps I should have specified "special relativity"?

What are the two possible interpretations of B that you are referring to?

It has to do about whether or not electric fields undergo relativistic aberration of electric fields.

If there is relativistic aberration of electric fields, it will make a difference. That difference will be found in the integration of the electric field with respect to a Gaussian surface.

The perception I had when I opened this thread is that light-time correction would make a difference in this integral. After some objections to the idea, I also noticed that relativistic aberration would make a difference, if applicable.

I am still pondering on the likelihood that both relativistic aberration and light-time correction will make distortions to this integral that are simply not accounted for by a Lorentz-transformation.

Recently I have even found that it is assumed in physics that the electric field doesn't aberrate at all. I have only found information concerning the lack of aberration of the gravitational field, but to find such information is not equivalent to verifying that electric field is not subject to relativistic aberration, though finding that information has apparently left the impression in some that relativistic aberration does not apply to 1/r^2 forces.

 

[Dale]

From your response I am not sure if you agree or disagree that "the system obeys relativity" should mean "the physical laws governing the system do not change their form under the Lorentz transform" or if you have a different meaning in mind.

The relativistic behavior of a system is certainly not solely determined by abberation. So, please respond clearly, what is your meaning when you say "the system obeys relativity" in general?