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

1. Jan 23, 2012

### 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?

2. Jan 23, 2012

### clem

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

3. Jan 23, 2012

### kmarinas86

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.

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• ###### Lienard-Wiechert Potential of a Relativistically Moving Point Charge Bounded by a Shape.gif
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Last edited: Jan 23, 2012
4. Jan 24, 2012

### Matterwave

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

5. Jan 24, 2012

### clem

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.

6. Jan 24, 2012

### kmarinas86

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:

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

Last edited: Jan 24, 2012
7. Jan 24, 2012

### kmarinas86

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.

Last edited: Jan 24, 2012
8. Jan 24, 2012

### Bill_K

Forget the ε0, I use Gaussian units. Integrate over a sphere centered at rs. 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

9. Jan 24, 2012

### clem

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

10. Jan 24, 2012

### kmarinas86

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.

Last edited: Jan 24, 2012
11. Jan 24, 2012

### Matterwave

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).

12. Jan 24, 2012

### kmarinas86

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.

Last edited: Jan 24, 2012
13. Jan 24, 2012

### 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.

14. Jan 24, 2012

### kmarinas86

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.

Last edited: Jan 24, 2012
15. Jan 24, 2012

### Staff: Mentor

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.

16. Jan 24, 2012

### kmarinas86

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

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).

17. Jan 24, 2012

### Staff: Mentor

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

18. Jan 24, 2012

### kmarinas86

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?

19. Jan 25, 2012

### Staff: Mentor

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.

20. Jan 25, 2012

### kmarinas86

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.