# Time-retarded E-field and Gaussian surface integral

1. ### kmarinas86

If I have an oscillating charge inside of a sphere, will the integral of E(t), where t=proper time of the sphere, over the sphere's surface area result in a value of electric flux equal to the value of the charge?

2. ### Matterwave

3,859
Yes, Gauss's law is fully compatible with relativity and is quite general.

3. ### kmarinas86

Are there examples in literature where I can find the derivation for the solution of E(t) on a sphere with an arbitrary continuous functions x(t), y(t), and z(t) for an oscillating charge moving close to the speed of light, including motion in a preferred side of a sphere? If not, I will not see why the above statement is correct, and instead I will sense parroting.

The time-dependent equations for the E-field do not look like simple Lorentz transforms at all.

http://webphysics.davidson.edu/applets/retard/Retard_FEL.html
http://farside.ph.utexas.edu/teaching/em/lectures/node41.html

I need to know where to look to find the proof (mathematical derivation) that any arbitrary sphere (at rest with the observer) surrounding the charge like that in the above picture (also in the attachment below) would have a constant integral of the E-field (value of electric flux) on its surface (based on the proper time in the frame shared by the observer and the sphere upon which the surface integral is computed) under these particular circumstances.

#### Attached Files:

• ###### Oscillating Electric Field (Time-Dependent).png
File size:
4.8 KB
Views:
157
Last edited: Feb 16, 2012
4. ### Matterwave

3,859
Certainly what you are looking for can be found, but doing it that way completely loses the point of the power of something like Gauss's law. The major power of Gauss's law is that we DON'T HAVE TO carry out that ridiculous integral, we already know the answer. If we had to carry out that ridiculous integral (where no symmetries are present), then we've completely missed the point and power of Gauss's law.

Think of it in terms of electric field lines in the same line of thought as Faraday's. No matter how that electron moves around, the number of field lines crossing a closed surface must remain constant as long as the charge itself does not move outside that surface. This is because although the field lines can get wavy or distorted, they cannot ever cross each other, they will not "multiply" (in the sense of the creation of more field lines), and they must start on the charge and end at infinity (since you have only 1 charge around). The number of field lines crossing the surface is exactly the electric flux.

A lot of people get bogged down by the specifics of what the field looks like "on the surface".

An alternative way to see this, if you are not satisfied with Faraday's picture of field lines, is to see simply that Maxwell's equations in differential form is completely in accord with special relativity, and the integral forms of Maxwell's equations can be obtained from the differential form by the generalized Stoke's theorem (which is also completely general).

One last point is that the Wigner Eckhart potentials, which are THE solution to E&M (in the sense that you can get the E&M fields from them for an arbitrary distribution of arbitrarily moving charges), are explicitly derived from Maxwell's equations. So, even if you get "time retarded potentials" that you are looking for, etc., they come from he Wigner-Eckhart potentials which come from Maxwell's equations!

5. ### kmarinas86

If I move charge quickly from rest the center of the sphere to the right side of a sphere (with respect to the viewer) at 0.999c, I would expect that the flux integral on the right side to be higher than that on the left side. Furthermore, while the charge being on the right side of sphere is closer to the right inner wall than the left inner wall at all times after t=0, where acceleration begins at t=0.

Just consider the following:

If the charge is at the center, the electric field intensity should be constant at all parts of the sphere. So 50% of the total integrated value is on the left hemisphere, and the other 50% of the total integrate value is on the right hemisphere.

If the charge is at very close to the azimuth of the right hemisphere, the electric field intensity should vary at all parts of the sphere. For example, if 10% of the total integrated value is on the left hemisphere, and the other 90% of the total integrate value is on the right hemisphere.

The problem is:
* Most of the left hemisphere will receive the field intensity from that charge when it was closer to the center than the azimuth of the right hemisphere.
* Most of the right hemisphere will receive the field intensity from that charge when it was closer to the the azimuth of the right hemisphere.

6. ### Matterwave

3,859
Yes, but the point is that the surface is closed...and so the inhomogeneities cancel.

The TOTAL flux out of a closed surface is equal to the charge.

7. ### kmarinas86

For our purposes, let's say that the E-field is recorded from all points on the sphere using sensors (fixed onto the sphere) whose clocks are synchronized. There is no "single observer" who makes the observations, but a set of observers in the same inertial frame as that of the sphere.

Last edited: Feb 16, 2012
8. ### Matterwave

3,859
The intensities may go up on that side, but the vector directions will tend to change. Remember that a moving charge's electric field tends to get stronger in the perpendicular directions and weaker in the parallel direction. In this case, the vectors, even though they are longer, rotate away from the normal of the sphere so the flux is conserved.

Just take a look at the picture you posted. No matter what circle I draw on that picture, even if I deform that circle, the same number of lines will pierce my circle.

Gauss's law is really only a statement that electric field lines must start and end on charges. This is one of the great utilities of Faraday's "electric field lines" picture. His picture makes this argument very transparent, whereas vector fields make this argument very complicated. That integral that you want to do is ridiculous, but we already know the answer!

Remember that Gauss's law holds not only for a surface of a sphere centered at my charge, it holds for ANY surface with my charge enclosed. I can draw any deformed, amorphous blob shape I wanted. The integral would be ridiculous, but the answer is the same! Why do you not question this aspect of Gauss's law? It is, as far as I'm concerned, completely analogous.

I think I outlined this in my last post already.

Also, didn't we have this same exact thread like 2 weeks ago or something.

Last edited: Feb 16, 2012
9. ### kmarinas86

Doesn't the angle between the line and the circle affect the component of the E-field that is normal to the differential barrier of that circle?

An area flux integral for the electric flux (units of V*m) is computed by the integration of the dot product of the local field vectors (units of V/m) with the corresponding vectors normal to the differential surfaces (vectors that represent those differential surfaces) (units of m^2) through which those local field vectors apply.

Last edited: Feb 16, 2012
10. ### kmarinas86

In a proper response, I expect to see the percentages 10%, 90%, and 50% used in context.

11. ### Matterwave

3,859
Don't confuse electric field LINES and electric field VECTORS. These are two compatible, but distinct, models for electromagnetism. The flux out of that sphere is EITHER the integral of the VECTORS (which depends on angle, and intensity and whatnot, and is, in general, very hard to calculate directly), or simply the number of field lines that pierce the sphere (which is very easy to see).

I did address your previous post directly. I told you that the angles for the VECTORS rotated away from the normal(s) to the sphere, and thereby compensate for their increased intensities. Your numbers 10%, 90%, and 50% are just example numbers, I don't get why you want me to address these 3 numbers directly?

You don't have 90% of the flux going in one sphere and 10% going the other. You may have something like 60% in one and 40% in the other (which, by the way, still adds up to 100%, so you haven't increased net flux at all) because the vectors turning has compensated for this effect. The lagged behind vectors don't rotate as much, but they don't grow, while the vectors in front of the electron grow but rotate away from the normal.

12. ### kmarinas86

I never knew that!

In what way are the field "lines" oriented differently than field vectors? Is the pattern of the time-dependent vector field hairy (in that their might be something to cancel in some way when finding the continuous electric field "lines" from that vector field?)

Ok.

Is 60% vs. 40% the maximum? It cannot go to, say, 61% and 39%?

Ok, now you've got me thinking about what the solution to the problem would look like. Thanks for your help so far!

Last edited: Feb 16, 2012
13. ### Matterwave

3,859
1) Electric field lines and electric field vectors are 2 fundamentally different (although compatible) models for the electric field.
Electric field vectors are always tangent to field lines, but the method by which they represent the STRENGTH of the electric field differ.
The field vectors represent STRENGTH by the vector's MAGNITUDE (longer=stronger), whereas the electric field lines represent STRENGTH by the DENSITY of lines (denser=stronger).

2) 60% and 40% are merely example numbers just like your 10% and 90%. All I was trying to get at is that the vectors where the electric field is stronger will tend to be rotated away from the normal of the surface.

14. ### kmarinas86

I'll look at it more closely. Again, thanks so far.
http://webphysics.davidson.edu/physlet_resources/bu_semester2/c02_field.html

15. ### kmarinas86

Let's say we have a sphere with radius R=1 centered on the origin.

When evaluating field intensity over the sphere, let's do it for each angle phi over the domain [0,pi]. phi represents degrees of latitude from the top of the sphere. From the viewer frame, phi=0 is at the top of the sphere, and phi=pi is at the bottom. The line connecting them is situated on what we will call the z-axis.
Then we will weigh the intensity at each phi according to the circumference of the circle at each phi, which is 2*pi*cos(phi).
Then after that, we then compute the integral over phi.

Let's say we have a charge moving along the z-axis and obeying the following system of equations. All variables here are treated a dimensionless numbers:
x=0
y=0
z=t if t>0, otherwise z=0.

t is the time variable.
The domain of t begins at t=0 and ends at t=1.

The electric field is measured by clock-synchronized sensors on the surface of the sphere. The reading from them is taken at t=1, when the electron reaches the sphere.

c, the speed of the field, will be set to a dimensionless number, unity, or 1.
c is therefore equal to the radius of sphere, which itself is represented as a dimensionless number.
v will be a dimensionless number less than 1, representing the velocity of that charge.

For each latitude phi, there exists a time t which matches the z position of the charge as seen from phi. This is due to the delay of the field to that position. The distance from that position to the latitude phi =
sqrt((cos(phi)-t)^2+sin(phi)^2) =
sqrt(cos(phi)^2-2t*cos(phi)+tt+sin(phi)^2) =
sqrt(1-2t*cos(phi)+tt)

The speed of the field divided by the speed of the charge =
The distance traveled by the field divided by the distance traveled by the charge.

The scenario under consideration assumes 0 acceleration of the charge being detected. Therefore v is constant.

c/v =
sqrt(1-2t*cos(phi)+tt)/(1-t)

To calculate t for a given phi, we must rearrange the formula. First, let's multiply both sides by (1-t)
(1-t)*c/v=sqrt(1-2t*cos(phi)+tt)

That operation is valid in the case that t is not equal to 1.
The equation returns meaningful values as long as t<1.
This is suitable for our problem.

We square both sides:
(1-t)^2*(cc/vv)=1-2t*cos(phi)+tt

Since c=1:
(1-t)^2/vv=1-2t*cos(phi)+tt

Let's multiply by v^2:
(1-t)^2=vv-2t*cos(phi)vv+ttvv

Now we put everything on the RHS:
0=vv-2t*cos(phi)vv+ttvv-(1-t)^2

And we expand a term:
0=vv-2t*cos(phi)vv+ttvv-(1-2t+tt)
0=vv-2t*cos(phi)vv+ttvv-1+2t-tt

Then we group things by order of t:
0=vv-1 +2t-2t*cos(phi)vv +ttvv-tt
0=(vv-1) +t(2-2*cos(phi)*vv) +tt(vv-1)

Therefore:
t= ( sqrt((2-2*cos(phi)*vv)^2-4*(vv-1)^2) - (2-2*cos(phi)*vv) ) / (2 * (vv-1))
t= ( sqrt(4*(1-cos(phi)*vv)^2-4*(vv-1)^2) - (2-2*cos(phi)*vv) ) / (2 * (vv-1))
t= ( 2*sqrt((1-cos(phi)*vv)^2-(vv-1)^2) - (2-2*cos(phi)*vv) ) / (2 * (vv-1))
t= ( sqrt((1-cos(phi)*vv)^2-(vv-1)^2) - (1-cos(phi)*vv) ) / (vv-1)

Despite the above aforementioned operation of multiplying by (1-t), t=1 is indeed a solution.
There's not a particular reason why it shouldn't be one. After all, we could have began with the quadratic formula, then proceeded to determine the same proportionality in the form of (c/v)*(1-t)=sqrt(1-2t*cos(phi)+tt) without ever dividing by zero.

v is a given, so when phi is chosen, the only remaining variable is t.
So t can be solved purely in terms of phi.

Let r equal the distance from the z-axis.

The r distance between the latitude phi (r=sin(phi)) and the charge (r=0) is:
sin(phi)-0
or simply
sin(phi)

The z distance between the latitude phi (z=cos(phi)) and the charge (z=t) is:
cos(phi)-t

The angle, nu, of the field's trajectory away from the +z direction can also be calculated by taking the inverse tangent of r/z:
nu=atan(sin(phi)/(cos(phi)-t))

Aside from:
phi, the angle of the line connecting latitude phi to the origin
nu, the angle of the line connecting the sphere to the charge's position at time t
Both of which exist in the domain from [0,pi]
We also have other angles:
pi/2 radians, the angle of the r component of the electric field
pi radians, the angle of the z component of the electric field
phi-nu, the angle, f, between the outward-facing normal at latitude phi and the direction from the point source of the electron at z=t to the receiver at latitude phi
phi-pi/2, the angle, g, between the outward-facing normal at latitude phi and the r component of the electric field
phi-0, the angle, h, between the outward-facing normal at latitude phi and the z component of the electric field

Due to the angle between phi and nu, only some fraction of the electric field, |cos(f)|*100% of it, will contribute to the field normal to the surface at latitude phi.
Of the r component of the electric field, which is sin(nu)*100% of the total non-relativistic electric field, |cos(g)|*100% of it will contribute to the field normal to the surface at latitude phi.
Of the z component of the electric field, which is |cos(nu)|*100% of the total non-relativistic electric field, |cos(h)|*100% of it will contribute to the field normal to the surface at latitude phi.

In the non-relativistic case, the electric field is inverse with the square of distance.
In the non-relativistic case, the electric field is proportional to 1/(sin(phi)^2+(cos(phi)-t)^2).

In the relativistic case, there is an increase of the lateral electric field by the factor gamma.
gamma=1/sqrt(1-vv/cc)
Given that c=1, this reduces to:
gamma=1/sqrt(1-vv)

In the relativistic case, the r component of the electric field normal to the surface at latitude phi=
|cos(g)|*(1/sqrt(1-vv))*(1/(sin(phi)^2+(cos(phi)-t)^2))*100% of the total non-relativistic electric field.

The increase does not apply to the z component of the electric field, which has strength of=
|cos(h)|*(1/(sin(phi)^2+(cos(phi)-t)^2))*100% of the total non-relativistic electric field.

The sum of these two gives the total relativistic electric field normal to the surface at latitude phi, with a strength of=
(|cos(g)|*(1/sqrt(1-vv))+|cos(h)|)*(1/(sin(phi)^2+(cos(phi)-t)^2))*100% of the total non-relativistic electric field=
(|cos(phi-pi/2)|*(1/sqrt(1-vv))+|cos(phi)-0|)*(1/(sin(phi)^2+(cos(phi)-t)^2))*100% of the total non-relativistic electric field=
(|cos(phi-pi/2)|*(1/sqrt(1-vv))+|cos(phi)-0|)*(1/(sin(phi)^2+(cos(phi)-(( sqrt((1-cos(phi)*vv)^2-(vv-1)^2) - (1-cos(phi)*vv) ) / (vv-1)))^2))*100% of the total non-relativistic electric field

This will allow us to determine the relativistic electric field strength that is normal to the surface at phi recorded at synchronized time t=1. Then it is simply a matter of integrating the electric field over the surface to get the electric flux detected at the sphere.

I will spare that integration for another day. Instead, I will display the following results that I got for the formula I used to solve for t as a function of phi (in degrees from the +z direction).

Notes for data below:
At v=0, the charge did not move. It was at z=1, and it stays at z=1. So for v=0, z=1 all phi.
For other v, at phi=0, the sample of the electric field is taken at the same place and time where and when the electron reaches z=1. Therefore, no matter the v selected, the reading at phi=0 comes from a field source of the electron that satisfies t=z.

Code (Text):

top row     v/c
left column phi in degrees from the +z direction
all other cells z position particle according sensor at phi a time t

0   0.05    0.1 0.15    0.2 0.25    0.3 0.35    0.4 0.45    0.5 0.55    0.6 0.65    0.7 0.75    0.8 0.85    0.9 0.95

0.0     1.00    1.00    1.00    1.00    1.00    1.00    1.00    1.00    1.00    1.00    1.00    1.00    1.00    1.00    1.00    1.00    1.00    1.00    1.00    1.00
10.0        1.00    0.99    0.98    0.97    0.97    0.96    0.95    0.94    0.93    0.92    0.90    0.89    0.88    0.86    0.84    0.82    0.79    0.76    0.70    0.59
20.0        1.00    0.98    0.97    0.95    0.93    0.91    0.90    0.88    0.86    0.84    0.82    0.80    0.77    0.74    0.71    0.68    0.63    0.58    0.50    0.36
30.0        1.00    0.97    0.95    0.92    0.90    0.87    0.85    0.82    0.80    0.77    0.74    0.71    0.68    0.64    0.61    0.56    0.51    0.44    0.36    0.24
40.0        1.00    0.97    0.93    0.90    0.87    0.84    0.81    0.78    0.74    0.71    0.68    0.64    0.60    0.56    0.52    0.47    0.41    0.35    0.27    0.16
50.0        1.00    0.96    0.92    0.88    0.84    0.80    0.77    0.73    0.69    0.66    0.62    0.58    0.54    0.49    0.45    0.40    0.34    0.28    0.21    0.12
60.0        1.00    0.95    0.90    0.86    0.82    0.77    0.73    0.69    0.65    0.61    0.57    0.52    0.48    0.44    0.39    0.34    0.29    0.23    0.16    0.09
70.0        1.00    0.94    0.89    0.84    0.79    0.74    0.70    0.65    0.61    0.57    0.52    0.48    0.43    0.39    0.34    0.29    0.24    0.19    0.13    0.07
80.0        1.00    0.94    0.88    0.82    0.77    0.72    0.67    0.62    0.57    0.53    0.48    0.44    0.39    0.35    0.30    0.26    0.21    0.16    0.11    0.06
90.0        1.00    0.93    0.87    0.81    0.75    0.70    0.64    0.59    0.54    0.50    0.45    0.41    0.36    0.32    0.27    0.23    0.19    0.14    0.10    0.05
100.0   1.00    0.93    0.86    0.79    0.73    0.67    0.62    0.57    0.52    0.47    0.42    0.38    0.33    0.29    0.25    0.21    0.17    0.13    0.08    0.04
110.0   1.00    0.92    0.85    0.78    0.72    0.66    0.60    0.55    0.50    0.45    0.40    0.36    0.31    0.27    0.23    0.19    0.15    0.11    0.07    0.04
120.0   1.00    0.92    0.84    0.77    0.70    0.64    0.58    0.53    0.48    0.43    0.38    0.34    0.29    0.25    0.21    0.18    0.14    0.10    0.07    0.03
130.0   1.00    0.91    0.83    0.76    0.69    0.63    0.57    0.51    0.46    0.41    0.37    0.32    0.28    0.24    0.20    0.17    0.13    0.10    0.06    0.03
140.0   1.00    0.91    0.83    0.75    0.68    0.62    0.56    0.50    0.45    0.40    0.35    0.31    0.27    0.23    0.19    0.16    0.12    0.09    0.06    0.03
150.0   1.00    0.91    0.82    0.75    0.68    0.61    0.55    0.49    0.44    0.39    0.34    0.30    0.26    0.22    0.19    0.15    0.12    0.09    0.06    0.03
160.0   1.00    0.91    0.82    0.74    0.67    0.60    0.54    0.49    0.43    0.38    0.34    0.30    0.25    0.22    0.18    0.15    0.11    0.08    0.05    0.03
170.0   1.00    0.91    0.82    0.74    0.67    0.60    0.54    0.48    0.43    0.38    0.33    0.29    0.25    0.21    0.18    0.14    0.11    0.08    0.05    0.03
180.0   1.00    0.90    0.82    0.74    0.67    0.60    0.54    0.48    0.43    0.38    0.33    0.29    0.25    0.21    0.18    0.14    0.11    0.08    0.05    0.03

#### Attached Files:

• ###### IC412528.png
File size:
2.7 KB
Views:
53
Last edited: Feb 18, 2012