Talk:Relativistic electromagnetism

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1 Regrets (I have a few!)
2 Request for help
3 Cut from article to be worked on
4 The physical basis of electromagnetic radiation
- 4.1 Strength of the radiation field.
- 4.2 The Larmor Formula
  - 4.2.1 Energy radiated by a charged particle
5 Applications of radiation
- 5.1 Electromagnetic Waves
6 Diagrams

[edit] Regrets (I have a few!)

I really wish Id not started this page now as it is too complex a subject to illustrate without diagrams that I cant draw! Suggestions on how to get out of this mess will be most welcome!--Light current 17:51, 30 April 2006 (UTC)

Diagrams will be inserted to illustrate the text ASAP--Light current 16:03, 21 October 2005 (UTC)

Having trouble getting the equations right. Please bear with me (or help!)--Light current 17:26, 21 October 2005 (UTC)

Lots more work needed on article especially with diagrams & equations. Please bear with me.--Light current 01:46, 22 October 2005 (UTC)

Started work on diagrams. Any help from persons with drawing packages gratefully recieved.--Light current 21:45, 22 October 2005 (UTC)

[edit] Request for help

I really need someone to help on improving my very basic (and not ver satisfactory) text based diagrams in this article. Any offers? You will be rewarded with thanks!!--Light current 00:30, 23 October 2005 (UTC)

[edit] Cut from article to be worked on

Using angle brackets to denote an average over all points on the shell,

�sin2 θ� = �1 − cos2 θ� = 1− �x2� R2 .

Now since the origin is at the center of the sphere, the average value of x2 is the same as the average value of y2 or z2: �x2� = �y2� = �z2�. (4.9)

But this implies that �x2� = 1 3�x2 + y2 + z2� = 1 3�R2� = R2 3

since x2 +y2 +z2 = R2 and R is constant over the whole shell. Combining equations (4.8) and (4.10) gives

�sin2 θ� = 1− R2 3R2 = 2 3. (4.11) 4.3

So the average energy per unit volume stored in the transverse electric field is

q2a2 48π2�0c4R2.

[edit] The physical basis of electromagnetic radiation

                \       wall 
                 \     |
                  \    |
                   \   |
                    \  |
                     \ |
                      \|B 
                      /|\
                     / | \
                    /  |  \ 
                  A/   |   \C

Consider a positively charged particle, initially traveling to the right at 1/4 the speed of light. It bounces off a wall at point B. The particle is now at point A, but if there had been no bounce it would now be at C. An imaginary circle (actually a cross section of a sphere) encloses the region of space where news of the bounce has already arrived; inside this circle the electric field points directly away from A. Outside the circle the news has not yet arrived, so the field points directly away from C. As time passes, the circle expands outward at the speed of light, and points A and C move away from B at 1/4 the speed of light. The field at your location points away from where the particle would be now if there had been no bounce. We know from special relativity that no information can travel faster than the speed of light

Assume that the information travels at precisely the speed of light, but no faster. This assumption, together with Gauss’s law, is enough to determine the electric field everywhere around the accelerated charge. The complete map of the electric field of an accelerated charge turns out to be quite complicated. An abstract representation in terms of field lines will be used instead. Field lines are continuous lines through space that run parallel to the direction of the electric field. A drawing of the field lines in a region therefore indicates immediately the direction of the electric field.

A map of the field lines for the situation of figure is shown in figure . No field lines through the gray spherical shell in figure are shown, since this is the region that is just in the midst of receiving the news of the particle’s acceleration. To determine the direction of the field here, imagine a curved Gaussian “pillbox”, indicated by the dashed line in the figure, which straddles the gray shell. This surface is symmetrical about the line along which the particle is moving; but viewed from along this line, it would be circular.

The Gaussian surface encloses no electric charge, so Gauss’s law tells us that the total flux of �E through it must be zero.

The direction of the field within the gray spherical shell can be found be considering the flux through the curved Gaussian “pillbox” indicated by the dashed line. Now consider the flux through various parts of the surface. On the outside (right-hand) portion there is a positive flux, while on the inside (left-hand) portion there is a negative flux. But these two contributions to the flux do not cancel each other, since the field is significantly stronger on the outside than on the inside. This is because the field on the outside is that of a point charge located at C, while the field on the inside is that of a point charge located at A, and C is significantly closer than A. The net flux through the inside and outside portions of the surface is therefore positive. To cancel this positive flux, the remaining edges of the pillbox must contribute a negative flux. Thus the electric field within the gray shell must have a nonzero component along the shell, in toward the center of the Gaussian surface. This component is the transverse field, since it points transverse (i.e., perpendicular) to the purely radial direction of the field on either side. To be more precise about the direction of the field within the gray shell, consider the modified Gaussian surface shown in figure.

Since the flux along segment cd must be zero, the electric field within the gray shell must be parallel to this segment. until it subtends the same angle, as viewed from C, that the inner surface ab subtends as viewed from A. Now the fluxes through ab and ef do indeed cancel. Segments bc and de are chosen to be precisely parallel to the field lines in their locations, so there is no flux through these portions of the surface. In order for the total flux to be zero, therefore, the flux must be zero through segment cd as well. This implies that the electric field within the gray shell must be parallel to cd. If you start at A and follow any field line outward, you will turn a sharp corner at the gray shell’s inner edge, then make your way along the shell and slowly outward, turning another sharp corner at the outer edge. The thickness of the gray shell is determined by the duration of the acceleration of the charge.

A complete drawing of the field lines for this particular situation is shown in figure. The transverse portion of the electric field of an accelerated charge is also called the radiation field, because as time passes it “radiates” outward in a sphere expanding at the speed of light. If the acceleration of the charged particle is sufficiently great, the radiation field can be quite strong, affecting faraway charges much more than the ordinary radial field of a charge moving at constant velocity. The radiation field can also store a relatively large amount of energy, which is carried away from the charge that created it.

[edit] Strength of the radiation field.

A complete sketch of the electric field lines for the situation shown in the preceding figures, including the transverse radiation field created by the acceleration of the charge.

Consider a somewhat simpler situation, in which a positively charged particle, initially moving to the right, suddenly stops and then remains at rest. Let v0 be the initial speed of the particle, and let the deceleration begin at time t = 0 and end at time t = t0. Assume that the acceleration is constant during this time interval; the magnitude of the acceleration is then a = |�a| = v0 t0. Assume that v0 is much less than the speed of light, so that the relativistic compression and stretching of the electric field discussed above is negligible. Figure shows the situation at some time T, much later than t0. The “pulse” of radiation is contained in a spherical shell of thickness ct0 and radius cT. Outside of this shell, the electric field points away from where the particle would have been if it had kept going; that point is a distance v0T to the right of its actual location. (The distance that

R = cT v0T sin θ ct0 θ v0T θ

For clarity, only a single field line is shown here. it traveled during the deceleration is negligible on this scale. A single field line is shown in the figure, coming out at an angle θ from the direction of the particle’s motion. There is a sharp kink in this line where it passes through the shell, as discussed in the previous section.

The radial component Er of the kinked field can be found by applying Gauss’s law to the pillbox shown. How strong is the electric field within the shell? Break the kinked field up into two components: a radial component Er that points away from the location of the particle, and a transverse component Et that points in the perpendicular direction. The ratio of these components is determined by the direction of the kink; See that Et Er = v0T sin θ ct0 = aT sin θ c

We can find the radial component Er by applying Gauss’s law to a tiny pillbox that straddles the inner surface of the shell. Let the sides of the pillbox be infinitesimally short so that the flux through them is negligible. Then since the net flux through the pillbox is zero, the radial component of �E (that is, the component perpendicular to the top and bottom of the pillbox) must be the same on each side of the shell’s inner surface. But inside the sphere of radiation the electric field is given by Coulomb’s law. Thus the radial component of the kinked field is Er = 1 4π�0 q R2 , where q is the charge of the particle. Combining equations and using the fact that R = cT , it can be shown that Et = qa sin θ 4π�0c2R.

Looking back at the figure, we see that the size of the kink in the field is a qualitative indication of the field strength. Also, the strength of the transverse field is proportional to a, the magnitude of the particle’s acceleration. The greater the acceleration, the stronger the pulse of radiation. This pulse of radiation carries energy. The energy per unit volume stored in any electric field is proportional to the square of the field strength. This implies Energy per unit volume ∝ a2 R2.

Since the volume of the spherical shell (the shell itself, not the region it encloses) is proportional to R2, the total energy it contains does not change as time passes and R increases. Thus when a charged particle accelerates, it loses energy to its surroundings, in an amount proportional to the square of its acceleration. This process is the basic mechanism behind all electromagnetic radiation: visible light and radio waves to gamma rays.

[edit] The Larmor Formula

[edit] Energy radiated by a charged particle

In any electric field, the energy store per unit volume is epsilon_0/2 E^2

The total energy in the pulse is:

$Ep = \frac{e^2 a^2 t}{6 \pi \epsilon_0 c^3}$ where t is the duration of the acceleration

If we divide both sides by the duration t of the particles acceleration, we obtain the power: P.

$P = \frac{e^2 a^2}{6 \pi \epsilon_0 c^3}$

where a is the acceleration, e is the electronic charge, epsilon0 is the permittivity of free sapce and c is the speed of electromagnetic radiation This is the Larmor formula

If the direction in which the energy goes is not important, we can average equation (4.7) over all directions. Using a mathematical device, introduce a coordinate system with the origin at the center of the sphere and the x axis along the particle’s original direction of motion. Then for any point (x, y, z) on the spherical shell, cos θ = x/R.

To obtain the total energy stored in the transverse electric field, we must multiply equation by the volume of the spherical shell. The surface area of the shell is 4πR2 and its thickness is ct0, so its volume is the product of these factors. Therefore the total energy is Total energy in electric field = q2a2t0 12π�0c3 . The total energy is independent of R; that is, the shell carries away a fixed amount of energy that is not diminished as it expands. There is also a magnetic field, which carries away an equal amount of energy. Many details about magnetic fields have been omitted. A factor of 2 would need putting in. Thus the total energy carried away by the pulse of radiation is twice that of the previous equation or

When a charged particle accelerates, part of its electric field breaks free and travels away at the speed of light, forming a pulse of electromagnetic radiation. Often, in practice, charged particles oscillate back and forth continuously, sending off one pulse after another in a periodic pattern.

An example of the electric field around an oscillating charge is shown in figure. A map of the electric field lines around a positively charged particle oscillating sinusoidally, up and down, between the two gray regions near the center. Points A and B are one wavelength apart. If you follow a straight line out from the charge at the center of the figure, you will find that the field oscillates back and forth in direction. The distance over which the direction of the field repeats is called the wavelength. For instance, points A and B in the figure are exactly one wavelength apart. The time that it takes the pattern to repeat once is called the period of the wave, and is equal to the time that the source charge takes to repeat one cycle of its motion. The period is also equal to the time that the wave takes to travel a distance of one wavelength. Since it moves at the speed of light, we can infer that the wavelength and the period are related by

speed = wavelength period or c = λ T ,

where λ (“lambda”) is the standard symbol for wavelength, T is the standard symbol for period, and c is the speed of light.

[edit] Applications of radiation

[edit] Electromagnetic Waves

If you follow a straight line out from the charge at the center of figure you will find that the field oscillates back and forth in direction. The distance over which the direction of the field repeats is called the wavelength. For instance, points A and B in the figure are exactly one wavelength apart.

If you sit at a fixed point and watch the electric field as it passes by, you will again find that its direction oscillates. The time that it takes the pattern to repeat once is called the period of the wave, and is equal to the time that the source charge takes to repeat one cycle of its motion. The period is also equal to the time that the wave takes to travel a distance of one wavelength. Since it moves at the speed of light, we can infer that the wavelength and the period are related by

speed =wavelength.period

or c = λT

where λ (“lambda”) is the standard symbol for wavelength, T is the standard symbol for period, and c is the speed of light. The frequency of an oscillation or a wave is the reciprocal of the period.

extracted/modified from Daniel V Schroeder paper