larmor radiation
TRANSCRIPT
Radiation from an Accelerated Charge[see also Longair, High Energy Physics, 2nd edition, vol. 1, p. 64, and Wikipedia]
Maxwell's equations imply that all classical electromagnetic radiation is ultimately generated byaccelerating electrical charges. It is possible to derive the intensity and angular distribution of theradiation from a point charge (a charged particle) subject to an arbitrary but small acceleration
via Maxwell's equations (and retarded potentials), but the complicated math obscuresthe physical interpretation that remains clear in J. J. Thomson's simpler derivation.
If a particle with electrical charge is at rest, Coulomb's law implies that its electric field lines willbe purely radial: . Suppose a charged particle is accelerated to some small velocity
in some short time . At time later, there will be a perpendical component ofelectric field
;
where is the angle between the acceleration vector and the line from the charge to theobserver.
This figure shows the electric fields from an acclerated electron.
Coulomb's law for the radial component of the electric field (electric force per unit charge)a distance from a charge is
E
Áv=Át
q E = Er
Áv Ü c Át t
Er
E? =cÁt
Áv t sin Ò
Ò
E rr q
r =q
r2(2D1)
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in Gaussian cgs units. Substituting gives
and
E
This equation is valid for any small acceleration, not just a sinusoidal one. The delay time hasdropped out; the transverse field instantaneously reflects the applied acceleration. Thus a
sinusoidal acceleration would result in a sinusoidal variation with the same frequency in . Note
that in contrast to . Far from the charged particle (large ), only willcontribute significantly to the radiation field. From the observer's point of view, only the visibleacceleration perpendicular to the line of sight contributes to the radiated electric field; theinvisible component of acceleration parallel to the line of sight does not radiate—what you see iswhat you get.
How much power is radiated in each direction? In a vacuum, the Poynting flux, or power per unitarea (erg s cm ) is
In cgs units so
jSj E
The charge radiates with a dipolar power pattern that looks like a doughnut whose axis isparallel to .
t =c = r
E ? =q
r2
Ò
Át
ÁvÓ
c2r sin Ò
? =rc2
qv_ sin Ò(2D2)
t E ?
E ?E ? / rÀ1 E r / rÀ2 r E ?
À1 À2
S E ~ =c
4Ù~ ÂH~
jEj H j ~ = j ~
~ =c
4Ù2 (2D3)
jSj ~ =c
4Ù
Ò
rc2qv_ sin Ò
Ó2
jSj ~ =1
4Ù c3q v2 2_
r2sin2 Ò
v _
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The power pattern of Larmor radiation from a charged particle shown for an acceleration vector tilted from the line of sight. The power received in any direction is proportional to the
component of perpendicular to the line of sight.
Coordinate system used to calculate the total power emitted by an accelerated charged particle.
v _ 60 Î
v _
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The total power emitted is obtained by integrating over all directions:
Evaluating this integral gives so the total power emitted is
P
This result is called Larmor's equation. It states that any charged particle radiates whenaccelerated and that the total radiated power is proportional to the square of the acceleration.Since the greatest astrophysical accelerations are usually electromagnetic, the acceleration isusually proportional to the charge/mass ratio of the particle. Thus radiation from electrons istypically stronger than radiation from protons, which are times moremassive.
Larmor's equation will be the basis for our derivations of radiation from a short dipole antennaas well as for free-free and synchrotron emission from astrophysical sources. Beware thatLarmor's formula is nonrelativistic; it is valid only in frames moving at velocities withrespect to the radiating particle. To treat particles moving at nearly the speed of light in theobserver's frame, we must use Larmor's equation to calculate the radiation in the particle's restframe and then transform the result to the observer's frame in a relativistically correct way. Also, Larmor's formula does not incorporate the constraints of quantum mechanics, so it shouldbe applied with great caution to microscopic systems such as atoms. For example, Larmor'sequation incorrectly predicts that the electron in a hydrogen atom will quickly radiate away all ofits kinetic energy and fall into the nucleus.
P SjdA =
Z
sphere
j~
P dÒ d¶ =q v2 2_
4Ùc3
Z 2Ù
¶=0
Z Ù
Ò=0 r2sin2 Ò
r sin Ò r
P dÒ =2c3q v2 2_
Z Ù
Ò=0
sin3 Ò
dÒ =3 R Ù
Ò=0sin3 Ò = 4
=3
2
c3q v2 2_
(2D4)
Ù 0 4Â 1 6 Ù 0 2Â 1 3
v Ü c
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