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Larmor's Formula

If one is far (enough) away from the an accelerating charge in the right direction, the field is given by primarily by the second (acceleration) term. This is the ``usual'' transverse EM field. If the particle is moving slowly with respect to $c$ (so $\beta « 1$), then

$\displaystyle \mbox{\boldmath$E$}$ $\textstyle =$ $\displaystyle \frac{e}{4\pi\epsilon_0}\frac{1}{c}
...{\boldmath$n$}}\times \dot{\mbox{\boldmath$\beta$}})}{R} \right\vert _{\rm ret}$ (18.32)
$\displaystyle \mbox{\boldmath$B$}$ $\textstyle =$ $\displaystyle \frac{e}{4\pi\epsilon_0}\frac{1}{c^2}
\left.\frac{\hat{\mbox{\boldmath$n$}}\times \dot{\mbox{\boldmath$\beta$}}}{R} \right\vert _{\rm ret}$ (18.33)

The energy flux is given by the (instantaneous) Poynting vector:

$\displaystyle \mbox{\boldmath$S$}$ $\textstyle =$ $\displaystyle \frac{1}{\mu_0}(\mbox{\boldmath$E$}\times \mbox{\boldmath$B$})$  
  $\textstyle =$ $\displaystyle \frac{e^2}{16\pi^2\epsilon_0 R^2}
\frac{1}{\mu_0\epsilon_0} \fra...
...math$n$}}\times \dot{\mbox{\boldmath$\beta$}})\vert^2 \hat{\mbox{\boldmath$n$}}$  
  $\textstyle =$ $\displaystyle \frac{e^2}{16\pi^2\epsilon_0 R^2}
\frac{1}{c^3} \vert\hat{\mbox{...
...h$n$}}\times c^2\dot{\mbox{\boldmath$\beta$}})\vert^2 \hat{\mbox{\boldmath$n$}}$  
  $\textstyle =$ $\displaystyle \frac{e^2}{16\pi^2\epsilon_0 R^2}
\frac{1}{c^3} \vert\hat{\mbox{...
...boldmath$n$}}\times \dot{\mbox{\boldmath$v$}})\vert^2 \hat{\mbox{\boldmath$n$}}$ (18.34)

As always, the power cross-section (energy per unit solid angle) is

$\displaystyle \frac{dP}{d\Omega}$ $\textstyle =$ $\displaystyle \mbox{\boldmath$S$}\cdot \hat{\mbox{\boldmath$n$}}R^2$  
  $\textstyle =$ $\displaystyle \frac{e^2}{16\pi^2\epsilon_0}
\frac{1}{c^3} \vert\hat{\mbox{\boldmath$n$}}\times (\hat{\mbox{\boldmath$n$}}\times \dot{\mbox{\boldmath$v$}})\vert^2$  
  $\textstyle =$ $\displaystyle \frac{e^2}{16\pi^2\epsilon_0}
\frac{1}{c^3} \vert\dot{\mbox{\boldmath$v$}}\vert^2 \sin^2(\Theta)$ (18.35)

where $\Theta$ is the angle between $\hat{\mbox{\boldmath$n$}}$ and $\dot{\mbox{\boldmath$v$}}$.

Aha! we say. The characteristic $\sin^2 \Theta$! Aha again! Inspecting the vector products, we see that the radiation is polarized in the plane of ${\bf n},\dot{\bf v}$, perpendicular to n. Finally, the integral over angles yields $8\pi/3$, so that

P = \frac{e^2}{6\pi \epsilon_0 c^3} \left\vert \dot{\bf v} \right\vert^2.
\end{displaymath} (18.36)

This is the Larmor formula for the power radiated from a nonrelativistic accelerated point charge. This has a covariant generalization that is valid for any velocity of charge. First we factor out an $m^2$ and convert this to momentum coordinates. Then we realize that the energy carried by this field (per unit time) is indeed related to the momentum by a factor of $1/c$ and convert the whole thing to 4-vector form. Last, we convert $t$ into $\tau$:
$\displaystyle P$ $\textstyle =$ $\displaystyle \frac{e^2}{6\pi\epsilon_0 c^3} \frac{1}{m^2}
  $\textstyle =$ $\displaystyle \frac{e^2}{6\pi\epsilon_0 m^2 c^3} \left\vert\frac{d(m\mbox{\boldmath$v$})}{\gamma
d\tau }\right\vert^2$  
  $\textstyle =$ $\displaystyle \frac{e^2}{6\pi\epsilon_0 m^2 c^3} (1 - \beta^2)
\left\vert\frac{d\mbox{\boldmath$p$}}{d\tau }\right\vert^2$  
  $\textstyle =$ $\displaystyle \frac{e^2}{6\pi\epsilon_0 m^2 c^3}
\left( \frac{d\mbox{\boldmath$p$}}{d\tau }\right)^2 -
\left( \frac{1}{c^2}\frac{dE}{d\tau}\right)^2$  
  $\textstyle =$ $\displaystyle - \frac{e^2}{6\pi\epsilon_0 m^2c^3} \left( \frac{dp_\alpha}{d\tau}
\frac{dp^\alpha}{d\tau} \right)$ (18.37)

This can be written one more way, (substituting $E = \gamma m c^2$ and ${\bf
p} = \gamma m {\bf v}$ and using some vector identities) due to Liénard:

P = \frac{e^2}{6\pi \epsilon_0 c^3} \gamma^6 [(\dot{\mbox{\...
...ox{\boldmath$\beta$}\times \dot{\mbox{\boldmath$\beta$}})^2 ]
\end{displaymath} (18.38)

We are all better people for knowing this.

Why, you may ask, is this torture necessary? Because quite a few of you will spend unreasonable amounts of your lives calculating things like radiative losses in accelerators. After all, if we could build GeV accelerators in a little bitty ten foot ring it would be a whole lot cheaper than 6 billion bucks, plus inflation. Unfortunately, nature says that if you try it the nasty thing will give off synchrotron radiation! Let us see that tanstaafl18.1.

The radiated power is proportional to the acceleration. The work is proportional to the tangential force times the velocity. Light particles accelerate the most for a given tangential force and have the highest velocity for a given energy; radiative losses are thus the most important for those particles at all energies. We will evaluate the radiative power loss for an electron in a linear accelerator.

We begin with

P = \frac{e^2}{6\pi \epsilon_0 m^2c^3} \left( \frac{dp}{dt} \right)^2
\end{displaymath} (18.39)

where $-e$ is now really the charge on the electron. Since the accelerator is linear, we can find the force directly from the rate at which work is done on the electron (otherwise we would have to include the force bending it in a curved path, which does no work). It is related to the ``gradient'' of the total energy,
P = \frac{e^2}{6\pi\epsilon_0 m^2c^3} \left( \frac{dE}{dx} \right)^2 .
\end{displaymath} (18.40)

For linear acceleration we don't care what the actual energy of the particle is; we only care how that energy changes with distance.

We will turn this into a rate equation by using the chain rule:

P_{\rm rad} = \frac{e^2}{6 \pi \epsilon_0 m^2c^3} \frac{dE}{dx}
\frac{dE}{dt} \frac{dt}{dx}
\end{displaymath} (18.41)

Thus the ratio of power radiated to power supplied by the accelerator $P_{\rm acc} = dE/dt$ is:
\frac{P_{\rm rad}}{P_{\rm acc}} = \frac{e^2}{6\pi\epsilon_0 ...
\frac{e^2/mc^2}{mc^2} \frac{dE}{dx}
\end{displaymath} (18.42)

where the latter form is valid when the electron is travelling at $v \approx

This quantity will be less than one while the gain in energy in a distance $e^2/mc^2 = 2.82 \times 10^{-13}$ cm is of the order of $mc^2 =
.5$ MeV. That would require a potential difference (or other force) on the order of $10^{14}$ MV/meter.

Maybe at the surface of a positron. Come to think of it, falling into a positron there comes a point where this is true and at that point the total mass energy of the pair is radiated away. But nowhere else. We can completely neglect radiative losses for linear acceleration simply because the forces required to produce the requisite changes in energy when the particle is moving at nearly the speed of light are ludicrously large. For a charged particle moving in a straight line, radiative losses are more important at low velocities. This is fortunate, or radios and the like with linear dipole antennas would not work!

However, it is incovenient to build linear accelerators. That is because a linear accelerator long enough to achieve reasonable energies for electrons starts (these days) at around 100-500 miles long. At that point, it is still not ``straight'' because the earth isn't flat and we don't bother tunnelling out a secant. Also, it seems sensible to let a charged particle fall many times through the ``same'' potential, which is possible only if the accelerator is circular. Unfortunately, we get into real trouble when the accelerator is not straight.

In a circular accelerator, there is a non-zero force proportional to its velocity squared, even when little or no work is being done to accelerate the particle! In fact, the centripetal force on the particle is

\left\vert \frac{d{\bf p}}{d\tau} \right\vert = \gamma \omega \left\vert \bf p \right\vert » \frac{1}{c}
\end{displaymath} (18.43)

all of which increase as the speed of the particle increases. If we completely neglect the radiative loss due to tangential acceleration (which is completely negligible once relativistic velocities have been reached) we see that
P = \frac{e^2}{6\pi\epsilon_0 m^2 c^3} \gamma^2 \omega^2 \l...
...\vert^2 =
\frac{e^2 c}{6\pi\epsilon_0 r^2} \beta^4 \gamma^4
\end{displaymath} (18.44)

where we have used $\omega = (c\beta/r)$. The loss per revolution is obtained by multiplying by $T$ (the period of a revolution). This yields
\Delta E = \frac{2 \pi r}{c\beta} P = \frac{e^2}{3\epsilon_0 r}
\beta^3 \gamma^4
\end{displaymath} (18.45)

which is still deadly if $r$ is small and/or $\gamma$ and $\beta$ are large.

If one does some arithmetic (shudder), one can see that for high energy electrons (where $\beta \approx 1$), this is

\Delta E {\rm (MeV)} = 8.85 \times 10^{-2} \frac{[E {\rm (GeV)}]^4 }{r
{\rm (meters)}} .
\end{displaymath} (18.46)

At around 1 GeV, one needs roughly $1/(10 r)$ of that energy gain per cycle in order to turn (heh, heh) a net profit. That is not so bad, but the power of 4 says that at 10 GeV, one needs a gain per cycle of $1000/r$ GeV (!) in order to turn a profit. Now, it is true that the bigger the radius the longer the circumference (linearly) and the longer the circumference the more work one can do with a given fixed potential in a cycle. So in terms of force this relation is not as bad as it seems. But it is bad enough, because you still have to do the work, which costs you the same no matter how hard you have to push to do it. Clearly even at 10 GeV, an orbit of radius $\sim 100$ meters or better is necessary. In electron-positron storage rings, work must be done at this general rate just to keep the particles moving.

Those of you who need to know can read section 14.3 on your own. The results are straightforward but algebraically tedious, and are of use only if you plan on studying accelerator design or neutron stars. Don't get me wrong. Nobel prizes have been won for accelerator design and may be again. Go for it.

Ditto for 14.4. This is highly readable and contains no algebra. In a nutshell, a particle moving in a synchrotron emits its radiation in its instantaneous direction of motion (which is indeed perpendicular to the acceleration). Since it moves in a circle, a stationary observer in the plane of motion sees short bursts of radiation at the characteristic frequency $c/r$. The length (in time) of the pulses is $L/c$ in time, and thus will contain frequencies up to $c/L \sim (c/r) \gamma^3 $ in a fourier decomposition of their ``wave packet'' where $L \approx r/(2 \gamma^3)$ is the length of the pulse in space. For highly relativistic particles moving in big circles, the characteristic frequency can be many orders of magnitude smaller than the high frequency cut off, as in AM radio frequencies to X-rays or worse. Synchrotron radiation is a potential source of high frequency electromagnetic energy.

Of course, it isn't tunable or coherent (in fact, its highly incoherent since the spectrum is so wide!) and we'd love to use the same kind of trick to make coherent, tunable, high frequency light. Some of you probably will use the same kind of trick before you leave, since free electron lasers produce energy from a similar principle (although with a totally different spectrum!). Section 14.6 deals with the spectrum, and we will blow that off, too. Suffice it to say that it can be calculated, and you can learn how, if you need to. You really should remember that $\omega_c \approx \omega_0 \gamma^3$, and should take a peek at the distribution curves. These curves let one detect synchrotron radiation from cosmological sources. These sources are generally charged particles falling into dark stars, radiation belts around planets, sunspots, or anyplace else that relativistic electrons are strongly accelerated in a circular, or helical, path. Finally, we will neglect 14.5 too, which analyzes radiation emitted by particles moving in wierd ways. Jackson is encyclopaediac, but we needn't be.

We will come back into focus at section 14.7, Thomson Scattering of Radiation. This is scattering of radiation by charged particles and is closely related to Compton scattering. It is important, as it is a common phenomenon.

next up previous contents
Next: Thomson Scattering of Radiation Up: Radiation from Point Charges Previous: Radiation from Point Charges   Contents
Robert G. Brown 2007-12-28