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Next: Radiation losses in charged Up: Relativity and electromagnetism Previous: The electromagnetic field due

The Lamor formula

Let us transform to the inertial frame in which the charge is instantaneously at rest at the origin at time $t=0$. In this frame $u\ll c$, so that ${\bfm r}_u \simeq{\bfm r}$ and $s\simeq r$, for events which are sufficiently close to the origin at $t=0$ that the retarded charge still appears to travel with a velocity which is small compared to that of light. It follows from the previous section that
$\displaystyle {\bfm E}_{\rm rad}$ $\textstyle \simeq$ $\displaystyle \frac{e}{4\pi\epsilon_0 c^2} \frac{{\bfm r}\wedge
({\bfm r}\wedge\dot{\bfm u})}{r^3},$ (354)
$\displaystyle {\bfm B}_{\rm rad}$ $\textstyle \simeq$ $\displaystyle \frac{e}{4\pi\epsilon_0 c^3} \frac{\dot{\bfm u}
\wedge{\bfm r}}{r^2}.$ (355)

Let us define spherical polar coordinates whose axis points along the direction of instantaneous acceleration of the charge. It is easily demonstrated that
$\displaystyle E_\theta$ $\textstyle \simeq$ $\displaystyle \frac{e}{4\pi\epsilon_0 c^2} \frac{\sin\theta}{r} \,\dot{u},$ (356)
$\displaystyle B_\phi$ $\textstyle \simeq$ $\displaystyle \frac{e}{4\pi\epsilon_0 c^3} \frac{\sin\theta}{r} \,\dot{u}.$ (357)

These fields are identical to those of a radiating dipole whose axis is aligned along the direction of instantaneous acceleration. The Poynting flux is given by
\begin{displaymath}
\epsilon_r = \frac{E_\theta B_\phi}{\mu_0} = \frac{e^2}
{16\pi^2 \epsilon_0 c^3}\frac{\sin^2\theta}{r^2} \,\dot{u}^2.
\end{displaymath} (358)

We can integrate this expression to obtain the instantaneous power radiated by the charge
\begin{displaymath}
P = \frac{e^2}{6\pi\epsilon_0 c^3} \,\dot{u}^2.
\end{displaymath} (359)

This is known as Lamor's formula. Note that zero net momentum is carried off by the fields (2.289).

In order to proceed further it is necessary to prove two useful theorems. The first theorem states that if a 4-vector field $T^\mu$ satisfies

\begin{displaymath}
\partial_\mu T^\mu = 0,
\end{displaymath} (360)

and if the components of $T^\mu$ are non-zero only in a finite spatial region, then the integral over 3-space,
\begin{displaymath}
I = \int T^4\,d^3 x,
\end{displaymath} (361)

is an invariant. In order to prove this theorem we need to use the 4-dimensional analog of Gauss's theorem, which states that
\begin{displaymath}
\int_V \partial_\mu T^\mu\, d^4 x = \oint_S T^\mu\,dS_\mu,
\end{displaymath} (362)

where $dS_\mu$ is an element of the 3-dimensional surface $S$ bounding the 4-dimensional volume $V$. The particular volume over which the integration is performed is indicated in Fig. 1. The surfaces $A$ and $C$ are chosen so that the spatial components of $T^\mu$ vanish on $A$ and $C$. This is always possible because it is assumed that the region over which the components of $T^\mu$ are non-zero is of finite extent. The surface $B$ is chosen normal to the $x^4$-axis whereas the surface $D$ is chosen normal to the $x^{4'}$-axis. Here, the $x^\mu$ and the $x^{\mu'}$ are coordinates in two arbitrarily chosen inertial frames. It follows from Eq. (2.294) that
\begin{displaymath}
\int T^4\,dS_4 + \int T^{4'}\,dS_{4'} = 0.
\end{displaymath} (363)

Here, we have made use of the fact that $T^\mu\,dS_\mu$ is a scalar and, therefore, has the same value in all inertial frames. Since $dS_4=-d^3 x$ and $dS_{4'} = d^{3}x'$ it follows that $I = \int T^4\,d^3 x$ is an invariant under a Lorentz transformation. Incidentally, the above argument also demonstrates that $I$ is constant in time (just take the limit in which the two inertial frames are identical).

Figure 1: The region of integration for proving the theorem associated with Eq. (2.293)
\begin{figure}\epsfysize =2in
\centerline{\epsffile{fig1.eps}}\end{figure}

The second theorem is an extension of the first. Suppose that a 4-tensor field $
Q^{\mu\nu}$ satisfies

\begin{displaymath}
\partial_\mu Q^{\mu\nu}=0,
\end{displaymath} (364)

and has components which are only non-zero in a finite spatial region. Let $A_\mu$ be a 4-vector whose coefficients do not vary with position in space-time. It follows that $T^\mu = A_\nu Q^{\mu\nu}$ satisfies Eq. (2.292). Therefore,
\begin{displaymath}
I = \int A_\nu Q^{4\nu} \,d^3 x
\end{displaymath} (365)

is an invariant. However, we can write
\begin{displaymath}
I = A_\mu B^\mu,
\end{displaymath} (366)

where
\begin{displaymath}
B^\mu = \int Q^{4\mu}\,d^3 x.
\end{displaymath} (367)

It follows from the quotient rule that if $A_\mu B^\mu$ is an invariant for arbitrary $A_\mu$ then $B^\mu$ must transform as a constant (in time) 4-vector.

These two theorems enable us to convert differential conservation laws into integral conservation laws. For instance, in differential form the conservation of electrical charge is written

\begin{displaymath}
\partial_\mu J^\mu =0.
\end{displaymath} (368)

However, from Eq. (2.293) this immediately implies that
\begin{displaymath}
Q = \frac{1}{c} \int J^4\,d^3 x= \int \rho\,d^3 x
\end{displaymath} (369)

is an invariant. In other words, the total electrical charge contained in space is both constant in time and the same in all inertial frames.

Suppose that $S$ is the instantaneous rest frame of the charge. Let us consider the electromagnetic energy tensor $T^{\mu\nu}$ associated with all of the radiation emitted by the charge between times $t=0$ and $t=dt$. According to Eq. (2.249) this tensor field satisfies

\begin{displaymath}
\partial_\mu T^{\mu\nu} = 0,
\end{displaymath} (370)

apart from a region of space of measure zero in the vicinity of the charge. Furthermore, the region of space over which $T^{\mu\nu}$ is non-zero is clearly finite, since we are only considering the fields emitted by the charge in a small time interval, and these fields propagate at a finite velocity. Thus, according to the second theorem
\begin{displaymath}
P^\mu = \frac{1}{c}\int T^{4\mu}\,d^3 x
\end{displaymath} (371)

is a 4-vector. It follows from Section 2.23 that we can write $P^\mu = (d{\bfm p}, dE/c)$, where $d{\bfm p}$ and $dE$ are the total momentum and energy carried off by the radiation emitted between times $t=0$ and $t=dt$, respectively. As we have already mentioned, $d{\bfm p} = 0$ in the instantaneous rest frame $S$. Transforming to an arbitrary inertial frame $S'$ in which the instantaneous velocity of the charge is $u$, we obtain
\begin{displaymath}
dE^{'} = \gamma(u) \left(dE + u\,dp^1\right) = \gamma \,dE.
\end{displaymath} (372)

However, the time interval over which the radiation is emitted in $S'$ is $dt'=\gamma dt$. Thus, the instantaneous power radiated by the charge,
\begin{displaymath}
P' = \frac{dE'}{dt'} = \frac{dE}{dt} = P,
\end{displaymath} (373)

is the same in all inertial frames.

We can make use of the fact that the power radiated by an accelerating charge is Lorentz invariant to find a relativistic generalization of the Lamor formula (2.291) which is valid in all inertial frames. We expect the power emitted by the charge to depend only on its 4-velocity and 4-acceleration. It follows that the Lamor formula can be written in Lorentz invariant form as

\begin{displaymath}
P = -\frac{e^2}{6\pi\epsilon_0 c^3} \,A_\mu A^\mu,
\end{displaymath} (374)

since the 4-acceleration takes the form $A^\mu =
(\dot{\bfm u}, 0)$ in the instantaneous rest frame. In a general inertial frame
\begin{displaymath}
-A_\mu A^\mu = \gamma^2\left(\frac{d\gamma}{dt}\,{\bfm u} + ...
...m u}\right)^2 - \gamma^2 c^2\left(\frac{d\gamma}{dt}\right)^2,
\end{displaymath} (375)

where use has been made of Eq. (2.85). Furthermore, it is easily demonstrated that
\begin{displaymath}
\frac{d\gamma}{dt} = \gamma^3\frac{{\bfm u}\!\cdot\dot{\bfm u}}{c^2}.
\end{displaymath} (376)

It follows, after a little algebra, that the relativistic generalization of Lamor's formula takes the form
\begin{displaymath}
P= \frac{e^2}{6\pi\epsilon_0 c^3}\, \gamma^6\, \left[ \dot{\bfm u}^2 -
\frac{({\bfm u}\wedge\dot{\bfm u})^2}{c^2}\right].
\end{displaymath} (377)


next up previous
Next: Radiation losses in charged Up: Relativity and electromagnetism Previous: The electromagnetic field due
Richard Fitzpatrick 2002-05-18