particle physics - Does an electron accelerated in a linear accelerator lose any energy?

In a circular one, it would lose energy due to bremsstrahlung (synchrotron emission), but if it is accelerated from rest to ~GeV energies, does it lose any energy in this acceleration process?

Answer

According to Larmors' formula in Lorentz invariant form the power loss $P$ of a accelerated particle is given by (cgs-units are used, $e$ is the electron charge, $c$ speed of light, $m_0$ the rest mass of the electron):

$$ P = - \frac{2}{3}\frac{e^2 c}{(m_0 c^2)^2}\left[ \frac{dp^\mu}{d\tau} \frac{dp_\mu}{d\tau}\right] $$

where $\tau$ is the proper time, and $p^{\mu}= (E/c, \vec{p})$ are the 4 components of the 4-momentum (E is the energy of the electron and $\vec{p}$ its 3-momentum). If the 4-momentum change rate is written out in its components we get:

$$ P = \frac{2}{3}\frac{e^2c}{(m_0 c^2)^2}\left[ \left(\frac{d\vec{p}}{d\tau}\right)^2 - \frac{1}{c^2} \left( \frac{d E}{d\tau}\right)^2\right] $$

Using $\frac{dE}{d\tau} = v \frac{dp}{dt}$ and replacing the proper time of the particle by the observer time $t$ by $d\tau = \frac{1}{\gamma}dt $ gives ( $\frac{1}{\gamma} = \sqrt{1-\beta^2}$ with $\beta =\frac{v}{c}$):

$$ P = \frac{2}{3}\frac{e^2c \gamma^2}{(m_0 c^2)^2}\left[ \left(\frac{d\vec{p}}{dt}\right)^2 - \frac{v^2}{c^2} \left( \frac{d p}{dt}\right)^2\right]\,\,\, (*)$$

In case of a linear acceleration we can set $\vec{p} = p\cdot \vec{e}_p$ where $p$ is the length of the momentum vector and $\vec{e}_p$ its time-independent (!) unit direction vector ($\frac{d \vec{e}_p}{dt}=0$). Then we get:

$$ P = \frac{2}{3}\frac{e^2c}{(m_0 c^2)^2}\left[ \gamma^2(1-\beta^2) \left( \frac{d p}{dt}\right)^2\right]$$

The term $ \gamma^2(1-\beta^2) =1$.

The parameter which in linear accelerators measures the acceleration typically is the energy gain per length, i.e. $\left(\frac{dE}{dx}\right)$ which is equal to: $\frac{dE}{dx} = \frac{dp}{dt}$. With this in mind we get:

$$ P = \frac{2}{3}\frac{e^2c}{(m_0 c^2)^2} \left(\frac{dE}{dx}\right)^2 $$

The energy gain per meter in a typical linear accelerator is $15$MeV/m, with this value one gets a radiation power in order of $10^{-17}$W (changing the formula to SI-units). So the answer is: Yes, principally there is energy loss in a linear accelerator, but it is so small, that it can be neglected right away.

In the opposite case one is only interested in the radiation on a circular path, term $\frac{dp}{dt}$ can be set to zero in formula (*), as the length of the momentum vector does not change. However we can use:

$$ \left(\frac{d\vec{p}}{dt}\right)^2 = \vec{F}^2 = (m\cdot a)^2 = \left(\frac{mv^2}{R}\right)^2 $$

where for $a$ the centripetal acceleration $a= \frac{v^2}{R}$ was used.
If however the centripetal acceleration is plugged in we will finally get the well-known formula for the radiation power in a circular accelerator which is tremendously much larger than in case of the linear accelerator.