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?
 A: 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.
