An accelerated charge (say an electron for simplicity) emits photons. Changing direction of movement is an acceleration. Electron-positron collisions are preferably done in linear colliders. But why is this the case? Also in a linear collider the charges are accelerated and should therefore emit photons (which means that they lose energy, something you don't want).

An answer including

  • Feynman graph

  • cross section (why should it be more likely to emit a photon on a bent track than on a straight one?)

  • QED

would be highly appreciated.

  • $\begingroup$ (I have a sense that QED isn't involved/necessary here.) $\endgroup$ – Helen Apr 30 at 18:23

I heard in my particle physics lectures that the reason against circular $e^{\pm}$ accelerators were the huge synchrotron radiation energy losses.
For a particle rotating around a magnetic field line with intensity $B$, one can derive from the Larmor formula \begin{equation} \frac{d E}{d t} = \frac{16 \pi}{3} \left(\frac{Q^2}{m c^2} \right)^2 c \left(\frac{B^2}{8 \pi} \right) \gamma^2 \beta^2 \sin^2\theta \end{equation} where Q is the charge, $m$ the mass of the particle, $\gamma$ and $\beta$ its Lorentz factor and velocity, respectively, $\theta$ is the pitch angle with respect to the direction of the magnetic field lines ($\sin\theta=1$ if the motion is in the plane orthogonal to the magnetic field lines).

Following the suggestion of @Mark H, let us simplify the formula further. Recalling $E = \gamma m c^2$: \begin{equation} \left( \frac{d \gamma}{d t} \right)_{\rm syn} = \frac{16 \pi}{3} \frac{Q^4}{m^3 c^5} \, U_B \gamma^2 \beta^2 \sin^2\theta \end{equation} where I have introduced the magnetic field energy density as $U_B = B^2 / 8\pi$.

As you can see $d\gamma/ dt \propto 1/m^3$, therefore the ratio of energy losses by synchrotron radiations for electrons will be $\approx 6\times10^{9}$ higher than for protons and this poses a severe limit to the circular acceleration of electrons, while the effect is negligible for protons.
My answer does not contain all the specifics you required, but it shows you can derive an explanation from simple electrodynamics (Larmor formula).

  • 2
    $\begingroup$ It's actually worse than you say. The $\gamma^2$ factor gives another $1/m^2$ factor, making the rate of energy loss proportional to $1/m^4$. $\endgroup$ – Mark H Apr 29 at 8:23
  • 1
    $\begingroup$ Thanks, @MarkH, I'll edit the answer later including your comment. $\endgroup$ – cosimoNigro Apr 29 at 8:44

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