# Why is it better to accelerate electrons in a linear collider?

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).

• 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.

• (I have a sense that QED isn't involved/necessary here.) – 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 $$$$\frac{d E}{d t} = \frac{16 \pi}{3} \left(\frac{Q^2}{m c^2} \right)^2 c \left(\frac{B^2}{8 \pi} \right) \gamma^2 \beta^2 \sin^2\theta$$$$ where Q is the charge, $$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$$: $$$$\left( \frac{d \gamma}{d t} \right)_{\rm syn} = \frac{16 \pi}{3} \frac{Q^4}{m^3 c^5} \, U_B \gamma^2 \beta^2 \sin^2\theta$$$$ where I have introduced the magnetic field energy density as $$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.
• 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 $1/m^4$. – Mark H Apr 29 at 8:23