Is it right that circular particle accelerators use magnetic fields to deflect the particle beam?

Using the simple equation of "a charged particle in a magnetic field":

  • $\vec{f}=q\vec{v}\times\vec{B}$

Say the particle accelerators is currently used for a proton beam. $\vec{f}$ will be directed to the center of the particle accelerators (centripetal force).

The proton itself is constituted of two up quarks and one down quark, which have electric charge:

  • $\frac{2}{3}e$ for one up quark.
  • $-\frac{1}{3}e$ for one down quark.

Now if I look at the effect of the magnetic field at quark level. The force is:

  • $\frac{2}{3}\vec{f}$ for one up quark (centripetal force).
  • $-\frac{1}{3}\vec{f}$ for one down quark (centrifugal force).

I imagine it as a centrifugation of the proton, the down quarks are pushed to the exterior side of the accelerator when the up ones are attracted to the center.

Is this effect real and taken into account in particle accelerators?

What about the mass ? The down quark is at least 2 times heavier than the up quark, does the rotation of the proton "separate" the quarks of the proton?

  • $\begingroup$ You have a neat imagination, but it's completely wrong in this case. Nothing even remotely related to that happens at an accelerator facility. Indeed, the protons won't even notice the acceleration in the accelerator, it's much too gentle to do anything at all about the inner workings of a proton, maybe with exception of beam polarization. $\endgroup$
    – CuriousOne
    Sep 21, 2015 at 8:19

1 Answer 1


The centripetal acceleration that the protons feel as they circulate in the LHC is roughly:

$$ a = \gamma^2 \frac{v^2}{r} $$

This is the usual equation for centripetal acceleration but multiplied by a factor of $\gamma^2$ to allow for the time dilation the protons experience. The speed $v$ is approximately $c$. The radius of the LHC is about 4.3km but the bending radius is sharper thna this because the protons are deflected only at certain points around the ring. The bending radius is about 2.8km. Finally, the Lorentz factor is about 7000 (the ratio of the proton enery to its rest mass). Feed all these into my equation and you get:

$$ a \approx 1.6 \times 10^{21} \,\text{m}\,\text{s}^{-2} $$

This seems an awfully large number, but what we're really interested in is the potential energy change across the width of a proton, and this is given by:

$$ U = m\, d\, a $$

where $m$ is the mass of the object, $d$ is the width of the proton and $a$ is the acceleration we've just calculated. The width of a proton is easy as that's around $1.7 \times 10^{-15}$ m. The mass is harder because we have to decide what object we're moving. Let's take the mass to be 1% the mass of a proton or about $1.7 \times 10^{-29}$kg. This is a bit heavier than the bare mass of the up and down quarks but it's of the same order of magnitude. Anyhow, if we feed these into the equation above, along with our calculated value for $a$, we get:

$$ U \approx 5 \times 10^{-12} \,\text{J} \approx 0.0002 \,\text{eV} $$

And 0.2meV is utterly insignificant compared to the energies inside the proton, so we can be confident that the protons don't get centrifuged by their passage round the ring.

A footnote: honesty compels me to admit I'm not sure if the equation for $a$ should contains a factor of $\gamma^2$ or $\gamma^3$ (though it makes no difference to the conclusion anyway). For linear acceleration you need $\gamma^3$ as you get a factor of $\gamma$ for each power of $t$ due to the time dilation and another factor of $\gamma$ due to the length contraction. For motion in a circle I'm not sure whether length contraction would introduce another factor of $\gamma$. Any contributions on this point are gratefully received.

  • $\begingroup$ Thanks the response and to regroup the needed equations to solve this. I just realize that I can't just integrate the acceleration to get the variation of speed between 2 points. I'm stuck with newtonian physics. Also, that probably explain why only a small energy is needed to deflect a high energy particle. (In my mind, to deflect a 1TeV particle so it perform a half of a turn in the LHC you need 2TeV of energy). $\endgroup$
    – Orace
    Sep 21, 2015 at 12:25
  • $\begingroup$ @Orace: You don't always need energy to alter an object's course. Take the ISS, it's also in a circular orbit. It's orbital energy is quite a bit more than 1 TeV ;). Gravity makes it orbit, but clearly there's no energy source powering gravity. $\endgroup$
    – MSalters
    Sep 21, 2015 at 14:32
  • $\begingroup$ If it helps, I got $\gamma^2$ too. $\endgroup$
    – Javier
    Sep 21, 2015 at 15:52
  • $\begingroup$ @MSalters But the ISS is clearly moving in a straight line (in a curved spacetime). $\endgroup$
    – Kyle Oman
    Sep 21, 2015 at 16:59
  • $\begingroup$ Would it be reasonable to say that the accelerator does indeed cause a centrifuge like effect due to magnetic forces, it's just far too small to be anything but negligible? When it comes to "no" answers in physics, I'm always interested in the difference between "no, we'd never reasonably measure such a thing," and "no, the equations don't actually do what you think they do." $\endgroup$
    – Cort Ammon
    Sep 21, 2015 at 17:45

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