2
$\begingroup$

So the inner tracker system of a particle detector (say CMS) can detect charged particles because they ionize particles in the detector. These inner tracker systems are made of silicon pixels and silicon strips. As pointed out by the quantum diaries blog, this is the same technology as is used in a digital camera.

Only, a digital camera can see neutral particles, crucially photons, otherwise it's not a very good camera. It does this via indirect ionization, from Compton scattering (edit; for future readers, this very confident statement is incorrect, see https://physics.stackexchange.com/a/646780/147600). By contrast, photons are not visible in the inner tracker of a particle collider, see here.

I'm sure that if an inner tracker could be built to detect photons it would be, because the inner tracker greatly improves the accuracy of the vertex finding. So there is some reason that isn't possible. We know the inner tracker is being hit by some very hard photons, so it's not a question of the photons being less ionizing than those detected by a camera.

Perhaps whatever a camera does to make photons detectable is not radiation hard, and so cannot be used here. It seems unlikely that it is too bulky, the inner tracker is measures in cm, and a phone camera is measures in mm. It could have too long a deadtime between successive hits. Alternatively, whatever a camera does to detect photons has a high stopping power, it's very opaque, and it would shield the rest of the detector from radiation. But these ideas are just my speculation.

It appears the answer to this is so obvious that nobody bothers to put it in their review/report/paper, which makes it a little embarrassing to ask, by why can't inner trackers see photons? If you had a citation for the cause I'd be very grateful to have that too.

$\endgroup$

2 Answers 2

1
$\begingroup$

The interaction cross-section for charged particles in matter is generally larger than the interaction cross-section for photons. (This is why it's much easier to shield against charged-particle radiation like alpha and beta particles than it is to shield against gamma radiation with the same energy.) So if you have a detector which is optimized to collect a reasonable number of ion pairs when charged radiation passes through it, that same detector will have a much smaller charge from a photon on the same trajectory.

Furthermore, above a few mega-eV, the primary interaction mechanism for photons is the production of electron-positron pairs by scattering from the electric fields near nuclei. The fate of a high-energy photon in dense matter is to turn into a shower of charged secondary radiation, in a Molière cone. A "photon calorimeter" is a detector which is long enough and wide enough to collect all of the energy from this shower; any thin tracking detector is going to interact much more with the charged-particle component of the shower than with the photon component.

For what it's worth, "Compton scattering" usually refers to the interaction between a photon and a free (or quasi-free) electron. The cross-section from Compton scattering for photons in matter is largest when the photon energy is around 100 keV. Compton scattering is different from the photoelectric effect, where the photon interacts with the entire ocean of electrons in the crystal lattice. (As a handwaving argument that collective behavior matters, consider that the work function for a silicon crystal is only about half the ionization energy for a single silicon atom.) What's happening in a CCD is yet again a different thing, where an eV-scale photon moves a single electron across the energy gap from the valence band to the conduction band.

I recommend you read the Particle Data Group's reviews on the passage of particles through matter and on the various kinds of particle detectors. I wrote this answer looking at figure 34.15. I re-read those reviews every couple of years, and I learn something new every time.

$\endgroup$
1
  • $\begingroup$ So if I'm understanding this correctly, a detector that was likely to interact with photons would lead to lots of showering in the detector? I can see why that would be unhelpful in the tracker. Also, thanks for pointing out my confusion around the Compton effect/photo-electric effect, I didn't know that. $\endgroup$
    – Clumsy cat
    Jun 20, 2021 at 15:15
1
$\begingroup$

inner tracker system of a particle detector

Trackers are designed in HEP experiments in order to measure tracks of charged particles coming from the main interaction. In CMS this is achieved among other things of having many pixels with known geometry as close to the interaction point as possible with a large enough volume to be able to see the bending of tracks in the superposed magnetic field, in order to measure the momenta of the charged tracks.

Note the emphasis on charged. A charged track through a medium of atoms ,interacts with the electrons of the atoms ,ionizing them consecutively,leaving a footprint of its passing. A photon coming from the same interaction region, has very small probability of even one scatter with an atom, exactly because it is neutral and should really hit the atom to interact with it, whereas the field of a charged track gives it a much larger interaction probability.

This is clear in the bubble chamber photo of an interaction of much lower energy than the LHC, but the tracks are formed by the ionisation of the hydrogen atoms in the chamber , by the charged particles. The neutral $K^0$ leaves no track, but its decay products do.

kaon enter image description here

Production and decay of $K^0$ in the reaction $K^- p$ to $K^0$ to $π^- p$

This event shows an interaction of an 8.2 GeV/c particle with a proton in the CERN 2 metre hydrogen bubble chamber.

A neutral particle is produced which decays into one positive and one negative particle with a characteristic 'vee’ pattern. Measurements are required to show that this is a a $K^0$.

It is educational to peruse the stored images in the link, there is a picture for photons but not very clear.It shows no "track" until they pair produce an electron positron.

So the answer is that the detector is designed in such a way that charged particles will leave a pixel footprint (instead of the ionisation dots in the bubble chanmber) and the mass of the detector is too small for photons to have more than a very very small probability to interact, even in one pixel, so they cannot leave a track footprint.

btw the photons of the digital camera are in the range of eV,(not GeV as measured in the CMS detector), where single photons interact with the lattice.

See how complicated, compared to low energy bubble chamber events , the CMS LHS events are.

$\endgroup$
3
  • $\begingroup$ So if I am reading you correctly, you don't think it would be impossible to record photons/neutral particles in the inner tracker, you think they chose to design it not too? I understand the importance of the tracker on measuring charged tracks but it also makes big difference for vertex reconstruction accuracy, which would be a great advantage for photons too. I'm not convinced this is a design choice. $\endgroup$
    – Clumsy cat
    Jun 20, 2021 at 15:10
  • $\begingroup$ I thought I make clear that neutral particles cannot leave footprings like charged ones. If they interact they interact drasitcally at a single point with a lot of debris if it is high energy neutrals, so cannot leave a "track" $\endgroup$
    – anna v
    Jun 20, 2021 at 15:17
  • $\begingroup$ No sorry, that bit didn't come through. I suppose it should have been clear from the behavior of photons in the hadron calorimeter. $\endgroup$
    – Clumsy cat
    Jun 20, 2021 at 15:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.