Is it feasible to measure the energy of cosmic ray muons with a consumer Digital Single Lens Reflex camera?

Question

I have read this article

SIBBERNSEN, Kendra. Catching Cosmic Rays with a DSLR. Astronomy Education Review, 2010, 9: 010111.

and it talks about estimating the muon cosmic ray flux by means of a DSLR camera.

Is it possible to measure the muon's energy with the same apparatus?

If yes, how can I "calibrate" (?) the camera in order to measure the energy?

This post imported from StackExchange Physics at 2014-04-25 16:59 (UCT), posted by SE-user uvts_cvs

Answer 1

The short answer is you can't, or at least not at all easily.

Your detector has only a single detection plane, and almost all muons are minimum ionizing, so you get essentially the same energy deposition from every muon (well, there is a factor from the angle of incidence the detection plane).

The usual mechanism for measuring the energy of a particle are

• For charged particles, radius of curvature in a magnetic field. This will work for muons, but you need multiple detection planes and a high field. (This really gets momentum, so you need some kind of PID as well.)

• Calorimetry of stopping particles. You need a detector with a lot of mass-per-unit-area in the direction of motion (and preferably segmented). For muons that'll have to be a lot of mass-per-unit-area.

• Time of Flight between multiple detection planes. You need multiple detection planes. (Gives you velocity, so you need PID.)

• Ring imaging of Cerenkov radiation (RICH). I suppose you could try this. You'll need a modified lens and a very sensitive CCD. Not sure how you are going to trigger it, though. (Velocity again.)

• There are a few special tools like transition radiation. Not something I know much about.

With cosmic rays the bulk of the impinging particles are muons, so you can just assume the species instead of doing a proper job of PID.

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A RICH modification would presumably have a single element bi-planer lens made out of some very clear glass or plastic and have the lens cover permanently attached (because you don't want outside light). You'll choose the index of refraction and distance from the focal plane according to the size of the focal plane detector and the slowest muon you want to image.

I don't know enough about modern digital cameras to know if they can be self-triggering. Maybe it is enough to simply take timed exposures without triggering and do a counting experiment.

This is interesting enough that I'm going to look into it further.

This post imported from StackExchange Physics at 2014-04-25 16:59 (UCT), posted by SE-user dmckee

Comments

What is a PID in the context you are using it?

This post imported from StackExchange Physics at 2014-04-25 16:59 (UCT), posted by SE-user Chris Mueller

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@Chris Particle IDentification. Determining the species of the particle responsible for a track.

This post imported from StackExchange Physics at 2014-04-25 16:59 (UCT), posted by SE-user dmckee

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@dmckee Thank you for your answer! I am afraid I can't afford a custom lens for a home made RICH detector :-)

This post imported from StackExchange Physics at 2014-04-25 16:59 (UCT), posted by SE-user uvts_cvs

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@dmckee What about a rough energy measure? After all I have very dark pixels (due to dark current) and the bright pixels (not saturated) due to muons... I already have a sort of relative measure, isn't it?

This post imported from StackExchange Physics at 2014-04-25 16:59 (UCT), posted by SE-user uvts_cvs

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"What about a rough energy measure?" @uvts_cvs Over a fairly wide range of energies muon deposit roughly the same amount of energy in ionization per unit mass penetrated ($\approx 2 \,\mathrm{MeV/g/cm^2}$. To a reasonable approximation low and high energy muons will excite the pixels by the same amount. So, mostly no. Sorry. You can find some useful background in the chapter on "Passage of radiation Through Matter" in the Particle Data Book (including what you'd need to design a RICH lens).

This post imported from StackExchange Physics at 2014-04-25 16:59 (UCT), posted by SE-user dmckee