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Cosmic Ray Detector - Data Center Guided Tour Stop 4

A model of cosmic ray production in the upper atmosphere

The cosmic rays we measure at sea level are not really "cosmic". In fact, they are mostly muons created in the upper atmosphere by "real" cosmic rays -- protons from outer space. These protons themselves hardly ever make it to the surface of the earth, because they interact with the nuclei of air molecules while they are still high above the earth's surface. When a high-energy proton interacts with a nucleus, a great many different processes may take place, almost all of which result in the creation of "secondary" particles. With high probability, many of the secondaries will be pions, a type of meson with a very short life time, so they decay almost immediately. Pions can be neutral (p°) or charged (p+ and p-, one being the antiparticle of the other). The neutral pions decay preferentially into a pair of photons, which in turn can create other particles. The charged pions decay preferentially into muons through the reactions:

p+ -> m+ + nm
(positively charged pion decays into an antimuon and a muon-like neutrino)

and

p- -> m- + `nm
(negatively charged pion decays into a muon and muon-like antineutrino).

The resulting "relativistic" muons have two properties that are of interest.

  1. They have a much longer life time than pions. A muon at rest has a mean life time of about 2 msec.
  2. They do not interact very much at all, and can penetrate through the air fairly easily and arrive at the surface of the earth.

The "cosmic rays" we measure at sea level are almost all muons. The neutrinos do not have electric charge and interact only through the weak force, and therefore they cannot be detected in most  detectors (including ours). In the following we will try to predict the angular distribution of cosmic rays (that is, the relative number of cosmic rays arriving from different directions in the sky), using a simple model. We will then use the actually measured values to check this prediction.

diagram of the earth, earth's crust, h, R, x, and theta.

The model

To make the calculation manageable without the use of computers, we will make a few simplifying assumptions. They are (see the diagram):

  1. The earth is a sphere of radius R.
  2. The atmosphere is a spherical shell of air which extends to a height h above the surface of the earth.
  3. Protons from outer space come in equally from all directions.
  4. When they reach the outer edge of the earth's atmosphere, they interact immediately.
  5. The (final) interaction products are all muons.
  6. The muons all go in the same direction as the initial proton.
  7. The muons all have the same energy Em.
  8. The muons do not interact at all with the air: they do not change direction by scattering and they do not lose energy.

None of these assumptions are exactly correct, but most of them are approximately correct. The last two of them, however, need some further discussion. The muon energies are, of course, not all the same. In fact, there are some muons with very large energies, but they are rare. There is a definite peak in the muon energy spectrum between 2 and 3 GeV. Furthermore, muons do lose energy in the atmosphere, in fact, they lose quite a bit. The energy loss is a second reason (the first is that they decay) why there are more muons at high altitudes than at sea level. But this primarily affects the lowest energy muons: the higher the muon energy, the less difference it makes whether they lose some energy. Muons also change direction when they scatter off nuclei in the atmosphere, although for the most part the change in direction is small.

This demonstrates something about models: a model is a simplified version of a physical system, constructed to gain insight into the physical processes that may cause a certain phenomenon. Often models need to be adjusted later to incorporate more detail or new knowledge to give a better description of the system. We will see where this particular model gets us.

Let us now consider this model. If muons didn't decay, then equal numbers of muons would come in from all directions. But muons do decay, and so muons that travel further have a higher probability to decay before reaching our detector than ones with less distance to travel. Muons decay mostly into electrons or positrons and neutrinos as described in the following reactions:

m+ -> e+ + `nm + ne

and

m- -> e- + nm + `ne.

The electrons and positrons are absorbed almost immediately in the atmosphere and do not reach our detector. As said before, the neutrinos are not detectable by most detectors.

We can make some guess as to what should happen. It is clear from the picture above that a muon arriving at our detector at some angle (q) travels further (distance x) than a muon arriving from directly above (distance h). Therefore, muons arriving at an angle will have had more time to decay, so there will be fewer of them.

Distance of travel

Using the law of cosines, we can derive the following formula for the distance x that a muon travels after being created in the collision of a proton with the atmosphere:

Formula: x(theta)=R cos(90 degrees + theta) + square root of (Rsquared sin squared theta + 2Rh + h squared)

which we can rewrite (since cos(90º+q)=-sin(q)) as:

Formula: x(theta) = R -sin theta + square root of (R squared sin squared Theta+2Rh+h squared)

To derive this, consider the triangle made up of the legs x, R and R + h, and use leg R + h as leg "a" in the law of cosines (the angle between legs R and x is 90º+ q).

Relativity

The muons have a lot of energy, in fact they have a lot more energy than their own rest energy mm c2. According to the Particle Data Tables, the mean energy, Em, of cosmic ray muons at sea level is about 2 GeV. The muon rest mass, mm, is 105.6 MeV/c2.

Using the well-known relation from relativity theory:

Em = g mm c2

it follows that the typical value of g for these muons is roughly 20 [see the calculation]. With such a high g, muons travel at a velocity v which is very close to the speed of light (Formula: v/c = square root of (1-1/gamma squared)).

Now we can use the relation for time dilation:

t = gt

where t is the "proper" time (i.e. the time as measured in the rest frame of the muon) to calculate the "apparent" mean life time of a cosmic ray muon in the earth's rest frame as 20 x 2 = 40 msec. Earlier, though, we measured that the apparent life time of a muon was only 12 msec. Something is not quite right, either we told you the wrong value for the muon energy or our model has a significant flaw. We will return to this point later.

Muon decay and angular distribution

Muons decay just like any other radioactive substance. For this reason, the number of muons left over after an elapsed time T is given by:

Nm(T) = Nm(T = 0)e(-T / t)

Using the above it follows that in our model the angular distribution for cosmic rays, relative to q= 90°, is given by:

N(q)/ N(90°) = e(-[x(q) - h] / [v g t])

Comparison with measured data 

To evaluate our model, we need to plug in some numbers. Using h = 20 km and R = 6378 km, we can calculate the value of the angular distribution given above for the angles 45° and 0°relative to the value at 90°:

N(45°)/N(90°) = 0.503  (formula)

N(0°)/N(90°) = 2.7 x 10-18 (formula)

In Tour Stop 1, we measured the following values:

N(90°) = 15.0 +/- 0.5

N(45°) = 7.23 +/- 0.35

N(0°) = 0.515 +/- 0.065

where we divided the 0 degree number by a factor of two to account for the fact that it measures muons coming in from both sides. We can convert these numbers to the following ratios:

N(45°)/N(90°) = 0.482 +/- 0.028 (measured)

N(0°)/N(90°) = 0.0345 +/- 0.0045 (measured)

Discussion

At first sight, it seems that these values compare pretty well to the values we calculated from the formula: the value at 45 degrees is within the measured uncertainty the same as the calculated value, and the 0 degree value is close to zero. But really, the value at 0 degrees is nowhere near as small as it should have been according to the calculation. One reason for this is that at angles near 0 degrees, the flux rapidly changes and because of the size of our scintillator panels, we accept muons from angles other than (but close to) 0 degrees. For example, if we repeat the calculation for 15 degrees, we get:

N(15°)/N(0°) = 0.0096 (formula)

which is fifteen orders of magnitude bigger than at 0 degrees but it still is not in very good agreement with our measurement.

So we conclude, then, that our simple model accounts for the data pretty well for angles between 90 and 45 degrees, but fails when the angle is near horizontal. It is possible that our model could be improved by taking into account the fact that muons lose energy while they travel through the atmosphere. Remember in Tour Stop 3, we measured a muon apparent mean life time of 12 msec, where we expected the mean life time to be closer to 40 msec. If this difference is entirely due to energy loss it would likely make a big difference in our model! Also, muons scatter in the atmosphere and change direction. Therefore, muons produced from protons coming in at larger angles may travel horizontally by the time they get to the detector.

But there are some other factors that should be taken into account. Our model has one obvious parameter: the height of the atmosphere h. We put it at 20 km, without much ado, but this choice really needs some justification. From the measurements aboard the KAO, it seems clear that there are a lot of muons at 41,000 ft, so it makes no sense to have h be less than about 12 km. We also know that the air pressure changes rapidly with altitude, and we know that there is little air left above 100 km, but that still leaves a wide range of choices for h. Of course, in reality, muons are not produced at a fixed distance in altitude at all, so it would probably be more productive to attempt to modify the model to take that (and the varying air pressure with altitude) into account.

Another factor is that muons come with a wide range of energies, not just the one fixed energy we assumed (2 GeV/c2, giving a g of 20). And since muons do lose energy, the value of g actually decreases as they travel through the atmosphere.

It is clear that if we want to take all these facts into account, our model would become a bit complicated, and more suited for a computer-based simulation. Computer simulations of cosmic rays are, in fact, done for a large number of experiments, although most concentrate on very-high energy cosmic rays. See, for example, the following link : Milagro Animations.