Jon Bender
Alan West
Atmospheric Absorption of Muons from Cosmic Rays
A cosmic ray shower occurs when a particle, typically a proton, enters the atmosphere at a speed close to the speed of light. Interaction with an atom or molecule in the atmosphere creates a spray of particles. The new particles rush onward toward the surface of the earth, interact with more atoms and molecules in the atmosphere, and produce even more particles. The process continues and grows until the energy for making new particles is exhausted. After that, the earth’s atmosphere attenuates the number of particles rushing toward the ground. The attenuation of high energy particles depends on two factors: the nature of the particle and the material through which the particle is passing.
The CROP detectors consist of scintillator panels on which photomultiplier tubes (PMTs) have been mounted. The panels produce light which the PMTs turn into an electrical signal whenever a charged particle passes through one of the panels. A muon is the most likely particle in a cosmic ray shower to trigger a CROP detector. Muons do not readily interact. So even if created high in the earth’s atmosphere, it is possible for muons to reach the surface of the earth. On the other hand, since muons are charged particles, they do slow because of Coulomb forces. Given enough material, a muon will eventually stop and be absorbed. To readily compare the absorbing ability of materials of different densities, the interaction length is defined as the density of a material multiplied by the depth of the material through which a particle passes. Lead for example has a large density, while air has a small density. So for a muon that is absorbed in a couple of centimeters of lead, there is a comparable muon which is absorbed only after passing through meters of air. Muons detected at the surface of the earth generally only have to pass through air. But the density of the air in the earth’s atmosphere is not constant. Fortunately, it is still possible to determine the interaction length for materials which have a changing density.
In a fluid model of the earth’s atmosphere, the weight of any disk of air is supported by the pressure difference between the bottom surface of the disk and the top surface of the disk.
r Vdisk g = (Plower - Pupper) A (1)
In the equation r is the density of the air, V is the volume of the disk and A is the surface are of the lower or upper surface – assumed equal. For a disk of thickness dy, dividing the volume of the disk by its area gives:
r g dh = - dP (2)
Using the ideal gas law, the density of the air is given as:
r = P M / ( R T ) (3)
where M is the molar mass of air, R is the ideal gas constant and T is the temperature. Assuming that the temperature and the acceleration due to gravity are roughly constant, combining equations 2 and 3 gives an exponential expression for the pressure or equivalently for the density of the earth’s atmsophere with ro being the atmospheric density at the surface of the earth.
r = ro exp[- M g h /(R T)] (4)
Towards the bottom of the earth's atmosphere the density is great. As you climb up, the density decreases exponentially. To determine the interaction length for the earth’s atmosphere, it is necessary to sum the product of the density times some fixed distance over which the density changes little, beginning from the surface of the earth until one reaches the edge of the atmosphere. For the purposes of this paper, the edge of the earth’s atmosphere was assumed to be 640 km. A BASIC program was written to numerically calculate this sum. (Listing 1) At sea level, the computation yields 1032 g/cm2. As the muons pass through the atmosphere, they lose energy primarily by ionization. Ionizing particles passing through matter such as the air lost about 2 MeV/gm-cm-2 as given by the Particle Data Group. (Figure 1) When that rate of energy loss is multiplied by the interaction length, the result is 2 GeV. Consequently, one would not expect that many muons with an energy below 2 GeV could reach the surface of the earth. In fact measurements indicate that the average energy of muons reaching the surface of the earth is 4 GeV.
Not all cosmic ray showers come from directly overhead. Cosmic rays can come from large angles to the zenith. However the showers must then pass through more of the earth’s atmsophere. The BASIC program was extended to caclulate the interaction length for angles away from the zenith. (Listing 2) The interaction length rises dramatically above 60 degrees. (Figure 2) One notes that the interaction length for showers coming along the horizon is 35000 g/cm2. When that interaction length is multiplied by the rate of energy loss, the result is 70 GeV. That means that muons reaching the detectors from the horizon need significantly more energy than muons from the zenith to reach the detectors.
The graph suggests that the ratio of interaction length at the zenith to the interaction length at the horizon will determine the ratio of cosmic rays arriving from the zenith as compared to those arriving from the horizon. (Figure 3) From December 2000 through February 2001, three detectors stacked in a vertical telescope measured the count rate for triple coincidences. The top detector was 95 cm above the bottom detector. The third was midway between the other two. Over the three month time period during which measurements were made, the detectors counted an average of one thousand six hundred fifty-one counts in five minutes. Equipment problems caused the loss of one detector. So in April and May of 2001, the remaining two detectors were aligned so that they measured cosmic rays arriving from the horizon. The separation distance between the two detectors was two meters. The count rate for cosmic rays from the horizon averaged eighty-nine counts in ten minutes. The count rates for zenith and for horizon cannot be compared directly since the separation distance was different for the two, and the time interval was different for the two. However it is possible to compare the flux ratios. To determine flux, the count rate is multiplied by detector separation squared and divided by the product of the areas of the detectors.
F = r d2/(A1A2) (5)
The zenith flux was 828 counts/(minute-m2-steradian). The horizontal flux was 99 counts/(minute-m2-steradian). The ratio of zenith flux to horizontal flux was found to be 8.4. This result bore checking for many reasons. The zenith measurement used three detectors and the horizontal measurement used only two detectors. The separation distance between detectors was different for the two experiments. The time interval for collecting results was different for the two experiments. The experiments were done at different times of year. There is no guarantee that cosmic ray activity should be the same at different times of the year. To overcome these limitations, a new experiment was designed with two detectors looking towards the zenith and two detectors looking towards the horizon. The time interval for collecting results was the same for each pair. Each pair was separated by the same distance, 1.4 m. Counts were taken simultaneously.
Results yielded an average of 407 counts in 30 minutes for the pair directed toward zenith. The detector pair aimed towards the horizon collected an average of 68 counts in 30 minutes. To ensure there was no bias to one set of detectors, the detectors were switched. The pair facing the horizon were now aimed at the zenith. Similarly, the pair facing the zenith was now directed toward the horizon. When this was done, an average of 548 counts in 30 minutes was recorded for the pair facing the zenith. An average of 59 counts in 30 minutes for the pair facing the horizon. For purpose of comparison with the previous data, the flux was determined for the pairs in the new arrangement. The ratio of zenith to horizon is the first case was 6.4. The ratio for the second case was 8.7. These results compare favorably with the result from the previous experiment which gave a zenith to horizon ratio of 8.4.
The ratio of interaction length for horizon to zenith appears to be thirty five to one. However, the observed ratio is obviously not close. How to account for the discrepancy? Several reasons suggest themselves. The detector pair aimed toward the horizon actually responds to cosmic rays coming from either direction. They can detect cosmic rays coming from both sides. However, the detectors looking towards the zenith can only receive cosmic rays coming from straight above them. Cosmic rays can't travel through the earth. So the horizontal count is too large by a factor of two. That makes the ratio closer to sixteen to one. A second factor is the angle of acceptance. Given that the detectors are 60 cm tall and separated by a distance of 1.4 m, the tangent function gives an angle of acceptance of 23 degrees. (Figure 4) This means that cosmic rays are not just striking the detectors from the horizon, but the detectors are also seeing cosmic rays coming from an angle of up to 23 degrees above the horizon. This again makes the count rate for the horizontal pair higher than it should be. It would be possible to reduce the angle of acceptance by separating the detector pairs by a greater distance. The drawback is that the count rate becomes smaller and the experiment has to run for a longer period of time to give roughly the same number of counts. A third factor concerns the energy spectrum of the cosmic rays. As mentioned previously, most muons arriving at the surface of the earth have an average energy of 4 GeV. Muons from the zenith need 2 GeV to reach the zenith detector pair. Muons which come from the horizon need an energy of 70 GeV to reach the horizon detector pair. The ratio of high energy muons to low energy muons thus affects the count rate. As there are fewer 70 GeV muons than 2 GeV muons, the count rate for the zenith pair will be higher than it would be if equal numbers of muons were created at these two energies. An experiment which accounts for these factors systematically could check the muon energy spectrum for cosmic rays in this manner.
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Listing 1
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| CLS | ‘Clear screen |
| H0 = 8000 | ‘Scale height of atmosphere |
| D0 = 1.29 | ‘Atmospheric density at sea level |
| S = 0 | ‘Set sum to zero |
| FOR h = 0 TO 400000 | ‘Step from sea level to atmosphere edge |
| D = D0 * EXP(-h / H0) | ‘Find interaction length |
| S = S + D | ‘Add to sum |
| NEXT h | ‘Continue |
| PRINT "Sum :"; S | ‘Print result |
| END | |
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Listing 2
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| CLS | ‘Clear screen |
| s = 0 | ‘Set sum to zero |
| pi = 3.141592654# | ‘Define pi |
| d = 1.225 | ‘Density at sea level |
| R = 6378000 | ‘Horizon distance to atmosphere edge |
| h = 400000! | ‘Zenith distance to atmosphere edge |
| FOR a = 0 TO 90 STEP 10 | ‘Step through angles from 0 to 90 degrees |
| theta = a * pi / 180 | |
| top = -R * COS(theta) + SQR(R ^ 2 * COS(theta) ^ 2 + (2 * R * h + h ^ 2)) | ‘Find distance to atmosphere edge at angle |
| PRINT top | |
| FOR x = 0 TO top STEP 10 | ‘Step by 10 meters to top |
| y = SQR(R ^ 2 + x ^ 2 + 2 * R * x * COS(theta)) - R | ‘Find height |
| l = d * EXP(-y / 8000) * 10 | ‘Find interaction length |
| s = s + l | ‘Add to sum |
| NEXT x | |
| PRINT "angle"; a, "interaction length"; s | ‘Print angle and interaction length |
| s = 0 | ‘Reset sum |
| NEXT a | ‘Next angle |
| END | |
Figure 1
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Figure1: The energy loss for muons passing through different absorbers. Taken from Particle Data Group.
Figure 2
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Figure 2: Absorption depth of the atmosphere as a function of angle measured from the zenith.
Figure 3
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Figure 3: The ratio of atmospheric depth at zenith to atmospheric depth at angle measured away from zenith as a function of angle.
Figure 4
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Figure 4: Angle of acceptance for detectors aimed toward horizon. Note that detectors will record coincidences coming from either direction.