# Sound from cosmic rays

1. Sep 15, 2007

### SpitfireAce

this isn't HW, it is more for myself, I don't want answers or links to arvix papers... im stuck and I just need something to help me proceed... a 10^20 ev proton strikes the atmosphere a km up and creates a vertical down to the ground, the energy is evenly distributed along this vertical, assume all the energy is converted to sound, up to what distance will this sound be audible... here's where I am... I think I want to find the intensity and see what it is at some distance away from the vertical... Intensity=Power/Area... for area I use (2r)(pi)(h)+(2pi)(r^2)... is this right? now I don't know how to get the power, do I just do Energy over time where the time is the time it takes the particle to travel a km, or do I use some other equation for power (I looked through some... I found it in terms of sound pressure, and then the sound pressure is in terms of particle velocity, angular momentum, or displacement, im assuming this refers to the air particles, and im not sure how to get any of these from energy) Any help would be greatly appreciated, it doesn't have to be eloquent, anything that I can use! I wanted to do this by myself but searching for reliable explanations of formulas on the internet is so frustrating =(

Last edited: Sep 15, 2007
2. Sep 15, 2007

### Loren Booda

An air shower caused by one cosmic ray, can be "many kilometers wide."

The cosmic ray of energy E=1020 eV equates to 1.6 x 108 ergs. Take the area impinging on Earth to be A=100 km2, i.e., 1012 cm2. The pressure on Earth, having been mediated through interactions in an atmospheric cone (V=Ah/3, where V is its volume and h is its effective height, ~10 km=106 cm), is much less even than the the following consideration, which assumes an efficiency of 100% for energy cascading:

p=3E/Ah=5 x 10-10 dyne/cm2

3. Sep 15, 2007

### SpitfireAce

Thanks a lot Loren =)
according to you're calculations, the sound should still be audible (by ear) to someone about 5.5 km away from the vertical.... which sounds too good, even with the 100% energy to sound transfer... is there some way to make this more exact?

4. Sep 16, 2007

### Loren Booda

At f=1000 Hz, defining zero decibels as z=10-12 W/m2 and with p=5 x 10-10 dyne/cm2, I figure the sound intensity in decibels to be:

log10(pf/z)=27 dB

barely audible, even at 100% efficiency and in the most audible range.

5. Sep 16, 2007

### Loren Booda

Just one correction, the last equation should read

10log10(pf/z)=27 dB

6. Sep 17, 2007

### SpitfireAce

wouldn't the particles be going faster than sound and emitting sonic booms and such... I can't find one equation that describes this

7. Sep 18, 2007

### Loren Booda

A guess.

The slight density gradient of the upper atmosphere would dissipate particles' energy gradually since their equivalent phonon wavelength is negligible to intermolecular distances there.