# Cosmic ray flux data in steradians

1. Jul 11, 2010

### PBurke1985

Hi guys,

i have previously posted this question in the 'maths' section but had no reply, so i thought i would try my luck here. I'm having problems understanding the units of a cosmic ray flux graph. The cosmic ray flux is given in 'Counts/cm^2 s str', or the counts per cm squared, per second, per steradian. I understand that i have to multiply this graph by the solid angle of my detector, to get the flux in 'Counts/cm^2 s', but this is where i'm having problems. The flux values represent the flux of cosmic rays at the earthâ€™s surface. To calculate the solid angle, i need to divide the area of my detector A, by r^2, which is the radius of the radiating sphere (distance from my detector on the earth's surface to the cosmic ray point source). Since the flux is for cosmic rays, how do i calculate the distance, or radius of the sphere? Am i going about this the wrong way? I also need to do this calculation for solar particles.

Can someone please shed some light on this for me please?

Thanks,

Pete.

2. Jul 11, 2010

### mgb_phys

Not quite
if your detector could only pickup events at the normal then it's solid angle couldn't be calculated. It wouldn't be the radius of the Earth because there is nothing special about the radius of the earth.

The actual solid angle of your detector (assuming you have no telescope) will depend on the structure and thickness of the detector layers.
In the simplest case it would be 180deg in each direction = pi sterad

3. Jul 11, 2010

### Bob S

If your detector were a single thin scintillator, it would count cosmic rays from all directions; i.e., 2pi coming from above. In actuality, your cosmic ray "telescope" is probably an array of maybe four scintillators in tight coincidence (10 or 20 ns). Suppose the scintillators were 10 cm by 10 cm each, and separated by perhaps 30 cm of lead to eliminate accidental coincidences from low energy stuff. So now the "view" seen by your telescope is limited to the cosmic rays that can hit all four scintillators, each of area 100 cm2, separated by 30 cm. Obviously the cosmic rays have to be within about atan(10/30) of the axis of your telescope, or about 0.32 radians of the axis of your telescope. What is the solid angle of that? In actuality, the effective solid angle would have to be determined by an integration of the effective area and the angle of incidence, because only cosmic rays from directly overhead would "see" the full 100 cm2 area.

Bob S

4. Jul 13, 2010

### PBurke1985

Hi guys,
my detector is just a flat CCD (i am trying to determine the cosmic ray background), so based on your replies, the solid angle covered by the CCD, is 2*pi*steradians.

Thanks for the replies.

Pete.