Compute the Flux of Neutrinos

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tony873004
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Compute the flux of neutrinos arriving at Earth, i.e. the number of neutrions that land on each square meter of Earth's surface each second.

The question is referring to neutrinos created in the photon-photon chain in the Sun.

I already computed that the Sun releases [tex]1.78*10^{38}[/tex] neutrinos per second in the photon-photon chain.

So 1 square meter, if face-on to the Sun at Earth's distance should intercept[tex]\frac{1}{4 \pi r^2}*1.78*10^{38} [/tex] neutrinos per second, where r is 149598000000 the radius of Earth's orbit in meters.

So, 1 square meter, if face-on to the Sun at Earth's distance should intercept [tex]6.33*10^{14} neutrinos/s[/tex]

I have a feeling that this is the answer the teacher will consider correct. However, not all square meters on Earth's surface are face-on to the Sun. Only square meters where the Sun is directly overhead will receive a full dosing of [tex]6.33*10^{14} neutrinos/s[/tex]. Square meters of Earth's surface where the Sun is setting or rising should receive 0 neutrinos (consider the Sun a point or a whole new can of worms is opened!)

All points inbetween will receive anywhere between 0 and [tex]6.33*10^{14} neutrinos/s[/tex].

But the change from minimum to maximum not a linear function, so I can't just average it. What would I use to compute the average flux? Can I avoid integrating? This class is supposed to avoid Calculus, but I have a feeling the integration in this case is easy since it is related only to SIN.

Any thoughts?...
 

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  • #2
dextercioby
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I think it's correct.A square meter of surface situated on avg at 1 AU gets each second [itex]6.33\cdot 10^{14} [/itex] electronic neutrinos.

Daniel.
 
  • #3
tony873004
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Thanks, dextercioby. Do you have any idea about figuring out the average for a square meter on Earth, being that not all square meters are directly facing the Sun. Some are nearly edge-on and would receive no neutrinos.
 
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dextercioby
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It's a pretty complicated problem indeed.What u did was to assume both Earth & Sun as pointlike object,which is a pretty darn good approximation...

Daniel.
 
  • #5
jtbell
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tony873004 said:
Do you have any idea about figuring out the average for a square meter on Earth, being that not all square meters are directly facing the Sun. Some are nearly edge-on and would receive no neutrinos.
The average number of neutrinos per square meter equals (the total number of neutrinos that strike the earth) / (number of square meters on the side of the [spherical] earth that faces the sun).

To calculate the numerator from the number/m^2 that you've already calculated, imagine what the earth "looks like" from the sun's vantage point.
 
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dextercioby
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Yeah,Jtbell,it looks like an ellipse.

Daniel.
 
  • #7
tony873004
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jtbell said:
The average number of neutrinos per square meter equals (the total number of neutrinos that strike the earth) / (number of square meters on the side of the [spherical] earth that faces the sun).

To calculate the numerator from the number/m^2 that you've already calculated, imagine what the earth "looks like" from the sun's vantage point.
I thought about doing it that way. Treat the Earth as a circle and not a sphere, then compute the circle's cross section against the surface area of a sphere of radius 1AU. Then divide by the surface area of a sphere with radius 1 Earth radii. I think that will give me the right answer.

Neutrinos also strike the side of the Earth that faces away from the Sun since they easily pass through the Earth. So I'm wondering whether or not I should multiply my answer by 2?
 
  • #8
jtbell
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tony873004 said:
Neutrinos also strike the side of the Earth that faces away from the Sun since they easily pass through the Earth.
Those are just the same neutrinos as entered on the side facing the sun, of course. You might want to ask your instructor for clarification, but if it were me, I wouldn't want to count the same neutrinos twice.
 

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