Angular diameter of neutrino emission

[martinhiggs]

Homework Statement

If the Superkamiokande neutrino detector could accurately determine the direction of detected neutrinos, what would the angular diameter, in arcminutes, be of the neutrino emission as seen from Earth.

Assume all the neutrinos are $$^{8}$$B neutrinos with energies <14.02MeV.

Homework Equations

$$\delta = 2tan^{-1}(\left \frac{0.5 d}{D} \right)$$

Where d = visual diameter
D = distance to the object.

The Attempt at a Solution

D = The distance from the sun to the Earth = 1.5x10^8 km.

I can't work out what the value of d would be, and how I can use the value of neutrino energies in working this out.

Is there a different equation that I can use?

 

[mark726]

I would first clarify some details about the problem. Are we assuming that the Superkamiokande detector is located on Earth, and that it is detecting neutrinos emitted from the sun? If so, we can use the equation provided to calculate the visual diameter of the neutrino emission as seen from Earth.

To do this, we need to know the distance from the sun to the Superkamiokande detector, which I will assume is negligible compared to the distance from the sun to Earth. We also need to know the distance from the sun to the neutrino emission point, which I will assume is also negligible compared to the distance from the sun to Earth.

Using the equation, we can calculate the angular diameter of the neutrino emission as seen from Earth:

d = 2tan^{-1}(\left \frac{0.5 (1.5x10^8 km)}{1.5x10^8 km} \right) = 2tan^{-1}(0.5) = 53.13 arcminutes

Therefore, the angular diameter of the neutrino emission as seen from Earth would be approximately 53.13 arcminutes.

As for the value of d, it represents the visual diameter of the neutrino emission, which is the diameter of the circle that would encompass the entire emission. Since we don't have any specific information about the size of the neutrino emission, we can't accurately calculate this value. However, we can assume that it would be very small compared to the distance from the sun to Earth, hence why we can ignore it in our calculations.