# Measuring Neutrino Mass from a Supernova

• touqra
In summary, the speed of a neutrino is given by:10 MeV neutrinos: Speed = \frac{100 000}{(t+10)}50 MeV neutrinos: Speed = \frac{100 000}{t}

#### touqra

I tried and I got a ridiculous large number for the neutrino mass. I basically used $$E=\gamma m c$$ and then, for 10 MeV neutrinos, time taken is t + 10 seconds to reach Earth, for 50 MeV neutrinos, time taken is t seconds.
$$Speed = \frac{100 000}{(t+10)}$$ for 10 MeV and $$Speed = \frac{100 000}{t}$$ for 50 MeV. Here's the question:

Neutrinos were detected from a supernova (distance ~ 100 000 light years) over a time period of ~ 10 seconds. The neutrinos have an energy range from 10 MeV to 50 MeV.
What's the mass of the neutrino, assuming all neutrinos were produced at the same time and their mass is small compared to their energy.

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Your speed is in LY/sec. Did you convert units?
Speed in LY/Y is so close to one that you have to use a Taylor expansion.

Meir Achuz said:
Your speed is in LY/sec. Did you convert units?
Speed in LY/Y is so close to one that you have to use a Taylor expansion.

Yes, I did unit conversions. A light year = $$9.46 \times 10^{15} meters$$

Here's what I did:

for 10 MeV neutrinos, $$10\;Me = \frac{m\;c}{\sqrt{1 - (\frac{d}{t+10})^2/c^2}}$$

and hence, $$t+10 = \frac{10\;M\;e\;d}{10\;M\;e\;c}(1-\frac{m^2\;c^2}{10^2\;M^2\;e^2})^{-1/2}$$

Binomial expansion, keep only first two terms yield:
$$t+10 = \frac{d}{c}(1+\frac{m^2\;c^2}{2\times10^2\;\M^2\;e^2})$$

do the same for 50 Mev neutrinos,

for 50 MeV neutrinos, $$50\;Me = \frac{m\;c}{\sqrt{1 - (\frac{d}{t})^2/c^2} }$$

simultaneous equations, cancelling t, yield:

$$10 = \frac{d\;m^2\;c}{2\;M^2\;e^2}(\frac{1}{10^2}-\frac{1}{50^2})$$

Solving for m (neutrino mass), I got 7.7 GeV !?

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I get: 10 sec=(10^5 yr X m^2/2)(1/100-1/2500), with m in MeV.
I don't see where your e^2 came from, why your c is i the numerator.
Use 1LY/c=1 year, keep all masses in MeV.
Use t=(L/c)\sqrt{m^+p^2}/p~L/c-(m^2/2E)(L/c).

You also must take into account how long the neutrino sees itself moving as compared to what the Earth sees as how long the neurtino is moving from a to b

## 1. How can neutrino mass be measured from a supernova?

Neutrino mass can be measured from a supernova by observing the neutrinos emitted during the explosion. Neutrinos have a very small but nonzero mass, and this mass affects the way they travel through space. By analyzing the patterns of neutrino interactions with detectors on Earth, scientists can infer the mass of these elusive particles.

## 2. Why is it important to measure neutrino mass from a supernova?

Measuring neutrino mass from a supernova can help us better understand the fundamental properties of these particles and the processes involved in a supernova explosion. It can also provide insights into the nature of dark matter and the evolution of the universe.

## 3. What techniques are used to measure neutrino mass from a supernova?

The most common technique is to use large underground detectors, such as the Super-Kamiokande detector in Japan, to observe the neutrinos emitted from a supernova. These detectors use a variety of methods, such as scintillation and Cherenkov radiation, to detect and measure the energy and direction of the neutrinos.

## 4. Can neutrino mass be directly measured from a supernova?

No, neutrino mass cannot be directly measured from a supernova. Neutrinos have a very small mass and do not interact strongly with other particles, making them difficult to detect and measure. Instead, scientists must use indirect methods to infer the mass of neutrinos from a supernova.

## 5. Have scientists successfully measured neutrino mass from a supernova?

Yes, scientists have successfully measured neutrino mass from a supernova. In 1987, a supernova explosion called SN1987A was observed in the nearby Large Magellanic Cloud galaxy. The neutrinos emitted from this explosion were detected by multiple detectors on Earth, providing valuable data for measuring neutrino mass.