# Can we observe proton decay?

I know that the half life of a proton is more than $6.6\cdot 10^{33}$ years (antimuon decay). I have found this data on Wikipedia proton decay but I do not know the probability distribution that leads to this data.

Moreover I know that the universe life has been extimated $t_{univ}\simeq13\cdot10^9$ years.

The universe mass is quoted as $10^{53}$ kg [wikipedia]. And from the stable nuclides table I can extimate that the ratio between neutrons and protons in the universe is about $1.5$

Then assuming that the universe mass is principally given by neutron and protons we have that, approximating neutrons and protons' masses equal to $1.67\cdot10^{-27}$ kg there should be

$$1.5\cdot n_{p^+} +1 \cdot n_{p^+}=\frac{10^{53}}{1.67\cdot10^{-27}}\simeq 5.98\cdot10^{80}$$ so the number of $p^+$ in the universe ($n_{p^+}$) should be $$n_{p^+}\simeq 2.39\cdot 10^{80}$$

Now, applying to the proton's decay the radiactive decay law:

$$N(t) = N_0\,e^{-{\lambda}t}$$

where $N_0=2.39\cdot 10^{80}$ and $\lambda$ can be determined from the half life time:

$$\lambda= \frac{\ln(2)}{t_{1/2}}$$

Since the exponent becomes

$$\frac{\ln(2)}{t_{1/2}}\cdot t_{univ}\simeq 1.36\cdot10^{-24}$$

and so I obtain a number of proton that decaded during universe life equal to $3.26\cdot10^{56}$

This number is about $1.36\cdot10^{-24}$ the number of $p^+$ in the universe so it should be not impossible for us to observe the $p^+$ decay during the total universe life.

This reasoning is very very primitive (too primitive). Has anyone a suggestion to improve it?

• "I know that the half life of a proton is $6.6\cdot10^{33}$ years (antimuon dacay)." Er ... I believe you know that the current lower experimental limit on the proton lifetime by that mode has that value. – dmckee --- ex-moderator kitten Oct 14 '14 at 3:14

1mol of protons is $6\times10^{23}$ particles. If proton decay has a half life of $6\times10^{33}$ years, then there should be one proton decaying per year per $10^{10}$mol. Since one mol of protons weighs one gram, that's a mere $10^7$kg of protons. Even at the density of liquid hydrogen (70kg/m$^3$), that's only $1.4\times10^6$m$^5$ of liquid volume. Take the third root and you end up with a tank of roughly 50m dimensions. That's not an easy experiment to instrument, but certainly nowhere close to what we have already done for neutrino measurements and cosmic ray showers.
The number you cite has, by the way, been measured experimentally in water Cherenkov detectors, so it has already been done. The current limit seems to be above $10^{34}$ years and I would expect us to push that by probably another three or four orders of magnitude over the next century, if necessary.
• @Caos: I was just doing a quick back of the envelope. We could write $N(t)=N_0exp(-t/{\tau})$. Then we get $dN(t)/dt=-N_0/{\tau}exp(-t/{\tau})$. For t=0 this is $dN(0)/dt=N_0/{\tau}$. $\tau=t_{1/2}/\ln{2}$, so I was neglecting a factor of $\ln{2}\approx 0.69$. – CuriousOne Oct 13 '14 at 21:00