Exponential decay question. fundamental

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Muons decay with a mean lifetime of about 2 microseconds, leading to an exponential decay plot rather than a Gaussian distribution. This is because the decay process is random, and the exponential function accurately describes the probability of decay over time. After 2 microseconds, a significant portion of the muons will have decayed, but some will still remain due to the nature of the mean lifetime. The discussion emphasizes the importance of understanding probability theory in predicting the number of remaining muons. Overall, the exponential decay model effectively captures the behavior of muon decay over time.
rjsbass
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this is just something has been bugging me for the last few days. it seems like it has a very basic solution.

Muons decay randomly, but have a mean lifetime of about 2 us. If I plot the # of muons that decay vs. time (say the axis spans from 0 to 20 us), why is the plot exponential decay? shouldn't it be a gaussian distribution centered around the mean lifetime?
 
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got it. textbooks ftw
 
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Suppose you had a pile pile of newly minted muons. After 2 μs, how many would you expect to have left?
 
the majority of them will be gone but some will still remain?
 
You should be able to give a numerical answer from probability theory. Remember, the lifetime is a mean value.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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