Re: HOW IS HALFLIFE OF A SUBATOMIC PARTICLE MEASURED

Halflife is a measure of the amount of time it takes for half the number of
particles in a given sample to disintegrate.  The disintegration process
happens exponentially in time via the following equation:

N(t) = N(0) * exp(-Rt)

where N(0) is the number of particles you started with, N(t) is the number
of particles at time t, R is the disintegration rate, and t is time.  To
figure out the halflife, you say that half the sample has disintegrated,
rearrage the equation in the following way and solve for the time:

N(t)/N(0) = 1/2 = exp(-Rt[halflife]), or ln(1/2) = -R t[halflife]

so, 

t[halflife] = 0.693/R

So ideally, you have a sample of particles that you watch, and count the
number of disintegrations as a function of time.  You plot them on a
semilog plot, and figure the slope.  The slope is the rate of
disintegration R, and you plug it into the above formula for halflife.  

But, you ask, how do you measure the half life of a particle that won't sit
still (like the muon)?  Muons are energetic little electron-like particles
that don't interact with matter easily.  They stream from outer space and
hit us continuously (thousands of them a second).  Luckily, they don't
interact with us much, and pass right through mostly.  So how do we even
get to see them?  And how do they get to a detector on Earth if they only
last for 2x10^-6 seconds anyway?  Einstein's theory of relativity states
that things that go really fast experience time dilation.  So if the
particle is going fast enough, its lifetime can be stretched on
indefinately (remember the "twin paradox" problems stated in most physics
text books).  So the muons that we see are going really fast, and have
stretched out lifetimes.  That means that they can last until they hit
something and interact with it, slowing the muon down.  As soon as the muon
slows down, its lifetime is no longer stretched, and it quickly decays.  So
how do we measure the muon half life?  

	 
	/
       /
      /
      muon
 ___________
(__________() <-scintillator 1
(__________() <-scintillator 2
XXXXXXXXXXXXX
XXXXXXXXXXXXX
XXXXXXXXXXXXX <-stopping crystal scintillator
XXXXXXXXXXXXX
XXXXXXXXXXXXX
 ____________
(___________() <-scintillator 3

We set up a detector called a coincidence-anticoincidence NaI stopping
crystal detector.  The coincidence-anticoincidence (scintillators 1, 2, and
3) part ensures that the muons are coming from one direction (up).  When a
muon hits the upper coincidence scintillator set (1 and 2), a timer is
started.  The muon then goes into the stopping crystal.  Some number of the
muons will stop inside the crystal.  The crystal is also a scintillator, so
when the muon finally stops, a flash of light will occur.  The timer stops
when the muon stops in the crystal.  The time is recorded, and the timer is
reset.  A typical setup registers 1 muon per second.  The apparatus records
the times for several thousand events.  Then, the data are plotted
semilogarythmically; number of events versus time.  The slope of this line
gives you the muon decay rate.  You then can solve for the muon half life,
and you get the published number, 2.2x10^-6.  

You might wonder about the time the muon spent getting to your apparatus. 
As far as the muon is concerned, practically no time at all had passed
since its creation and its eventual stop in your detector.  Particles from
space hit our atmosphere, and kick out pions.  The pions usually decay into
muons high in our upper atmosphere.  The muons produced are very fast, and
therefore their time spent going close to the speed of light adds little to
the total lifetime of the particle.  The experimental errors are larger
than this small effect, so it is usually ignored anyway. 

I hope this answers your question!

-Fred Niell