Not sure on topic - relativity if I had to guess

[pat666]

Homework Statement

Pions have a half-life, at rest, of 1.77 x 10-8 s. A collimated beam of pions leave an accelerator at 0.95c. What distance from the accelerator will the intensity be half?

Homework Equations

The Attempt at a Solution

as you can probably tell from the title I really need some help getting started (not even sure on the topic), so that's problem 1. One thing I have been able to ascertain is that a pion is anyone of three subatomic particles, thanks google&wiki! The question gives the half life at rest, that suggests to me that that(half life) is when the energy will be half of the original but there are two reasons why I can't see any use for that:
1. they are not at rest. does that change the half life?
2. I don't want to know when the energy is half I want to know when the Intensity is half and remembering sound stuff (not sure if it translates to this?) the two are not on a linear scale, ie. if energy goes down by half intensity does not necessarily?

Intensity is watts/m^2 ( I think) so I=E/t/A so if I knew the initial energy I suspect I could easily find intensity.

Sorry about rambling a bit but I find that that helps people understand where I'm coming from and then help be "better".
Thanks for any Help...

 

[Doc Al]

In the rest frame of the pions, their half-life is 1.77 x 10-8 s. What would an observer in the lab frame, who sees the pions moving at 0.95c, measure as the half-life? (Hint: Time dilation!)

 

[pat666]

AHHHHHHHHHHH, thanks Doc!
I found gamma to be 3.20256
then delta t0 =1.77*10^-8/gamma
so the "new" half life is 5.5268*10^-9 s that I think is the relativity part done but I am still unsure of what to do next regarding intensity?

 

[Doc Al]

I found gamma to be 3.20256

Good.

then delta t0 =1.77*10^-8/gamma
so the "new" half life is 5.5268*10^-9 s

Careful! The 'rest-frame' half-life is the proper time. Think of the pions as little moving clocks.

To an observer in the lab, will the pions take a longer or a shorter time to decay, compared to what you'd see in the rest frame of the pions?

that I think is the relativity part done but I am still unsure of what to do next regarding intensity?

Once you have the correct half-life as observed in the lab, you can figure out how much time it takes for the beam to reach half intensity (how many half-lives is that?). Then it's just a kinematics problem: Distance = speed X time.

 

[pat666]

I actually had gamma*1.77*10^-8 = 5.66*10^-8s first but then I changed it - I figured that if the half life was a at rest then it would have to get faster if it were accelerated.
Is the half intensity x half lifes that I could just look up which I did and I get alpha and beta half lifes and nothing simple and relevant to this?

Thanks.

 

[Doc Al]

I actually had gamma*1.77*10^-8 = 5.66*10^-8s first but then I changed it - I figured that if the half life was a at rest then it would have to get faster if it were accelerated.

No, moving clocks run slow.

Is the half intensity x half lifes that I could just look up which I did and I get alpha and beta half lifes and nothing simple and relevant to this?

You don't need to look anything up--they give you the half life. Hint: After how many half lives will the intensity be 1/2 the original?

 

[pat666]

Hint: After how many half lives will the intensity be 1/2 the original?

The only thing that I can think of is one half life as that would be half the energy? but as I said in my 1st post I don't think that 1/2 energy means 1/2 intensity?

 

[Doc Al]

In this context, by 'half the intensity' they just mean half the particles are left. (Don't confuse this with the intensity and amplitude of an electromagnetic wave.)

You're thinking way too much!

 

[pat666]

ok so it will just be .95c/5.66*10^-8s = 0.5m ?

 

[Doc Al]

ok so it will just be .95c/5.66*10^-8s = 0.5m ?

Distance = speed X time.

 

[pat666]

so it does - 16.3m?

 

[Doc Al]

so it does - 16.3m?

That looks better.

 

[pat666]

Thanks for all your help Doc - really appreciate it.

 

[CompuChip]

Note that the "classical" calculation 0.95c x 1.77 10^-8 would only give a distance of about 5 meters, so the relativistic effect is really important here.
In fact, because of it, we can detect particles (like muons) that are created in our atmosphere from cosmic rays, that "classically" should have decayed long before they reach the surface of the Earth due to their short half-life time.