Calculating Muon Speed Using Lorentz Transformations | Relativity Help

[vsage]

A muon is an unstable elementary particle with an average lifetime of 2.20×10-6 seconds (from the moment of creation until it decays) as measured by an observer at rest with the muon. If an average muon travels a distance of 900 meters during one lifetime, according to an observer in the laboratory, what is the muon's speed (in m/s)?

I have the Lorentz Transformations sitting right in front of me but I can't seem to make the logical jump again. If O is the observer's frame and O' is the muon's frame of reference I have the following data:

x = ?
t = 2.2*10^6s
x' = 900m
t' = ?

I really don't remember where to go from here. I tried taking the ratio of x to t and x' to t' the Lorentz transformations but it didn't yield anything relevant. Thanks in advance.

 

[vsage]

I tried using the fact that t' = $$\gamma$$t and x' = x/$$\gamma$$ and v = x/t to yield v = x'$$\gamma$$/t but the answer came out imaginary. What am I doing so wrong?

 

[Doc Al]

Think in terms of the time and distance intervals between the birth and death of the muon. Here's what you are given:
$$\Delta t = ?$$
$$\Delta x = 900$$m
$$\Delta t' = 2.2*10^6$$s
big hint:
$$\Delta x' = 0$$

Now use the LT to find $\Delta t$, at least in terms of v. Then realize that $\Delta x/\Delta t = v$. Solve for v.

 

[Doc Al]

I tried using the fact that t' = $$\gamma$$t and x' = x/$$\gamma$$ and v = x/t to yield v = x'$$\gamma$$/t but the answer came out imaginary. What am I doing so wrong?

Careful of those "facts". While it's true in this case that $\Delta t = \gamma \Delta t'$, $\Delta x \ne \Delta x' / \gamma$. Use the full LT; don't take shortcuts (until you have more experience and know when to use them).

 

[vsage]

Well I got the right answer with your advice (thanks btw) but I don't understand why $$\Delta$$x' = 0 still. It should be obvious but I apparently forgot all of relativity.

 

[Doc Al]

Well I got the right answer with your advice (thanks btw) but I don't understand why $$\Delta$$x' = 0 still. Does that mean it was observed to have barely moved in the time interval or what?

Remember that the primed coordinates are observations made in the "moving" frame attached to the muon. How far does the muon move in its own frame?

 

[vsage]

Argh I see now. I completely switched the perspectives. For some reason I thought the decay time was measured from an observer standing on Earth or something. Thank you!