Time dilation and length contraction help

dswatson
Messages
36
Reaction score
0
The average lifetime of a muon is 2.2 microsec. What is the average distance the muon would travel in free space before decaying according to a stationary observer watching the muon travel with a speed of (.60c).

I know how to find the proper time.

2.2/Sqrt[1-(.6c/c)^2] = 2.2/Sqrt[1-.6^2] = 2.2*1.25 =2.75 microsec

thank you in advance for any help.
 
Physics news on Phys.org
Once you've found the proper time you're basically done. You don't have to worry about length contraction in this case because the observer isn't traveling at any speed. The objects agree on how fast they are traveling with respect to each other.

~Lyuokdea
 
It also may be instructive to draw a 1+1 D space-time diagram and see what is the x-value after you shift to the appropriate point on the hyperbola.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top