Probability of finding a particle at X?

[Chuckstabler]

Hey all,

Here's my question : Given X is the position of a particle at time T, how would I go about finding the probability of finding said particle at any given position if i randomly pick a time out of the interval T1 to T2?

Let's say that my X(t) = cos(t). How can I find the probability of observing the value of X(t) to be equal to some value X if i randomly select a value for time from my interval T1<T<T2?

This is kind of inspired by quantum physics; in particular the time independent solutions to the shrodinger wave equation for a harmonic oscillator. Thanks :)

 

[mfb]

Integrate from T1 to T2?
For a classical motion with a non-zero velocity, this leads to a sum over $\displaystyle \frac{1}{v_i}$ where v_i are the velocities at times i where the particle crosses the position you are looking at.

 

[Chuckstabler]

Integrate what from T1 - T2? X(t)? 1/V(t)? What I'd ultimately want would be a function of X that gives me the probability.

Kinda confused, sorry :/

 

[mfb]

Integrate the probability to find the particle at position x at time t from T1 to T2 (and divide by (T2-T1) to normalize it properly). With an exact position that needs distributions to do it properly, but with the result I posted above.

It is probably easier with the example:
X(t) = cos(t)
X'(t) = -sin(t)
Let's say T1=0 and T2=2 pi and we are interested in the probability to find the particle at x=0. The particle crosses this point twice, at 1/2 pi and 3/2 pi. The derivative there is -1 and 1 respectively. The sum I suggested (forgot to take the absolute value) gives $\frac{1}{|-1|} + \frac{1}{|1|}$, dividing it by 2 pi gives 1/pi.
The probability to find $x=\sqrt{3/4}$, following the same steps, is $\frac{1}{2\pi} \left( \frac{1}{|-1/2|} + \frac{1}{|1/2|} \right) = \frac{2}{\pi}$.