# Classical position/velocity probability

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1. Nov 3, 2015

### j1m1

Probability to find a particle in some region of space is inversely proportional to velocity particle has in that region of space.
Let's say we have two cases: one particle has velocity given by v(t)=v0*Cos(w*t), and other by v(t)=v0-v1*Cos(w*t), (v0>v1).
Since particle spends more time in regions of low velocity this should imply that probability to find a particle with low velocity is bigger than to find it with high velocity . For the first case probability to find a particle with velocity around v0 should be equal to probability to find a particle with velocity around -v0. In the second case the probability to find a particle with velocity around v0-v1 should be much bigger that to find it with velocity v0+v1, but on the other hand v(t) is distribution of velocity of a particle in time, and from this it looks that probability to find a particle with velocity around v0-v1 should be equal to probability to find a particle with velocity around v0+v1.
All opinions appreciated.

2. Nov 3, 2015

### andrewkirk

It will be bigger, but not necessarily much bigger.

I don't follow your reasoning here. Can you try to explain in more detail why you think this?

3. Nov 4, 2015

### j1m1

I think probability in low velocity region should depend on v0-v1 (the closer they are together, the longer the time particle spends in low velocity region).
My reasoning is: if you look at v(t)=v0-v1*Cos(w*t), particle's velocity oscillates around v0, and Cos function is symmetric in t, meaning in one full period it will be as much positive as it will be negative,and in the same way. If I were to choose random time t, just by looking at this velocity distribution, I would have equal probability that particle has velocity around v0+v1 as v0-v1.

4. Nov 5, 2015

### andrewkirk

The cos function is symmetric, but the function $t\mapsto v(t)$ is not symmetric because of the asymmetric velocity v0, and it is that function that determines the likelihood of finding a point in a region, not just the Cos function.

5. Nov 5, 2015

### j1m1

Ok I think I understand now. Thank you for clarifying.