Probability is inverse of velocity?

  • #1

Main Question or Discussion Point

Dear all,

Intuitively, the probability of finding a trapped particle at a given location x should be proportional to the inverse of the velocity of the particle. But for even the simplest of systems, such as 1D harmonic potential SHM, the velocity is zero at the two end points (+A & -A) of the motion. This makes 1/v(x) not defined for x=+A & -A. How is this intuitively possible according to classical mechanics only?

On a side note, I am actually trying to find rdot as a function of r or t for a central force problem with potential proportional to r^3. Anyone tackled such a problem? :)

Cheers.
 

Answers and Replies

  • #2
jtbell
Mentor
15,484
3,248
Intuitively, the probability of finding a trapped particle at a given location x should be proportional to the inverse of the velocity of the particle.
More precisely, the probability density at x is proportional to 1/v(x).

But for even the simplest of systems, such as 1D harmonic potential SHM, the velocity is zero at the two end points (+A & -A) of the motion. This makes 1/v(x) not defined for x=+A & -A. How is this intuitively possible according to classical mechanics only?
In order to get an actual probabilty of finding the particle in the range a < x < b, you must integrate the probability density between x = a and x = b. I expect you will find that even though the probability density approaches infinity as x approaches +A or -A, an integral which includes one or both of those points as a limit nevertheless converges.
 
  • #3
412
2
Dear all,

Intuitively, the probability of finding a trapped particle at a given location x should be proportional to the inverse of the velocity of the particle.
What tells you this? I'm not sure if there is some principle or anything as such.. but if you look at it, why do you consider that the probability of a desired quantity needs to be inversely proportional to only it's first order rate of change. In the case you presented, the places where velocity was zero, the second order rate of change i.e. acceleration is maximum.

Other than that, the case is of a continuous probability distribution. In that case, the probability that the object will be at any point is always zero. It is only between an interval that you can find the probability of occurrence.

P.S: I like ur nickname :D
 
  • #4
In order to get an actual probabilty of finding the particle in the range a < x < b, you must integrate the probability density between x = a and x = b. I expect you will find that even though the probability density approaches infinity as x approaches +A or -A, an integral which includes one or both of those points as a limit nevertheless converges.
That makes sense. :) It might end up that I do not have an analytic expression for the velocity. I wonder if a numerical integration of the inverse velocity would yield converging results... guess I will need to play around in Matlab.

Thanks for your help.
 
  • #5
What tells you this? I'm not sure if there is some principle or anything as such.. but if you look at it, why do you consider that the probability of a desired quantity needs to be inversely proportional to only it's first order rate of change. In the case you presented, the places where velocity was zero, the second order rate of change i.e. acceleration is maximum.
The probability density most definitely depends only on the velocity. There are no higher order terms. Think of the radial distance probability density function of a circular motion problem. It is of course a delta-function at a radius R, this is because the magnitude of the radial velocity is zero while the radial acceleration is finite.

P.S: I like ur nickname :D
Thanks. It came to me in my sleep. :)
 
  • #6
111
2
On a side note, I am actually trying to find rdot as a function of r or t for a central force problem with potential proportional to r^3. Anyone tackled such a problem?
[tex]
dr/dt = \sqrt{2/m}\left(E - V(r) - \frac{L^2}{2mr^2}\right)^{1/2}
[/tex]

Put in whatever potential you wish for V(r). L is the angular momentum.
E is the total energy.
 
Last edited:
  • #7
[tex]
dr/dt = \sqrt{2/m}\left(E - V(r) - \frac{L^2}{2mr^2}\right)^{1/2}
[/tex]

Put in whatever potential you wish for V(r). L is the angular momentum.
E is the total energy.
Yes I have gotten that far. But as you can see from the equation, there are regions where dr/dt is imaginary. Physically, the particle is forbidden to enter these regions. That is, there exists a [tex] r_{min}[/tex] and a [tex]r_{max} [/tex] given by solving:

[tex]
E - V(r) - \frac{L^2}{2mr^2} = 0
[/tex]

Trying to find these has been the problem. I used the symbolic toolbox in matlab and I can get solutions where the potential V(r) is proportional to r^-1, r, r^2, r^4. But when I put in the case I am interested in, r^3, it says that a closed form solution can't be found. Do you have idea of how one goes about solving this? Maybe I should make a new post on it.
 
  • #8
111
2
Do you have idea of how one goes about solving this?
Yes indeed. I have done this thousands of times for many different potentials V(r) in connection with a program that computes classical trajectories to simulate chemical reactions. For most potentials there is no closed-form solution, so you have to solve for the classical turning points numerically. For each turning point you need to do the following:

1) You need a rough estimate of the root of the equation.
2) Then you need to bracket the root.
3) Once the root is localised, you must use an algorithm that homes in on the solution. I used Newton's method in those cases where the derivative of V(r) wrt r was known; otherwise, I used the bisection method (binary chop), which is slower but safer. There are also canned programs using combined strategies.

Look at the book "Numerical Recipes" by Press, Flannery, Teukolsky, and Vetterling.
 

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