# Motion of a particle from its momentum distribution

• WWCY
In summary, the conversation discusses the concept of using momentum distributions to study the probabilistic motion of a classical free particle. The classical equation of motion for a free particle is given, and the use of Cumulative Distribution Functions and Probability Density functions is explored. The assumption of independence between the distributions of ##pt## and time is questioned, and the definition of ##x_0## as a random variable is clarified.
WWCY
Hi all,

I recently learned the concept of Maxwell's speed distribution and became interested in how to use similar momentum distributions to study the probabilistic motion of a classical free particle. I have done some of my own reading on probabilities and distributions (no formal lessons yet) and with some help, tried to derive a probability density for a classical particle. Could someone look through my work and point out mistakes in my understanding?

The classical equation of motion for a free particle is (##m = 1##),
$$X = x_0 + pt$$
where ##x_0## and ##pt## are random variables.
Let ##F## denote Cumulative Distribution Function and ##f## for Probability Density
$$F_X = \int_{-\infty}^{\infty} F_{pt} (a - y) f_{x_0}(y)dy$$
$$f_X = \int_{-\infty}^{\infty} f_{pt} (a - y) f_{x_0}(y)dy$$
Since this is a classical particle we talk about, then ##f_{x_0}(y) = \delta (y - x_0)## since it has ##P = 1## of being at ##x_0##
This gives,
$$f_X = f_{pt} (a - x_0)$$
If I know the momentum probability density ##f_p##, I could use the fact that
$$F_p = P(p<a) = \int_{-\infty}^{a} f_p(b) db$$
$$F_{pt} = P(pt<a) = P(p<\frac{a}{t}) = \int_{-\infty}^{\frac{a}{t}} f_p(c) dc$$
and do a quick change of variables, compare integrands (both with limits ##-\infty## to ##a##) that would give,
$$f_{pt}(b) = \frac{1}{t} f_p(b/t)$$
Finally,
$$f_X = f_{pt} (a - x_0) = \frac{1}{t} f_p( \frac{a - x_0}{t})$$
Which gives me a probability density of a free particle , ##f_X##

Does the above work? Many thanks in advance!

WWCY said:
The classical equation of motion for a free particle is (##m = 1##),
$$X = x_0 + pt$$
where ##x_0## and ##pt## are random variables.

What does it mean to say "##pt##" is a random variable?

If we treat ##pt## as a single random variable with a distribution independent of time then as time passes the distribution of ##p## would have to vary in order to keep the distribution of the product ##pt## the same. Is that the situation you want to model?

It amounts to saying that the position of the particle (at any given time whatsoever) has the same distribution but the distribution of its velocity (and momentum) changes with respect to time in a special way.

(I'm assuming you don't wish to treat time ##t## as a random variable.)
Let ##F## denote Cumulative Distribution Function and ##f## for Probability Density
$$F_X = \int_{-\infty}^{\infty} F_{pt} (a - y) f_{x_0}(y)dy$$
$$f_X = \int_{-\infty}^{\infty} f_{pt} (a - y) f_{x_0}(y)dy$$

I interpret this to mean:
##F_X(a) = \int_{-\infty}^{\infty} F_{pt}(a-y) f_{x0}(y) dy ##
##f_X(a) = \int_{-\infty}^{\infty} f_{pt} (a - y) f_{x_0}(y)dy##

which makes sense if the distribtion of ##pt## is independent of time and the disribution of velocity is not, but that strikes me as a strange assumption for physical process.
Since this is a classical particle we talk about, then ##f_{x_0}(y) = \delta (y - x_0)## since it has ##P = 1## of being at ##x_0##
What does "##x_0##" denote? Previously, you said ##x_0## denoted a random variable, but now you are treating ##x_0## as some constant.

Perhaps you are expressing the thought that if ##f_X(x_0)## is the probability density for the particle being at a position ##x_0## then it is also the probability density for the particle really being at ##x_0##. It is unnecessary to say this. That is implicit in defining the probability distribution ##f_X(x)## for position.

## 1. What is the momentum distribution of a particle?

The momentum distribution of a particle is a probability function that describes the likelihood of a particle having a certain momentum value. It is often represented graphically as a histogram, where the height of each bar represents the probability of finding a particle with a specific momentum value.

## 2. How is the momentum distribution related to the motion of a particle?

The momentum distribution is directly related to the motion of a particle. It provides information about the range of possible momenta that a particle can have, which in turn affects its velocity and overall motion.

## 3. Can the momentum distribution change over time?

Yes, the momentum distribution of a particle can change over time. This is due to various factors such as interactions with other particles or external forces acting on the particle.

## 4. How is the momentum distribution experimentally measured?

The momentum distribution of a particle can be experimentally measured using techniques such as scattering experiments or particle accelerators. These methods allow for the determination of a particle's momentum and thus its momentum distribution.

## 5. What does the shape of the momentum distribution tell us about the particle?

The shape of the momentum distribution can provide valuable information about the particle's properties, such as its mass and energy. It can also give insights into the interactions and dynamics of the particle within a system.

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