---
title: "Classical probability density"
chunk: 2/2
source: "https://en.wikipedia.org/wiki/Classical_probability_density"
category: "reference"
tags: "science, encyclopedia"
date_saved: "2026-05-05T13:41:34.829381+00:00"
instance: "kb-cron"
---

== Momentum-space distribution ==
In addition to looking at probability distributions in position space, it is also helpful to characterize a system based on its momentum. Following a similar argument as above, the result is

  
    
      
        P
        (
        p
        )
        =
        
          
            2
            T
          
        
        
          
            1
            
              
                |
              
              F
              (
              x
              )
              
                |
              
            
          
        
        ,
      
    
    {\displaystyle P(p)={\frac {2}{T}}{\frac {1}{|F(x)|}},}
  

where F(x) = −dU/dx is the force acting on the particle as a function of position. In practice, this function must be put in terms of the momentum p by change of variables.

=== Simple harmonic oscillator ===
Taking the example of the simple harmonic oscillator above, the potential energy and force can be written as

  
    
      
        U
        (
        x
        )
        =
        
          
            1
            2
          
        
        k
        
          x
          
            2
          
        
        ,
      
    
    {\displaystyle U(x)={\frac {1}{2}}kx^{2},}
  

  
    
      
        
          |
        
        F
        (
        x
        )
        
          |
        
        =
        
          |
        
        −
        k
        x
        
          |
        
        =
        
          
            2
            k
            U
            (
            x
            )
          
        
        =
        
          
            
              
                k
                m
              
            
            (
            2
            m
            E
            −
            
              p
              
                2
              
            
            )
          
        
        .
      
    
    {\displaystyle |F(x)|=|-kx|={\sqrt {2kU(x)}}={\sqrt {{\frac {k}{m}}(2mE-p^{2})}}.}
  

Identifying (2mE)1/2 = p0 as the maximum momentum of the system, this simplifies to

  
    
      
        P
        (
        p
        )
        =
        
          
            1
            π
          
        
        
          
            1
            
              
                p
                
                  0
                
                
                  2
                
              
              −
              
                p
                
                  2
                
              
            
          
        
        .
      
    
    {\displaystyle P(p)={\frac {1}{\pi }}{\frac {1}{\sqrt {p_{0}^{2}-p^{2}}}}.}
  

Note that this has the same functional form as the position-space probability distribution. This is specific to the problem of the simple harmonic oscillator and arises due to the symmetry between x and p in the equations of motion.

=== Bouncing ball ===
The example of the bouncing ball is more straightforward, since in this case the force is a constant,

  
    
      
        F
        (
        x
        )
        =
        m
        g
        ,
      
    
    {\displaystyle F(x)=mg,}
  

resulting in the probability density function

  
    
      
        P
        (
        p
        )
        =
        
          
            1
            
              m
              
                
                  8
                  g
                  h
                
              
            
          
        
        =
        
          
            1
            
              2
              
                p
                
                  0
                
              
            
          
        
        
           for 
        
        
          |
        
        p
        
          |
        
        <
        
          p
          
            0
          
        
        ,
      
    
    {\displaystyle P(p)={\frac {1}{m{\sqrt {8gh}}}}={\frac {1}{2p_{0}}}{\text{ for }}|p|<p_{0},}
  

where p0 = m(2gh)1/2 is the maximum momentum of the ball. In this system, all momenta are equally probable.

== See also ==

== References ==