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