Phase space

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ID:(1659, 0)



Diagram in momentum-position space $p-q$

Definition

An analytical technique to analyze motion involves representing momentum as a function of the position of a moving object. This representation enables us to study how momentum evolves with respect to the achieved position.

The representation of motion in the momentum-position space $p-q$ allows us to analyze the displacement's evolution, highlighting extremes in position and momentum.



In the case of periodic motion or when considering the journey both ways, it can be represented as:



Furthermore, it's evident that the area enclosed by the curve

$\displaystyle\int_{q_1}^{q_2} p dq = \displaystyle\int_{v_1}^{v_2} m v dv = \displaystyle\frac{1}{2} m v_2^2 - \displaystyle\frac{1}{2} m v_1^2$



corresponds to the system's energy.

The area enclosed by the curve in the momentum-position diagram $p-q$ corresponds to the system's energy.

ID:(1240, 0)



Phase space

Storyboard

Variables

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$E_k$
E_k
Energy of a spring system
J
$E_G$
E_G
Energy of a system with general gravity
J
$E_g$
E_g
Energy of a system with gravitational acceleration
J
$m_g$
m_g
Gravitational mass
kg
$k$
k
Hooke Constant
N/m
$m_i$
m_i
Inertial Mass
kg
$M$
M
Mass of the celestial body
kg
$p$
p
Moment
kg m/s
$p$
p
Momentum and Path
kg m/s
$\vec{s}$
&s
Posición (vector)
m
$s$
s
Posición camino aleatorio
m
$V$
V
Potential Energy
J
$K$
K
Space Time Position
J
$E$
E
Total Energy
J

Calculations


First, select the equation:   to ,  then, select the variable:   to 

Symbol
Equation
Solved
Translated

Calculations

Symbol
Equation
Solved
Translated

 Variable   Given   Calculate   Target :   Equation   To be used



Equations

Given that the kinetic energy is

equation=4425

and the potential energy is

equation=12551

we can express the energy with the radius denoted by the variable $q$ as

equation

In the case where the kinetic energy surpasses the potential energy at the starting radius and the energy is positive (indicating that the object can escape the planet), the equation can be written as

$1 = \left(\displaystyle\frac{p}{\sqrt{2mE}}\right)^2 - \displaystyle\frac{GmM}{Eq}$



which simplifies to

$y=\pm\sqrt{1+\displaystyle\frac{1}{x}}$



with

$x=\displaystyle\frac{q}{GmM/E}$

, and

$y=\displaystyle\frac{p}{2mE}$



In the case where the kinetic energy does not exceed the potential energy (indicating that the object cannot escape the planet's attraction), the energy is negative, and the expression is written as

$1 = -\left(\displaystyle\frac{p}{\sqrt{2mE}}\right)^2 + \displaystyle\frac{GmM}{Eq}$



where $E$ is the absolute value of the energy. With the definitions of $x$ and $y$, we then have

$y=\pm\sqrt{\displaystyle\frac{1}{x}-1}$

The kinetic energy as a function of momentum is given by

equation=4425

and the potential energy as a function of height is

equation=3246

So, if we express the elongation as the position

$x = q$



we obtain

equation

Since kinetic energy is equal to

equation=3244

and momentum is

equation=10283

we can express it as

$K_t=\displaystyle\frac{1}{2} m_i v^2=\displaystyle\frac{1}{2} m_i \left(\displaystyle\frac{p}{m_i}\right)^2=\displaystyle\frac{p^2}{2m_i}$



or

equation

As the kinetic energy as a function of momentum is

equation=4425

and the potential energy as a function of height is

equation=3245

so if the height is expressed as the position

$h = q$



we get

equation

Since in general, energy is the sum of kinetic energy

equation=4425

and potential energy U, we can express it as

$E=\displaystyle\frac{p^2}{2m}+U$



If we solve for momentum, we get the following expression:

equation


Examples

An analytical technique to analyze motion involves representing momentum as a function of the position of a moving object. This representation enables us to study how momentum evolves with respect to the achieved position.

The representation of motion in the momentum-position space $p-q$ allows us to analyze the displacement's evolution, highlighting extremes in position and momentum.



In the case of periodic motion or when considering the journey both ways, it can be represented as:

image

Furthermore, it's evident that the area enclosed by the curve

$\displaystyle\int_{q_1}^{q_2} p dq = \displaystyle\int_{v_1}^{v_2} m v dv = \displaystyle\frac{1}{2} m v_2^2 - \displaystyle\frac{1}{2} m v_1^2$



corresponds to the system's energy.

The area enclosed by the curve in the momentum-position diagram $p-q$ corresponds to the system's energy.

The kinetic energy of a mass $m$

equation=3244

can be expressed in terms of momentum as

kyon

If the energy is solved for the momentum, the expressions for positive and negative momentum are obtained:

kyon

For the case of a particle in the gravitational field of the Earth, the energy as a function of momentum $p$ and position $q$ is given by

kyon

The equation can be written in an adimensional form as

$y=\pm\sqrt{1-x}$



with

$x=\displaystyle\frac{q}{mg/E}$

, and

$y=\displaystyle\frac{p}{\sqrt{2mE}}$



which is represented below

image

For the case of a mass oscillating with a spring, the energy as a function of momentum $p$ and position $q$ is

kyon

The equation can be expressed in an adimensional form as

$1=y^2 + x^2$



where

$x=\displaystyle\frac{q}{\sqrt{2E/k}}$

, and

$y=\displaystyle\frac{p}{\sqrt{2m_iE}}$



solving for y, we obtain

$y=\pm\sqrt{1-x^2}$



Its representation in the xy plane is shown below

image

For the case of a mass in a gravitational field, the energy as a function of momentum p and position q is given by

kyon

The equation can be written in an adimensional form for the case of positive energy as (blue and green curves)

$y=\pm\sqrt{1+\displaystyle\frac{1}{x}}$



and for the negative energy case as (red and purple curves)

$y=\pm\sqrt{\displaystyle\frac{1}{x}-1}$



with

$x=\displaystyle\frac{q}{GmM/E}$

, and

$y=\displaystyle\frac{p}{2mE}$



which is depicted below

image


>Model

ID:(1659, 0)