# I Why are position and velocity independent variables in a phase space?

1. Aug 25, 2016

### Happiness

Why do we take a particle's position $x$ and its velocity $\dot{x}$ as independent variables in a phase space when they are dependent in the sense that given the function $x(t)$, we can get the function $\dot{x}(t)$?

I'm thinking they are independent variables but not independent functions. Does this make sense?

2. Aug 25, 2016

### Lucas SV

I would say the variables are independent by assumption. $x$ and $v$ are both defined to be functions of time, and the equation $\dot{x}=v$ becomes a constraint that must hold in order for $v$ to correspond to the physical velocity when $x$ is interpreted as the position of a particle.

3. Aug 25, 2016

### Happiness

If we drop the assumption that $x$ and $v$ are independent, would we run into any problems?

Say, suppose we define the variable $v$ as $v=2x$. We then plot the $x$-axis as the horizontal axis and the $v$-axis as the vertical axis. Is there anything wrong with plotting two dependent variables as two perpendicular axis?

We say $x$ and $v$ are independent, is it because $v=2x$ is true only for a particle and may not be true for another particle? In this case, is there a mathematical definition for independence, say, in the form of an equation, similar to the one below for the definition of linear dependence of 2 vectors?

Last edited: Aug 25, 2016
4. Aug 25, 2016

### Lucas SV

In the usual particle mechanics phase space is defined as a vector space. More precisely the phase space of a classical particle in 3 dimensionals is a 6 dimensional vector space. This is because the state of a particle at time $t$ is described by 3 position and 3 velocity components. In general phase space is the space of states or configurations of the system.

However a space is not sufficient to describe physics. One needs laws and interpretations. One interprets the motion of a particle as a path in phase space parametrized by time, where each point in the path is the state at a certain time. The laws governing the system are Hamilton's equations. They will tell you which path is the actual path of the particle in phase space.

It is true that the axis for position and momentum are orthonormal by definition. So yes, you can think of the 'position' unit vector and 'momentum' unit vector in phase space as linearly independent in the sense of linear algebra.

5. Aug 25, 2016

### Happiness

So can I say that the position unit vector and the momentum unit vector in a phase space are linearly independent, but the position function $x(t)$ of a particle and the velocity function $v(t)$ of a particle are not linearly independent? And that it is nonsensical/ambiguous to say position $x$ and velocity $v$ are dependent or independent?

6. Aug 25, 2016

### Lucas SV

Oh, I said position and momentum out of practise, since these are the fundamental variables in quantum mechanics. In classical physics they are prety much the same except for the mass factor. You can replace momentum by velocity in everything I said. Phase space is defined as the 6 dimensional vector space which contains the position and velocity unit vectors as an orthonormal basis. So in phase space position and velocity are linearly independent (the momentum-position and velocity-position phase spaces are equivalent by scaling).

7. Aug 25, 2016

### Happiness

Alright, what I get from your replies is that it makes no sense to just say $x$ and $v$ are dependent or independent. Because $\hat{x}$ unit vector and $\hat{v}$ unit vector in phase space are linearly independent, whereas $x(t)$ function and $v(t)$ function are dependent. So technically, we have to be clear and say $\hat{x}$ unit vector or $x(t)$ function. Am I right?

8. Aug 25, 2016

### Staff: Mentor

But we are not given $x(t)$. Otherwise we wouldn't be using phase space to solve for it. For each system there are states which go through each $x$ at any $\dot x$, so they are independent. Then we solve for $x(t)$

9. Aug 25, 2016

### Happiness

Could we make this explanation more formal? My attempt is as follows:

For each system there are states that go through each $x$ value at any $\dot x$ value, so the variables are independent, but the functions $x(t)$ and $\dot x(t)$ are dependent.

A variable can take on any value in a given set, whereas a function cannot. A function is a relation that maps a value in a given set to another value (which could be distinct or identical) in another given set. (Another is not a good word, but I can't think of a better word.)

Definition of independent variables
Two random variables X and Y are independent if and only if for every a and b, the events {Xa} and {Yb} are independent events.

Definition of independent events

Definition of independent functions
I can't find it. I can only find the definition of linearly independent functions. But the functions $x(t)$ and $\dot x(t)$ could be linearly independent but dependent.

Last edited: Aug 25, 2016
10. Aug 25, 2016

### Staff: Mentor

I think that the correct way to go about this is to say that $x(t)$ and $\dot x(t)$ are governed by a set of coupled differential equations.

11. Aug 25, 2016

### Lucas SV

Don't confuse phase space with a particle trajectory. The unit vectors I was talking about do not represent a particle. They are linearly independent. But the particle's actual position and actual momentum are not, in general. Unfortunately in physics, we use $x$ to mean both the particle trajectory and the coordinate in phase space/real space. So I will use a different notation for clarity's sake.

A phase space (for a classical 1 particle system in 3 dimensions) is $P=\mathbb{R}^6$. We denote an element of the phase space by $(x_1,x_2,x_3,v_1,v_2,v_3)$. The unit vectors I mentioned previously are nothing but
$$(1,0,0,0,0,0), (0,1,0,0,0,0), (0,0,1,0,0,0), (0,0,0,1,0,0), (0,0,0,0,1,0), (0,0,0,0,0,1)$$
The classical trajectory of a particle is a smooth map $\gamma:\mathbb{R} \rightarrow P$ such that $\dot{\gamma}_1=\gamma_4$, $\dot{\gamma}_2=\gamma_5$ and $\dot{\gamma}_3=\gamma_6$, where $\dot{\gamma}=(\dot{\gamma}_1,\dot{\gamma}_2,\dot{\gamma}_3,\dot{\gamma}_4,\dot{\gamma}_5,\dot{\gamma}_6)$ is the derivative of $\gamma=(\gamma_1,\gamma_2,\gamma_3,\gamma_4,\gamma_5,\gamma_6)$ and $\gamma_i:\mathbb{R}\rightarrow\mathbb{R}$, $i=1,2,3,4,5,6$. There is a unique particle trajectory $\gamma$ satisfying the initial condition $\gamma(t_0)=u$, for some $u\in P$ and $t_0\in\mathbb{R}$.

The right material for you to learn this from is Dynamical Systems/Classical Physics, in particular the Hamiltonian formalism.

12. Aug 25, 2016

### Happiness

In the derivation of the Euler-Lagrange equation from the principle of stationary action, Morin, in his book, wrote that nowhere do we assume that $x$ and $\dot x$ are independent variables. But in phase space, they are independent variables.

Am I right to say that he should have said "nowhere do we assume that $x$ and $\dot x$ are independent functions" instead? This is because in (5.16), we differentiate $x$ and $\dot x$, and variables cannot be differentiated. For example, we can differentiate a function $y(x)$ wrt $x$, but we can't differentiate a variable $x$ wrt $x$ or a variable $y$ wrt $x$. (If we do $\frac{dx}{dx}$, we are differentiating the function $y(x)=x$ wrt $x$.)

13. Aug 25, 2016

### Lucas SV

In the lagrangian formulation we do not use phase space. The word 'variable' is not used precisely in mathematics, in my experience. You do not need to talk about the variable if you want to talk about the funtion. Learning some logic, set theory and functions in a more systematic way may help you.

I think you are getting too bogged down on this dependent\independent issue. It doesn't matter if you want to understand things precisely, or not so much so, you will have to work hard anyway.

Do some exercises in the lagrangian formulation, and you will understand how to work with it. After you know the lagrangian formulation well, move on to the hamiltonian formulation. Don't worry about learning things too quickly.

When it comes to functionals (the action is a functional), the proper mathamatics can get quite technical. Also if you are rigid about notation you may have problems. Even mathematicians in differential geometry will use many different notations quite flexibly. Do not worry too much about the technicalities and details for now, learn the subject by doing exercises, and later you will have the opportunity to come back to this once you have more experience.

14. Aug 25, 2016

### Khashishi

x is frequently used to mean two related but different things.
The first is a coordinate. There is a set of coordinates for each point in space, or of phase space.
The second is the actual position of an object. This only occupies a curve in space much smaller than the whole space.

The Lagrangian uses the first meaning.

15. Aug 25, 2016

### Happiness

How do you reconcile Morin's comment that $x$ and $\dot x$ are dependent variables with the fact that they are independent variables in phase space?

16. Aug 26, 2016

### Khashishi

I'm not sure I agree with Morin. 5.17 and 5.16 are fine even if they are independent.

Euler-Lagrange equation is an optimization problem, so the solution is more constrained than the functional that is being optimized. So $x$ and $\dot{x}$ are independent in the functional, but not in the solution. I think it's analogous to this simpler example. Consider the line
y=mx+b
Clearly, x and y are not independent on the line. Either y is function of x, or x is a function of y.
But now consider the function
f(x,y) = mx+b-y
Now consider the optimization problem: f(x,y) = 0. Obviously the solution is y=mx+b.
But, the function f(x,y) is defined on all x and y. x and y are independent parameters to the function f. But the solution is a constraint between them.

17. Aug 26, 2016

### Happiness

Can I say there are two ways of viewing (5.16), depending on the vector space you use to visualize the optimization?

If we use the position-time vector space, then we vary the trajectory $x(t)$ of the particle. In this case, $\dot x(t)$ changes accordingly as $x(t)$ is varied. Then the $x(t)$ and $\dot x(t)$ in (5.16) are NOT independent.

On the other hand, if we use the velocity-position phase space to visualize the optimization, then we vary the phase-space "trajectory" $\begin{pmatrix}x(t)\\\dot x(t)\end{pmatrix}$ of the particle. In this case, $x(t)$ and $\dot x(t)$ are varied independently. Then the $x(t)$ and $\dot x(t)$ in (5.16) are independent.

However, $\frac{\partial x}{\partial a}$ and $\frac{\partial\dot x}{\partial a}$ are related by (5.17) as the latter is the derivative of the former. Does this imply that not all variations of the phase-space "trajectory" $\begin{pmatrix}x(t)\\\dot x(t)\end{pmatrix}$ are considered in the optimization? But only those variations that satisfy (5.17)? If so, then $x(t)$ and $\dot x(t)$ are NOT varied independently. And the function $x(t)$ and the function $\dot x(t)$ in (5.16) are NOT independent.

Also, consider the term $\frac{\partial L}{\partial x}\frac{\partial x}{\partial a}$ in (5.16). Can I say that the $x$ in $\frac{\partial L}{\partial x}$ and the $x$ in $\frac{\partial x}{\partial a}$ mean slightly different things, though they are both symbolized in the same way? The $x$ in the former is a variable (which can take on certain values), whereas the $x$ in the latter is a function (which is a map).

Last edited: Aug 26, 2016
18. Aug 26, 2016

### PeroK

As mentioned in post #11, you are confusing a specific particle trajectory with all the possible states of a particle.

in a sense, its no different from x and y being independent variables, in terms of the x-y plane, but not if you looking at a specific graph, like $y = x^2$.

19. Dec 25, 2016

### Sam Robertson

They are only independent in the sense that you need position and velocity to specify the initial state of a system (and thus the further time evolution of the system - you are completely right about this) because F = m a is a 2nd order differential equation and there is an initial velocity integration constant. If the law were different say F = m v you could specify all the initial states with just position because v = dx / dt would just be F(x) / m and there is no initial velocity freedom. Leonard Susskind talks about this at his first lecture at stanford (google) if I am not misunderstanding his lecture.

20. Dec 25, 2016

### Sam Robertson

What I mean is that they are only independent in the sense that you are free to specify them *independently* as initial conditions. Once you do that all you have to do is take some nice derivatives (no integration constants needed!) and you will have the evolution of your system for all time.