I Independent coordinates are dependent

[Kashmir]

(This is not about independence of $q$, $\dot q$)

A system has some holonomic constraints. Using them we can have a set of coordinates ${q_i}$. Since any values for these coordinates is possible we say that these are independent coordinates.

However the system will trace a path in the configuration space ,which means that actually all our independent coordinates are in fact dependent.

Why do we then say that we have independent coordinates?
For example,
Goldstein
>The fundamental problem of the calculus of variations is easily generalized to the case where $f$ is a function of many **independent variables** $y_{i}$, and their derivatives $\dot{y}_{i}$. (Of course, all these quantities are considered as functions of the parametric variable $x$.) Then a variation of the integral $J$,
$\delta J=\delta \int_{1}^{2} f\left(y_{1}(x) ; y_{2}(x), \ldots, \dot{y}_{1}(x) ; \dot{y}_{2}(x), \ldots, x\right) d x$

 

[PeroK]

The idea of Lagrangian mechanics is to analyse the functional form of the system. In the simplest case we have: $$L(x, \dot x) = \frac 1 2 m \dot x^2 - V(x)$$ That is a well-defined function of two variables. For the time being we ignore that those variables will turn out to be related along a specific trajectory.

The Euler-Lagrange equations are generated using this functional approach and minimising the action. All the time treating the Lagrangian as nothing more than a function of two independent variables.

The trick is that this approach leads to equations of motion that are satisfied by the system. There's always a point at which you switch from looking at the system in a generalised functional form, to one where you are looking at a specific trajectory.

 

[Kashmir]

Thank you. Please give me some time to fully think about it.

 

[vanhees71]

The problem is partially the physicists sloppy notation. On the one hand they denote with $x$ and $\dot{x}$ simply two independent variables of the Langrange function,
$$(x,\dot{x}) \mapsto L(x,\dot{x}).$$
Then you can take partial derivatives of this function wrt. these independent variables, i.e.,
$$\frac{\partial L}{\partial x}$$
is the derivative of $L$ with respect to $x$ with $\dot{x}$ kept constant and
$$\frac{\partial L}{\partial \dot{x}}$$
is the derivative of $L$ wrt. $\dot{x}$ with $x$ kept constant.

On the other hand you define the action functional, maps a trajectory in configuration space $t \mapsto x(t)$ with time as the parameter of this trajectory:
$$S[x]=\int_{t_1}^{t_2} \mathrm{d} t L[x(t),\dot{x}(t)],$$
where now
$$\dot{x}(t)=\frac{\mathrm{d} x}{\mathrm{d} t}.$$
If you now want to take the total time derivative of some function $f(x,\dot{x})$ along the trajectory you have
$$\frac{\mathrm{d}}{\mathrm{d} t} f[x(t),\dot{x}(t)]=\frac{\mathrm{d} x(t)}{\mathrm{d} t} \left (\frac{\partial f(x,\dot{x})}{\partial x} \right)_{x=x(t),\dot{x}(t)} + \frac{\mathrm{d}^2 x(t)}{\mathrm{d} t^2} \left (\frac{\partial f(x,\dot{x})}{\partial \dot{x}} \right)_{x=x(t),\dot{x}=\dot{x}(t)}.$$
Being lazy, for this the physicists simply write
$$\frac{\mathrm{d}}{\mathrm{d} t} f=\dot{x} \partial_x f + \ddot{x} \partial_{\dot{x}} f.$$

 

[Kashmir]

The idea of Lagrangian mechanics is to analyse the functional form of the system. In the simplest case we have: $$L(x, \dot x) = \frac 1 2 m \dot x^2 - V(x)$$ That is a well-defined function of two variables. For the time being we ignore that those variables will turn out to be related along a specific trajectory.

The Euler-Lagrange equations are generated using this functional approach and minimising the action. All the time treating the Lagrangian as nothing more than a function of two independent variables.

The trick is that this approach leads to equations of motion that are satisfied by the system. There's always a point at which you switch from looking at the system in a generalised functional form, to one where you are looking at a specific trajectory.

How can I assume an independence already knowing that they are dependent. It looks like some sort of cheating the math.

 

[Kashmir]

The problem is partially the physicists sloppy notation. On the one hand they denote with $x$ and $\dot{x}$ simply two independent variables of the Langrange function,
$$(x,\dot{x}) \mapsto L(x,\dot{x}).$$
Then you can take partial derivatives of this function wrt. these independent variables, i.e.,
$$\frac{\partial L}{\partial x}$$
is the derivative of $L$ with respect to $x$ with $\dot{x}$ kept constant and
$$\frac{\partial L}{\partial \dot{x}}$$
is the derivative of $L$ wrt. $\dot{x}$ with $x$ kept constant.

On the other hand you define the action functional, maps a trajectory in configuration space $t \mapsto x(t)$ with time as the parameter of this trajectory:
$$S[x]=\int_{t_1}^{t_2} \mathrm{d} t L[x(t),\dot{x}(t)],$$
where now
$$\dot{x}(t)=\frac{\mathrm{d} x}{\mathrm{d} t}.$$
If you now want to take the total time derivative of some function $f(x,\dot{x})$ along the trajectory you have
$$\frac{\mathrm{d}}{\mathrm{d} t} f[x(t),\dot{x}(t)]=\frac{\mathrm{d} x(t)}{\mathrm{d} t} \left (\frac{\partial f(x,\dot{x})}{\partial x} \right)_{x=x(t),\dot{x}(t)} + \frac{\mathrm{d}^2 x(t)}{\mathrm{d} t^2} \left (\frac{\partial f(x,\dot{x})}{\partial \dot{x}} \right)_{x=x(t),\dot{x}=\dot{x}(t)}.$$
Being lazy, for this the physicists simply write
$$\frac{\mathrm{d}}{\mathrm{d} t} f=\dot{x} \partial_x f + \ddot{x} \partial_{\dot{x}} f.$$

I'm sorry but I don't understand how it answers my doubt. :(

 

[wrobel]

Take a particle on a plane (x,y). You can provide the particle with arbitrary initial position coordinates and arbitrary initial velocities $x,y,\dot x, \dot y$. The independence is understood in this way.

 

[PeroK]

How can I assume an independence already knowing that they are dependent. It looks like some sort of cheating the math.

It's not cheating, it's using a clever mathematical idea. There's nothing to stop you calculating the numeric value of $$L(x, \dot x) = \frac 1 2 m \dot x^2 - V(x)$$ for any values of $x, \dot x$. And there's nothing to stop you plotting the surface in 3D for all values in the $x, \dot x$ plane. And, there's nothing to stop you analysing that surface mathematically. And, there's noting to stop you minimising the action integral for paths on that surface. And, there's nothing to stop you showing that those paths satisfy the Euler-Lagrange equations.

Those paths might have nothing to do with physics. But, those paths (found by mathematically analysing the Lagrangian and solving the E-L equations) turn out to be precisely the paths determined by Newton's laws (and this can be proved).

So, that gives us two ways to determine the motion of the system:

1) Use Newton's laws of motion directly.

2) Find paths on the surface defined by the Lagrangain that satisfy the E-L equations.

And, it can be proved that these two are indeed equivalent.

 

[Ibix]

How can I assume an independence already knowing that they are dependent. It looks like some sort of cheating the math.

The trick is to realize that $x$ and $\dot{x}$ are coordinates on a plane where you've defined a displacement-versus-velocity graph. In principle, a thing can be anywhere on that plane: it can be 10m to the left doing 30m/s or -100m/s. You don't know what the relationship between position and velocity is at this point - you haven't worked it out. The Lagrangian defines a function on that plane and the Euler-Lagrange equations give you a way to find the route across the plane that extremises its integral. That's what defines the relationship between the specific values of $x$ and $\dot{x}$ that form the trajectory of the object across your displacement-velocity graph.

On a standard Euclidean plane $x$ and $y$ coordinates can take any value. It's only once you've got a path specified by a relationship like $y=t^2,x=t$ that you have a dependency - Euler-Lagrange is a way of discovering physically interestibg dependencies.

 

[andresB]

A system has some holonomic constraints. Using them we can have a set of coordinates ${q_i}$. Since any values for these coordinates is possible we say that these are independent coordinates.

However the system will trace a path in the configuration space ,which means that actually all our independent coordinates are in fact dependent.

Why do we then say that we have independent coordinates?

In this context, "independent" means that they are not related to each other via a constraint.

Take for example a particle that is constrained to move on the surface of a sphere. You can take the polar and azimuthal angles as generalized coordinates.

Moving on the surface of a sphere creates no relation between the angles, they are, thus, said to be independent. This has nothing to do with the dynamics (Lagrangian).