Showing that the euler lagrange equations are coordinate independent

In summary, the conversation discusses the form invariance of the EL equations under coordinate changes. The participant provides a derivation of this concept and points out a mistake in the original equation. They use the chain rule and Einstein's summation convention to show that the EL equations are valid with respect to both the original coordinates and the transformed coordinates. The conversation also mentions the use of LaTeX for clearer presentation of mathematical equations.
  • #1
demonelite123
219
0
so i know for example that d/dt (∂L/∂x*i) = ∂L/∂xi for cartesian coordinates, where xi is the ith coordinate in Rn and x*i is the derivative of the ith coordinate xi with respect to time. L represents the lagrangian.

so using an arbitrary change of coordinates, qi = qi(x1, x2, ..., xn)

i have that ∂L/∂x_i = ∂L/∂qi * ∂q_i/∂x_i and that ∂L/∂x*i = ∂L/∂q*i * ∂q*i/∂x*i and substituting i get:

d/dt(∂L/∂q*i * ∂q*i/∂x*i) = ∂L/∂qi * ∂qi/∂xi and using the product rule i get:

d/dt(∂L/∂q*i) * ∂q*i/∂x*i+ ∂L/∂q*i * d/dt(∂q*i/∂x*i) = ∂L/∂qi* ∂qi/∂xi

using the fact that ∂q*i/∂x*i = ∂qi/∂xi,

d/dt(∂L/∂q*i) * ∂qi/∂xi + ∂L/∂q*i * d/dt(∂qi/∂xi) = d/dt(∂L/∂q*i) * ∂qi/∂xi + ∂L/∂q*i * (∂q*i/∂xi) = d/dt(∂L/∂q*i) * ∂qi/∂xi+ ∂L/∂xi = ∂L/∂qi * ∂qi/∂xi.

if i didn't have that ∂L/∂xi on the left side then i could just cancel out the ∂qi/∂xi on both sides and i would be done. have i made a mistake somewhere or am i missing something? thanks.
 
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  • #2
Why don't you use the nice LaTeX functionality of this forum? It's really hard to read, what you have written. So I don't want to decipher it.

Here is my derivation of the form invariance of the EL equations under coordinate changes (diffeomorphisms). Let [itex]x^j(q)[/itex] denote the transformation with [itex]j \in \{1,\ldots,f \}[/itex]. Further [itex]q=(q^k)_{k \in \{1,\ldots, f\}}[/itex].

First of all we need an auxilliary equation for the time derivatives. From the chain rule we have

[tex]\dot{x}^j=\frac{\partial x^j}{\partial q^k} \dot{q}^k.[/tex]

Here and in the following I use Einstein's summation convention, according to which one has to sum over any pair of repeated indices. From this equation we see

[tex]\frac{\partial \dot{x}^j}{\partial \dot{q}^k}=\frac{\partial x^j}{\partial q^k}. \quad (1) [/tex]

Now we have

[tex]\frac{\partial L}{\partial q^k}=\frac{\partial L}{\partial x^j} \frac{\partial x_j}{\partial q_k} + \frac{\partial L}{\partial \dot{x}^j} \frac{\partial \dot{x}^j}{\partial q^j} \quad (2)[/tex]

and

[tex]\frac{\partial L}{\partial \dot{q}^k}=\frac{\partial L}{\partial \dot{x}^j} \frac{\partial \dot{x}^j}{\partial \dot{q}^k} \stackrel{(1)}{=}\frac{\partial L}{\partial \dot{x}^j} \frac{\partial x^j}{\partial q^k}.[/tex]

Taking the time derivative of this, we find

[tex]\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{q}^k} = \frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{x}^j} \cdot \frac{\partial x^j}{\partial q^k} + \frac{\partial L}{\partial \dot{x}^j} \frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{\partial x^j}{\partial q^k} \right). \quad (3)[/tex]

Now we obviously have

[tex]\frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{\partial x^j}{\partial q^k} \right) = \frac{\partial \dot{x}^k}{\partial q^k}, \quad (4)[/tex]

and thus, if the EL equation are valid wrt. the coordinates, [itex]x^k[/itex] they are also valid wrt. to the coordinates, [itex]q^j[/itex], which can be seen by comparing (2) and (3) and making use of (4) and the EL equation wrt. [itex]x^k[/itex].
 
  • #3
demonelite123 said:
i have that ∂L/∂x_i = ∂L/∂qi * ∂q_i/∂x_i

This is one of the mistakes. Instead, write:

[itex]\frac{\partial L}{\partial x_i}=\sum_j \frac{\partial L}{\partial q_j}\frac{\partial q_j}{\partial x_i}+\sum_j \frac{\partial L}{\partial \dot{q}_j}\frac{\partial \dot{q}_j}{\partial x_i}[/itex]

You fergot the second piece. Remember:

[itex]q_j=q_j(x_1,\dots,x_n)[/itex]
[itex]\dot{q}_j=\sum_k\frac{\partial q_j}{\partial x_k}\dot{x}_k[/itex]

so

[itex]\frac{\partial\dot{q}_j}{\partial x_i}=\frac{\partial^2 q_j}{\partial x_i\partial x_k}\dot{x}_k[/itex]
 
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1. How do the Euler-Lagrange equations demonstrate coordinate independence?

The Euler-Lagrange equations are a set of mathematical equations that describe the behavior of a physical system. They are derived from the principle of least action, which states that a physical system will take the path between two points that minimizes the action (a measure of the system's energy). Since the equations are derived from this fundamental principle, they hold true regardless of the coordinate system used to describe the system.

2. What is the significance of coordinate independence in the context of the Euler-Lagrange equations?

Coordinate independence is significant because it allows us to describe a physical system using different coordinate systems without changing the underlying equations that govern its behavior. This is important because it allows us to choose the most convenient coordinate system for a given problem, without affecting the validity of our results.

3. Can you provide an example of how the Euler-Lagrange equations show coordinate independence?

One example is the motion of a particle in a conservative force field, such as a gravitational field. The equations of motion for this system, which are derived using the Euler-Lagrange equations, do not change when using different coordinate systems (e.g. Cartesian, polar, spherical coordinates). This demonstrates the coordinate independence of the equations.

4. How do the Euler-Lagrange equations ensure that the principle of least action is satisfied?

The Euler-Lagrange equations ensure that the principle of least action is satisfied by finding the path between two points that minimizes the action. This is done by taking the derivative of the action with respect to the position of the particle, and setting it equal to zero. The resulting equation is the Euler-Lagrange equation, which must be satisfied for the principle of least action to hold true.

5. Why is it important to show that the Euler-Lagrange equations are coordinate independent?

It is important to show that the Euler-Lagrange equations are coordinate independent because it confirms the validity and applicability of the equations in describing physical systems. It also allows us to use different coordinate systems to analyze the same system, which can provide valuable insights and simplify calculations.

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