Find the Conserved Quantity of a Lagrangian Using Noether's Theorem

[koil_]

Homework Statement

Find the conserved quantity of the Lagrangian $$L = (\frac {\dot{x}}{x})^2$$ associated with the invariance given by the transformation $$ x \to sx$$

Relevant Equations

$$ \frac {\partial L}{\partial \dot{Q}} \frac {\partial Q}{\partial s}$$

So Noether's Theorem states that for any invarience that there is an associated conserved quantity being:

$$ \frac {\partial L}{\partial \dot{Q}} \frac {\partial Q}{\partial s}$$
Let $$ X \to sx $$
$$\frac {\partial Q}{\partial s} = \frac {\partial X}{\partial s} = \frac {\partial (sx)}{\partial s}= x $$
This is the part that I'm now unsure about:

$$\frac {\partial L}{\partial \dot{Q}}=\frac {\partial (\frac {\dot{X}}{X})^2} {\partial \dot{X}} = \frac {2\dot{X}} {X^2} $$

This would make the conserved quantity therefore:

$$ \frac {2\dot{X}} {X^2} * x $$

However I'm not sure where to go from here as there X and x is present and 'reversing' the transformation produces an 's' in the quantity which wouldn't make sense as this quantity is supposed to be conserved.

I'm sure it is quite evident that this is the first piece we have been given on this so I may have the method completely wrong as, in some examples I have seen, the derivatives are not fully expanded. e.g:

$$\frac {\partial L}{\partial \dot{X}} * x$$ is some conserved quantity.

Many thanks in advance.

 

[PeroK]

I must admit that I'm not at all familiar with the notation or this representation of Noether's theorem.

First, we can find a conserved quantity directly from the Euler-Lagrange equation:
$$\frac{\partial L}{\partial x} = \frac{d}{dt}(\frac{\partial L}{\partial \dot x})$$ Which yields $$x\ddot x - \dot x^2 = 0$$ Now, if we note that $$\frac{d}{dt}(\frac{\dot x}{x}) = \frac 1 {x^2}(x\ddot x - \dot x^2) = 0$$ then we see that $$\frac{\dot x}{x}$$ is conserved.

And, indeed, that is the quantity that emerges from an application of Noether's theorem in the formulation that I am familiar with.

 

[PeroK]

PS the form of Noether's theorem I am familiar with is described here. See in particular the section on "Simple Form user Perturbations".

https://en.wikipedia.org/wiki/Noether's_theorem

 

[wrobel]

$x\mapsto sx$ is not a flow
the corresponding flow is $x\mapsto e^sx$.
thus the generating vector field is $$v(x)=\frac{d}{ds}\Big|_{s=0}g^s(x)=x,\quad g^s(x)= e^sx$$
Thus the first integral is
$$\frac{\partial L}{\partial\dot x}v(x)$$

 

[Orodruin]

Homework Statement:: Find the conserved quantity of the Lagrangian $$L = (\frac {\dot{x}}{x})^2$$ associated with the invariance given by the transformation $$ x \to sx$$
Relevant Equations:: $$ \frac {\partial L}{\partial \dot{Q}} \frac {\partial Q}{\partial s}$$

So Noether's Theorem states that for any invarience that there is an associated conserved quantity being:

$$ \frac {\partial L}{\partial \dot{Q}} \frac {\partial Q}{\partial s}$$
Let $$ X \to sx $$
$$\frac {\partial Q}{\partial s} = \frac {\partial X}{\partial s} = \frac {\partial (sx)}{\partial s}= x $$
This is the part that I'm now unsure about:

$$\frac {\partial L}{\partial \dot{Q}}=\frac {\partial (\frac {\dot{X}}{X})^2} {\partial \dot{X}} = \frac {2\dot{X}} {X^2} $$

This would make the conserved quantity therefore:

$$ \frac {2\dot{X}} {X^2} * x $$

However I'm not sure where to go from here as there X and x is present and 'reversing' the transformation produces an 's' in the quantity which wouldn't make sense as this quantity is supposed to be conserved.

I'm sure it is quite evident that this is the first piece we have been given on this so I may have the method completely wrong as, in some examples I have seen, the derivatives are not fully expanded. e.g:

$$\frac {\partial L}{\partial \dot{X}} * x$$ is some conserved quantity.

Many thanks in advance.

You need to evaluate $X$ for the value of $s$ that gives $x$.

$x\mapsto sx$ is not a flow
the corresponding flow is $x\mapsto e^sx$.
thus the generating vector field is $$v(x)=\frac{d}{ds}\Big|_{s=0}g^s(x)=x,\quad g^s(x)= e^sx$$
Thus the first integral is
$$\frac{\partial L}{\partial\dot x}v(x)$$

All that is required is a one-parameter symmetry of the Lagrangian and calling the scaling $s$ or $e^s$ is quite arbitrary. Indeed, you end up with the same conserved quantity in both cases

 

[wrobel]

All that is required is a one-parameter symmetry of the Lagrangian and calling the scaling or is quite arbitrary.

Sure and some formulations of the theorem stress its invariant nature while other ones darken it. That is why questions rise.