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

• koil_
In summary, the conserved quantity in Noether's theorem is given by:$$\frac {\partial L}{\partial \dot{Q}} \frac {\partial Q}{\partial s}$$
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.

PhDeezNutz
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.

PhDeezNutz
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

##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
$$\frac{\partial L}{\partial\dot x}v(x)$$

koil_ said:
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.

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

wrobel said:
##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

Orodruin said:
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.

## 1. What is Noether's Theorem?

Noether's Theorem is a fundamental principle in physics that states that for every continuous symmetry in a physical system, there exists a corresponding conserved quantity.

## 2. What is a conserved quantity?

A conserved quantity is a physical quantity that remains constant over time, even as the system undergoes changes or transformations. This can include quantities such as energy, momentum, and angular momentum.

## 3. How is Noether's Theorem used to find conserved quantities?

Noether's Theorem provides a method for identifying the conserved quantity associated with a particular symmetry in a physical system. This is done by first identifying the continuous symmetry, such as time translation or spatial rotation, and then using the corresponding mathematical equations to determine the conserved quantity.

## 4. Why is finding conserved quantities important?

Conserved quantities play a crucial role in understanding the behavior and dynamics of physical systems. They provide insights into the underlying symmetries and laws governing the system and can be used to make predictions and solve problems in physics.

## 5. What is the significance of finding the conserved quantity of a Lagrangian?

The Lagrangian is a mathematical function that describes the dynamics of a physical system. By using Noether's Theorem to find the conserved quantity associated with the Lagrangian, we can gain a deeper understanding of the symmetries and underlying principles that govern the system, and potentially make predictions about its behavior and evolution.

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