Lagrangians and Noether Theorem

[atomqwerty]

Homework Statement

Let be the lagrangian given by

$L(x,y,\dot{x},\dot{y})=\frac{m}{2}(\dot{x^2} +\dot{y^2})-V(x^{2}+y^{2})$

and

$L(x,y,\dot{x},\dot{y})=\frac{m}{2}(\dot{x^2} + \dot{y^2})-V(x^{2}+y^{2}) - \frac{k}{2}x^{2}$

and the transformation

$x'=\cos\alpha x - \sin\alpha y$
$y'=\sin\alpha x + \cos\alpha y$

Show the invariant observable with respect to the symmetry of this transformation (Noether's Theorem) for both Lagrangians.

Homework Equations

Noether Theorem says that for each variable x such as $\partial L/\partial x = 0$ then x is said to be invariant and $\partial L/\partial \dot{x} = 0$ is constant of motion for that system.

The Attempt at a Solution

First, the transformation from (x,y) to (x',y') doesn't change the Lagrangian, and since L does depend on $x, y, \dot{x}, \dot{y}$ on both Lagrangians, then the only non explicit variable is time t, and in that case the conserved quantity would be the Energy E = T + V. I must be wrong since this is a problem of an exam of Theoretical Mechanics in University...

Thanks a lot!

 

[dextercioby]

[...]

$L(x,y,\dot{x},\dot{y})=\frac{m}{2}(\dot{x}^2 +\dot{y}^2)-V(x^{2}+y^{2}) - \frac{k}{2} x^{2}$

and the transformation

$x'=cos\alpha x - sin\alpha y$
$y'=sin\alpha x +cos\alpha y$

Show the invariant observable with respect to the symmetry of this transformation (Noether's Theorem) for both Lagrangians.

[...]

You need to think about Noether's theorem. Can you show that the 2 Lagrangians are invariant under the mentioned transformations ?

 

[atomqwerty]

I could check that both $\dot{x^2}+\dot{y^2}$ and $x^2+y^2$ keep the same under that transformation, and so that there's no variable $x$ or $y$ that verifies $\partial L / \partial x =0$, $\partial L / \partial y =0$. That's what I thought of time instead...

 

[physicus]

Noether's theorem is something else. It states, that for every continuous symmetry of a theory there is a conserved charge. You simply try to find cyclic coordinates which does not give you a conserved quantity here.

What you need to do is to consider an infinitesimal transformation. So just choose $\alpha$ infinitesimal. Therefore:
$x'=x-\alpha y, y'=y+\alpha x$

The conserved quantity is then given by
$\frac{\partial L}{\partial \dot{x}}\frac{\delta x}{\alpha}+\frac{\partial L}{\partial \dot{y}}\frac{\delta y}{\alpha}$

You will see, for example, that it is angular momentum $m(\dot{y}x-\dot{x}y)$ for the first lagrangian.

 

[atomqwerty]

And by $\delta x$ do you mean $-\alpha y$, and by $\delta y$ do you mean $\alpha x$?

 

[physicus]

Yes, exactly.

 

[atomqwerty]

Why is the conserved quantity given by that expression? I've just been thinking about it via Poisson brackets and I can't derive it. Thanks

 

[physicus]

The Lagrangian is invariant under the transformation $\delta L=0$.

I just prove the conservation of the Noether charge for one variable x instead of x and y. The proof obviously works for more variables:
You want to show $\frac{\partial L}{\partial \dot{x}}\frac{\delta x}{\alpha}$ is conserved, i.e. $\frac{d}{dt} \left(\frac{\partial L}{\partial \dot{x}}\frac{\delta x}{\alpha}\right)=0$

$\frac{d}{dt} \left(\frac{\partial L}{\partial \dot{x}}\frac{\delta x}{\alpha}\right) = <br /> \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)\frac{\delta x}{\alpha}+\frac{\partial L}{\partial \dot{x}}\frac{\delta \dot{x}}{\alpha}=\frac{\partial L}{\partial x}\frac{\delta x}{\alpha}+\frac{\partial L}{\partial \dot{x}}\frac{\delta \dot{x}}{\alpha}= \frac{1}{\alpha}\delta L = 0$

In the second step I have used the Euler Lagrange equations to simplify the first term. This is actually important. The conserved charge is only conserved for solutions of the Euler Lagrange equations. Classically, all particles obey the equations. However, in QFT for example, there are so-called off-shell particles, that do not obey the equations of motion. For these, the Noether current is not necessarily conserved.

 

[atomqwerty]

I can see that the same quantity )angular momentum) is conserved also for the second Lagrangian, isn't it?

 

[physicus]

The second Lagrangian is not invariant under the given transformation. Are you sure it is right?

 

[atomqwerty]

Well, following

The conserved quantity is then given by
$\frac{\partial L}{\partial \dot{x}}\frac{\delta x}{\alpha}+\frac{\partial L}{\partial \dot{y}}\frac{\delta y}{\alpha}$

then the new term of the second Lagrangian does not depend on $\dot{x}$ or $\dot{y}$, right?

 

[physicus]

This formula for the conserved charge is only valid if the Lagrangian is invariant under the given transformation (you have seen that I used $\delta L=0$ in the proof). But I don't see why the second Lagrangian should be invariant. Pick for example $\alpha=\frac{\pi}{2}$, then $x \rightarrow -y, y \rightarrow x$. Then the tranformed Lagrangian is
$\frac{m}{2}(\dot{x^2} + \dot{y^2})-V(x^{2}+y^{2}) - \frac{k}{2}y^{2} \not= L$. There seems to be an error in the problem the way you have written it down in the beginning.