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Given the Lagrangian $$L(x,y,\dot{x},\dot{y},t)=y^2\dot{x}^2+x^2\dot{y}^2+x+y+\dot{x}t+\dot{y}t $$ in the $x,y$ coordinates, I want to find some conserved quantities. Since there are no cyclic coordinates, I am searching for a symmetry for the system but I'm not sure how to do that. Is there any method I could use?

Ginevra Carbone
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  • Just saying that this is a very very very big Lagrangian. – R.W Mar 16 '17 at 16:21
  • I am curious about this Lagrangian. May you tell if it's homework or whatever? And if homework, what's the solution provided. I know what's the coordinate change leading to that Lagrangian and I'd like to know if at least you got it and what you were said. – Rafa Budría Mar 19 '17 at 09:36
  • Yes, It's homework. I just had to say which of three given quantities was a constant of motion for the Lagrangian, but I was curious to find a symmetry. The solution provided is exactly the one I accepted. – Ginevra Carbone Mar 20 '17 at 10:02
  • @Rafael Wagner and magnetissimo, thanks! I am sure now what's the coordinate transformation leading from that lagrangian to the one for a free particle. I asked for someting very related and I got a nice complete answer here Somebody put the "Homework" tag, but it isn't :) – Rafa Budría Mar 22 '17 at 14:57

2 Answers2

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To find conserved quantities there is nothing better then derive the canonical equations of Hamilton. So I have the hint

Hint: Find the Hamiltonian and the canonical equations of motion for the hamiltonian system

$$\frac{\partial \mathcal{H}}{\partial q} = -\dot{p} \hspace{2cm} \frac{\partial \mathcal{H}}{\partial p} = \dot{q}$$

You should of course find the equations of motion for the Lagrangian, i.e., the Euler-Lagrange equations.

If you are searching specifically for a symmetry there is the most beatifull theorem of physics (my personal opinion), so the other hint should be

Hint: Use Noether's Theorem to find a symmetry. If you read the link you should see that, in the notation used,

$$\sum_\alpha \frac{\partial \mathcal{L}}{\partial \dot{q_\alpha}}\gamma_\alpha = \text{constant}$$

The theorem says that for a specific symmetry you always can find conserved quantities. See that if you change $x \rightarrow y$ you get the same Lagrangian.

R.W
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First of all, we can write the lagrangian in this form:

$$L(x,y,\dot{x},\dot{y},t)=y^2\dot{x}^2+x^2\dot{y}^2+x+y+\dot{x}t+\dot{y}t=$$

$$=y^2\dot{x}^2+x^2\dot{y}^2+\dfrac{\mathrm d}{\mathrm dt}(xt+yt)$$

We can drop the total derivative with respect to time because it doesn't affect the equations of motion:

$$\tilde L(x,y,\dot{x},\dot{y},t)=y^2\dot{x}^2+x^2\dot{y}^2$$

Now, as the lagrangian doesn't depend explicitly of time, the energy is conserved. So is,

$\dot x\dfrac{\partial\tilde L}{\partial\dot x}+\dot y\dfrac{\partial\tilde L}{\partial\dot y}-\tilde L=K$

$\dot x\dfrac{\partial\tilde L}{\partial\dot x}+\dot y\dfrac{\partial\tilde L}{\partial\dot y}-\tilde L=\dot xy^22\dot x+\dot yx^22\dot y-y^2\dot x^2-x^2\dot y^2=y^2\dot x^2+x^2\dot y^2$

$y^2\dot x^2+x^2\dot y^2=K$

I tink this lagrangian is simply the one for the free particle in dimension two, but in a weird coordinate system.

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Rafa Budría
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