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It is the first time I get in touch with classical mechanics and this stuff...

Suppose a mechanical system has $n$ degrees of freedom described by coordinated $q\in V\subset\mathbb{R}^n$, set $v=\dot{q}$. In the situation of particles under the influence of some forces we have the Lagrange function $$ L(v,q)=\frac{1}{2}vMv-U(q).~~~(*) $$ where $M$ is a positive diagonal matrix with the masses of the particles as entries and $U$ is the potential corresponding to the forces. Let $$ p(v,q)=\frac{\partial L}{\partial v}(v,q) $$ denote the momentum.

Theorem of Noether Let $\Phi(t,q)$ be the flow generated by $f(q)$. If $\Phi$ leaves the Lagrangian invariant, then $$ I(v,q)=p(v,q)\cdot f(q) $$ is a constant of motion.

Now there is the following task:

Consider $L(v,q)$ from $(*)$ in $\mathbb{R}^3$ with $M=m\mathbb{I}_3$ and suppose $U(q)=U(\lvert q\rvert)$ is rotation invariant. Show that the angular momentum $l= x\wedge p$ is conserved in this case. Here $\wedge$ denotes the cross product in $\mathbb{R}^3$.

I think I have to apply the theorem of Noether, but I do not know how to do so. The first problem for me already is to calculate $p(v,q)$.

M. Meyer
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  • For your first problem note that $L = \frac{1}{2}m\sum v_i^2 - U(q)$ so $\frac{dL}{dv_i} = m v_i \equiv p_i$ or in more compact notation $\frac{dL}{dv} = mv \equiv p$. You can use Noethers theorem to solve this, but it is just as easy to compute $\frac{d\vec{l}}{dt}$ directly applying the equaiton of motion and $\vec{q}\times \vec{q} = 0$ to simplify. – Winther Sep 10 '15 at 17:08
  • Isn't it $\frac{\partial L}{\partial v_i}=m_iv_i^2$? – M. Meyer Sep 10 '15 at 17:11
  • No. Note that $\frac{1}{2}vMv = \frac{1}{2} \sum_{i,j} v_i M_{ij} v_j$. Now since $M_{ij} = m\delta_{ij}$ (the identity matrix) we get $\frac{dL}{dv_k} = \frac{d}{dv_k}\frac{1}{2}m\sum_i v_i^2 = \frac{1}{2}m (2v_k) = m v_k$. This is nothing but $\frac{dx^2}{dx} = 2x$. – Winther Sep 10 '15 at 17:16
  • In order to apply the given theorem of Noether, I need to know what $f(q)$ in the formular is and the generated flow... how do I know/ get them? – M. Meyer Sep 10 '15 at 17:19
  • As far as I do understand, this is meant as the right side of $\dot{q}=f(q)$ (autonomous ODE) and the flow is meant to be generated by $f(q)$. – M. Meyer Sep 10 '15 at 17:28
  • It looks more like $f(q)$ are the generators of the symmetry that leave $L$ invariant. For rotations about $\vec{n}$ this would be something like $\vec{f}(\vec{q}) = \vec{n}\times \vec{q}$. – Winther Sep 10 '15 at 17:40
  • I think this is the right side of an ODE, since here http://www4.ncsu.edu/~schecter/ma_732_sp13/teschl_ode.pdf this seems to be the case. In particular, pp. 238. – M. Meyer Sep 10 '15 at 17:45

1 Answers1

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If we have a Lagrangain $L(\vec{v},\vec{q})$ and the (infinitesimal) transformation $\vec{q}\to \vec{q} + \epsilon \vec{\sigma}(t)$ leaves it invariant then Noether's theorem states that

$$I = \frac{dL}{d\vec{v}}\cdot \vec{\sigma}$$

is a conserved quantity. For rotations about a direction $\vec{n}$ we have $\vec{\sigma} = \vec{n}\times \vec{q}$ so

$$I = \vec{p} \cdot (\vec{n}\times \vec{q}) = \vec{n}\cdot (\vec{q}\times \vec{p}) = \vec{n}\cdot \vec{l}$$

Since $\vec{n}$ is arbitrary we get that $\vec{l}$ is conserved. For more information see for example this note.

One can also solve this problem explicitly. For the case $M = m I_n$ and $U(\vec{q}) = U(q)$ where $q=|\vec{q}|$ we find

$$\frac{dL}{d\vec{v}} = m\vec{v} \equiv \vec{p},~~~\frac{dL}{d\vec{q}} = -\frac{dU(q)}{dq}\frac{\vec{q}}{q}$$

so from the Euler-Lagrange equations the equation of motion is

$$\frac{d\vec{p}}{dt} = \frac{dU}{dq}\frac{\vec{q}}{q}$$

Using the equation of motion we can calculate the time derivative of $\vec{l} = \vec{q}\times \vec{p}$ to find

$$\frac{d\vec{l}}{dt} = \frac{d\vec{q}}{dt}\times \vec{p} + \vec{q}\times \frac{d\vec{p}}{dt} = \frac{\vec{p}\times \vec{p}}{m} + \frac{dU}{dq}\frac{\vec{q}\times\vec{q}}{q} = 0$$

and angular momentum is conserved.

Winther
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  • Without any doubt, this is a perfect answer. Nonetheless, my main question remains, what the formulation of the theorem means with $f(q)$ and the generated flow $\Phi(t,q)$. – M. Meyer Sep 10 '15 at 19:26
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    @M.Meyer As I said I'm not familiar with the language used, but it seems to be the exact same concept just in a different language. What I call the infinitesimal generator of the transformation (here the rotation) the theorem calls the generator of the flow. I'm only guessing but I would think that something like $\Phi(t,q) = R(t) q$ with $R$ being some rotation matrix would be the generated flow and the generator of this flow would be $f(q) \propto \vec{\sigma} = \vec{n}\times \vec{q}$. This choice for $f(q)$ would give the correct result. – Winther Sep 10 '15 at 19:39
  • This may be correct. I am just wondering, since there is no remark in the text that explains sth like this -and in the previous parts, flow always seems to be associated with an ODE: – M. Meyer Sep 10 '15 at 19:41
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    @M.Meyer Well there is an associated ODE here, the equation of motion for the system $\frac{d\vec{q}}{dt} = \frac{dU}{dq}\frac{\vec{q}}{q}$. Both $\vec{q}$ and $\Phi = R\vec{q}$ for any rotation matrix $R$ satisfy it (multiply by $R$ and use $|\vec{q}| = |R\vec{q}|$ to see this). So it might be that this is meant by the flow $\Phi(t,q)$ – Winther Sep 10 '15 at 20:00
  • How do you get $\frac{dL}{d\vec{q}}$? – M. Meyer Sep 10 '15 at 20:14
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    @M.Meyer Since $U$ is only a function of the norm $q = |\vec{q}| = \sqrt{q_1^2+ \ldots + q_n^2}$ we have $\frac{dU}{dq_i} = \frac{dU}{dq}\frac{dq}{dq_i} = \frac{dU}{dq}\frac{q_i}{q}$. I now see that I have a sign error in my answer, will correct it. – Winther Sep 10 '15 at 20:16
  • @Winther One question please. Can you indicate me please some information/papers/anything you have - regarding infinitesimal transformation? thanks! – Iuli Sep 10 '15 at 20:22
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    @Iuli This note has a nice introduction to this subject with regards to Noethers theorem. If you want to digg deeper try searching for 'Lie groups' or 'generators of Lie groups'. – Winther Sep 10 '15 at 20:27
  • Did not get it completely. The ODE here seems to be $\ddot{q}=-\frac{1}{m}\text{grad}U(q)$. So here, $f(q)=-\frac{1}{m}\text{grad}(U(q)$? If I write this as a system, I have $\dot{q}=v, \dot{v}=-\frac{1}{m}\text{grad}(U(q))$ and so the right side now is $f(q,v)=(v,-\frac{1}{m}\text{grad}(U(q))$. What is now the generated flow by $f(q)$ and $f(q,v)$, respectively? – M. Meyer Sep 10 '15 at 20:39
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    @M.Meyer Well $f(q)$ cannot be the right hand side of the Euler Lagrange ODE. To clear this up I checked the book you are using. Take a look at (8.51) on PDF-page 252 and try to reproduct the example below (for linear momentum). It seem the authors do exactly the same as I did above. It seems they consider the transformation $q⃗^s(t)=Φ(s,q(t))=q+sf(q)$. Thus in this case we would get $f(q)=n⃗ × q⃗$ which is what I called $σ$. – Winther Sep 10 '15 at 22:21