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I've read in a book, that in general case energy $E$, momentum $\textbf{p}$ and angular momentum $\textbf{M}$ of a closed system are the only additive constants of motion, that is, if I have $N$ closed systems that are not interacting with each other then

$$ E = \sum_{i=1}^N E_i, \;\;\;\;\; \textbf{p} = \sum_{i=1}^N \textbf{p}_i,\;\;\;\;\; \textbf{M} = \sum_{i=1}^N \textbf{M}_i $$

It is obvious that indeed they are additive constants of motion, but how can I prove that there are no other?

Qmechanic
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xaxa
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1 Answers1

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It is false in Newtonian physics just referring to the standard situation of systems of points, interacting with conservative forces only depending on relative distances.

Consider the so-called boost vector, $${\bf k}:= t{\bf p} - M {\bf x}_G\:,$$ where $M$ is the total mass of the system and ${\bf x}_G$ the position of the centre of mass. This quantity is conserved, is additive and it is not a function of the remaining constants of motions you mentioned.

For an isolated system $\frac{d{\bf p}}{dt}=0$, hence: $$\frac{d}{dt}(t{\bf p} - M {\bf x}_G)= {\bf p}-M \frac{d{\bf x}_G}{dt} =0\:.$$

All that is consequence of Noether theorem. Galileo's group has $10$ generators and they correspond to $10$ scalar conserved quantities long the evolution of an isolated system. They are the energy, the three components of the momentum,the three components angular momentum and the three components of the boost.

ADDENDUM. Notice that the boost is a function of the positions of the $n$ points, of the system and their momenta, but it is also a function of time in view of the factor $t$ in front of the total momentum:

$${\bf k} = {\bf k}(t, {\bf x}_1,\ldots, {\bf x}_n, {\bf p}_1, \ldots, {\bf p}_n )\:.$$

Explicitly,

$${\bf k} = t \sum_{k=1}^n {\bf p}_k - \sum_{k=1}^n m_k {\bf x}_k\:.$$

However, when evaluated on a motion of the system: $${\bf x}_i = {\bf x}_i(t)\:,\quad {\bf p}_i = {\bf p}_i(t)$$ it turns out to be a constant: $$\frac{d}{dt} {\bf k}(t, {\bf x}_1(t),\ldots, {\bf x}_n(t), {\bf p}_1(t), \ldots, {\bf p}_n(t) )=0$$ This is the definition of constant of motion.

Valter Moretti
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  • Good point! But if my understanding is correct, this is not a constant of motion. Those things should not depent on time explicitly and must be functions only of $p$s and $q$s. – xaxa Apr 30 '14 at 18:10
  • It is! For an isolated system $d{\bf p}/dt=0$ hence $d/dt(t{\bf p} - M {\bf x}_G)= {\bf p}-M d{\bf x}_G/dt =0$. – Valter Moretti Apr 30 '14 at 18:18
  • It is consequence of Noether theorem, Galileo's group has 10 generators and they correspond to 10 scalar conserved quantities. They are the energy, the three components of the momentum,the three components angular momentum and the three components of the boost. – Valter Moretti Apr 30 '14 at 18:21
  • Yes, it is conserved, but as it depends on $t$, it is not a constant of motion – xaxa Apr 30 '14 at 18:56
  • It is a constant of motion! A function $f(t,x,p)$ is a constant of motion when when evaluated on a motion of the system $x=x(t), p=p(t)$ is constant! Namely $f(t,x(t),p(t)) =$ constant. This is the case. It is irrelevant the fact that $f$ may depend explicitly on time as in this case. – Valter Moretti Apr 30 '14 at 19:16
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    It absolutely is a constant of the motion. Explicit dependence on $t$ does not mean that it has to be changing. – webb Apr 30 '14 at 19:34
  • Looks like I mixed up constant of motion and integral of motion... – xaxa Apr 30 '14 at 20:18