Landau-Lifshitz Mechanics says that there are $2s-1$ constants of a system with $s$ degrees of freedom (beginning of the second chapter on Conservation Laws). If this is true, for a single free particle moving in 3D space, there should be $5$ constants of the motion. However, I can just count $7$ of them: three components of linear momentum, three components of the angular momentum, and the kinetic energy. What is wrong here?
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2$\begingroup$ $E = p^2/2m$ reduces the number of independent quantities by one. Is there another such constraint? $\endgroup$– d_bCommented Jul 6, 2019 at 7:09
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1$\begingroup$ It's possible that knowledge of the three components of the linear momentum and two of the angular momentum is sufficient to recover the third. That needs careful analysis to confirm, though. $\endgroup$– Emilio PisantyCommented Jul 6, 2019 at 9:52
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$\begingroup$ Isn't $E=L^2/2I$ the other constraint? $\endgroup$– probably_someoneCommented Jul 6, 2019 at 9:52
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$\begingroup$ Emilio is right. Given the fact that three components of linear momentum and two components of the angular momentum don't change in time, it follows that the third component of angular momentum is also unchanged. $\endgroup$– SinghCommented Jul 6, 2019 at 12:42
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2 Answers
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It's true there are 7 conserved quantities. But for a free particle \begin{equation*} E=\frac{1}{2}m\dot{\mathbf{q}}^{2}=\frac{\mathbf{p}^{2}}{2m} \end{equation*} so $E$ can be written in terms of $\mathbf{p}$. So we have now 6 conserved quantities. The other relation you need is $\mathbf{p}\cdot\mathbf{L}=0$ where $\mathbf{p}$ is the linear momentum and $\mathbf{L}$ is the angular momentum, which implies that one component of them can be expressed in terms of the others. So the independent conserved quantities are back to 5.
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Landau and Lifshitz distinguish between constants of motion and conserved quantities--the former being mere mathematical results of integration and the latter being profound statements regarding unchanging quantities in physical systems (energy, momentum, etc.). The $2s-1$ quantity is a count of the former.
For a free particle, its motion can be given by $\vec{x}(t) = \vec{x}_0 + \vec{v}t$, where $\vec{x}_0$ is the position when time $t=0$ and $\vec{v}$ is the velocity. There are six constants of motion: three in $\vec{x}_0$ and three in $\vec{v}$. However, this is, in a sense, over-specified since time does not explicitly occur in the equations of motion (i.e., the original Lagrangian). Choosing another initial position, $\vec{x}(t) = \vec{x}_0' + \vec{v}t$ results in the exact same motion as long as $\vec{x}_0' = \vec{x}_0 + \vec{v}k$ for some value of $k$. The space of constants that don't make a difference is parameterized by a one-dimensional scalar $k$, so the space of constants of motion that do make a difference and are independent is reduced by one.
Since a translation in time is an unimportant transformation (time is homogeneous, so the choice of what $t=0$ means is arbitrary), we can create a constant of motion associated with the choice of the initial time coordinate $t_0$, leaving $2s-1$ constants that depend only on the position and velocity coordinates ($q_i$ and $\dot{q}_i$ for $i \in 1, 2, ... , s$ in the textbook notation). The latter set of constants contain all of the interesting physics.
Here's an example of a set of five ($2s-1$ where $s = 3$) constants of motion that completely define a free particle:
- Velocity along the x-axis,
- Velocity along the y-axis,
- Velocity along the z-axis,
- The distance of closest approach to the origin,
- The angle of the position vector from the positive z-axis at the closest approach to the origin.
These five numbers define the motion of the particle unambiguously in a timeless way.
