PDA

View Full Version : Special invariants with few constants of motion


wkb13
Oct28-09, 09:57 PM
Ordinarily, a system of N particles in d dimensions has 2Nd constants of motion, but there are certain invariants, like energy and angular momentum, that have a lot fewer. What's so special about these? Why do they have so few constants of motion?

Gerenuk
Oct29-09, 03:12 AM
Ordinarily, a system of N particles in d dimensions has 2Nd constants of motion, but there are certain invariants, like energy and angular momentum, that have a lot fewer. What's so special about these? Why do they have so few constants of motion?
Hope I understand this correctly. Not sure why you refer to everything as "constants".
The invariants are due to the special form of the interaction between particles which is \propto\vec{r}/r^3. If you have another fictional, crazy type of interaction, you wouldn't have energy and angular momentum conserved.

wkb13
Oct29-09, 03:25 AM
Yes, I guess "parameters of motion" would have been a more appropriate term. So, if I understand correctly, any inverse square interaction will have fewer than 2Nd parameters?

Gerenuk
Oct29-09, 03:30 AM
Yes, any inverse square law where action equals reaction (and I believe magnetism also doesn't harm), will conserve the sum of kinetic plus potential energy and also will conserve the the total angular momentum for the system and thus reduce the total number of parameters you need.

Bob_for_short
Oct29-09, 05:19 AM
Ordinarily, a system of N particles in d dimensions has 2Nd constants of motion, but there are certain invariants, like energy and angular momentum, that have a lot fewer. What's so special about these? Why do they have so few constants of motion?
The special feature of the total energy, momentum, and the angular momentum is that they are additive in particles. The inter-particle interaction potentials are not involved. It helps in certain simple cases (scattering, for example). The other integrals of motion are harder to find and they are not additive in particles.

Gerenuk
Oct29-09, 05:28 AM
The special feature of the total energy, momentum, and the angular momentum is that they are additive in particles.
What means additive? Every quantity can be added up. And we want to consider only those whose sum is constant over time.
And quantities like entropy are additive in the idealized case, but not conserved. So additivity doesn't play a role.


The inter-particle interaction potentials are not involved.

The interaction potential is the only determining feature. Just imagine a crazy unphysical force equation for the particles with index i:

F_i=\begin{cases}
C & \text{if }i=j\\
0 & \text{otherwise}
\end{cases}

This physics would make particle j fly away to infinity and there is no conservation of energy of angular momentum.

Bob_for_short
Oct29-09, 05:59 AM
What means additive?
It means that the total momentum is a sum of particle momenta, for example.
The interaction potential is the only determining feature. ...
I speak of inter-particle potentials, not of the external force. In presence of external force the additive conservation laws may not be valid.

Gerenuk
Oct29-09, 06:05 AM
It means that the total momentum is a sum of particle momenta, for example.

That also applies for the x-component of particles. The total "x-component" of particle set A and B together is equal to the sum of their individual x-component sums. Yet, the quantity is not conserved during motion.


I speak of inter-particle potentials, not of the external force. In presence of external force the additive conservation laws may not be valid.
Correct. I should refer to the total force on each particle which should be of the form

F_i=\sum_{j\neq i} \frac{a_{ij}\hat{r}_{ij}}{r_{ij}^2}


a_{ij}=a_{ji}

for energy and momentum and angular momentum to work. That is necessary and sufficient I believe. Well, almost. I guess a conservative velocity dependent force like a magnetic field can also be added and yet the derivations for the conserved quantities would work.

Bob_for_short
Oct29-09, 06:33 AM
That also applies for the x-component of particles. The total "x-component" of particle set A and B together is equal to the sum of their individual x-component sums. Yet, the quantity is not conserved during motion.

Component of what vector? If you speak of momentum, the total vector P is conserved:

dPx/dt = 0, dPy/dt = 0, dPz/dt = 0.

And Px = Σk(px)k, etc.

Gerenuk
Oct30-09, 05:45 AM
Component of what vector? If you speak of momentum, the total vector P is conserved
Oh come on. I forgot to say component of velocity, but it's really not hard to make up additive quantities that are not conserved. How about x+v_y\cdot 1\mathrm{s} where x is x coordinate and v_y the y component of the velocity.