Conservation of Momentum for system of particles

In summary, the momentum is conserved in a system of particles if there is no external force acting on them.
  • #36
Thank you.
 
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  • #37
Delta2 said:
Yes it is not needed but we got to refer component wise then. Because if say initially the momentum of body 1 is in x-direction and after a collision it exchanges momentum with body 2 in an angle 45 degrees with the x-direction, then I am not sure what we ll mean by saying its momentum increased or decreased as its momentum now will make an an angle with the x-direction. Momentum is a vector quantity.
Right. So you stop saying that [a] momentum [component] has increased or decreased and say instead that the vector sum of the change on body 1 and the change on body 2 is the zero vector.

The math does not care about the words you use. The equation that says that the vector sum is zero works regardless.
 
  • #38
Instead of saying that the momentum is conserved pairwise why not say the vector sum is conserved as a whole even for 2 body system. That solves the problem I guess.
 
  • #39
mark2142 said:
Instead of saying that the momentum is conserved pairwise why not say the vector sum is conserved as a whole even for 2 body system. That solves the problem I guess.
Its the exact same thing with different words whether you say momentum conserved or the vector sum is conserved. Neither momentum neither vector sum is necessarily conserved pairwise if you have more than 2 bodies.

I thought your point was to make some sort of intuitive conclusion of the style" Since momentum is conserved, if momentum is increased for somebody then for some other body must decrease in order for the sum to remain constant". Yes that's true but only component wise . That is you may say, if the x-momentum of body 1 is increased, then the x-momentum of at least one other body must decrease. You got to use x,y,z components for these because momentum is a vector that is it has magnitude and direction, so if it changes it can change either in magnitude or direction or both. But if it changes in direction then I just can't understand the meaning of the premise "If the momentum is increased". However since the x-component has always constant direction along the x-axis and conservation of momentum holds component wise, you can say that for the x (or y or z) component.
 
  • #40
mark2142 said:
Instead of saying that the momentum is conserved pairwise why not say the vector sum is conserved as a whole even for 2 body system. That solves the problem I guess.
It's conserved as a whole. That's it. I don't know, what you mean by "pairwise conserved". I've given the derivation of momentum conservation in #7 for the most common case that there are only pair interactions relevant. Is that what you mean by "pairwise conserved"?
 
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  • #41
Pairwise conserved means that for any two ##i,j## $$\dot p_i+\dot p_j=0$$

It holds if those are the two bodies in our system but in the case our system includes 3 or more bodies I don't think it necessarily holds.
 
  • #42
No, only total momentum is conserved.
 
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  • #43
Delta2 said:
Pairwise conserved means that for any two ##i,j## $$\dot p_i+\dot p_j=0$$

It holds if those are the two bodies in our system but in the case our system includes 3 or more bodies I don't think it necessarily holds.
That applies if the interactions are a sequence of discrete collisions involving two particles at a time. And it only applies to particles ##i## and ##j## when they interact.

The alternative model, e.g., a gravitational or EM system, is a continuous process of all particles interacting continuously with all others.
 
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  • #44
vanhees71 said:
Is that what you mean by "pairwise conserved"?
Delta2 said:
Pairwise conserved means that for any two ##i,j## $$\dot p_i+\dot p_j=0$$

It holds if those are the two bodies in our system but in the case our system includes 3 or more bodies I don't think it necessarily holds.
That is what I mean. I get it, It can only happen if there are only pair collisions(between 2 particles at a time) inside a system but that is not what happens.
jbriggs444 said:
The proviso about collisions being on the same line is not needed.

Newton's third law applies in the x direction, the y direction and the z direction separately. As a result, momentum is conserved in the x direction, the y direction and the z directions separately.

One does not need the lines of action of all of the x-direction forces/impulses to coincide in order to conserve momentum in the x direction. The position of the line of action of a force is irrelevant for the purposes of linear momentum. It is enough that for each internal impulse there is an equal and opposite internal impulse.
Sorry for replying so late. I was trying to understand. I have made two diagrams. Is that what you mean by no need for position of line of action of a forces to coincide?
(Since the impulses are equal and opposite so all the force components in x and y directions will cancel out)
I have not made components in second diagram since it was getting messy.
 

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  • #45
mark2142 said:
That is what I mean. I get it, It can only happen if there are only pair collisions(between 2 particles at a time) inside a system but that is not what happens.

Sorry for replying so late. I was trying to understand. I have made two diagrams. Is that what you mean by no need for position of line of action of a forces to coincide?
No, that is not what I had in mind.

In context, I believe that we are talking about three (or more) bodies. For simplicity, we could call them A, B and C. We have a number of interactions between pairs. For simplicity, we could reduce it to two interactions: AB and AC.

There was an assertion against which I was arguing. Well, not quite an assertion -- a caveat that I claim is not needed. That caveat is boldfaced here:

"If the interactions all have lines of action that line up on a single axis and conserve momentum pairwise then total momentum is conserved."

In fact, given nothing but pairwise momentum conservation for each interaction, linear momentum is still conserved. The caveat about lines of action is not needed.This is not to say that the lines of action of the pairwise interaction forces are irrelevant. They are important. One can make the following assertion:

If each pairwise interaction conserves momentum and has a line of action that passes through the center of mass of the two interacting bodies then angular momentum is conserved.

More advanced physics students may note that the caveat here about lines of action can be taken as a somewhat corrupted way of saying that "there is no preferred direction" or "the laws of physics are isotropic". This, in turn means that Noether's theorem assures us that there is a corresponding conserved quantity (i.e. angular momentum).
 
  • #46
vanhees71 said:
No, only total momentum is conserved.
Newton's formulation involves only pairwise interactions. Technically, we need some kind of handwave or incantation make the leap from momentum conservation for pairwise interactions to momentum conservation for all possible many-body interactions.

A more modern formulation would indeed take total momentum conservation as the given and pairwise conservation as a trivial special case.
 
  • #47
jbriggs444 said:
Newton's formulation involves only pairwise interactions. Technically, we need some kind of handwave or incantation make the leap from momentum conservation for pairwise interactions to momentum conservation for all possible many-body interactions.
Perhaps waving a few differentials and reciting the principle of superposition of vectors would do!
 
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  • #48
PeroK said:
Perhaps waving a few differentials and reciting the principle of superposition of vectors would do!
Just don't let anyone see you do it and remember to clean up after.
 
  • #49
jbriggs444 said:
Newton's formulation involves only pairwise interactions. Technically, we need some kind of handwave or incantation make the leap from momentum conservation for pairwise interactions to momentum conservation for all possible many-body interactions.

A more modern formulation would indeed take total momentum conservation as the given and pairwise conservation as a trivial special case.
Newton's mechanics or Noether's theorem (aka the symmetrie of the Newtonian spacetime model) does in no way restrict interactions to pair interactions. E.g., the interaction between nucleons contains generic ##n##-body forces. Even with only pair interactions the momentum conservation is not fulfilled for each particle pair but only for the total momentum (see the very simple derivation based on Newton's 3 postulates in #7).
 
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  • #50
vanhees71 said:
Newton's mechanics or Noether's theorem (aka the symmetrie of the Newtonian spacetime model) does in no way restrict interactions to pair interactions. E.g., the interaction between nucleons contains generic ##n##-body forces.
Yes indeed.

Newton's original formulation is not so broad, however:

Lex III: Actioni contrariam semper et �qualem esse reactionem: sive corporum duorum actiones in se mutuo semper esse �quales et in partes contrarias dirigi.

Translation:
  • All forces occur in pairs, and these two forces are equal in magnitude and opposite in direction.
 
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  • #51
Can I do it like this to conserve total P of say 3 body system in y-direction?
$$F_{aby}=-F_{bay}$$
$$F_{cby}=-F_{bcy}$$
$$F_{cay}=-F_{acy}$$
$$F_{aby}+F_{cby}+F_{cay}=-F_{bay}-F_{bcy}-F_{acy}$$
$$\frac d{dt}(p_{ay}+p_{cy}+p_{cy})=-\frac d{dt}(p_{by}+p_{by}+p_{ay})$$
$$\frac d{dt}(p_{ay}+p_{cy}+p_{cy}+p_{by}+p_{by}+p_{ay})=0$$
$$\frac d{dt}(p_{ay}+p_{by}+p_{cy})=0$$
$$p_{ay}+p_{by}+p_{cy}=constant$$
 
  • #52
mark2142 said:
Can I do it like this to conserve total P of say 3 body system in y-direction?
$$F_{aby}+F_{cby}+F_{cay}=-F_{bay}-F_{bcy}-F_{acy}$$
$$\frac d{dt}(p_{ay}+p_{cy}+p_{cy})=-\frac d{dt}(p_{by}+p_{by}+p_{ay})$$
Please explain the above step. It looks like hideous garbling of Newton's second law.

As I decrypt it, the idea is that (for instance) the derivative of the momentum of c is equal to the force of a on c and is also equal to the force of b on c. But Newton's second actually says that the derivative of the momentum of c is equal to the sum of the force of a on c and the force of b on c.

It's ##\sum F = ma##, not ##F_1 = ma = F_2##.
 
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  • #53
jbriggs444 said:
Newton's second actually says that the derivative of the momentum of c is equal to the sum of the force of a on c and the force of b on c.
Oh yes! It should be
$$(F_{aby}+F_{acy})+(F_{cby}+F_{cay})+(F_{bcy}+F_{bay})=0$$
$$dp_a/dt+dp_b/dt+dp_c/dt=0$$
$$p_a+p_b+p_c=constant$$
Thank you.
Also I understand that in any system if there are only internal transfers of momentum then the net P of the system can not change. Its just that the interaction is so complex between particles of different masses (atoms) that it seems impossible.
 
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