B Conservation of Momentum for system of particles

[mark2142]

We know that if we take two particles and assume no external force is applied then by Newtons third law total momentum gets conserved after collision. If we take three particles and there is collision between them and no external force then the momentum is again conserved for each pair like in figure below (arrows represent internal forces).
How can we do this for system of particles?

 

[Ibix]

I don't understand what it is you want to do for a system of particles. You just seem to be stating the law of conservation of momentum - it applies equally to a closed $n$-body system.

 

[mark2142]

I don't understand what it is you want to do for a system of particles

I am asking how is momentum conserved for system of n particles?

 

[Ibix]

Newton's third law applies to each interaction, so momentum is conserved.

 

[mark2142]

Newton's third law applies to each interaction, so momentum is conserved.

Can you show or prove it mathematically?

 

[Delta2]

the momentum is again conserved for each pair

No in the case of 3 bodies and more general for n bodies, the momentum is NOT necessarily conserved pairwise. It is conserved as a whole, that is "vector sum of momentum of all bodies before collision=vector sum of momentum of all bodies after collision". The full mathematical proof in the case of n bodies comes from Newton's 2nd and 3rd law and it is more an exercise in indexes and summation than a proof that has really something else to say besides the proof we do for 2 bodies.
I 'll do it for 3 bodies to see what I mean. By $F_{ij}$ I ll mean the force from body i to body j. And $P_i$ the momentum of the body i before collision and $P'_i$ its momentum after collision.

So we will have by Newtons 2nd law for each body (actually from impulse momentum theorem which is an immediate consequence of Newton's 2nd)
$$P'_1=P_1+\int F_{21} dt +\int F_{31} dt$$
$$P'_2=P_2+\int F_{12} dt+\int F_{32} dt$$
$$P'_3=P_3+\int F_{13} dt+\int F_{23} dt$$

By summing all the three equations above we get
$$P'_1+P'_2+P'_3=P_1+P_2+P_3+\int F_{21}dt+\int F_{12}dt+\int F_{31} dt+\int F_{13} dt+\int F_{32} dt+\int F_{23} dt$$

Now from Newton's 3rd Law we have $F_{ij}=-F_{ji}$ so $F_{21}=-F_{12}$, $F_{31}=-F_{13}$ and $F_{32}=-F_{23}$ so all the integrals are summed out to zero and what is left is $$P_1+P_2+P_3=P'_1+P'_2+P'_3$$.

PS. I should have put the vector sign in Ps ,P's and Fs so for example when i write $P_1$ i should have written $\vec{P_1}$ and for $F_{12}$ , $\vec{F}_{12}$ and so on but I did it that way for Latex convenience.

 

[vanhees71]

Can you show or prove it mathematically?

It's indeed much simpler to use mathematics than many words. In almost all cases the interaction forces in a many-body system are sums over two-body interactions, i.e., the force on particle $j$ is given by
$$\vec{F}_j= \sum_{k \neq j} \vec{F}_{jk}.$$
$\vec{F}_{jk}$ is the interaction force on particle $j$ due to particle $k$.

Newton's third law now says that
$$\vec{F}_{jk}=-\vec{F}_{kj}.$$
The equation of motion for particle $j$ is
$$\dot{\vec{p}}_j=m_j \ddot{\vec{x}}_j = \sum_{k \neq j} \vec{F}_{jk}.$$
Now sum this over $j$:
$$\dot{\vec{P}}=\sum_j \dot{\vec{p}}_j = \sum_j \sum_{k \neq j} \vec{F}_{jk} = \frac{1}{2} \sum_{j \neq k} (\vec{F}_{jk}+\vec{F}_{kj})=0 \; \Rightarrow \; \vec{P}=\text{const}.$$
QED.

 

[mark2142]

No in the case of 3 bodies and more general for n bodies, the momentum is NOT necessarily conserved pairwise. It is conserved as a whole, that is "vector sum of momentum of all bodies before collision=vector sum of momentum of all bodies after collision".

Oh. Ok!That theory would need to have even number of particles I guess + there are many forces acting on each particle?

The full mathematical proof in the case of n bodies comes from Newton's 2nd and 3rd law

Shouldn't we just prove $\frac{dP}{dt}= F_{ext}$ ?
(Where P is momentum of whole system).

 

[Delta2]

Shouldn't we just prove $\frac{dP}{dt}=F_{ext}$ ?
(Where P is momentum of whole system).

In a way that's what I prove. I assume $F_{ext}=0$ and prove that $$P=\text{constant}\Rightarrow \frac{dP}{dt}=0=F_{ext}$$
Maybe you should follow Vanhees71 approach and add a term $F_{i,ext}$ for the external force on body i. At the end of the day you ll get a + $\sum F_{i,ext}$ term which won't simplify to zero so at the bottom line you ll have $$\frac{dP}{dt}=\sum F_{i,ext}=F_{ext}$$ where $F_{ext}$ the total external force on the system of the bodies

 

[Delta2]

Oh. Ok!That theory would need to have even number of particles I guess.

No that is not the reason that the momentum is not conserved pairwise. The reason is that there would be forces and impulses from other "external" bodies to the 2-body system which will affect the momentum of the 2body system.

 

[mark2142]

In a way that's what I prove. I assume and prove that

Great! So for n particles there will be many forces on each particle. But by Newtons 3rd law each will get canceled and we will get the same total momentum as before.
(Sorry I was editing!)

 

[PeroK]

The following analogy works by effectively the same argument.

If we have $n$ bank accounts and a series of transactions between any two accounts where the total money in the two accounts remains the same then the total money in all $n$ accounts stays the same.

The total amount in the $n$ accounts can only be changed by an external input or output of money.

 

[Delta2]

The following analogy works by effectively the same argument.

If we have $n$ bank accounts and a series of transactions between any two accounts where the total money in the two accounts remains the same then the total money in all $n$ accounts stays the same.

The total amount in the $n$ accounts can only be changed by an external input or output of money.

This analogy seems to imply that the momentum is conserved pairwise in a system of n bodies which is not necessarily true.

 

[jbriggs444]

This analogy seems to imply that the momentum is conserved pairwise in a system of n bodies which is not necessarily true.

The analogy seems solid to me.

I transfer two gold coins to Joe. Joe transfers three gold coins to Brad. Brad takes one gold coin from Holly. If these are the only transactions that take place, what is the resulting change in the total number of gold coins held by the four of us?

There is no implication of pairwise conservation for the total held by Holly and myself.

 

[Delta2]

The analogy seems solid to me.

I transfer two gold coins to Joe. Joe transfers three gold coins to Brad. Brad takes one gold coin from Holly. If these are the only transactions that take place, what is the resulting change in the total number of gold coins held by the four of us?

There is no implication of pairwise conservation for the total held by Holly and myself.

I don't know the following sentence is confusing me

where the total money in the two accounts remains the same

 

[mark2142]

Thank You Guys.

 

[jbriggs444]

I don't know the following sentence is confusing me

If we have n bank accounts and a series of transactions between any two accounts where the total money in the two accounts remains the same

So @PeroK is contemplating a series of simple pairwise transactions. Each individual transaction involves only two accounts and conserves the total in those two accounts.

This much seems obvious:

"If each individual transaction conserves the total amount of money then a whole series of such transactions [even if not all between the same two individuals] must also conserve the total amount of money."

We could add some additional observations which seem pretty obvious:

[Finite Commutativity]

"[If we allow overdrafts] When performing a finite set of of pairwise transactions the sequence in which the individual transactions are performed is irrelevant to the final result."

and

[Superposition or Pyramid Schemes Don't Work]

"Any valid transaction among three or more participants.
can be achieved with a finite sequence of money-conserving pairwise transactions"

or

"No non-money-conserving transaction among three or more participants can be achieved with a finite sequence of money-conserving pairwise transactions".

 

[Delta2]

Nvm I think I got it, the total amount of money pairwise is conserved for a single transaction but this doesn't mean that it is conserved after many transactions.

Of course the total amount of money (not pairwise) is conserved after a single and after many transactions.

 

[jbriggs444]

Nvm I think I got it, the total amount of money pairwise is conserved for a single transaction but this doesn't mean that it is conserved after many transactions.

Mathematical induction would allow conservation over every finite sequence of transactions to be proven.

Edit: on re-parsing the quoted statement, I see what you are asserting and agree.

 

[Delta2]

Mathematical induction would allow conservation over every finite sequence of transactions to be proven.

If you are saying that the momentum is conserved pairwise, no sorry I can't agree to that.

 

[nasu]

Momentum is conserved for any pairwise interaction. But the momentum of the pair may change due to interactions with other particles. The momentum of the pair is not necessarily conserved even if the pairwise interaction conserves momentum.

 

[jbriggs444]

If you are saying that the momentum is conserved pairwise, no sorry I can't agree to that.

Fortunately, I am only saying that total money is conserved. Not that the sum of the money in every pair of accounts is conserved.

 

[Delta2]

Fortunately, I am only saying that total money is conserved. Not that the sum of the money in every pair of accounts is conserved.

Ok great I can agree to that.

 

[mark2142]

If you are saying that the momentum is conserved pairwise, no sorry I can't agree to that.

If total P doesn't change does that mean the individual p's of the particles also not change at all. In other words are the particles at rest because there are equal and opposite forces at work?

 

[Delta2]

Does this happen in system of particles as well or do we assume

Yes I believe it holds for a system of more than 2 bodies as well, if all the collisions are among the same line, then you can say that if the momentum of 1 body increases, then there must be at least one other body of the system whose momentum decreases.

 

[mark2142]

if all the collisions are among the same line

But its not in same line. Then?

 

[jbriggs444]

Yes I believe it holds for a system of more than 2 bodies as well, if all the collisions are among the same line, then you can say that if the momentum of 1 body increases, then there must be at least one other body of the system whose momentum decreases.

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.

 

[Delta2]

But its not in same line. Then?

You got to refer in the momentums component wise then. For example if the x-momentum of body 1 increases, then the x-momentum of some other body has to decrease (not necessarily by the same amount).

 

[Delta2]

The proviso about collisions being on the same line is not needed.

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.

 

[mark2142]

You got to refer in the momentums component wise then. For example if the x-momentum of body 1 increases, then the x-momentum of some other body has to decrease (not necessarily by the same amount).

How can you be so sure that these things work in the same way at microscopic level as it works at macroscopic level?
These principles are found out by experimenting with balls and stuffs. Is it because we don't see bodies move on their own? ( For that we need to apply force?)
I hope you understand me.

 

[Delta2]

How can you be so sure that these things work in the same way at microscopic level as it works at macroscopic level?
These principles are found out by experimenting with balls and stuffs. Is it because we don't see bodies move on their own? ( For that we need to apply force?)
I hope you understand me.

I am not sure what you asking. Deep down we can't be sure about anything but whenever we do experiments in the lab, either microscopic or macroscopic experiments we have found that conservation of momentum holds.

 

[mark2142]

I am not sure what you asking. Deep down we can't be sure about anything but whenever we do experiments in the lab, either microscopic or macroscopic experiments we have found that conservation of momentum holds.

But there is no microscope that exist with which we can see atoms at work.

 

[Delta2]

But there is no microscope that exist with which we can see atoms at work.

Hm maybe not with small atoms like Hydrogen, but with large atoms like that of Uranium , I think in nuclear fission experiments conservation of momentum has been experimentally confirmed.

 

[mark2142]

Hm maybe not with small atoms like Hydrogen, but with large atoms like that of Uranium , I think in nuclear fission experiments conservation of momentum has been experimentally confirmed.

Oh. Thats great.
We can also say the momentum has to be conserved or else we would see a ball moving on its own without applying any force. Yes?

 

[Delta2]

Oh. Thats great.
We can also say the momentum has to be conserved or else we would see a ball moving on its own without applying any force. Yes?

Yes.

 

[mark2142]

Thank you.

 

[jbriggs444]

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.

 

[mark2142]

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.

 

[Delta2]

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.

 

[vanhees71]

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"?

 

[Delta2]

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.

 

[vanhees71]

No, only total momentum is conserved.

 

[PeroK]

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.

 

[mark2142]

Is that what you mean by "pairwise conserved"?

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.

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.

 

[jbriggs444]

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).

 

[jbriggs444]

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.

 

[PeroK]

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!

 

[jbriggs444]

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.

 

[vanhees71]

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).

 

[jbriggs444]

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.