# Conservation of Momentum for system of particles

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• mark2142
In summary, the momentum is conserved in a system of particles if there is no external force acting on them.

#### 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?

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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.

Ibix said:
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?

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

Ibix said:
Newton's third law applies to each interaction, so momentum is conserved.
Can you show or prove it mathematically?

mark2142 said:
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.

mark2142
mark2142 said:
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.

malawi_glenn, mark2142 and Delta2
Delta2 said:
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?
Delta2 said:
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).

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mark2142 said:
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

mark2142 said:
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
Delta2 said:
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!)

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.

Nugatory and malawi_glenn
PeroK said:
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.

Delta2 said:
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.

jbriggs444 said:
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
PeroK said:
where the total money in the two accounts remains the same

Thank You Guys.

Delta2
Delta2 said:
I don't know the following sentence is confusing me
PeroK said:
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".

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PeroK
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.

Delta2 said:
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.

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Delta2
jbriggs444 said:
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.

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.

Delta2 said:
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
jbriggs444 said:
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.

jbriggs444
Delta2 said:
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?

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PeroK
mark2142 said:
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.

Delta2 said:
if all the collisions are among the same line
But its not in same line. Then?

Delta2 said:
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.

mark2142 said:
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).

jbriggs444 said:
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
Delta2 said:
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.

mark2142 said:
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.

Delta2 said:
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.

mark2142 said:
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.

Delta2 said:
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
mark2142 said:
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