Sum of internal forces equals zero

[Frank Castle]

This is probably a very trivial question, but my brain isn't "playing ball" today so I'm hoping someone can help me with this.

Suppose I have a system of $N$ mutually interacting particles, then the force on the $i$-th particle due to the other $N-1$ particles is given by $$\mathbf{F}_{i}=\sum_{j=1, i\neq j}^{N}\mathbf{F}_{ij}$$ If I then introduce a net external force, $\mathbf{F}^{ext}$ acting on the whole system, then the total force, $\mathbf{F}$ acting on the system is given by $$\mathbf{F}=\mathbf{F}^{ext}+\sum_{i=1}^{N}\sum_{j=1, i\neq j}^{N}\mathbf{F}_{ij}$$ Concentrating on the double sum one can expand this as $$ \sum_{i=1}^{N}\sum_{j=1, i\neq j}^{N}\mathbf{F}_{ij}=(\mathbf{F}_{12}+\mathbf{F}_{13}+\cdots)+(\mathbf{F}_{21}+\mathbf{F}_{23}+\cdots)+(\mathbf{F}_{31}+\mathbf{F}_{32}+\cdots)+(\mathbf{F}_{41}+\mathbf{F}_{42}+\cdots)+\cdots \\ =(\mathbf{F}_{12}+\mathbf{F}_{21}+\cdots)+(\mathbf{F}_{13}+\mathbf{F}_{31}+\cdots)+(\mathbf{F}_{14}+\mathbf{F}_{41}+\cdots)+\cdots \\ = \sum_{1<j}(\mathbf{F}_{1j}+\mathbf{F}_{j1})+\sum_{2<j}(\mathbf{F}_{2j}+\mathbf{F}_{j2})+\sum_{3<j}(\mathbf{F}_{3j}+\mathbf{F}_{j3})+\sum_{1<j}(\mathbf{F}_{4j}+\mathbf{F}_{j4})+ \cdots\\ =\sum_{i=1}^{N}\sum_{i<j}(\mathbf{F}_{ij}+\mathbf{F}_{ji})$$ From Newton's 3rd law we have that $\mathbf{F}_{ij}=-\mathbf{F}_{ji}$, and so clearly this whole term vanishes (term-by-term). Hence we are left with the known result that the total force acting on a system of $N$ particles is equal to the external force acting on the system, thus enabling us to treat the system (as a whole) as a point particle.

Now, I'm sure there must be a more elegant why to arrive at this result than the way I have above, i.e. manipulating the double sum without having to expand in the way I did, but I can't seem to see the wood for the trees at the moment.

[One thought I had was to write $$\sum_{i=1}^{N}\sum_{j=1, i\neq j}^{N}\mathbf{F}_{ij}=\sum_{i=1}^{N}\left(\sum_{i<j}\mathbf{F}_{ij}+\sum_{i>j}\mathbf{F}_{ij}\right)$$ but I'm a little unsure about this. I mean, can one legitimately write $\sum_{i>j}\mathbf{F}_{ij}=\sum_{i<j}\mathbf{F}_{ji}$?]

If anyone can provide a more elegant approach I'd much appreciate it.

 

[Fightfish]

There's a nice trick that makes use of the fact that $F_{ij}$ is antisymmetric with respect to an exchange in the indices. By making use of the antisymmetric property, we have
$$\sum_{i \neq j} F_{i j } = - \sum_{i \neq j} F_{ji } = -\sum_{j \neq i} F_{i j } = -\sum_{i \neq j} F_{i j }$$
where in going from the second to the third expression, we used the fact that $i$ and $j$ are dummy indices and from the third to the fourth, that the double summation is commutative.

 

[Frank Castle]

There's a nice trick that makes use of the fact that $F_{ij}$ is antisymmetric with respect to an exchange in the indices. By making use of the antisymmetric property, we have
$$\sum_{i \neq j} F_{i j } = - \sum_{i \neq j} F_{ji } = -\sum_{j \neq i} F_{i j } = -\sum_{i \neq j} F_{i j }$$
where in going from the second to the third expression, we used the fact that $i$ and $j$ are dummy indices and from the third to the fourth, that the double summation is commutative.

Would it be correct to express it as follows:

$$\sum_{i=1}^{N}\sum_{j=1, i\neq j}^{N}\mathbf{F}_{ij}=\sum_{i=1}^{N}\left(\sum_{i<j}\mathbf{F}_{ij}+\sum_{i>j}\mathbf{F}_{ij}\right)$$

and then use the fact that double summation is commutative (and that $i$ and $j$ are dummy indices) to write $$\sum_{i=1}^{N}\sum_{i>j}\mathbf{F}_{ij}=\sum_{i=1}^{N}\sum_{i<j}\mathbf{F}_{ji}$$ such that
$$\sum_{i=1}^{N}\sum_{j=1, i\neq j}^{N}\mathbf{F}_{ij}=\sum_{i=1}^{N}\left(\sum_{i<j}\mathbf{F}_{ij}+\sum_{i>j}\mathbf{F}_{ij}\right)=\sum_{i=1}^{N}\left(\sum_{i<j}\mathbf{F}_{ij}+\sum_{i<j}\mathbf{F}_{ji}\right)=\sum_{i=1}^{N}\sum_{i<j}\left(\mathbf{F}_{ij}+\mathbf{F}_{ji}\right)$$

 

[Fightfish]

Yes, it looks confusing but is technically correct. You might want to be a little careful about notation though; when you wrote $\sum_{i > j}$ you were actually summing over $j$ only, and so the choice of notation might be confusing, because to most people it either means summing over $i$ or an implicit double summation.

 

[Frank Castle]

Yes, it looks confusing but is technically correct. You might want to be a little careful about notation though; when you wrote $\sum_{i > j}$ you were actually summing over $j$ only, and so the choice of notation might be confusing, because to most people it either means summing over $i$ or an implicit double summation.

Ah ok. Would it be better to write it something like this:

$$\sum_{i=1}^{N}\sum_{j=1,\;i\neq j}^{N}\mathbf{F}_{ij}=\sum_{i=1}^{N}\left(\sum_{j= 1}^{i-1}\mathbf{F}_{ij}+\sum_{j= 1}^{i-1}\mathbf{F}_{ji}\right)=\sum_{i=1}^{N}\sum_{j= 1}^{i-1}\left(\mathbf{F}_{ij}+\mathbf{F}_{ji}\right)$$

What originally confused me was the notation used in David Tong's "Dynamics and relativity" notes. In section 5 (starting on page 67) he discusses this and I'm confused by his summation notation on page 68, in particular, How he ends up with $$\sum_{i}\sum_{i\neq j}\mathbf{F}_{ij}=\sum_{i<j}\left(\mathbf{F}_{ij}+\mathbf{F}_{ji}\right)$$ Should one interpret this as a sum over all pairs $(i,j)$ with the condition that $i<j$?

 

[Fightfish]

Ah ok. Would it be better to write it something like this:
$$\sum_{i=1}^{N}\sum_{j=1,\;i\neq j}^{N}\mathbf{F}_{ij}=\sum_{i=1}^{N}\left(\sum_{j= 1}^{i-1}\mathbf{F}_{ij}+\sum_{j= 1}^{i-1}\mathbf{F}_{ji}\right)=\sum_{i=1}^{N}\sum_{j= 1}^{i-1}\left(\mathbf{F}_{ij}+\mathbf{F}_{ji}\right)$$

Yup, its clearer that way.

What originally confused me was the notation used in David Tong's "Dynamics and relativity" notes. In section 5 (starting on page 67) he discusses this and I'm confused by his summation notation on page 68, in particular, How he ends up with $$\sum_{i}\sum_{i\neq j}\mathbf{F}_{ij}=\sum_{i<j}\left(\mathbf{F}_{ij}+\mathbf{F}_{ji}\right)$$ Should one interpret this as a sum over all pairs $(i,j)$ with the condition that $i<j$?

Yes, the only reasonable interpretation is that it is a double summation over both indices. Although this then means that there may be confusion or inconsistency in his earlier summation notation. I guess one can always tell from the context what it is that is being summed over, but personally I don't quite like that formalism haha.

 

[Frank Castle]

Yup, its clearer that way.

The only problem it maybe inconsistent, since in the case where $i=1$, the second sum becomes $\sum_{j=1}^{0}$ which doesn't make any sense since there is no "zeroth term"?!

I guess one can always tell from the context what it is that is being summed over, but personally I don't quite like that formalism haha.

Me neither, I feel that it causes unnecessary confusion.

 

[Fightfish]

The only problem it maybe inconsistent, since in the case where $i=1$, the second sum becomes $\sum_{j=1}^{0}$ which doesn't make any sense since there is no "zeroth term"?!

Well, it just means that nothing happens for that value of $i$ i.e. no contribution.