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I Sum of internal forces equals zero

  1. May 4, 2016 #1
    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.
     
  2. jcsd
  3. May 4, 2016 #2
    There's a nice trick that makes use of the fact that [itex]F_{ij}[/itex] is antisymmetric with respect to an exchange in the indices. By making use of the antisymmetric property, we have
    [tex]\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 } [/tex]
    where in going from the second to the third expression, we used the fact that [itex]i[/itex] and [itex]j[/itex] are dummy indices and from the third to the fourth, that the double summation is commutative.
     
  4. May 4, 2016 #3
    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
    $$\sum_{i=1}^{N}\sum_{j=1, i\neq j}^{N}\mathbf{F}_{ij}=\sum_{i=1}^{N}\left(\sum_{i<j}\mathbf{F}_{ij}+\sum_{i>j}\mathbf{F}_{ij}\right)=\sum_{i=1}^{N}\left(\sum_{i<j}\mathbf{F}_{ij}+\sum_{i<j}\mathbf{F}_{ji}\right)=\sum_{i=1}^{N}\sum_{i<j}\left(\mathbf{F}_{ij}+\mathbf{F}_{ji}\right)$$
     
  5. May 4, 2016 #4
    Yes, it looks confusing but is technically correct. You might want to be a little careful about notation though; when you wrote [itex]\sum_{i > j}[/itex] you were actually summing over [itex]j[/itex] only, and so the choice of notation might be confusing, because to most people it either means summing over [itex]i[/itex] or an implicit double summation.
     
  6. May 4, 2016 #5
    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##?
     
    Last edited: May 4, 2016
  7. May 4, 2016 #6
    Yup, its clearer that way.
    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.
     
  8. May 4, 2016 #7
    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"?!

    Me neither, I feel that it causes unnecessary confusion.
     
  9. May 4, 2016 #8
    Well, it just means that nothing happens for that value of ##i## i.e. no contribution.
     
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