Sum of internal forces equals zero

In summary: 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...In summary, the conversation discusses the calculation of the force on a system of mutually interacting particles and how it can be simplified using the antisymmetric property of the force. The conversation also touches upon the use of double summations and the need for clear notation to avoid confusion.
  • #1
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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.
 
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  • #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.
 
  • #3
Fightfish said:
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.

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
$$\sum_{i=1}^{N}\sum_{j=1, i\neq j}^{N}\mathbf{F}_{ij}=\sum_{i=1}^{N}\left(\sum_{i<j}\mathbf{F}_{ij}+\sum_{i>j}\mathbf{F}_{ij}\right)=\sum_{i=1}^{N}\left(\sum_{i<j}\mathbf{F}_{ij}+\sum_{i<j}\mathbf{F}_{ji}\right)=\sum_{i=1}^{N}\sum_{i<j}\left(\mathbf{F}_{ij}+\mathbf{F}_{ji}\right)$$
 
  • #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.
 
  • #5
Fightfish said:
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.

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##?
 
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  • #6
Frank Castle said:
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.
Frank Castle said:
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.
 
  • #7
Fightfish said:
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"?!

Fightfish said:
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.
 
  • #8
Frank Castle said:
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.
 

1. What does the concept of "Sum of internal forces equals zero" mean?

The concept of "Sum of internal forces equals zero" refers to the principle of static equilibrium in physics. It states that for an object to be in a state of rest or constant motion, the sum of all internal forces acting on the object must be equal to zero.

2. Why is it important to understand the concept of "Sum of internal forces equals zero"?

Understanding this concept is important because it helps us to analyze and predict the behavior of a system or structure under different forces. It also allows us to determine if an object is in a state of equilibrium or if it is experiencing unbalanced forces.

3. How is the concept of "Sum of internal forces equals zero" related to Newton's first law of motion?

The concept of "Sum of internal forces equals zero" is closely related to Newton's first law of motion, also known as the law of inertia. This law states that an object at rest will remain at rest and an object in motion will continue in motion with constant velocity unless acted upon by an external force. The concept of "Sum of internal forces equals zero" ensures that there are no unbalanced forces acting on an object, thus allowing it to maintain its state of rest or constant motion.

4. Is the concept of "Sum of internal forces equals zero" applicable to all types of forces?

Yes, the concept of "Sum of internal forces equals zero" is applicable to all types of forces, including gravitational, electromagnetic, and contact forces. It is a fundamental principle in physics and is used to analyze the equilibrium of both simple and complex systems.

5. How can the concept of "Sum of internal forces equals zero" be applied in real-world situations?

The concept of "Sum of internal forces equals zero" has many practical applications in engineering and everyday life. It is used to design and analyze structures, such as bridges and buildings, to ensure they can withstand external forces without collapsing. It is also used in the field of biomechanics to study human movement and determine the forces acting on the body during different activities.

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