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I came across the following in Goldstein's Classical Mechanics book, section 1.3.

In a system of particles, the equation of motion for the $i$'th particle is to be written $$ \sum_j F_{ji}+F_i^{(e)}= \frac{dp_i}{dt}$$ where $F_i^{(e)}$ stands for an external force and $F_{ji}$ is the internal force on the $i$'th particle due to the $j$'th particle.

My question is: what is the meaning of this equation? In the external force $F_i^{(e)}$ why don't we use $j$'th particle symbol? That is, why don't we use $F_{ji}^{(e)}$? If it is meaningless, then how? I'm new to physics, so I would appreciate an explanation.

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    $\begingroup$ Is there a particular reason you chose your first physics book to be a graduate/advanced undergraduate level text? $\endgroup$
    – J. Murray
    Aug 6, 2021 at 4:25
  • $\begingroup$ advanced undergraduate level sir $\endgroup$
    – user310684
    Aug 6, 2021 at 4:26

3 Answers 3

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$F_i^{(e)}$ is the external force on the $i^{th}$ particle - that is, the force on the $i^{th}$ particle which cannot be attributed to any other particle in the system - whereas $F_{ji}$ is the force on the $i^{th}$ particle due to the $j^{th}$ particle. The expression for the total force is then

$$\frac{d\mathbf p_i}{dt} = \underbrace{\mathbf F^{(e)}_i}_{\text{external}}+ \underbrace{\sum_j \mathbf F_{ji}}_{\text{internal}}$$


As an example, consider a system of 3 particles under the influence of gravity. The net force on particle $1$ is

$$\frac{d\mathbf p_1}{dt} = \underbrace{(-mg \hat y)}_{\text{external}} + \underbrace{(\mathbf F_{21} + \mathbf F_{31})}_{\text{internal}}$$

where $\mathbf F_{21}$ and $\mathbf F_{31}$ are the forces on particle $1$ due to particles $2$ and $3$, respectively.

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  • $\begingroup$ sir..could you please give an example of an object/particle and one external force acting on it ? $\endgroup$
    – user310684
    Aug 6, 2021 at 4:30
  • $\begingroup$ @LearneR I've updated my answer. $\endgroup$
    – J. Murray
    Aug 6, 2021 at 4:34
  • $\begingroup$ Thank you sir. (Im only have 3 reputation, so Im unable to upvote (It requires 15 rep) your answer. $\endgroup$
    – user310684
    Aug 6, 2021 at 4:45
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What Goldstein is referring to here is how one particle in an object responds to a force. $F_i^{(e)}$ would be something like gravity acting on the particle directly. In all likelihood $F^{(e)}$ also acts on particle j, but we don't need to know that in order to calculate $\frac{dp_i}{dt}$. Whatever it is that the external force does to particle j is accounted for in $F_{ji}$.

For example, a particle in a ball on a table feels pressure from the particle above it because gravity is acting as $F^{(e)}$ on it. That gives $F_{ji}$ its value.

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  • $\begingroup$ Thank you sir.. $\endgroup$
    – user310684
    Aug 6, 2021 at 4:47
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We are calculating change in moment of "i"th particle so it is sum of all forces applied by other particles and external forces acting on "i"th particle.

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