# What is the meaning of external force in the Newtonian force equation?

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.

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

$$F_i^{(e)}$$ 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.

• sir..could you please give an example of an object/particle and one external force acting on it ?
– user310684
Aug 6, 2021 at 4:30
• @LearneR I've updated my answer. Aug 6, 2021 at 4:34
• Thank you sir. (Im only have 3 reputation, so Im unable to upvote (It requires 15 rep) your answer.
– user310684
Aug 6, 2021 at 4:45

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.

• Thank you sir..
– user310684
Aug 6, 2021 at 4:47

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.