# Why is Ratio of $3$ particle is $1$?

Suppose that when $$P_2$$ interacts with a third particle $$P_3$$ the induced accelerations are $$a_{23}$$ and $$a_{32}$$, and when $$P_1$$ interacts with $$P_3$$ the induced accelerations are $$a_{13}$$ and $$a_{31}$$. Then the magnitudes of these accelerations satisfy the consistency relation∗ $$|a_{21}| /|a_{12}| × |a_{32}|/ |a_{23}| × |a_{13}| /|a_{31}| = 1$$. Why is multiplication of ratio of 3 particle's accleration is 1?

My thought is ratio of each particle accleration acting on each other is 1. But this is true for some case or they must be canceling each other in this equation. But the way they give the definition sounds like it is true for all.

Newton's Third Law says that $$\vec F_{ij} = -\vec F_{ji}$$ where $$\vec F_{ij}$$ is the force on $$j$$ due to $$i$$. Newton's Second Law says that, if this is the only force acting on that object $$\vec F_{ij} = m_j \vec a_{ij}$$ where $$m_j$$ is the mass of $$j$$ and $$a_{ij}$$ is the acceleration of $$j$$ resulting from interaction with $$i$$. This gives us $$\lvert \vec a_{ij} \rvert = \frac{\lvert \vec F_{ij} \rvert}{m_j} = \frac{\lvert \vec F_{ji} \rvert}{m_j}$$ Substituting this into your equation gives $$\frac{\lvert a_{21} \rvert}{\lvert a_{12} \rvert} \cdot \frac{\lvert a_{32} \rvert}{\lvert a_{23} \rvert} \cdot \frac{\lvert a_{13} \rvert}{\lvert a_{31} \rvert} = \frac{m_2}{m_1} \cdot \frac{m_3}{m_2} \cdot \frac{m_1}{m_3} = 1$$