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Source: An Introduction to modern Astrophysics

Source: An Introduction to Modern Astrophysics

This is from the proof of Virial Theorem. In the above picture's second equation, how the first term is zero due to symmetry from Newton's third Law? I'm more specifically concerned with $(r_{i}+r_{j})$ which are position vector of ith and jth particle. After multiplying by Fij to both of them, how it results to the condition in which we can use $F_{ij}=-F_{ji}$?

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    $\begingroup$ Are you asking why $\mathbf{F}_{ij}=-\mathbf{F}_{ji}$? It's Newton's third law. $\endgroup$
    – J.G.
    Jul 16 at 10:51
  • $\begingroup$ I meant to ask how the first term leads to that Newton's third law. $\endgroup$
    – Bkas
    Jul 16 at 11:16
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I think the simplest way is to take into account that the sum $$\frac{1}{2}\sum_i\sum_{j, j\ne i} {\bf F}_{ij}\cdot ({\bf r}_i + {\bf r}_j) $$ is a sum over all the values of $i$ and $j$, with the only constraint that $i \neq j$.

With respect to the exchange of $i$ and $j$, ${\bf F}_{ij}$ changes its sign (antisymmetric), by Newton's third law, while $({\bf r}_i + {\bf r}_j)$ is symmetric. The resulting scalar product is then antisymmetric. Therefore, each $i,j$ term of the sum is exactly canceled by the term $j,i$.

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  • $\begingroup$ I really like this approach! So if we define $A_{ij}:={\bf F}_{ij}\cdot ({\bf r}_i + {\bf r}_j)$ we have $A_{ij}=-A_{ji}$ and hence $\sum_i\sum_{j, j\ne i} {\bf F}_{ij}\cdot ({\bf r}_i + {\bf r}_j)=\sum_{i\neq j}A_{ij}=\sum_{i<j}A_{ij}+A_{ji}=0$. $\endgroup$
    – Filippo
    Jul 16 at 13:08
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Newton's second law states that the force the $i^{th}$ particle exerts on the $j^{th}$ is equal in magnitude, but opposite in direction to the force that the $j^{th}$ particle exerts back on to the $i^{th}$ particle. That is, $$\vec F_{ij}=-\vec F_{ji}$$

If we look at the sum, we see Newton's third law terms$^*$ multiplied by distance, so that $$\frac{1}{2}\sum_i\sum_{j, j\ne i} F_{ij}\cdot (r_i + r_j)=\frac{1}{2}\underbrace{\sum_i\sum_{j,j\ne i}(F_{ij}\cdot r_i+F_{ij}\cdot {r_j})}_{\text{*3rd law pairs}\ \ \large F_{ij}\cdot r_i=-\vec F_{ji}\cdot r_i}$$ that will continually cancel throughout the sum. Given that $i\ne j$ over the summation, we will get terms like $$F_{12}\cdot r_1 - F_{21}\cdot r_1 +F_{13}\cdot r_1 - F_{31}\cdot r_1 + \cdots +F_{21}\cdot r_2 - F_{12}\cdot r_2 + F_{23}\cdot r_2-F_{32}\cdot r_2 + \cdots=0$$

where we know that all the $F_{ij}=-F_{ji}$ by Newton's third law, and by the explicit symmetry of the sum, the entire term will vanish. This is a simple way of saying that since $F_{ij}$ is antisymmetric, while $(r_i+r_j)$ is purely symmetric, the dot product in the sum will also be antisymmetric, which once again means all of the terms in the original sum will vanish.

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  • $\begingroup$ How do we know that $\sum_{i\neq j}F_{ij}(r_i-r_j)=0$? $\endgroup$
    – Filippo
    Jul 16 at 11:58
  • $\begingroup$ That term should not equal zero. What makes you think it does? $\endgroup$
    – joseph h
    Jul 16 at 12:01
  • $\begingroup$ We want to show that $\frac{1}{2}\sum_i\sum_{i, j\ne i} F_{ij}\cdot (r_i + r_j)=0$, don't we? And $\frac{1}{2}\sum_i\sum_{j, j\ne i} F_{ij}\cdot (r_i + r_j)=\frac{1}{2}\sum_i\sum_{j,j\ne i}F_{ij}\cdot(r_i-r_j)$ according to your second equation. $\endgroup$
    – Filippo
    Jul 16 at 12:05
  • $\begingroup$ Or did I misunderstand the underbrace? $\endgroup$
    – Filippo
    Jul 16 at 12:06
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    $\begingroup$ Consider the second equation from the book, the equation above (41). It has the form $A=B+C$ and according to the book, $B=0$ such that $A=C$ (41). The LHS of the second equation in your answer equals $B$. So we need to show that $B=0$. Inserting Newtons Law this boils down to proving the equality $\sum_i\sum_{j,j\neq i}F_{ij}r_i=\sum_i\sum_{j,j\neq i}F_{ji}r_j$ $\endgroup$
    – Filippo
    Jul 16 at 12:21

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