I was watching a video describing a way to deduce Newton's Third law using symmetry considerations.
Consider a system of $N$ particles not acted upon by any external force (isolated system). Suppose the medium in which the system in placed is homogenous.
Now we are displacing each particle in the system simultaneously and in the same direction such that the configuration of the system remains the same.
Now it has been said that the displacement which we have done is instantaneous i.e., there is no time interval involved in making such displacement. So it is just a thought process. Such displacements are known as virtual dislacements. So, there is no question of what agency has brought out this displacement.
Now we are trying to find out the virtual work done by the internal forces (as external force = $0$)
It has been said that $\text{Virtual work done}\;=\;0\tag{1}$
So, $(\sum_i\vec F_i).\vec{\delta s}=0$
As the above equation holds for all $\delta\vec s$
By Newton's second law,
$\frac{d}{dt}\sum_i\vec{p_i}=0\implies\frac{d}{dt}\vec P=0$
So, total momentum is conserved.
If we have two particle system, then
$\sum_{i=1}^2\vec F_i=0\implies \vec{F_1}+\vec{F_2}=0$
Thus, proving Newton's Third Law.

I have a doubt that the proof relies on $(1)$. I am not able to understand how $(1)$ is true?
In Goldstein's Classical Mechanics, momentum conservation has been proved through virtual work but by considering $(1)$ as a consequence of Newton's Third Law.
But here, I am not able to understand how we are considering $(1)$ to be true? I am not abe to understand that.

  • $\begingroup$ Similar questions have been asked here before several times, check out for example physics.stackexchange.com/questions/12122/…. It isn't strictly enough to assume that translation (what you called "virtual displacement") doesn't change the forces. You also need to assume that Lagrangian stationary-action is a principle of the universe, and other stuff (like only two-body interactions, rotational invariance, etc.) to strictly reach the third law. $\endgroup$ May 8, 2023 at 9:22

1 Answer 1


By the work-energy theorem, the amount of work done to a system equals the increase in its mechanical energy. However, after performing a virtual displacement, the system is in exactly the same state, only at a different point in space. Thus its mechanical energy must be the same before and after the virtual displacement (assuming only the homogeneity of space). This can only be so if the work done in displacing the system is zero.

  • $\begingroup$ Thanks a lot for the answer. I have one doubt is it meaningful to evaluate the work done by internal force. Because the translation described in the question is a mental construct, translation is happening at an instant. So, it is not even meaningful to ask the agency behind the translation. So, how it is meaningful to even calculate the work done by the internal force, which leads to the derivation of conservation of linear momentum? $\endgroup$
    – Iti
    May 10, 2023 at 12:29
  • $\begingroup$ @Iti Think of the virtual work as a virtual change in the energy of the system after you perform your virtual displacement: if your system is supposed to have the same energy after the displacement you have zero virtual work. If for instance your system is in a gravitational well, virtual displacements in the vertical direction do not have zero virtual work. In fact you can think of momentum conservation as translational invariance. In a potential well your space is not translational invariant hence you have virtual work hence no momentum conservation $\endgroup$
    – LolloBoldo
    May 10, 2023 at 21:45
  • $\begingroup$ Yeah this is fine. Basically my doubt is that how quantity like virtual work (which is actually a mathematical construct) is used to prove the principles which we can observe physically. In reality, we can't do virtual displacement because this happen at an instant so how this is used to prove which we observe in reality (conservation of momentum) $\endgroup$
    – Iti
    May 11, 2023 at 7:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.