Here are the proofs regarding the center of mass motion as reported on my book.

$$\vec{r_{cm}}=\frac{\sum\vec{r_i} m_i}{\sum m_i}$$

$$\vec{v_{cm}}=\frac{d{\vec{r_{cm}}}}{dt}=\frac{1}{M}\sum \frac{d}{dt} m_i \vec{r_i}=\frac{1}{M} \sum m_i \vec{v_i}=\frac{1}{M} \vec{P} \tag{1}$$

$$\vec{a_{cm}}=\frac{d{\vec{v_{cm}}}}{dt}=\frac{1}{M}\sum \frac{d}{dt} m_i \vec{v_i}=\frac{1}{M} \sum m_i \vec{a_i}=\frac{1}{M} \vec{F^{(EXT)}}=\frac{1}{M} \frac{d\vec{P}}{dt} \tag{2}$$

Both in $(1)$ and $(2)$ derivatives this assumption was made: the mass of the system $M$ and the mass of each point $m_i$ are constant. Otherwise the derivatives would have been much more complicated.

But I also read that $$\vec{F^{(EXT)}}= \frac{d\vec{P}}{dt}\tag{3}$$

Holds true also if the mass is not constant.

Nevertheless to prove $(2)$ (and so $(3)$) the assumption of constant mass was used, so how can $(3)$ be true without that assumption?

And if $(3)$ holds which mass can vary? The mass of the system $M$ or the mass of the single points $m_i$?

  • 1
    $\begingroup$ Generally mass is conserved unless there are sources and/or sinks in a mass flux continuity equation or some form of nuclear interaction has occurred (e.g., fusion or fission). At high speeds, there is a Lorentz factor that would be included in $\vec{P}$ which is why it is written in the form of (3). Sometimes, the inclusion of the Lorentz factor is treated as though a particle's mass varies (though I am not fond of that description...). $\endgroup$ – honeste_vivere Apr 20 '16 at 16:40
  • $\begingroup$ @honeste_vivere: +1. I would add that, in an open system it is tricky to apply Newton's 2nd law directly, usually one will look at a closed system (closed for a duration $dt$ at least) to apply it safely. $\endgroup$ – L. Levrel Apr 20 '16 at 20:31

For Newton's Laws to hold, mass must not vary. Wherever you read otherwise was mistaken.

Take a look at this answer. It contains a description of why mass must be constant in Newton's Laws in the context of the rocket equation ... but the analysis applies generally. Newton's Laws are not valid for variable mass systems.

  • 1
    $\begingroup$ Why the down vote? Is it because I refer to an answer rather than spelling it out here? If so, I am guilty and perhaps deserve the down vote. But my answer is correct. To answer the OP explicitly: neither mass can vary. Newton's laws are not valid in systems of variable mass. See this from Wikipedia $\endgroup$ – garyp Apr 23 '16 at 16:04
  • $\begingroup$ well, I dont agree... at least it depends on what you mean by Newtons laws. The equation (3) in the question is valid (in the relativistic case, where mass varies). The question is: can we generalize the proof, or is the necessary proof conceptually different. $\endgroup$ – Ilja Apr 27 '16 at 14:34
  • $\begingroup$ @Ilja Mass does not vary in the relativistic case in the sense that atoms are added or removed. One needs to use the relativistic expression for momentum, not $dm/dt$. But perhaps the better way to say it is that Newton's Laws are not valid for open systems, where mass can enter or leave the system. $\endgroup$ – garyp Apr 27 '16 at 15:50
  • $\begingroup$ yes, that's right. I wanted to say, that probably the formula (3) was mentioned in the context of the relativistic distinction between $ma$ and $\dot p$ $\endgroup$ – Ilja Apr 27 '16 at 15:58

One could not defend a book you are most probably misquoting. I strongly suspect the book says "the mass distribution is not constant", that is M is constant but the distribution and number of constituent $m_i$s may vary, i.e. they may split or agregate, a common feature in astrophysics. You are confusing yourself with symbols and definitions and proofs.

The problem is trivial if you start with one particle of mass M which splits into two equal particles of mass m each, by a small explosion, and you find the cm position, velocity, and acceleration before and after the split, so you convince yourself (1), (2) and (3) are preserved. Then, and only then, bother with generic splits, constituent $m_i$s and positions.


Just a note, you can rewrite (3) as $$\vec{F} = \frac{d\vec{P}}{dt} = \frac{dm}{dt}\vec{v} + m\frac{d\vec{v}}{dt}$$ If you have a system where the total mass is not changing, then $\frac{dm}{dt}=0$, which brings us back to the form we're more commonly used to.

  • 1
    $\begingroup$ That's looks good mathematically, but it is invalid physically because Newton's Second Law has as a condition that $dm/dt = 0$. Your if must be satisfied before you start. $\endgroup$ – garyp Apr 27 '16 at 16:08
  • $\begingroup$ My mistake. I thought that was the derivation of the 'Ideal Rocket Equation', but it's actually derived in a slightly different manner. Thank you $\endgroup$ – Tweej Apr 27 '16 at 18:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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