# Violation of Newton's second law if the mass if changing?

I learned some thing called Galilean principle of relativity which says that two inertial frames are equivalent and the laws of physics are the same in both inertial frames.

However here comes the problem with Newton's second law: suppose inertial frame 2 is moving at a constant velocity $$\textbf{u}$$ with respect to inertial frame 1. There is an object with non-constant mass, $$m(t)$$, as a function of time, $$t$$, moving with a velocity $$\textbf{v}_{1}(t)$$, also as a function of $$t$$, in inertial frame 1. Therefore, the object has a vecolity $$\textbf{v}_{2}(t) = \textbf{v}_{1}(t) - \textbf{u}$$ in inertial frame 2. Apply Newton's second law in inertial frame 1 to find the net force, $$\textbf{F}_{1}$$, on the object in inertial frame 1: $$$$\textbf{F}_{1} = \frac{dm(t)\textbf{v}_{1}(t)}{dt} = \frac{dm(t)}{dt}\textbf{v}_{1}(t) + m\frac{d\textbf{v}_{1}(t)}{dt}.$$$$

Now, apply Newton's second law to the object to find the net force, $$\textbf{F}_{2}$$, on the object in inertial frame 2: $$$$\textbf{F}_{2} = \frac{dm(t)\textbf{v}_{1}(t) - \textbf{u}}{dt} = \frac{dm(t)}{dt}\textbf{v}_{1}(t) + m\frac{d\textbf{v}_{1}(t)}{dt} - \frac{dm}{dt}\textbf{u}.$$$$

Clearly, $$\textbf{F}_{1} \neq \textbf{F}_{2}$$. How to explain the difference?

• Newton's 2nd law $F=dp/dt$ is meant for constant mass bodies, thus $m$ is assumed constant. For bodies that lose/acquire mass, it does not hold. Commented Mar 6 at 18:18
• You mean violation of Newton's third law? Commented Mar 7 at 9:03

Let me just slightly restate your problem to make sure I'm being clear. Let's say we have frame A and frame B, where $$\vec{v}_{BA} = \vec{u}$$, meaning the "velocity of B with respect to A" is $$\vec{u}$$. Then if we have an object, C, with velocity $$\vec{v}_{CA} = \vec{v}$$, its velocity with respect to frame B is (using the relative velocity equation): $$\vec{v}_{CB} = \vec{v}_{CA} + \vec{v}_{AB} = \vec{v} + \left( - \vec{v}_{BA} \right) = \vec{v} - \vec{u} \,.$$ Now you are assuming that Newton's Second Law says that the net force on an object over time can be found kinematically from the momentum: $$F_C = \frac{d \vec{p}_C(t)}{d t} \quad \text{with} \quad \vec{p}(t) = m(t) \, \vec{v}(t)\,.$$ Then, for frame A, we have: \begin{align} \vec{F}_C^{(A)} &= \frac{d \vec{p}^{(A)}_C(t)}{d t}\\ & = \frac{d m_C}{dt}\, \vec{v}_{CA} + m_C \, \frac{d \vec{v}_{CA}}{dt} = \frac{d m_C}{dt}\, \vec{v} + m_C \frac{d \vec{v}}{dt}\,, \end{align} and for frame B: \begin{align} \vec{F}_C^{(B)} &= \frac{d \vec{p}^{(B)}_C(t)}{d t} \\ & = \frac{d m_C}{dt}\, \vec{v}_{CB} + m_C \, \frac{d \vec{v}_{CB}}{dt} \\ & = \frac{d m_C}{dt} \left(\vec{v} - \vec{u} \right) + m_C \frac{d \left(\vec{v} - \vec{u} \right)}{dt} = \frac{d m_C}{dt} \left(\vec{v} - \vec{u} \right) + m_C \frac{d \vec{v}}{dt}\,, \end{align} where the last equality follows because $$\vec{u}$$ is a constant. So we find that: $$\left(\vec{F}_C^{(A)} - \vec{F}_C^{(B)}\right)(t) = \frac{dm_C(t)}{dt} \, \vec{u}$$ Is that your conclusion?

I think the issue is that your statement of Newton's Second Law is incorrect. The law is the equation of motion for a particle of mass $$m$$. In that way we can write Newton II in all of these equivalent ways: $$\vec{F}_\text{net} = m \vec{a} = m \frac{d\vec{v}}{dt} = \frac{d \left( m \vec{v}\right)}{dt} = \frac{d \vec{p}}{dt}$$ exactly because we assume that the particle has constant mass.

If you would like to write down a "Newton's Second Law for an object with changing mass", then you need to derive an effective equation for a system of particles. See, for example, my recent answer here. The result of summing many Newton II equations for many (say, $$N$$) particles is this: $$\vec{F}_\text{net, external} = M_\text{tot} \frac{d \vec{V}_\text{com}}{d t}\,,$$ where the left-hand side is the net external force on the system (all internal forces have cancelled out via Newton III); $$M_\text{tot}$$ is the sum of all the particle masses; and the center-of-mass velocity is defined as: $$\vec{V}_\text{com} = \frac{d}{dt} \vec{R}_\text{com} = \frac{d}{dt} \left( \frac{m_1 \vec{r}_1 + \cdots + m_N \vec{r}_N}{M_\text{tot}} \right)\,.$$

I think you need to frame your question in terms of that extension of Newton's Second Law. I'm not sure how to precisely state such a question, but it would be a nice exercise. I don't think, though, that in doing so you will arrive at such contradictions, because to have a contradiction you'll need to show it happening at the level of the equation of Newton's Second Law for a particle.

In your original statement, which I summarized at the beginning of the answer, there is simply no contradiction because it is impossible, within the theory, for a particle $$m$$ to have changing mass.

Actually $$F_1$$ does equal $$F_2$$ because even for an inertial reference frame, the mass/weight (apparent weight) remains constant, so the $$\frac{dm}{dt}$$ terms are just 0. This is not a violation of Newtons second law by any means, things get interesting for rockets though ($$\frac{dm}{dt}\neq 0$$ because the propellant carries quite a bit of the mass of the rocket and is being flushed out at a high rate so there is some mass flow rate).

• $\frac{dm}{dt} = 0$ because $m$ is a constant. But if $m$ is changing, you get my result above Commented Mar 6 at 15:23
• Yeah but then if mass flow rate = 0 the argument that newtons laws are invalid doesn't hold because the only way this is true is if the frame is non-inertial Commented Mar 6 at 15:47
– Community Bot
Commented Mar 6 at 15:51

The acceleration is not really central to your question; we could ask essentially the same thing by considering any object that is not accelerating, but whose mass is changing.

To restate your point: in the object's own rest frame, there is no net force on the object, since the velocity and acceleration are zero. In a frame where the object is moving, its momentum is changing, despite there being no force on it, leading to an apparent paradox.

I think the resolution is that, if things could magically gain or lose mass, Newton's laws would not obey Galilean invariance. Of course, there would also be no energy conservation, no momentum or angular momentum conservation, etc.

In Newtonian mechanics, if a cart gains mass, the mass is coming from somewhere, e.g. someone pouring sand into it.

So imagine a cart rolling along the ground to the right at constant speed $$v,$$ and someone pouring sand into it from above. We can think about two scenarios:

1. The sand is also moving to the right at $$v$$ as it falls.
2. The sand has no left/ right movement as it falls.

In scenario 1, there is no external push needed to keep the cart moving at constant speed. As seen from the ground frame, the system of the cart plus any sand in it gains momentum to the right, but this momentum came in from the sand as it entered the system. As seen in the cart's own frame, the sand has no left/right momentum and the momentum of the system isn't changing. Both perspectives tell us we don't have to push on the cart.

In scenario 2, both perspectives say we do have to push on the cart. In the ground frame, the cart is gaining momentum without the sand bringing that momentum in, so we have to push with a force $$F_{\rm push} = \dot{m} v$$ where $$\dot{m}$$ is the rate that the sand adds mass to the system of the cart and any sand in it. As seen in the cart frame, the sand is moving to the left at $$v,$$ bringing in momentum per unit time $$-\dot{m} v,$$ and we need the same $$F_{\rm push}$$ to keep the momentum of the system at zero.

In either scenario, Galilean relativity holds. Observers in any reference frame can apply a Newtonian analysis and agree about whether there must be an external push on the cart, and on the magnitude of that push.

What remains to be resolved is the exact semantics of "force". In scenario 1, in the ground frame, the sand is bringing momentum into the system of the cart plus any sand in it. If we define "force" to mean any flux of momentum into a system and include systems where mass is flowing in or out, then the sand is "exerting a force" on the system as it falls in the cart, and from the cart's own reference frame, the sand is not "exerting a force" on the system. In that sense "force" would not be a Galilean invariant. I don't think this is usually what people mean when they say "force", though. Instead, they usually mean whether or not someone needs to push on the cart to keep the velocity constant.

You can not simply change your mass with magic since the mass of a particle is a conserved quantity, but if you could, Newton's laws would not apply since they only apply to physics. If the mass of your rocket changes due to the material leaving the exhaust then Newton's laws do apply, but that is trivial.

Newton's second law is:

$$\,m\,\frac{d\vec v}{dt}=\vec F$$

the mass m is constant

$$\Rightarrow$$ $$\frac{d}{dt}\left(m\,\vec v\right)= \vec F\quad,\text{or}\\ \int d\left(m\,\vec v\right)=\int \vec F\,dt$$

hence if $$~\vec v(t)\mapsto \vec v(t)-\vec u ~$$ you obtain

$$m\,\vec v=\int \vec F\,dt$$