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I have a question regarding the meaning of the force in the equation of motion for a variable mass system. I will begin by deriving the equation, in order to better explain my reasoning.

Thus consider an arbitrary control volume, and let the mass inside this volume be $m$ and let the total momentum of the particles inside the control volume be $\mathbf{p}$. Suppose that at time $t+\Delta t$ a parcel of mass $\Delta m_o$ ('o' for out) leaves the system with absolute velocity $\mathbf{v}_o$ in some inertial frame. Similarly, assume that at time $t$ a parcel of mass $\Delta m_i$ ('i' for in) enters the system with absolute velocity $\mathbf{v}_i$.

By the impulse equation, we have $$(\mathbf{p}(t+\Delta t)+\Delta m_o\mathbf{v}_o)-(\mathbf{p}(t)+\Delta m_i\mathbf{v}_i) = \int_{t}^{t+\Delta t}\mathbf{F}dt. $$ Here, the force $\mathbf{F}$ must be the force not just on the control volume, but on all the particles, even those which may leave or enter the volume, in order that we may use Euler's first law.

Dividing through by $\Delta t$, taking the limit $\Delta t \to 0$, invoking the mean value theorem, putting $q_o = \lim_{\Delta t \to 0} \frac{\Delta m_o}{\Delta t}$, similarly putting $q_i = \lim_{\Delta t \to 0} \frac{\Delta m_i}{\Delta t}$ and noting that $\dot{m}=q_i-q_o$, we obtain

$$\mathbf{F}+q_i\mathbf{v}_i-q_o\mathbf{v}_o = \dot{\mathbf{p}} $$

or, if you will,

$$\mathbf{F}+q_i(\mathbf{v}_i-\mathbf{v})-q_o(\mathbf{v}_o-\mathbf{v})=m\dot{\mathbf{v}},$$ in order to get an expression that's explicitly Galielei invariant.

My question then is about the force $\mathbf{F}$. In the derivation, we remarked that this has to be the total external force on all the particles under consideration, not just the particles inside the control volume itself. However, this seems quite unreasonable from another perspective. If I fire a machine gun in space, I will experience some recoil and accelerate as a result of it. Neglecting gravitational influence, we might be tempted to set $\mathbf{F}=\mathbf{0}$. But now suppose that someone puts a plate in the way of my bullets, so that the bullets eventually hit this plate and experience some recoil as a result. This would amount to a nonzero average force on the bullets and thereby on the entire system under consideration, affecting my acceleration according to the equation of motion. However this seems quite absurd. The plate might be pretty far away, and my acceleration is evidently not going to change when the bullets hit the plate.

What am I missing here?

EDIT

We can see the problem even easier in the following way. Say we are still in space and let out collection of particles be, say some nice asteroid. As the control volume, pick a region of empty space, thus with $\mathbf{p}=\mathbf{0}$ and such that no mass passes through it. Our equation of motion is then $\mathbf{F}=\mathbf{0}$. Again, this is absurd since e.g. gravitational forces will be acting on the asteroid, which is part of our collection of particles.

It seems then, that the equation of motion only makes sense if $\mathbf{F}$ refers to the net external force on the control volume. However, in using the impulse law and thereby Euler's first law, we must assume that $\mathbf{F}$ is the net external force on the entire collection of particles.

EDIT

Alright, so I guess the issue is in the original impulse equation. If there are forces acting one the mass outside of the control volume it is not correct, since the LHS does not account for the momentum of this mass. I suppose then that we can only use the above equation for variable mass systems if there is no force on particles which have left the control volume.

Étienne Bézout
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  • I'm not sure about your 1st part, but for your second part, just look at the equation $\vec{F} = \frac{d\vec{P}}{dt}$ where $\vec{P}$ is the total momentum of a system of particles. If all internal forces are equal and opposite, $\vec{F}$ is the net external force summed over all particles. For your 2nd scenario, $\vec{F}$ is nonzero. This just means that some part of the system is changing it's momentum. You won't contribute to the nonzero $dP/dt$, but the bullets will contribute. The equation doesn't tell you which part of the system will have a changing momentum, just that some part will – DWade64 May 16 '18 at 13:35
  • (but we of course know that the nonzero external force F is on the "bullet part of the system" so we know that's where the changing momentum will take place) – DWade64 May 16 '18 at 13:40
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    @DWade64Yes, but my "system" is the control volume, say myself plus gun, and the momentum p refers to this system. – Étienne Bézout May 16 '18 at 15:33
  • I think I see your confusion. From a physics perspective, everything I said was okay. I've only used "control volume" in a fluid dynamics class. Anyways in my previous comments, I had some system, and I was keeping track of everything. That "everything" (which was the person and bullets) was my system. If you want to put a box around the person and gun, this is your control volume, in which bullets (mass) is leaving the control volume. The only difference between this case and the previous description is language. It's just language. $F$ in your derivation should be worded – DWade64 May 16 '18 at 16:17
  • "external force on control volume." $p$ is the momentum of the control volume. If something leaves the control volume and experiences an external force, that force doesn't change the momentum of whatever is in the control volume. So in both cases, $F$ is $0$. In both cases, you recoil out 1 side and the bullets fly out the other. Therefore in both cases $dp/dt$ will be equal to the same value (whatever the outflow rate is) – DWade64 May 16 '18 at 16:22
  • Oh that's interesting. Control volumes are actually a lot more practical. In physics, you always keep track of everything. In reality, you can only put up measuring devices, at say, a particular location. You instruct the devices to keep track of everything in that "control volume". If you could keep track of everything, the equations would be simpler. Keeping track of just a section (a control volume) is much more practical and easier. The downside is that in our equations, we need "flow terms through the surface." – DWade64 May 16 '18 at 16:37
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    @DWade64 Well the whole point of my question is that F in the derivation is not the force on the control volume, but the force on the collection of particles. Otherwise we may not use Euler's first law (basically Newton's second law), which is implicit in the impulse equation. – Étienne Bézout May 17 '18 at 08:35
  • It you and the bullets are part of the same system, there is nothing in the equation which says that a force on the bullets must cause you to accelerate. It says the total change in momentum equals the impulse of an external force. In your example you do not change momentum, only the bullets do, and this change equals the impulse on them. – sammy gerbil May 18 '18 at 14:56
  • @sammygerbil Again, please note my distinction between collection of indestructible particles (here we can apply Euler's laws) and the control volume (system). – Étienne Bézout May 20 '18 at 14:36
  • Please do not make your post look like a revision table, instead just seamlessly integrate the new material into the post. There is an edit history button at the bottom of the post for those interested in seeing what changed. – Kyle Kanos May 22 '18 at 10:18

1 Answers1

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Let us consider a scenario where particles are only leaving the control volume. The vector $\mathbf F(t)$ in your derivation is sum of all external forces acting on constant number of particles relevant for the interval $t,t+\Delta t$: the particles that remain inside plus the particles that leave in this time interval. External forces acting on other particles that left earlier than those are not included in $\mathbf F$.

Since the number of particles that leave in time $\Delta t$ is negligible compared to number of particles inside the control volume, $\mathbf F$ is almost the same as force acting on the particles that remain inside the control volume, and in the limit $\Delta t\rightarrow 0$ they become the same.

This explains your worry - the external forces that occur only some finite time after the subject particles leave the control volume, are no longer relevant and they are not included in $\mathbf F$.

Your derivation is thus quite alright, except for one detail: the forces the leaving particles outside the control volume exert on the particles inside the control volume (I would not count those as external forces). For a rocket, the resultant of these forces can be made negligible if the control volume boundary is chosen far from the compact body inside it - the exhaust gas is rarified there, its pressure low so the force is negligible. See also my answer here:

Second law of Newton for variable mass systems

Ján Lalinský
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  • The constant mass system under consideration in $\mathbf F_{ext} + \mathbf F_{parts} = \sum_k m_k \mathbf a_k$ is not the control volume, nor the super-system, but rather a system of particles entirely within the control volume at $t$ but not entirely in the control volume at $t + dt$ (since some particles in this system have crossed the boundary in time $dt$). Might it be clearer to include a note on this, e.g. $(\mathbf{F}{parts} + \mathbf{F}{ext})dt =(d\mathbf{p}{lost} + \mathbf{p}{cv}(t+dt)) - \mathbf{p}_{cv}(t)$? – 13509 May 17 '20 at 11:33
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    @JamesWirth It would be better to put that comment here https://physics.stackexchange.com/a/142773/31895 – Ján Lalinský May 17 '20 at 11:54
  • Thanks for the advice, I've moved the comment across! – 13509 May 17 '20 at 12:00