Discontinuity of velocity in a problem on mass flow and momentum

[Leo Liu]

In this question, the particles are constantly transmitting their momentum to the rocket. The force required to keep the rocket stable can be express as $\vec F=(\vec v-\vec u)\dot m$.
However, when I tried to solve this question using the Newton's 2nd law, I found that the infinitesimal change in the velocity vector in the first derivative is discontinuous, as shown below.
$$\vec F=\frac{dm{\vec v}'}{dt}=m\frac{\overbrace{d{\vec v}'}^{\text{disc.}}}{dt}+{\vec v}'\frac{dm}{dt}$$
After getting stuck, I tried to express the differentiation in terms of limit.
$$\vec F=\lim_{\Delta t\to 0}\frac{\Delta P}{\Delta t}=\lim_{\Delta t\to 0}\frac{\Delta m\vec{v}'}{\Delta t}=\lim_{\Delta t\to 0}\frac{\Delta m}{\Delta t}(\vec v-\vec u)=\dot m(\vec v-\vec u)$$
And it is the same as the solution in my book.

All in all, I would like to know why I have failed to find the solution by applying the chain rule to the 2nd law. Thank you.

 

[etotheipi]

like to know why I have failed to find the solution by applying the chain rule to the 2nd law. Thank you.

Newton's second law only applies to constant mass systems. When the system is changing mass, you need to be very careful to account for how the mass is entering or leaving. For instance, if mass is accumulating with relative velocity $\vec{v}_{rel} = \vec{u} - \vec{v}$ to the spacecraft , then the 'modified form' of Newton's second law you need is$$\vec{F}_{ext} + \vec{v}_{rel} \frac{dm}{dt} = m\frac{d\vec{v}}{dt}$$Notice that in this example, $m\frac{d\vec{v}}{dt} = \vec{0}$ so it reduces to$$\vec{F}_{ext} = -\vec{v}_{rel} \frac{dm}{dt} = (\vec{v} - \vec{u}) \frac{dm}{dt}$$which is the result stated in the book. Notice that they were careful in their derivation to only apply NII to a system of constant mass!

$$\vec F=\lim_{\Delta t\to 0}\frac{\Delta P}{\Delta t}=\lim_{\Delta t\to 0}\frac{\Delta m\vec{v}'}{\Delta t}=\lim_{\Delta t\to 0}\frac{\Delta m}{\Delta t}(\vec v-\vec u)=\dot m(\vec v-\vec u)$$

I don't know what you did here, but it's not a correct manipulation. For instance, even ignoring that it is incorrectly equated to external force (because the system is of variable mass), the RHS should instead be$$\lim_{\Delta t \rightarrow 0}\frac{\Delta \vec{p}}{\Delta t} = \lim_{\Delta t \rightarrow 0} \frac{\Delta (m\vec{v})}{\Delta t} = \lim_{\Delta t \rightarrow 0} \frac{(\Delta m)\vec{v} + m(\Delta \vec{v}) + \Delta m \Delta \vec{v}}{\Delta t} = \lim_{\Delta t \rightarrow 0} \left( \frac{\Delta m}{\Delta t}\vec{v} + m \frac{\Delta \vec{v}}{\Delta t} \right)$$

 

[Leo Liu]

When the system is changing mass, you need to be very careful to account for how the mass is entering or leaving. For instance, if mass is accumulating with relative velocity $\vec{v}_{rel} = \vec{u} - \vec{v}$ to the spacecraft

Hi James. Thanks for your reply.
In your answer, you mentioned that the velocity in the equation is the relative velocity between the particles and the rocket. Yet isn't it true that the 2nd law is always true as long as the reference is an inertial frame?

 

[etotheipi]

Yet isn't it true that the 2nd law is always true as long as the reference is an inertial frame?

Yes but for a constant mass system, i.e. Newton's second law doesn't apply to a variable mass system.

 

[Leo Liu]

Yes but for a constant mass system, i.e. Newton's second law doesn't apply to a variable mass system.

Can you please show me the intuition here ? Why does a variable system behave so differently than a constant system?

 

[etotheipi]

Can you please show me the insight here ?

I don't know if I can do better than to point out that applying NII in it's standard form to an open system gives you contradictory results. When you have a body which is e.g. losing mass at relative velocity $\vec{v}_{rel} = \vec{u} - \vec{v}$, then you can consider the change in momentum of the closed (to mass) system containing both the rocket and the parcel of mass $-dm$ that is lost in the time $dt$,$$d\vec{p} = \left[(m+dm)(\vec{v} + d\vec{v}) + \vec{u} (-dm) \right] - m\vec{v} = (dm)\vec{v} + m(d\vec{v}) - (dm)\vec{u}$$ $$\frac{d\vec{p}}{dt} = \frac{dm}{dt} \vec{v} + m \frac{d\vec{v}}{dt} - \frac{dm}{dt}\vec{u} $$Because this system is of constant mass, and because we are working in an inertial frame, Newton II applies and $\vec{F} = \frac{d\vec{p}}{dt}$ where $\vec{F}$ is the total external force on the system,$$\vec{F} = (\vec{v} - \vec{u})\frac{dm}{dt} + m\frac{d\vec{v}}{dt} = - \vec{v}_{rel} \frac{dm}{dt} + m\frac{d\vec{v}}{dt}$$You'll notice that's the same equation as before, for variable mass systems. It includes the $\vec{v}_{rel}\frac{dm}{dt}$ term that you would not get if you erroneously applied NII to the variable mass system.

 

[etotheipi]

Also, there is another nice way at looking at this problem that involves the Reynold's Transport Theorem. The RTT states that if the boundaries of a system (of constant mass) and a control volume (through which mass can flow into and out of) coincide at some time $t$, then the following relation holds with any extensive quantity $A$, with the corresponding intensive quantity $a = A/m$:$$\frac{dA_{sys}}{dt} = \frac{dA_{cv}}{dt} + \int a \rho \vec{V} \cdot \hat{n} dS$$where $A_{sys}$ and $A_{cv}$ are the amounts of $A$ inside the system and control volume respectively, and $\vec{V}$ is the velocity of the contents w.r.t. the boundary of the control volume. If you consider the momentum $\vec{p}$, you see that$$\frac{d\vec{p}_{sys}}{dt} = \frac{d\vec{p}_{cv}}{dt} + \frac{d\vec{p}_{out}}{dt}$$where $\frac{d\vec{p}_{out}}{dt} = -\frac{dm}{dt} \vec{u}$ is the momentum carried away by the lost mass per time $dt$. The final piece of the puzzle is to realize that the total external force on the system (constant mass) satisfies $\vec{F} = \frac{d\vec{p}_{sys}}{dt}$. Noting that $\vec{p}_{cv} = m\vec{v}$, you have$$\vec{F} = \frac{d(m\vec{v})}{dt} - \frac{dm}{dt} \vec{u} = (\vec{v} - \vec{u})\frac{dm}{dt} + m\frac{d\vec{v}}{dt} = -\vec{v}_{rel} \frac{dm}{dt} + m\frac{d\vec{v}}{dt}$$which is again the variable mass equation!

 

[DrStupid]

However, when I tried to solve this question using the Newton's 2nd law, I found that the infinitesimal change in the velocity vector in the first derivative is discontinuous, as shown below.
$$\vec F=\frac{dm{\vec v}'}{dt}=m\frac{\overbrace{d{\vec v}'}^{\text{disc.}}}{dt}+{\vec v}'\frac{dm}{dt}$$

That is the total force acting on the rocket but not the external force. If you use this expression for force, than you need to do it for both, rocket and dust. Than there is a force acting on the dust and therefore a corresponding counter force acting on the rocket. Let me show that with different symbols:

The force acting on the rocket is equal to the sum of the external force and the counter force from the dust:

$F_R = m_R \cdot \dot v_R + \dot m_R \cdot v_R = F_{ext} - F_D$

The force acting on the dust is

$F_D = m_D \cdot \dot v_D + \dot m_D \cdot v_D$

The dust is moving with constant velocity ($\dot v_D = 0$) and the total mass is conserved ($\dot m_D = - \dot m_R$). This results in

$F_{ext} = m_R \cdot \dot v_R + \dot m_R \cdot \left( {v_R - v_D } \right)$

and with constant velocity of the rocket in

$F_{ext} = \dot m_R \cdot \left( {v_R - v_D } \right)$

 

[Leo Liu]

That is the total force acting on the rocket but not the external force. If you use this expression for force, than you need to do it for both, rocket and dust. Than there is a force acting on the dust and therefore a corresponding counter force acting on the rocket.

Aha, this makes a lot more sense. Thank you for your clear explanation.

the total mass is conserved \dot m_D = - \dot m_R

Could you please explain why this relation is negatively proportional? Is it because we don't consider the particles attached to the rocket a separate system?

Also, I like you username.

 

[etotheipi]

Yes sometimes the $\vec{v}_{rel} \frac{dm}{dt}$ term is identified as the thrust. I think this not the best way to think about it, because although that term has dimensions of force, it is not a force so much as a correction term to Newton II. To think of that term as a force acting on the rocket is a somewhat slippery concept. Please refer to Kleppner and Kolenkow:

Recall that $\vec{F} = \frac{d\vec{p}}{dt}$ was established for a system composed of a certain set of particles... it is essential to deal with the same set of particles throughout the time interval... Consequently, the mass of the system can not change during the time of interest.

An introduction to mechanics; Kleppner, Kolenkow

Also in Halliday and Resnick:

It is important to note that we cannot derive a general expression for Newton's second law for variable mass systems by treating the mass in $\vec{F} = \frac{d\vec{p}}{dt} = \frac{d(m\vec{v})}{dt}$ as a variable. [...] We can use $\vec{F} = \frac{d\vec{p}}{dt}$ to analyze variable mass systems only if we apply it to an entire system of constant mass having parts among which there is an interchange of mass.

Physics; Halliday, Resnick

 

[DrStupid]

Could you please explain why this relation is negatively proportional? Is it because we don't consider the particles attached to the rocket a separate system?

You may consider the attached particles as a separate system. But that would make the derivation more complicate without any benefit. It is better to distinguish between the free dust and the rocket with the dust particles attached. That's what I did above. Than the rocket gains the mass that the dust loses.

 

[DrStupid]

Yes sometimes the $\vec{v}_{rel} \frac{dm}{dt}$ term is identified as the thrust. I think this not the best way to think about it, because although that term has dimensions of force, it is not a force so much as a correction term to Newton II.

Yes, it is not (always) a force. But isn't that the very reason for the term "thrust" instead of "force"?

Please refer to Kleppner and Kolenkow:

I never understood their argumentation.

 

[etotheipi]

@DrStupid my problem with your approach is that whilst it gives the correct answer this case, it is physically misleading and even incorrect. The formula you used $$\vec{F} = m\frac{d\vec{v}}{dt} + \frac{dm}{dt} \vec{v}$$is not Galilean invariant, i.e. $\vec{F}$ changes under a Galilean boost, which is obviously wrong. The resolution is that Newton II is not valid for variable mass systems, and the updated form that I derived in #6 and #7 is required.

 

[DrStupid]

The formula you used $$\vec{F} = m\frac{d\vec{v}}{dt} + \frac{dm}{dt} \vec{v}$$is not Galilean invariant, i.e. $\vec{F}$ changes under a Galilean boost, which is obviously wrong.

Why is that wrong? The change of momentum obviously is frame dependent for open systems.

 

[etotheipi]

Why is that wrong? The change of momentum obviously is frame dependent for open systems.

Force is a frame independent quantity in classical physics, and since that expression for force is not Galilean invariant, it cannot be valid.

 

[DrStupid]

Force is a frame independent quantity in classical physics

Not according to F=dp/dt.

 

[etotheipi]

Not according to F=dp/dt.

I have no idea what you mean by this. $\vec{F} = \frac{d\vec{p}}{dt}$ only applies to a constant mass system, for reasons I outline above. Force is a frame-invariant quantity in classical physics. All the fundamental interactions are functions of differences in positions, which are invariants.

 

[DrStupid]

I have no idea what you mean by this.

In the second law Newton defined force to be proportional to the change of momentum. The change of momentum is frame dependent. Thus, force is frame dependent as well according to this definition.

$\vec{F} = \frac{d\vec{p}}{dt}$ only applies to a constant mass system, for reasons I outline above.

Above I see your claim that force need to be frame-independent in classical physics. But I do not see the reasons. You are just repeating this claim with different wordings. Here once again:

Force is a frame-invariant quantity, as is acceleration, in classical physics.

Yes, acceleration is frame-invariant in classical mechanics but with F=dp/dt force is not. You can simply say that you refuse to use the second law in this form. That's your personal choice. But if you say it is wrong than you need to provide a proper justification.

All the fundamental interactions are functions of differences in positions, which are invariants.

And they remain frame independent with F=dp/dt. I do not see the problem.

 

[etotheipi]

In the second law Newton defined force to be proportional to the change of momentum. The change of momentum is frame dependent. Thus, force is frame dependent as well according to this definition.

Since the momentum is linear in velocity, a change in momentum is Galilean-invariant (edit: I mean in cases where the mass is fixed). Then force is Galilean-invariant. And Newton's laws are defined for systems, which are by definition constant in mass.

You can simply say that you refuse to use the second law in this form. That's your personal choice. But if you say it is wrong than you need to provide a proper justification.

My justification is that $\dot{m} \vec{v} + m\dot{\vec{v}}$ is not Galilean invariant. Surely that's good enough for you!

 

[DrStupid]

a change in momentum is Galilean-invariant.

is contradicted by

$\dot{m} \vec{v} + m\dot{\vec{v}}$ is not Galilean invariant.

Either it is Galilean-invariant or it is not Galilean-invariant.

 

[PeterDonis]

If acceleration is Galilean-invariant then so is force, they're related by a constant of proportionality.

This doesn't seem consistent with your statement that Newton's second law only applies to constant mass systems. Can you elaborate?

 

[PeterDonis]

Since the momentum is linear in velocity

Only for a constant mass system. In a variable mass system, this is no longer true.

 

[etotheipi]

This doesn't seem consistent with your statement that Newton's second law only applies to constant mass systems. Can you elaborate?

Yes, that was a lapse in focus on my part (I had constant mass systems in mind when I wrote that). What I meant to say was this:

The quantity $\frac{d\vec{p}}{dt} = m\dot{\vec{v}} + \dot{m}\vec{v}$ is not Galilean invariant in cases where $\dot{m} \neq 0$. Hence, in the cases where $\dot{m} \neq 0$, you can't equate $\frac{d\vec{p}}{dt}$ with $\vec{F}$; you could say this is because fundamental forces in classical mechanics are Galilean invariant by virtue of them being functions of position. The fact that you can't equate a Galilean invariant quantity with a Galilean variant quantity is enough to deduce that Newton's second law cannot hold in variable mass systems.

In the second law Newton defined force to be proportional to the change of momentum. The change of momentum is frame dependent. Thus, force is frame dependent as well according to this definition.

I disagree here. In the second law, Newton defined force to be proportional to the change in momentum, assuming the system is of constant mass.

In any case, I wouldn't take my word for it. But I am putting forward a viewpoint that is widely shared in the literature, e.g. I attached two such examples in a previous post.

 

[PeterDonis]

In the second law, Newton defined force to be proportional to the change in momentum, assuming the system is of constant mass.

So how is force defined when the system is not of constant mass?

 

[A.T.]

In the second law, Newton defined force to be proportional to the change in momentum, assuming the system is of constant mass.

The most general case:
Force is the rate of momentum transfer.

A special case:
Net force is the rate of momentum change.

An even more special case:
For a constant mass system net force is proportional to acceleration.

 

[etotheipi]

So how is force defined when the system is not of constant mass?

I'm not sure, but I don't think it's an issue, because you can always work with systems of constant mass if you're careful and use e.g. the approach in the OP's textbook, or the momentum balance approach in #6, or the Reynold's transfer theorem in #7. (I think this is a bit similar in nature to asking something like how force is defined in a non-inertial frame, the answer being it's not directly defined...)

But I think I do see your point... are you saying that, for a variable mass system, $\vec{F} = \dot{\vec{p}} = \dot{m} \vec{v} + m\dot{\vec{v}}$ is not valid (for reasons of non-invariance of the RHS under Galilean boosts, given that force must be frame-invariant) however the form $\tilde{F} = m\vec{a} = m\frac{d\vec{v}}{dt}$ is valid? The latter would I guess be valid so long as you re-interpret $\tilde{F}$ as a force which contains, in addition, a Galilei-varient 'thrust' force $\dot{m}\vec{u}$. If you like, this can constitute a definition of 'force' of sorts for a variable mass system (although I'd argue $\dot{m} \vec{u}$ is not a force so much as a term with dimensions of force).

That's okay, I guess, but it seems a bit messy and is not as elegant IMO as the momentum balance or RTT arguments. I think I personally will stick with the view of Kleppner/Kolenkow and Resnick/Halliday for my studies.

(N.B. for the record, I still completely disagree with @DrStupid's derivation in #8. His $\vec{F}_D = \dot{m} \vec{v}_D$ is not this thrust term mentioned above, and furthermore I have no idea why he writes the rate of change of the dust velocity is zero, $\dot{\vec{v}}_D= 0$. To me, that 'derivation' is a series of unfortunate events that somehow lands at the right answer...)

 

[DrStupid]

In the second law, Newton defined force to be proportional to the change in momentum, assuming the system is of constant mass.

Can you support that with a reference? I'm not aware of any sources where Newton explained what he assumed when he defined force that way. A restriction to closed systems is neither included in his wording of the second law nor is it reqired. Such a restriction is necessary for the first law and the term "body" can indeed be read as closed systems. Newton also used this term in the third law. However, due to conservation of momentum this law works for open systems as well.

(N.B. for the record, I still completely disagree with @DrStupid's derivation in #8. His $\vec{F}_D = \dot{m} \vec{v}_D$ is not this thrust term mentioned above

Of course not. I clearly stated that $F_D$ is the force acting on the dust.

and furthermore I have no idea why he writes the rate of change of the dust velocity is zero, $\dot{\vec{v}}_D= 0$.

I did it because the dust has the velocity $v_D = u$ by definition. The momentum of the dust doesn't changes by acceleration of dust particles but by removing particles from the dust only. Thus, $F_D$ is limited to the convective momentum transfer.

To me, that 'derivation' is a series of unfortunate events that somehow lands at the right answer...)

That means you finally hope for a series of coincidences? I'm afraid I have bad news for you: If the second law is used in the form F=dp/dt than F is just another symbol for dp/dt. That means that my derivation is essentially based on conservation of momentum. That always works because conservation of momentum is an universal law. There are no unfortunate events.

 

[etotheipi]

Can you support that with a reference? I'm not aware of any sources where Newton explained what he assumed when he defined force that way.

Plucked directly from the Wikipedia page on Newton's laws of motion, and references therein...

"Since Newton's second law is valid only for constant-mass systems..."
https://en.wikipedia.org/wiki/Newton's_laws_of_motion#Newton's_second_law

Your approach does not work in the general case. Consider a mine-cart rolling along a frictionless track, filled with sand, and with a very small hole punctured in the bottom of the cart such that sand falls out at zero initial relative velocity w.r.t. the cart, and as such the leaving sand exerts no force on the cart or its contents

If you were to naively apply Newton's second law, in the way that you are suggesting, to the variable mass system consisting of the cart and its contents at a time $t$, you would find that $$\vec{F}_{ext} = \vec{0} = \frac{dm}{dt} \vec{v} + m\frac{d\vec{v}}{dt}$$and because $\frac{dm}{dt} < 0$, you would end up with non-zero $\frac{d\vec{v}}{dt}$, i.e. the cart would accelerate! Clearly this is nonsense, because it violates conservation of energy (remember that the sand that has just exited still carries the horizontal kinetic energy it had when it was inside the cart).

If you use the corrected version that I have derived in two different ways, you will correctly see that since $\vec{v}_{rel} = \vec{0}$, that $\frac{d\vec{v}}{dt} = \vec{0}$, and that the cart carries on with constant velocity.

 

[etotheipi]

Of course not. I clearly stated that $F_D$ is the force acting on the dust.

and is not Galilei invariant, which means something is wrong

 

[vanhees71]

It's always safe to use the conservation laws, i.e., momentum conservation of a closed system. E.g., to derive the rocket equation just consider the rocket as a "control volume" and use Reynold's transport theorem. This takes into account the correct momentum balance including the exhausted fuel which gives the correct equation given in #6.

 

[DrStupid]

Plucked directly from the Wikipedia page on Newton's laws of motion, and references therein...

That does not show what Newton was assuming. It just reperesents what the autors of the page assume.
At the end of the day it doesn't matter what Newton or somebody else assume. Everybody may decide whether to use Newton's second law in the form F=dp/dt for open systems or not. Both options are working.

Consider a mine-cart rolling along a frictionless track, filled with sand, and with a very small hole punctured in the bottom of the cart such that sand falls out at zero initial relative velocity w.r.t. the cart, and as such the leaving sand exerts no force on the cart or its contents

With F=dp/dt it does. Again: You do not need to use F=dp/dt for open systems. It is your personal choice. But if you do it (even in the attempt to falsify it) then you need to do it consistent. You must not mix it with assumptions that hold for closed systems only.

and is not Galilei invariant, which means something is wrong

Yes, it is not Galilean invariant but no, it doesn't mean it is wrong - no matter how often you repeat it. dp/dt obviously is frame-dependent for open systems and with F=dp/dt the force acting on an open system is frame-dependent as well. That's quite trivial. You decided not to use it for open systems. But that doesn't make my derivation wrong. It just doesn't fit to your personal preferences.

 

[etotheipi]

I didn't understand anything of what you just said, but that's okay. I've got better things to do than argue this. Please read this paper: http://articles.adsabs.harvard.edu/full/seri/CeMDA/0053//0000227.000.html

As for the mine-cart, there is no force acting on the leaving sand. And this is a frame independent observation. You can't just add one in by saying "with F=dp/dt it does", whatever that means.

 

[DrStupid]

It's always safe to use the conservation laws, i.e., momentum conservation of a closed system. E.g., to derive the rocket equation just consider the rocket as a "control volume" and use Reynold's transport theorem. This takes into account the correct momentum balance including the exhausted fuel which gives the correct equation given in #6.

Yes, it always works. However, when you ask for a force (like in case we are discussing here) than you need to include it into the momentum balace. With F=dp/dt for open systems this is quite easy. But if the second law is limited to closed systems than the derivation becomes ugly, with intersections into many small closed systems, difference quotients, limits etc. I do not see the benefit of limiting F=dp/dt to closed systems and than to do all these steps in order to make it working for open systems again.

 

[vanhees71]

You get the forces for free when doing the momentum balance right.

 

[etotheipi]

With F=dp/dt it does. Again: You do not need to use F=dp/dt for open systems. It is your personal choice. But if you do it (even in the attempt to falsify it) then you need to do it consistent. You must not mix it with assumptions that hold for closed systems only.

@DrStupid I have a conceptually easier example. Now consider a mine-cart with a large-ish hole in the lower face, with no sand this time. Instead, you are standing inside the mine-cart, holding loads of metal ball bearings in your hand.

The mine-cart rolls along the frictionless tracks at a constant velocity, and then you begin to release the ball bearings from rest (in the cart frame) through the hole in the base of the mine-cart, at a rate of one ball per second.

Now there is absolutely no way in which there can be a horizontal force between the ball bearings and the mine-cart/you, since you're just dropping them through the hole, imparting no horizontal impulse. Consequently, the mine-cart undergoes no acceleration.

You can now imagine taking the continuum-limit of this process.

Now, if you were to apply your equation, $\vec{F} = \vec{0} = \dot{m}\vec{v} + m\dot{\vec{v}}$, to the mine-cart, you would find that $\dot{\vec{v}}$ would be non-zero. How would you explain this? How would you also explain that $\dot{\vec{v}}$ would depend on your observational frame of reference?

The resolution is that $\vec{v}_{rel}$ has zero component parallel to the direction of motion, so then I use the modified equation $\vec{F} + \vec{v}_{rel} \dot{m} = m\dot{\vec{v}}$ to deduce that $\dot{\vec{v}} = \vec{0}$ (external constraint forces with the ground prevent vertical acceleration).