Discontinuity of velocity in a problem on mass flow and momentum

[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).

 

[DrStupid]

Please read this paper: http://articles.adsabs.harvard.edu/full/seri/CeMDA/0053//0000227.000.html

They claim

When F is zero, in particular, Equation (2) implies that the particle will remain at rest in a system where it is originally at rest, but it will be accelerated by the "force" -v dm/dt in a system where the particle moves with the velocity v!

Let's check that:

Equation (2) results in the acceleration

$\dot v = \frac{{F - \dot m \cdot v}}{m}$

With Galiean transformation we have

$v' = v - u$
$\dot v' = \dot v$
$m' = m$
$\dot m' = \dot m$

and due to the principle of relativity also

$F' = \dot m' \cdot v' + m' \cdot \dot v' = F - \dot m \cdot u$

That means for the acceleration resulting from (2)

$\dot v' = \frac{{F' - \dot m' \cdot v'}}{{m'}} = \dot v$

Everything remains consistent (obviously). The resulting acceleration is frame-independent as expected. If the particle is at rest in one frame then there is no other frame where it is accelerated. Thus, the claim above appears to be wrong. That makes the paper pointless.

As for the mine-cart, there is no force acting on the leaving sand.

If you split mine-cart and sand into an open system containing the mine-cart and the sand that did not yet leave the cart and another open systems containing the sand that already leaved the cart than the momentums of both systems are changing (if the cart has non-zero velocity). With F=dp/dt this means that there are forces acting on both systems.

And this is a frame independent observation.

Acceleration, mass and mass transfer are frame-independent. Velocity, momentum and F=dp/dt are not.

 

[DrStupid]

Now, if you were to apply your equation, $\vec{F} = \vec{0} = \dot{m}\vec{v} + m\dot{\vec{v}}$, to the mine-cart

F=dp/dt is frame-dependent! If it is zero in the rest-frame of the cart than it is not zero in a frame of reference where the cart is moving with v>0. You again made the mistake to use F=dp/dt for an open system together with an assumption that holds for closed systems only.

 

[etotheipi]

To be honest, I really have a hard time trying to interpret what you are saying.

If you split mine-cart and sand into an open system containing the mine-cart and the sand that did not yet leave the cart and another open systems containing the sand that already leaved the cart than the momentums of both systems are changing (if the cart has non-zero velocity). With F=dp/dt this means that there are forces acting on both systems.

F=dp/dt is frame-dependent! If it is zero in the rest-frame of the cart than it is not zero in a frame of reference where the cart is moving with v>0.

This is completely confused. If the external force on the 'mine-cart + me holding the ball bearings' system in my example in #35 is non-zero, then what is the physical origin of this supposedly present force?

 

[DrStupid]

If the external force on the 'mine-cart + me holding the ball bearings' system in my example in #35 is non-zero, then what is the physical origin of this supposedly present force?

It comes from the ball bearings that leaved the system. It is exactly the same situation as with the sand: You have an open system containing the cart, you and some mass (e.g. sand, ball bearings or whatever) and you have another open system containing the exhausted mass that already dropped out of the cart. There is a mass transfer $\dot m$ from one system to the other. If the systems are moving with $v$ than there is also a momentum transfer $\dot m \cdot v$ because the exchanged mass carries its momentum from one system to the other. According to F=dp/dt the momentum transfer corresponds to a force between the two systems. The cart system exerts the force $\dot m \cdot v$ on the exhaust system and the exhaust system exterts the corresponding counter force $-\dot m \cdot v$ on the cart system.

Maybe you don't understand it because you expect something difficult. But it is in fact very trivial.

 

[etotheipi]

According to F=dp/dt the momentum transfer corresponds to a force between the two systems.

No, that is wrong. The correct answer is that 'there is no such force'. You can't just magic up a force from somewhere, real forces arise due to physical interactions and there is no such physical interaction that you can identify in my example.

You are sitting in the mine-cart, and you drop the ball bearings from rest through the hole in the floor at a rate of one ball per second. You impart no horizontal impulse to the ball bearings, and they impart no horizontal impulse to you.

The point is that you can obtain a change in momentum without a force, due to mass leaving the system. That was the whole point of my example, to explain that naively applying $\vec{F} = \frac{d\vec{p}}{dt}$ to a variable mass system (like the 'mine-cart + me with ball bearings' system) gives contradictory results.

You have a seriously flawed interpretation of the equation $\vec{F} = \dot{\vec{p}}$, and your vague and deflective responses only go to show that you cannot justify your position.

Please, please, consult any fluid dynamics textbook that deals with control volumes, or any introductory mechanics textbook that deals with variable mass systems. It will become clear to you that the only fully consistent picture is that $\vec{F} = \dot{\vec{p}}$ can only be applied to systems of constant mass.

 

[DrStupid]

real forces arise due to physical interactions and there is no such physical interaction that you can identify in my example.

Are you really telling me that mass transfer is no physical interaction? That is ridiculous.

You impart no horizontal impulse to the ball bearings, and they impart no horizontal impulse to you.

I didn't claim that there is a force between me and the ball bearings. In #39 I defined the systems that I am talking about. Can I at least expect you to read my postings before you reply?

The point is that you can obtain a change in momentum without a force, due to mass leaving the system.

Not with F=dp/dt. Just for the case that you don't understand it: It means that the force F is equal to the change of momentum p with time t. With this relationship there is no change of momentum if there is no force. If you come to another conclusion than you did something wrong.

 

[berkeman]

Thread is closed for Moderation...

 

[PeterDonis]

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

To wrap up the discussion in this thread, I'm going to answer this question and then give some brief follow-up. The total force on the spacecraft in the scenario given in the OP of this thread is

$$
\vec{F} = \vec{F}_\text{external} + \left( \vec{u} - \vec{v} \right) \frac{dm}{dt} = m \frac{d \vec{v}}{dt} = m \vec{a}
$$

The total force here is frame invariant and has the obvious physical interpretation of the force exerted by the dust on the spacecraft , plus the "magical" external force exerted to oppose the dust force and keep the spacecraft moving uniformly (i.e., so that $\vec{a} = 0$).

This also shows that Newton's Second Law, in the form $\vec{F} = m \vec{a}$, with the total force $\vec{F}$ properly defined, works just fine for systems with time-varying mass.

What about the equation $\vec{F} = d \vec{p} / dt$? We can obtain that from the above equation by rearranging it as follows:

$$
\vec{F}_\text{external} + \vec{u} \frac{dm}{dt} = m \frac{d \vec{v}}{dt} + \vec{v} \frac{dm}{dt} = \frac{d \vec{p}}{dt}
$$

The LHS is the sum of a frame invariant term and a frame-dependent term, so the LHS as a whole is frame-dependent; thus, the RHS must also be frame-dependent. So this equation, unlike the equation above, does not express a relationship between frame-invariant quantities.

Furthermore, the "force" appearing on the LHS of this equation is not the total force, since the term $\vec{v} \, d \vec{m} / dt$, which is part of the total force, at least if "total force" has the obvious physical meaning described above, is included in the RHS, not the LHS. Defining the LHS of this equation as "force" is giving a different, frame-dependent meaning to the term "force".

It is true that, mathematically, both equations are equivalent in terms of what they will say about actual experimental predictions, and what one chooses to mean by the term "force" is a question of choice of words, not physics. The first equation seems to me to be a clearer representation of the actual physics, since it relates invariant quantities. But if someone wants to use the second equation, as long as they are able to get the right answers with it, it's up to them.

 

[PeterDonis]

And with that, this thread will remain closed.