Why not use the product rule to expand Newton's 2nd law?

[Prem1998]

Hi! I was reading the Wikipedia article on Newton's laws of motion. I read there that when mass is a variable function of time as well as velocity, one cannot use the product rule of derivatives to expand d/dt(mv)
It said that d/dt(mv)=mdv/dt+vdm/dt is WRONG
I don't know why that is wrong. The article said that this formula does not respect Galilean invariance: a variable-mass object with F = 0 in one frame will be seen to have F ≠ 0 in another frame.
Can anyone briefly explain the above line in simple language? I'm again saying: simple language, please.

 

[Simon Bridge]

Do you know what Galilean invariance is?
We cannot put it in terms you understand unless we know what terms you understand - however, you will be better off expanding your vocabulary: we have these "complicated terms" for a reason.

There is also:
http://physics.stackexchange.com/qu...the-product-rule-to-expand-Newtons-second-law

 

[Prem1998]

Will you explain in short what Galilean invariance is? I won't put any restriction to your language now.
EDIT: I googled Galilean invariance. What I get is: It just says that if Newton's laws are valid in one inertial frame then they're valid in all others. And, the equation of motion of a particle in other inertial frame can be obtained by Galilean transformation.
Now, can you explain how this happens if we expand using product rule:
A variable-mass object with F = 0 in one frame will be seen to have F ≠ 0 in another frame.

 

[Simon Bridge]

Galilean invariance is something you can look up for the detail.
Basically it means that all inertial frames are indistinguishable (think of inertial frames as boxes each moving with different constant velocity):
Put a skilled and clever experimenter, Alice, in a box (she cannot see out) with all the lab equipment she could want. Then there is no experiment she can do that will determine her motion though she can detect changes in her motion (ie. she can detect her acceleration but not her velocity).

In maths, this means the laws of physics must be unchanged when you replace v with v+Δv.

In Newtons laws - if Alice were to make an object that loses mass (ie a rocket) and measure the force needed to accelerate it across the lab, she could determine what speed she was doing by measuring the discrepancy from ##F=m\dot v + v\dot m## ... this would be a violation of Galilean invariance, so it is not true.

Now why do we not just say that Galilean invariance is wrong?
Well - that would create issues since it would imply there is some absolute resting frame ... but the equation has to be wrong because it does not account for the change in mass ... writing it out like that the mass just pops in or out by magic. IRL the mass has to come from somewhere and go somewhere. The process by which the mass changes affects the outcome. So this is tied up with the big conservation laws: energy and momentum.

This is how we handle things whose mass is changing - like rockets.

 

[DrStupid]

In Newtons laws - if Alice were to make an object that loses mass (ie a rocket) and measure the force needed to accelerate it across the lab, she could determine what speed she was doing by measuring the discrepancy from ##F=m\dot v + v\dot m## ... this would be a violation of Galilean invariance, so it is not true.

I can't follow. It is clear that F=dp/dt becomes frame dependent for variable mass systems, but I do not see how that violates Galiean invariance.

 

[vanhees71]

The OP Is not clear to me either, but maybe the problem is that one has to distinguish, case by case, how the variation of mass occurs?

One typical example is the rocket, where the loss or mass comes from exhausting the fuel, i.e., the fuel moves in the rest frame of the rocket and thus carries of momentum, and one has to take into account this in the momentum balance, i.e. Take the standard problem of the rocket in free space: In the rest frame of the rocket the velocity of the streaming gas ##v_{\text{gas}}## may be constant and also the exhaust per time, ##\mu##. Then the equation of motion reads
$$\dot{p}=m \dot{v} -\mu v=\mu (v_{\text{gas}}-v).$$
This is, because on the left-hand side we have the change of momentum with time, which is the force, which is due to the momentum carried away by the gas per unit time, which is ##\mu v_{\text{gas}}## in the rest frame of the rocket, but we have to give it in the "lab frame", which makes it the right-hand side of the equation.
To integrate it we also need ##m=m_0-\mu t##, i.e., the equation of motion reads
$$(m_0-\mu t)\dot{v}=\mu v_{\text{gas}}.$$
The solution is
$$v(t)=v_0+v_{\text{gas}} \ln \left (\frac{m_0}{m_0-\mu t} \right).$$
So the naive equation
$$\dot{p}=0$$
is wrong in this case, because it neglects the recoil of the rocket from the outstreaming gas, which of course is the main point of shooting a rocket in the first place :-).

 

[PeroK]

I can't follow. It is clear that F=dp/dt becomes frame dependent for variable mass systems, but I do not see how that violates Galiean invariance.

If you have an object that could spontaneously lose mass, then (assuming it does not change its velocity):

In its rest frame, there is no change in momentum: it remains ##0##

In any other frame, the momentum changes due to the change of mass.

I must confess, the conclusion I would draw from this is that (in classical physics) mass must be conserved. I see no reason to conclude that a basic theorem of calculus breaks down as a result!

 

[vanhees71]

It depends! In my rocket example in the rest frame of the rocket (which is a non-inertial frame by the way) there's change of momentum because of the recoil of the outstreaming gas. That's why you shoot a rocket in the first place, namely to move, i.e., to gain momentum :-). Of course, no basic rule (the product rule) of differentiation is violated anywhere here. In fact, I used it in my previous posting in this thread to derive the equation of motion of the rocket.

Also it's of course common that in a description, where you consider an open system, not the full Galileo invariance is manifest. If you consider the entire system, i.e., the rocket and the outstreaming gas as a whole, everything is Galileo invariant as it must be in a Newtonian treatment. Last but not least the full Newtonian dynamics follows from Galileo invariance thanks to Emmy Noether's theorems!

 

[DrStupid]

I must confess, the conclusion I would draw from this is that (in classical physics) mass must be conserved.

Yes, mass is conserved (not only in classical mechanics). That's well known.

I still do not see why F=m·a+v·dm/dt should be wrong and/or violate Galilean invariance.

 

[Simon Bridge]

I must confess, the conclusion I would draw from this is that (in classical physics) mass must be conserved. I see no reason to conclude that a basic theorem of calculus breaks down as a result!

It is not that the theorem of calculus "breaks down" - it is that this application results in something that does not work well in Nature.
This happens a lot and it is usually a sign that something was left out of the description. The power rule still works.

Yes, mass is conserved (not only in classical mechanics). That's well known.

Mass is not always conserved though - it can be exchanged for other kinds of energy.

It is clear that F=dp/dt becomes frame dependent for variable mass systems, but I do not see how that violates Galiean invariance.

And yet, "becomes frame dependent" is just another way of saying "violates Galilean invariance". The two are the same.

Do you know how Galilean relativity works?
Let's say there is some object at rest in the lab, and it's mass is decreasing.
... then there must be forces acting on it so that ##F=0## (since v=0 and a=0)
For Galilean invariance to hold, then this should be the same in all reference frames
.. ie. frame S', with relative speed u, must have F'=0 too.

So let's work this out - in frame S', the object moves at constant speed ##v'=v+u=u## (v=0),
* it does not accelerate, ##a'=\dot v' = \frac{d}{dt}(v+u) = 0## (v and u are constant);
* but it's mass changes ##\dot m < 0##
... then there must be forces acting on it in that frame so that ##F'=ma' + v'\dot m = u\dot m## (since v'=u and a'=0 from above)
##F' \neq F## therefore Galilean invariance fails.

What is that equation is telling us ... ##F=u\dot m## remember, u is the relative speed of you wrt the object ... so anyone moving past an object that is losing mass must apply a force to it, in the direction they are moving, to keep going at a constant speed?? Is there any way that result makes sense?

 

[PeroK]

It is not that the theorem of calculus "breaks down" - it is that this application results in something that does not work well in Nature.
This happens a lot and it is usually a sign that something was left out of the description. The power rule still works.

I would suggest an alternative explanation:

1) ##F = \frac{dp}{dt}##

Applies to any particle and, as explained above, in classical physics a particle cannot change its mass.

2) ##F = \frac{dp}{dt}##

Applies, therefore, to any fixed collection of particles.

But, if you change the particles under consideration over time, then clearly this is invalid. Using finite numbers of particles this would be more obvious because the change in mass would simply be the removal of certain particles from consideration. For example, if you started with two particles of mass ##m##, both traveling at the same velocity ##v##, then the initial momentum would be ##2mv##. If you drop one particle from your consideration, then the momentum of the remaining particle is ##mv##, but nothing has changed for this particle (except that it is no longer associated with the other particle).

The same applies to a "rocket". How do you define the "rocket"? At one moment the "rocket" is ##n## particles; the next moment the "rocket" is ##n-1## particles, as you are no longer considering a particle of fuel that has been ejected. But, these two rockets are not the same thing; they are not the same collection of particles. Therefore:

##F = \frac{dp}{dt}##

Applies to each and every particle that was and remains part of the rocket, but it cannot apply to a changing set of particles.

Most succinctly: if there is no external force, momentum is not conserved if you change the particles under consideration, e.g. by changing the number of particles that comprise something you define as a "rocket".

 

[Nugatory]

A very long digression driven by the ambiguities between mass, mass-energy, and total energy and mass within a closed system has been removed from this thread. Seeing as how this post is in the Classical Physics section and asks specifically about Newton's second law, posters should assume that we're working in a domain in which the energies are small enough that their contribution to the mass of the system can be ignored.

All posters are also reminded that we mentors are NOT paid by the number of posts we moderate, so giving us more work to do instead of less is NOT appreciated.

 

[DrStupid]

Do you know how Galilean relativity works?
[…]
... then there must be forces acting on it in that frame so that ##F'=ma' + v'\dot m = u\dot m## (since v'=u and a'=0 from above)
##F' \neq F## therefore Galilean invariance fails.

That doesn’t necessarily mean that ##F = \dot p## is wrong for variable mass. A limitation of the first law to constant mass would solve the problem.

What is that equation is telling us ... ##F=u\dot m## remember, u is the relative speed of you wrt the object ... so anyone moving past an object that is losing mass must apply a force to it, in the direction they are moving, to keep going at a constant speed?? Is there any way that result makes sense?

Yes, there is. Let’s assume a homogeneous medium which is at rest in the lab as well as a body with the mass m moving through this medium with the velocity v and accumulating mass with the rate ##\dot m##. With ##F = \dot p## there are no forces. But in a frame of reference, which moves with the velocity u relative to the lab system the force acting on the body turns into

[itex]F'_{body} = - u \cdot \dot m[/itex]

and the force acting on the medium into

[itex]F'_{medium} = + u \cdot \dot m[/itex]

That’s exactly what the third law requires for interactive forces and that makes sense because the mass transfer from the medium to die body actually is an interaction.

This also works for the general case of interacting bodies with variable mass in an isolated system. With ##F = \dot p## the sum of all forces is

[itex]\sum\limits_i {F_i } = \sum\limits_i {m_i } \dot v_i + \sum\limits_i {\dot m_i } v_i = 0[/itex]

Galilean transformation and conservation of mass result in

[itex]\sum\limits_i {F'_i } = \sum\limits_i {m_i } \dot v_i + \sum\limits_i {\dot m_i } v'_i = \sum\limits_i {F_i } - u \cdot \sum\limits_i {\dot m_i } = 0[/itex]

In case of open systems the second and third law of motion are full consistent with Galilean invariance. The problem is the first law because the body is accelerated in the lab system even though there is no force acting on it.

With ##F = m \cdot a## the first law can be saved for variable mass systems. But that results in an violation of the second law (because the force would be not proportional to the change of momentum) and the third law (because there would be forces without counter forces in intertial systems).

That means that the decision between ##F = \dot p## and ##F = m \cdot a## is equivalent to the decision between the violation of the first law and the violation of the second and third law. I prefer the violation of the first law (and therefore ##F = \dot p##) because it is not required for equations of motion and the original wording allows a limitation to constant mass (if the word "corpus" is translated as "closed system").

 

[dextercioby]

For what it's worth, there's a small comment in the wiki article with reference to the literature (Section "Common misconceptions") and also a discussion in the "Talk" page behind it. https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation

 

[DrStupid]

For what it's worth, there's a small comment in the wiki article with reference to the literature (Section "Common misconceptions") and also a discussion in the "Talk" page behind it. https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation

The article just repeats the usual claim that F=dp/dt is wrong for variable mass systems without proper justification. There is even a major mistake in the referenced literature: A.R. Plastino & J.C. Muzzio “On the use and abuse of Newton’s second law for variable mass probems” claims

If we consider the simple case of a variable mass, and we write Newton’s second law as:

[itex]\bar F = m\frac{{d\bar v}}{{dt}} + \bar v\frac{{dm}}{{dt}}[/itex] (2)

we can easily see it violates the relativity principle under Galilean transformation. When ##\bar 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 will be accelerated by the “force” ##- {\bar v} dm / dt## in a system where the particle moves with velocity ##{\bar v}##!

That's obviously wrong. Equation (2) results in the acceleration

[itex]a = \frac{{d\bar v}}{{dt}} = \frac{{\bar F - \bar v \cdot \dot m}}{m}[/itex]

and Galilean transformation in

##m' = m##
##\dot m' = \dot m##
##\bar v' = \bar v – u##
##\bar F' = \bar F - u \cdot \dot m##

and therefore in

##a' = \frac{{\bar F' - \bar v' \cdot \dot m'}}{{m'}} = a##

This error makes the entire paper obsolete because there is no problem that needs to be solved.

 

[Simon Bridge]

Is there any way this makes sense?

Yes, there is.

OK - however, the question was directed at a particular person and was rhetorical - for pedagogical purposes. ie. the person who needs the additional learning was supposed to reply.
I was not asking because I did not know the answer!
Sheesh!

 

[vanhees71]

A very long digression driven by the ambiguities between mass, mass-energy, and total energy and mass within a closed system has been removed from this thread. Seeing as how this post is in the Classical Physics section and asks specifically about Newton's second law, posters should assume that we're working in a domain in which the energies are small enough that their contribution to the mass of the system can be ignored.

All posters are also reminded that we mentors are NOT paid by the number of posts we moderate, so giving us more work to do instead of less is NOT appreciated.

Well, in some sense it was a digression. However, the discussion about "relativistic mass" came up, and in some textbooks it's indeed treated in the chapter about systems with varying mass, which is misleading and should be corrected right away. Also the discussion was entirely about classical physics. Nowhere quantum theory was invoked. We were just discussing relativistic energy-momentum conservation. Nevertheless it's fine that you have deleted the posts since there's enough said in the relativity forum about the idiosyncratic use of non-covariant mass concepts anyway.

 

[DrStupid]

I was not asking because I did not know the answer!

That means you think that F=dp/dt is wrong even thogh you know that it makes sense?

 

[stevendaryl]

That means you think that F=dp/dt is wrong even thogh you know that it makes sense?

Yes, [itex]F = \frac{dp}{dt}[/itex] is wrong, when applied to objects with variable mass.

Consider the case of a rocket whose mass is decreasing due to burning fuel and throwing the burnt fuel out the rear. The total external force is zero, so according to you, that means

[itex]\frac{dp}{dt} = M \frac{dv}{dt} + v \frac{dM}{dt} = 0[/itex]

So according to [itex]F= \frac{dp}{dt}[/itex], we would have

[itex]M \frac{dv}{dt} = -v \frac{dM}{dt}[/itex]

But the actual equation for a rocket, under the assumption that the burnt fuel is thrown behind the rocket at a speed of [itex]-v_{rel}[/itex] relative to the rocket, is:

[itex]M \frac{dv}{dt} = -v_{rel} \frac{dM}{dt}[/itex]

(This can be derived by conservation of momentum by considering an infinitesimal time period [itex]\delta t[/itex], and an infinitesimal amount of fuel, [itex]\delta m = - \frac{dM}{dt} \delta [/itex].)

So [itex]F = M \frac{dv}{dt} + v \frac{dM}{dt}[/itex] doesn't seem to give the right answer, except in the special case in which [itex]v_{rel} = v[/itex].

 

[DrStupid]

The total external force is zero

Really? The momentum of an accelerating rocket is always constant? Please rethink this assumption and correct your calculation accordingly.

 

[stevendaryl]

Really? The momentum of an accelerating rocket is always constant?

The external force is zero. The momentum is not constant. Therefore, [itex]F = \frac{dp}{dt}[/itex] is false for objects with changing mass.

I'm sure there is a way to define "the force on the object" that makes [itex]F = \frac{dp}{dt}[/itex] true, even for objects with changing mass, but doing so would be an exercise in tautologies, and not very helpful. Writing

[itex]F = M \frac{dv}{dt} + v \frac{dM}{dt}[/itex]

is at best useless.

 

[stevendaryl]

The external force is zero. The momentum is not constant. Therefore, [itex]F = \frac{dp}{dt}[/itex] is false for objects with changing mass.

I'm sure there is a way to define "the force on the object" that makes [itex]F = \frac{dp}{dt}[/itex] true, even for objects with changing mass, but doing so would be an exercise in tautologies, and not very helpful. Writing

[itex]F = M \frac{dv}{dt} + v \frac{dM}{dt}[/itex]

is at best useless.

The results I gave for a variable-mass system is the standard way to think about it. It's the way described in this Wikipedia article: https://en.wikipedia.org/wiki/Variable-mass_system (Except that I chose [itex]v_{rel}[/itex] with the opposite sign.)

The only way I can imagine making your [itex]F = M \frac{dv}{dt} + v \frac{dM}{dt}[/itex] to work is if you are being creative about what counts as a "force" on the system. Basically, if you know [itex]\frac{dv}{dt}[/itex] and [itex]\frac{dM}{dt}[/itex], then you can work your way backwards to figure out [itex]F[/itex] to make things work out.

 

[DrStupid]

The external force is zero.

That needs to be explained!

 

[stevendaryl]

That needs to be explained!

To whom? I think that I explained how variable mass systems work pretty well. If you have a rocket, the external force would be anything affecting the rocket from some source outside the rocket, such as gravity, friction, tractor beams, whatever.

The rule [itex]F = M \frac{dv}{dt} + \frac{dM}{dt}[/itex] is what needs a lot more explanation.

 

[DrStupid]

If you have a rocket, the external force would be anything affecting the rocket from some source outside the rocket, such as gravity, friction, tractor beams, whatever.

There is a mass transfer between the rocket and it's envoronment. If this mass carries momentum both the momentum of the rocket and of the environment change. As the force is per definition proportional to the change of momentum (second law) there must be forces acting on both systems.

You need to explain why you assume something else.

The rule [itex]F = M \frac{dv}{dt} + \frac{dM}{dt}[/itex] is what needs a lot more explanation.

Definitions do not need explanation.

---

Edit (in order to be more specific):

The mass ##{\dot m}## with the momentum ##\dot p = \dot m \cdot v## is transferred out of the rocket and into the environment. According to the second law the corresponding forces are

[itex]F_R = m_R \cdot \dot v_R + v_R \cdot \dot m_R = - \dot m \cdot v[/itex]
[itex]F_E = m_E \cdot \dot v_E + v_E \cdot \dot m_E = + \dot m \cdot v[/itex]

This already includes the third law. Conservation of mass requires

[itex]\dot m_R = - \dot m_E = - \dot m[/itex]

With the relative velocity

[itex]v_{rel} = v - v_R[/itex]

between the mass flow and the rocket this finally results in the rocket equation

[itex]\dot v_R = \frac{{v_{rel} \cdot \dot m_R }}{{m_R }}[/itex]

As you can see ##F = \dot p## works for rockets if you use it correctly. You got a wrong result because you started with wrong assumptions (##F = 0## and ##F = \dot p## for ##\dot p \ne 0##) and not because ##F = \dot p## is wrong. Garbage In, Garbage Out.

 

[stevendaryl]

There is a mass transfer between the rocket and it's envoronment. If this mass carries momentum both the momentum of the rocket and of the environment change. As the force is per definition proportional to the change of momentum (second law) there must be forces acting on both systems.

I went through the argument for figuring out the acceleration of a rocket due to burning fuel. [itex]F = M \frac{dv}{dt} + v \frac{dM}{dt}[/itex] played no role. You haven't given any indication why it would ever play a role. I think it's very bad pedagogy and worthless for solving problems.

 

[vanhees71]

Please, see #6. Of course, Newton's law is correct also in this case, ##\dot{\vec{p}}=\vec{F}##, where ##\vec{F}## is the force on the rocket. You only have to make sure to include the force due to the repulsion by the exhausted fuel. That's why one uses rockets for space travelling. Of course, Newton's force law is just the momentum balance equation, describing the conservation of total momentum in the closed system of rocket + streaming fuel.

 

[stevendaryl]

Please, see #6. Of course, Newton's law is correct also in this case, ##\dot{\vec{p}}=\vec{F}##, where ##\vec{F}## is the force on the rocket. You only have to make sure to include the force due to the repulsion by the exhausted fuel. That's why one uses rockets for space travelling. Of course, Newton's force law is just the momentum balance equation, describing the conservation of total momentum in the closed system of rocket + streaming fuel.

The issue is: what is the force on the rocket?

As I said, for the case of the rocket, the equation is indisputably:

[itex]- v_{rel} \frac{dM}{dt} = M(t) \frac{dv}{dt}[/itex]

(where I've defined [itex]v_{rel}[/itex] is the velocity of the rocket, relative to the fuel, rather than the other way around)

So the real issue is how to identify the force on the rocket plus remaining fuel, [itex]M(t)[/itex]. I would say that the force on the rocket [itex]M(t)[/itex] due to burning fuel is [itex]F_{fuel-on-rocket} = - v_{rel} \frac{dM}{dt}[/itex]

But with that identification,

[itex]F_{fuel-on-rocket} = M_{rocket} \frac{dv_{rocket}}{dt}[/itex]

not

[itex]F_{fuel-on-rocket} = M_{rocket} \frac{dv_{rocket}}{dt} + v_{rocket} \frac{dM_{rocket}}{dt}[/itex]

which is what the product rule would give you for [itex]\frac{d p_{rocket}}{dt}[/itex]

 

[vanhees71]

All I have to say is said in #6. The force on the rocket is due to the momentum per unit time transported away by the exhausted fuel (repulsion). You can of course write down the same equation in terms of total-momentum conservation
$$\frac{\mathrm{d}}{\mathrm{d} t} (p_r+p_{f})=m_r \dot{v}_{r}+\dot{m}_r v_{r} + \mu(v_{fr}+v_{r})=m_r \dot{v}_r +\mu v_{fr}=0.$$
This is the same as with the "force argument" in #6 (with ##v_{fr}=-v_{\text{gas}}## the velocity of the fuel relative to the rocket).

 

[stevendaryl]

All I have to say is said in #6. The force on the rocket is due to the momentum per unit time transported away by the exhausted fuel (repulsion).

My point is that that force is not [itex]v \frac{dM}{dt}[/itex], it's [itex]v_{rel} \frac{dM}{dt}[/itex]. Your analysis uses [itex]\frac{dp}{dt} = 0[/itex] for a constant-mass system, the fuel + the rocket. It isn't using [itex]F = \frac{dp}{dt}[/itex] for a variable-mass system. That's what I am advocating--that whatever you want to know about a variable-mass system is better understood in terms of a larger constant-mass system.

 

[mridul]

Its as simple as this .actually mass of the body does not change with time so it is a constant .Now the constant comes out of the factor that's why.sorry for bad eng

 

[Stephen Tashi]

This note is given in the current Wikipedia article on "Newton's laws of motion":

Halliday; Resnick. Physics. 1. p. 199. ISBN 0-471-03710-9. 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 F = dP/dt = d(Mv) as a variable. [...] We can use F = dP/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.

However, don't we put a similar restriction on applying any equation of classical physics?

Technically a "mass" is not "an object". "Mass" is a property of an object (or system) and, as far as I can see, in classsical mechanics, "object" and "system" have no rigorous definition. They are left as terms of common speech.

Our habit is to use the phrase "a mass" to designate an object. If that is to be an unambiguous designation, when "a mass" breaks into 5 pieces, all those 5 pieces must be considered part of the designated object and designated mass.

I we allow ourselves to define "a system" in a way not related to mass, we could make definitions like "System B: consists of all things having a mass that are in my bedroom". By that definition, a person who walked out of the bedroom would no longer be part of System B. So there would be no conservation of mass in System B. Likewise there need be no conservation of momentum in System B.

So the restriction that Halliday and Resnick state: "only if we apply it to an entire system of constant mass" must also be enforced on other equations of mechanics that apply to "a system".

It's interesting to ask whether "conservation of mass" (in classical physics) is merely a definition! - i.e. mass is conserved "in a system" because we require that "a system" be specified in such a way that it always includes a constant amount of mass".

 

[conscience]

@haruspex , do you mind giving your views on this topic .

 

[haruspex]

@haruspex , do you mind giving your views on this topic .

If you insist.
##\vec F=\frac{\vec{dp}}{dt}## is correct, where ##\vec p## is the momentum of a system of particles and ##\vec F## is the net external force acting on the system.
##\frac{\vec{dp}}{dt}=m\frac{\vec {dv}}{dt}+\vec v\frac{dm}{dt}## is correct, but normally we would take the mass of a particle to be invariant. A nonzero dm/dt term means mass is (particles are) being added to or removed from the system represented by ##\vec p##.
If you consider a rocket plus its fuel (used or unused) as the system, there is no external force and no net change in momentum.
If you consider the rocket as one system and all its fuel as another, neither is changing mass, so for each there is only the m dv/dt term, and this corresponds to the force each exerts on the other.
If you consider the rocket plus unused fuel as one system and spent fuel as the other, ##\vec F=\frac{\vec{dp}}{dt}## no longer applies, because we are no longer considering a consistent system of particles. We can illustrate this with a simple experiment. I throw a ball to which there is a particle of dirt attached. The particle falls off in flight. If we define a system as "the ball plus anything currently attached to it", the momentum suddenly dropped, yet there is no evident force.

A possible approach is to extend the concept of force to include the rate of momentum change resulting from addition or removal of mass.

 

[vanhees71]

I usually don't like forces, because they are a murky concept to me. The true laws of nature are all describable with the action principle, but here, I'd say it's pretty simple to see at the rocket example that Newton's law holds true at least in some cases of systems with changing mass. The force on the rocket is the recoil of the streaming gas on the rocket, and of course the product rule is correct. It's a mathematical theorem and has nothing to do with physics except that you can use it in physic-related calculations ;-). Also in your ball example you can use this concept: When the dust particle falls of it gains momentum relative to the ball and thus the ball also gets a recoil, which is a force acting on it (in addition to other forces obviously involved here). Of course, for the entire closed system of ball+dust particle momentum is conserved, and you can derive the equation of motion of the ball and the dust particle also by using momentum conservation, as in the example with the rocket.