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

  1. Oct 29, 2016 #1
    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.
  2. jcsd
  3. Oct 29, 2016 #2

    Simon Bridge

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  4. Oct 29, 2016 #3
    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.
  5. Oct 29, 2016 #4

    Simon Bridge

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    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.
  6. Oct 29, 2016 #5
    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.
  7. Oct 29, 2016 #6


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    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
    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 :-).
  8. Oct 29, 2016 #7


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    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!
  9. Oct 29, 2016 #8


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    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!
  10. Oct 29, 2016 #9
    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.
  11. Oct 29, 2016 #10

    Simon Bridge

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    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.

    Mass is not always conserved though - it can be exchanged for other kinds of energy.
    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 lets 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?
  12. Oct 30, 2016 #11


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    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 travelling 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".
  13. Nov 4, 2016 #12


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    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.
    Last edited: Nov 4, 2016
  14. Nov 4, 2016 #13
    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.

    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").
  15. Nov 4, 2016 #14


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  16. Nov 4, 2016 #15
    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

    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.
    Last edited: Nov 4, 2016
  17. Nov 5, 2016 #16

    Simon Bridge

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    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!
  18. Nov 5, 2016 #17


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    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.
  19. Nov 5, 2016 #18
    That means you think that F=dp/dt is wrong even thogh you know that it makes sense?
  20. Nov 6, 2016 #19


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    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].
    Last edited by a moderator: Nov 6, 2016
  21. Nov 6, 2016 #20
    Really? The momentum of an accelerating rocket is always constant? Please rethink this assumption and correct your calculation accordingly.
  22. Nov 6, 2016 #21


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    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.
  23. Nov 6, 2016 #22


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    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.
  24. Nov 6, 2016 #23
    That needs to be explained!
  25. Nov 6, 2016 #24


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    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.
  26. Nov 6, 2016 #25
    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.

    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.
    Last edited: Nov 6, 2016
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