From Goldstein:

... The mechanics of the particle is contained in Newton's second law of motion, which states that there exist frames of reference in which the motion of the particle is described by the differential equation $$\mathbf{F} = \frac{d\mathbf{p}}{dt} \equiv \dot{\mathbf{p}},$$ or $$\mathbf{F} = \frac{d}{dt}\left(m\mathbf{v}\right).$$ In most instances, the mass of the particle is constant and [the last equation] reduces to $$\mathbf{F} = m\frac{d\mathbf{v}}{dt} = m\mathbf{a}\textrm{...}$$

What Goldstein is saying troubles me for it implies that $\mathbf{F} = \dot{\mathbf{p}}$ works for a particle of time-varying mass and that contradicts Ján Lalinský's answer to this question: Second law of Newton for variable mass systems.

What's going on here?

Edit: Some people think that a particle that loses mass that isn't going anywhere--it simply disappears---isn't a useful fiction to solve problems. Consider a ball that is emitting mass isotropically whilst being pushed by some force. In analyzing the motion of this ball there is no need to consider the fact that material is indeed being emitted, all that is needed is the fact that the ball is losing mass. Here we can just use $\mathbf{F} = m\mathbf{a}$ instead of $\mathbf{F} = \dot{\mathbf{p}}$, no?

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    $\begingroup$ I'm not sure I see the contradiction per se. Jan's statement is that $\mathbf{F} = d/dt(m\mathbf{a}) = \mathbf{F}_{ext} + \mathbf{F}_{parts} = \sum m \mathbf{a}$. If you use the chainrule for Goldstein's expression, you get $\mathbf{F} = m\mathbf{a} + \mathbf{v}dm/dt$. If you let $ -\mathbf{v}dm/dt = \mathbf{F}_{parts}$, then the two equations are the same. And that matches the equation on the Wikipedia page. Can you clarify what you see as the contradiction? Or clarify if Goldstein says additional things that are relevant? $\endgroup$ – tpg2114 Mar 7 '19 at 23:50
  • $\begingroup$ I interpret Goldstein to mean that the mass of the particle is lost without going anywhere---it just disappears. $\endgroup$ – PiKindOfGuy Mar 8 '19 at 0:33
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Mar 8 '19 at 18:17
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    $\begingroup$ @tpg2114 it is not possible to let $-\mathbf vdm/dt = \mathbf F_{parts}$, because that expression depends on velocity of the body $\mathbf v$, not relative velocity of the parts leaving the body, as it should. The correct expression in case of ideal rocket is $\mathbf F_{parts} = \frac{dm}{dt}\mathbf c$, where $\mathbf c$ is relative velocity of the gas leaving the rocket with respect to the rocket. $\endgroup$ – Ján Lalinský Mar 8 '19 at 21:56
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    $\begingroup$ In the case you describe in your last paragraph, you are correct, $F=ma$ works. The rocket equation reduces to exactly that, with the exhaust showing up as an external force. $F=ma$ involves the instantaneous value of the mass, not its rate of change. $\endgroup$ – garyp Mar 11 '19 at 16:56

I think that there are basically four pieces to the puzzle for why this frequent error (of course, I may be missing something):

  • it is customary since old times to state traditional Newton's second law in terms of change of momentum, as he originally did (quantity of motion). This catched on despite this traditional formulation only applies to systems of constant mass and there is no more generality to it than in stating the law in terms of changes of velocity of the body. Perhaps one advantage of the momentum formulation is that it applies more directly to extended bodies which have no single velocity, but do have single momentum. But even for such bodies second law can be described using changes of velocity: if the body has no single velocity, the velocity of center of mass can be used.

  • since in special relativity $\mathbf F = m\mathbf a$ does not hold in any obvious sense, it was necessary to find some valid equivalent and in the decades after Einstein's 1905 publication it was generally accepted that the preferred way to do that is to try to give the traditional expression $$ \mathbf F = \frac{d\mathbf p}{dt} $$ a more non-trivial role - by allowing for the $m$ in $\mathbf p = m\mathbf v$ be a variable quantity that is a function of body speed. Later in 20th century and today, this become widely discouraged by particle physicists, for some valid reasons - while the method is internally logically consistent, for some people explanation of special relativity gets easier and more clear when not relying on the concept of relativistic mass.

  • Most courses on mechanics do not do justice to analysis and examples of variable mass systems, this area is often skimped over in courses and textbooks for physicists.

  • Given the above situation in both non-relativistic and relativistic mechanics teaching, it is likely that people then reconstruct the actual logic of Newton's second law in this incorrect way:

because we write it in a way that suggests differentiating mass by time can take place ($F = dp/dt$), this equation probably applies even if mass $m$ changes in time. (WRONG)

Anybody who ever derived the rocket equation of motion knows that variable mass systems need careful analysis in terms of interaction of constant mass systems, and $\mathbf F = \frac{d\mathbf p }{dt}$ is no more general than $\mathbf F = m\mathbf a$ - it applies only to constant mass systems.

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  • $\begingroup$ What do you mean by "nowhen"? $\endgroup$ – Acccumulation Mar 8 '19 at 23:06
  • $\begingroup$ So you're saying that Goldstein is plain wrong? $\endgroup$ – PiKindOfGuy Mar 8 '19 at 23:48
  • $\begingroup$ @Acccumulation bad choice of word, fixed that. $\endgroup$ – Ján Lalinský Mar 8 '19 at 23:59
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    $\begingroup$ @PiKindOfGuy I believe so. Or he means something else that is eluding us, but I think that is unlikely. $\endgroup$ – Ján Lalinský Mar 9 '19 at 0:00

That equation, as described by Goldstein and by many others, is wrong, because it's not Galilean invariant when the mass is variable. As an example, consider a boxcar full of sand with a hole in the bottom, and analyse its motion using Goldstein's equation from a reference frame where the boxcar is initially at rest and from a reference frame where the boxcar is moving horizontally. What happens?

Two remarks:

  1. The relationship between the two equations given by Goldstein is right the other way round: if the mass is constant, then you can write $\boldsymbol{F} = m\boldsymbol{a}$ as $\boldsymbol{F} = \dot{\boldsymbol{p}}$.
  2. In Newtonian mechanics, there's no variable mass: mass just moves from one place to another, and variable mass problems are better analysed by considering the whole system with total constant mass (as is done when deriving the correct rocket equation). Considering just a portion of the system with variable mass is a recipe for mistakes (it can be done, but why going that slippery slope?).

For more on the problem of variable mass in Newtonian mechanics, see e.g. Plastino (1990), Pinheiro (2004) and Spivak's book Physics for Mathematicians, Mechanics I.

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  • $\begingroup$ Do you mean that $\boldsymbol{F}$ is not Galilean invariant? Why must it be? $\endgroup$ – md2perpe Mar 8 '19 at 19:02
  • $\begingroup$ @md2perpe I mean that the whole equation is not Galilean invariant, and it should be because all Newtonian mechanics is expected to be Galilean invariant. $\endgroup$ – Massimo Ortolano Mar 8 '19 at 19:15
  • $\begingroup$ @MassimoOrtolano It's quite remarkable that people who do physics professionally misapply basic principles. Perhaps there's a problem with physics education? $\endgroup$ – PiKindOfGuy Mar 8 '19 at 19:23
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    $\begingroup$ @MassimoOrtolano Could you please address why this paper seems to think you're wrong: iopscience.iop.org/article/10.1088/1361-6404/aac751/… $\endgroup$ – PiKindOfGuy Mar 8 '19 at 22:31
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    $\begingroup$ @PiKindOfGuy You need to read that paper carefully. It is correct, but instead of saying that the equation is wrong they say it is correct as long as you "redefine" what the force is to account for the varying mass. $\endgroup$ – BioPhysicist Mar 16 '19 at 11:25

I think that the question is all about the statement

In most instances, the mass of the particle is constant . . .

Here Goldstein is talking explicitly about a particle in the realm of Classical Physics (the title of the book although later chapters include the special relativity) and Section $1.1$ has the title Mechanics of a particle.

Goldstein define the position of his particle with a single radius vector from some origin $\vec r$ which to me implies that the particle is a point mass.

Later on Goldstein reinforces my idea that a particle is a point in that he defines the angular momentum as $\vec L = \vec r \times \vec p$, ie $\vec r$ and $\vec p= m \vec v$ are well defined.

In Section $1.2$ Mechanics of a system of particles Goldstein then starts by distinguishing between external forces and internal forces which again implies that a particle is a point?

This leads me to believe that Goldstein has chosen poor wording, In most instances, the mass of the particle is constant, or I hesitate to write that Goldstein is wrong, a conclusion which is reinforced by the wording of Derivations $1$ in that same chapter:

Show that for a single particle with constant mass the equation of motion implies the following differential for kinetic energy [$T$]:
$$\dfrac {dT}{dt} = \vec F \cdot \vec v$$ while if the mass varies with time the corresponding expression is $$\dfrac {d(mT)}{dt} =\vec F \cdot \vec p$$

Finally in Exercise $13$ the correct form of the rocket equation with the velocity of the escaped gases relative to the rocket has to be derived which requires a number of particles to be consider together with there being internal and external forces.

In most instances, the mass of the particle is constant . . . should be replaced with something like Given that the mass of a particle is constant . . .

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    $\begingroup$ The task given to the student in Derivations, part 1. is quite misleading, as it suggests that there is a general relation between rate of loss of mass and force and momentum. That is obviously not true, the equation is valid only in special case where change of momentum of lost particles is perpendicular to momentum of the body (or zero). This is the case if the body loses parts isotropically in observer's frame, but not for a rocket where the mass ejected in preferred direction. $\endgroup$ – Ján Lalinský Mar 16 '19 at 11:56
  • $\begingroup$ Two remarks. First, it's a pity that Goldstein doesn't report the solutions of the exercises, because I'm pretty sure that the authors haven't solved exercise 13 by using the equation they postulated for variable mass. Second, don't hesitate to say that someone is wrong when they are: that statement has propagated by for too long time in classical physics. $\endgroup$ – Massimo Ortolano Mar 17 '19 at 11:27

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