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Wikipedia states that Newton's second law of motion is:

At any instant of time, the net force on a body is equal to the body's acceleration multiplied by its mass or, equivalently, the rate at which the body's momentum is changing with time.

In other words, it is either

$$ F = ma $$

or

$$ F = \frac{d}{dt}(mv) = v\frac{dm}{dt} + m\frac{dv}{dt}. $$

These are definitely not equivalent. The first says that a changing mass does not affect the force. The second says it does.

So, which is it?

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4 Answers 4

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Your product rule is incorrect without being handled properly. The expression $F=v\frac{\text dm}{\text dt}+m\frac{\text dv}{\text dt}$ does not follow Galilean relativity; $v$ is not the same for inertial observers moving relative to each other, but we should have force being the same for all inertial observers.

You have to consider the rate of change in momentum of the entire system (the "original" system and the mass that causes the change in mass of the "original" system). If you handle this correctly for, say, a system of mass $m$ with velocity $\mathbf v$ colliding with a change in mass $\text dm$ with velocity $\mathbf u$, then we end up with

$$\mathbf F_\text{ext}+\mathbf v_\text{rel}\frac{\text dm}{\text dt}=m\frac{\text d\mathbf v}{\text dt}$$

where $\mathbf v_\text{rel}$ is the relative velocity $\mathbf u-\mathbf v$ between the system and added mass.

I think we should defend $F=ma$ more here though. If two masses collide and stick together, for example, you can either treat it as

  1. a variable-mass system where the mass of one object is changing due to the mass of the other object, or
  2. Two constant-mass systems that are interacting, where $F=ma$ applies to each system individually

Of course, this breaks down / becomes unreasonable when approximating the change in mass as continuous, such as in rocket propulsion, but still, saying $F=ma$ is incorrect because we can have variable-mass systems would be like saying Newtonian mechanics is incorrect because of special relativity. Sure, you are technically correct, but practically it's not too relevant.

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  • $\begingroup$ I didn’t say $F=ma$ is incorrect. I said they are not equivalent and asked which is the actual second law. But, in saying the product rule is incorrect, does this mean that F proportional to rate of change in momentum itself is incorrect? $\endgroup$
    – liah
    Apr 10 at 6:28
  • $\begingroup$ @liah Sorry, I was looking at your comment in the main post about it being "generally incorrect" $F=\dot p$ is not incorrect, you just need to be careful with the "p" you are using. You have to take the whole system into account $\endgroup$ Apr 10 at 12:52
  • $\begingroup$ The physics that survives the quantum revolution and actually works exactly even today is the conservation of momentum and energy, whereas neither forces nor accelerations are well-defined. It makes much more sense to try to fix the momentum time derivative to agree with Galilean relativity by easily subtracting a constant representing the centre of momentum frame. Or better, special relativity. $\endgroup$ Apr 11 at 1:11
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Although mathematically inequivalent, the two definitions are equivalent within the scope of Newtonian Mechanics.

To understand the previous statement, one must understand the meaning of the quantities appearing in Newton's Second Law. In particular, definitions of momentum and mass are important.

Within Classical Mechanics, a point-like body (i.e. a body whose dynamical state is described only by its position and velocity) is characterized by a positive constant called mass. The operative definition of the mass can be based on the ratio of two bodies' accelerations or on the conservation of momentum. From the formal point of view, mass can be defined either from the analysis of the effect of the interaction between two isolated bodies (non-interacting with other bodies): $$ \frac{m_1}{m_2} = \frac{a_2}{a_1}, $$ where $a_i$ is the modulus of the acceleration of the $i$-th body, or from the requirement that the quantity $$ m_1 {\bf v}_1+m_2 {\bf v}_2, $$ for the same system, is constant. In both cases, there is an implicit assumption of constant mass. After the definition of mass, momentum is defined as the product of mass by velocity.

Such apparent limitation is not as severe as it could seem because, starting from the formulation of the equations of motion for constant-mass bodies, it is possible to extend the theory to treat systems of bodies if some of these compound systems vary their total mass. This is the only safe way to extend Newtonian mechanics to the cases of so-called mass-varying systems. A blind use of $$ \frac{{\mathrm d}{\bf p}}{{\mathrm d}t} = \frac{{\mathrm d}{ (m{\bf v}})}{{\mathrm d}t} = m \frac{{\mathrm d} {\bf v}}{{\mathrm d}t} + \frac{{\mathrm d}{m}}{{\mathrm d}t} {\bf v} $$ is well known to be wrong in many cases of mass-varying systems.

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  • $\begingroup$ This "operative definition of mass" occurs in some textbooks of mechanics, but it is a bad idea. It is pedagogically inappropriate use of Newton's 2nd law, or conservation of momentum, because it redefines simple concept in terms of more complicated ones, and degrades them into definitions, instead of laws of nature. This is not how mass of a body is found in practice, except for astronomic bodies, where this method (really an extrapolation of laws found on Earth) is usable due to lack of friction. A good definition of mass in teaching basic mechanics is "number of mass units" or "weight". $\endgroup$ Apr 10 at 21:17
  • $\begingroup$ On the other hand, I agree that there is the tacit assumption in Newton's laws in that mass of bodies referred to does not change. $\endgroup$ Apr 10 at 21:25
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    $\begingroup$ @JánLalinský What you call a bad idea goes back to Mach's formulation of mechanics and has been carefully scrutinized by many good physicists. It is not the only possibility, but "number of mass units" looks like a circular definition, and "weight" is too dangerous because it does not help to distinguish between a force (weight) and mass. $\endgroup$ Apr 10 at 23:09
  • $\begingroup$ Number of mass units is not circular, because it is just the natural number of identical copies of standardized masses (cubes, balls), whose mass is one. Distinguishing between mass and weight is a subtle topic and can be done later, after the three Newton's laws are understood, e.g. at the beginning of teaching of Newton's law of universal gravity. $\endgroup$ Apr 10 at 23:21
  • $\begingroup$ While I totally agree that number of mass units is a superior definition, the original Newton's Principia resorted to operative definition of mass, so I don't know what it is you are on about. $\endgroup$ Apr 11 at 1:07
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They are equivalent, because it is assumed (in both) that the body does not lose or gain any massive parts.

In case the body is losing massive parts, such as a rocket, both formulations seem inadequate, because it is not clear what the system considered when using words "mass" or "momentum" and "net force" is - whether the force of the expelled gases on the rocket motor nozzles is to be accounted in the net force, and whether $m$ is mass of and inside the rocket, or mass of the whole system, including the expelled gases.

But in fact, the $\mathbf F=m\mathbf a$ formulation is correct and applicable to the rocket, provided $m$ is the mass of and inside the rocket, not including the expelled gases (thus $m$ is changing in time), and $\mathbf F$ includes the force on the nozzles due to expelled gases. This force can be expressed as $\frac{dm}{dt} \mathbf c_0$ where $\mathbf c_0$ is the final velocity of the expelled gases in the frame of the rocket. So the equation of motion for the rocket which experiences external force $\mathbf F_{ext}$ (e.g. due to atmosphere) is

$$ m\mathbf a = \mathbf F_{ext} + \frac{dm}{dt} \mathbf c_0 $$

and this equation holds in all reference frames.

The other formulation $\mathbf F = d\mathbf p/dt$, if $\mathbf p$ means momentum of the rocket $\mathbf p=m\mathbf v$, is incorrect in general, because due to the term $\frac{dm}{dt}\mathbf v$, it depends on the frame of reference. This is not acceptable, because force does not depend on frame of reference - the "$dp/dt$ formulation" would lead to

$$ m\mathbf a + \frac{dm}{dt}\mathbf v = \mathbf F_{ext} + \frac{dm}{dt} \mathbf c_0. $$

This equation is correct only in the inertial frame where the rocket has zero velocity, where the term $\frac{dm}{dt}\mathbf v$ is zero.

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  • $\begingroup$ "because force does not depend on frame of reference" is false. Maybe in Galilean, but in SR it transforms rather non-trivially. I don't think its $c\to\infty$ limit is the Galilean behaviour either. $\endgroup$ Apr 11 at 1:05
  • $\begingroup$ @naturallyInconsistent we're discussing a question from Newtonian mechanics, in this theory, the force vector does not depend on the frame of reference, meaning its components transform as spatial coordinates of a point under rotations, so we can say there is only one vector. In special relativity, such statement is still true for four-vector force, it also does not depend on the reference frame. This is in the sense the four-vector force $F^\mu$ obeying $F^\mu = dp^\mu/d\tau$ transforms between inertial frames as the spacetime coordinates do, so we can say there is only one four-vector. $\endgroup$ Apr 11 at 11:56
  • $\begingroup$ It is also important and illuminating to note that for the entire mass of the rocket, including the fuel, Newton's law in its original form holds beautifully: The center of mass of a rocket in space without gravity does not move at all even when the rocket motors are on full throttle since all forces (if we include the fuel) are inner forces. The rocket actually never moves even a single inch. (All we do is increase the entropy of the system ;-) ). $\endgroup$ Apr 11 at 12:14
  • $\begingroup$ @Peter-ReinstateMonica Center of mass of rocket+fuel in space free of gravity does not accelerate, but it can move rectilinearly due to inertia. I don't think this is important or illuminating, it's a trivial consequence of conservation of momentum with no practical use (where even is the center of mass and why should we care). The important thing is how the rocket without the expelled gases moves, and that is described by the equation above. $\endgroup$ Apr 12 at 7:47
  • $\begingroup$ @JánLalinský Yeah, obviously (to me) the "does not move" was relative to the rocket. Trivial or not, it was an interesting realization in high school for me. $\endgroup$ Apr 12 at 7:56
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For a collection of particles, you have the definition of momentum $$ \boldsymbol{p} = \sum_i m_i \boldsymbol{v}_i \tag{1}$$

for which you can show that for a rigid body there exists a point C riding along the body, the center of mass, such that $$ \boldsymbol{p} = m \boldsymbol{v}_C \tag{2}$$ with $m = \sum_i m_i$ and the assumption that mass is constant.

Now Netwton's 2nd law relates the net force acting on a body with the change in momentum

$$ \boldsymbol{F}_{\rm net} = \tfrac{\rm d}{{\rm d}t} \boldsymbol{p} \tag{3} $$

The above is always the definition and if you do not assume a rigid body then (1) leads to

$$ \boldsymbol{F}_{\rm net} = \tfrac{\rm d}{{\rm d}t} \left( \sum_i m_i \boldsymbol{v}_i \right) = \sum_i \tfrac{\rm d}{{\rm d}t} \left( m_i \boldsymbol{v}_i \right) = \sum_i \left( (\tfrac{\rm d}{{\rm d}t} m_i) \boldsymbol{v}_i + m_i \boldsymbol{a}_i \right) \tag{4} $$

But if you only have one particle, or all the particles belong to the same rigid body you get

$$ \boldsymbol{F}_{\rm net} = \tfrac{\rm d}{{\rm d}t} \left( m \boldsymbol{v}_C \right) = m \tfrac{\rm d}{{\rm d}t} \left( \boldsymbol{v}_C \right) = m \boldsymbol{a}_C \tag{5} $$

You can say that both (4) and (5) are correct within their own assumptions, as they both derive from the 2nd law of motion (3).

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