Using the product rule to expand Newton's Second Law?

Question

Newton's second law says that $$F=\frac{\text{d}p}{\text{d}t},$$ where $F$ is the net force on a body. My question is, why can't the product rule be used to yield $$F=v\frac{\text{d}m}{\text{d}t}+m\frac{\text{d}v}{\text{d}t}$$ I've read that it doesn't respect Galilean Invariance, but can somebody explain why this is so? I've seen a respectable textbook (Morin) doing this without any problems, so I was wondering if this were correct.

Answer 1

As wikipedia says about the second law,

As Newton's second law is only valid for constant-mass systems, mass can be taken outside the differentiation operator by the constant factor rule in differentiation

This is expanded in the article about variable mass systems. You derived the formula $$ F = m\dot v + \dot m v. $$ where $\dot v \equiv {\text{d}v}/{\text{d}t}$ etc. This violates Galilean invariance because it is not invariant under $v\to v + \Delta v$. If, for example, the force is zero, $$ \dot v = -\frac{\dot m}m v, $$ an object with a velocity $v$ and changing mass accelerates. The value of $v$, however, is dependent on an observer's frame of reference, and thus so is the acceleration in this case.

This is resolved by considering that a changing mass implies that mass is being ejected from an object (or accrued). The correct equation in these circumstances is $$ F_{\text{external}} = m\dot v - \dot m u $$ where $u$ is a the relative velocity with which mass is ejected from the centre of mas of the object with mass $m(t)$ and velocity $v$. You can derive this formula by simple considerations about the momentum of a system that emits/accrues a mass $\text{d}m$ in time $\text{d}t$.

Because this equations depends on a relative velocity $u$ rather than an absolute velocity, it is invariant under Galilean transformations, i.e. the relative velocity does not change $u=v_1 - v_2 \to u$, even if the absolute velocities are shifted $v_i\to v_i + \Delta v$.

Answer 2

So, $F = \frac{dp}{dt} = \frac{d}{dt}(mv)=m\frac{dv}{dt}+v\frac{dm}{dt} = m\frac{dv}{dt} =\frac{dp}{dv} \times \frac{dv}{dt} = \frac{dp}{dt} = F $ nope, no problem. Note that $\frac{dm}{dt}$ is assumed zero because the situation described by the equation requires that the mass is constant, and hence time invariant. This is of course inapplicable if somehow the mass is changing. The most definite equation for the law should be $ F = ma = m\frac{dv}{dt} $ which works even if the mass is changing.