Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Is there ever a situation where the distinction between $F = m \frac{dv}{dt}$ and $F = \frac{dp}{dt}$ is important? I can't think of a situation where one is true and not the other (assuming only conservation of momentum).

Edit: Obviously it is important to take a changing mass into account (e.g. for a rocket) when you're considering a full time evolution, i.e. $F(t) = m(t) \frac{dv}{dt}$ (or in the relativistic case, perhaps something like $F(t) = m(t) \frac{d}{dt} \left( \frac{p}{m} \right)$ with $m$ the rest-mass). And perhaps there is a nontrivial relationship between the rate of change of mass, and the forces being exerted (again, e.g. with a rocket --- where the mass loss is tied to the propulsion). What is not clear is that there should ever be a $F = v\frac{dm}{dt}$ term.

Edit 2: My understanding of the solution:
There should not be a $dm/dt$ term, as pointed out by @garyp. The change in momentum expression is, however, more accurate because $p \neq mv$ in general (e.g. in relativistic cases, or when considering massless systems). It would seem that either one must take the caveat that $dp/dt$ cannot be used for mass-varying systems, or take the much less conceptual or aesthetically pleasing expression that $F = m \frac{d (\gamma v)}{dt}$ (which still only applies to classical systems).

share|cite|improve this question

4 Answers 4

up vote 11 down vote accepted

In classical mechanics Newton's second law applies only to constant mass systems. In those cases there's no difference between $F=ma$ and $F=\mathrm{d}p/\mathrm{d}t$. However, in special relativity the latter is valid, but the former is not. A relativistic definition of momentum is required: $p=\gamma m v$.

Some details

[rework of my original answer] Some of the responses given so far answer the OP, but have weaknesses that might lead to misconceptions about Newton 2. I'll try to address the issue.

Some of the answers so far are not entirely correct if by $p$ is meant $mv$ of the rocket. Newton's second law is valid only for constant mass systems. $F=\mathrm{d}p/\mathrm{d}t$ leads to the rocket equation by accident if the propellent is exhausted in a direction opposite to the direction of motion. If you are very careful about what you mean by $p$ a correct analysis can be made, but $p=m_\mathrm{rocket}v_\mathrm{rocket}$ does not work.

This Wikipedia entry. is the clearest statement of that fact that I've found.

To see why, consider a system comprising the rocket plus its remaining fuel and remaining propellant, which is what I think the other responders intend. (I may be wrong, but if so, they should clarify what exactly is their system.)

Imagine the system moving at constant velocity, and having two thrusters pointing perpendicular to the direction of motion, and directly opposite each other, and producing identical constant thrust by expelling exhaust gas. The thrust force of the two are equal and opposite. So the net force on the rocket is zero. Then $F_{net}=0$ and $\mathrm{d}v/\mathrm{d}t = 0$, and $v \neq 0$ and $\mathrm{d}m/\mathrm{d}t \neq 0$. Blindly applying $F=\mathrm{d}p/\mathrm{d}t$ leads to $0=v\,\mathrm{d}m/\mathrm{d}t$, a contradiction which can only be resolved if the mass is unchanging.

Comments on some comments

The system is losing momentum by losing mass, but it's velocity is not changing: there is no net force on the system. Momentum is conserved in the closed system consisting of the rocket and the exhausted propellant. The system consisting of the rocket plus the fuel and propellant not yet exhausted (fuel remaining in the tank) is an open system. Conservation of momentum does not apply to open systems.

Careful analysis of a variable mass system lead to $$ \vec{F}_\mathrm{ext}=\vec{u}\frac{\mathrm{d}m}{\mathrm{d}t} + m\frac{\mathrm{d}\vec{v}}{\mathrm{d}t}$$ where $\vec{u}$ is the velocity of the mass leaving the system relative to the velocity of the system, and $\vec{F}_\mathrm{ext}$ is the external force on the system. For a rocket $\vec{F}_\mathrm{ext}=0$, and $$ 0=\vec{u}\frac{\mathrm{d}m}{\mathrm{d}t} + m\frac{\mathrm{d}\vec{v}}{\mathrm{d}t}$$ This is not the same as $$ 0=\vec{v}\frac{\mathrm{d}m}{\mathrm{d}t} + m\frac{\mathrm{d}\vec{v}}{\mathrm{d}t}$$ where $\vec{v}$ is the velocity of the rocket.

Trying to write $F = m\,\mathrm{d}v/\mathrm{d}t + v\,\mathrm{d}m/\mathrm{d}t$ for a variable mass system is not correct. here's a nice discussion whose first sentences are "In mechanics, a variable-mass system is a collection of matter whose mass varies with time. Newton's second law of motion cannot directly be applied to such a system because it is valid for constant mass systems only."

Aeronautical engineers know all this. It's only physicists who are confused. I checked some books. Symon's and John R. Taylor's classical mechanics texts get it right, as does Halliday, Resnick, and Walker.

share|cite|improve this answer

Yes, and a rocket is a good example. In $$F=m\left( \frac{dv}{dt} \right)$$ you assume mass is constant. If mass is variable, like a rocket burning off fuel, you need to account for it, $$F=\frac{dp}{dt}=\frac{d}{dt}\left( mv \right)\; =\; m\frac{dv}{dt}\; +\; v\frac{dm}{dt}$$ where the m is decreasing. If you fiddle with this you might get the Rocket Equation. Start by examining the motion with m replaced by $$\left( m+\Delta m \right)$$ where the delta m can be negative, and use some ideas from conservation of momentum.

share|cite|improve this answer

In electrodynamics, the Lorentz force on a moving charge $q$ $$ \vec F = q \left( \vec E + \vec v \times\vec B \right) $$ can be rewritten as the force per unit volume $dV$ on charge density distribution, $\rho$, with current density $\vec J$: $$ \vec F = dV\ \vec f = dV \left( \rho\vec E + \vec J \times \vec B \right). $$ You can use Maxwell's source-term equations $$ \begin{align*} \vec\nabla\cdot\vec E &= \frac{\rho}{\epsilon_0} & \vec\nabla\times\vec B - {\epsilon_0\mu_0}\frac{\partial\vec E}{\partial t} = {\mu_0 \vec J} \end{align*} $$ to eliminate the charge and current distributions and write this force density $\vec f$ solely in terms of the fields. It takes some doing, but you wind up with $$ \vec f = \vec\nabla\cdot\mathbf{T} - \epsilon_0\mu_0 \frac{\partial \vec S}{\partial t} $$ where $\vec S$ is the Poynting vector and $\mathbf T$ is the Maxwell stress tensor. This shows that you may have momentum entering and leaving the fields in a region of space, even when there isn't any mass around to accelerate.

share|cite|improve this answer

Just a little bit of addition:

Newton's 2nd Law is F = ma = m dv/dt

This is correct for constant mass and in non-relativistic context.

However since mass can be assumed constant in Newton's original formulation is equivalent to:

F = dp/dt

And this form is correct for both variable mass AND in a relativistic context (of course with the appropriate interpreations)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.