Momentum

For a nonrelativistic particle of mass $m$ and velocity $v$, the momentum is $p=mv$.

For a charged particle minimally coupled to a magnetic vector potential, the mechanical momentum is $\mathbf{\pi }=\mathbf{p}-q\mathbf{A}$, with $\mathbf{p}$ the canonical momentum. It is the mass times velocity in nonrelativistic mechanics.

Momentum conservation states that the total momentum of an isolated system is constant. In a translation-invariant field theory, integrating over the spacetime zero mode enforces momentum conservation at interactions and scattering amplitudes.

For a body of varying mass $M$ ejecting material at fixed relative velocity, conservation of momentum gives a differential relation between $dM$ and the body's change $dv$. Integrating it produces a logarithm of the mass ratio.

When matter leaves a moving body with zero instantaneous relative velocity, it carries the body's current momentum per unit mass but creates no additional recoil term. A short-interval momentum balance including the newly released material gives $q\dot{v}=F_{\mathrm{ext}}$ for the remaining body's mass $q(t)$. The expression $\frac{d}{dt}(qv)=F_{\mathrm{ext}}$ omits the outgoing momentum flux. Nonzero relative ejection velocity instead produces the thrust term of the rocket equation.

A classical rocket ejecting fuel at rate $\alpha >0$ and relative exhaust speed $u>0$, while collecting stationary dust of mass $\beta dx$ per traveled distance, obeysThese are mass conservation and momentum conservation for the rocket, incoming dust and outgoing gas. The dust-loading term supplies the momentum needed to accelerate incoming stationary material. Put $\lambda =\sqrt{\alpha /(\beta u)}$ and $V=\lambda u$. While the engines operate and $0\le v<V$, an initially stationary rocket satisfiesFor $\lambda >1$, continuing to $V$ would drive total mass to zero. For $0<\lambda <1$, total mass instead grows without bound and $V$ is approached only after infinite time under indefinitely maintained thrust. In both cases a finite fuel supply runs out before that limiting speed. Total stored mass must not be confused with remaining fuel.

For collinear motion without external forces, a rocket ejecting material backwards at constant speed $u$ in its instantaneous rest frame obeys $m{\gamma }^{2}dv+udm=0$. Here $m$ is the rocket's rest mass, including stored internal energy, and $\gamma$ is its Lorentz factor in the chosen inertial frame. The equation follows from momentum conservation, conservation of relativistic energy, and the relativistic velocity-addition formula; expelled rest mass is not simply the decrease in rocket rest mass.

For zero initial velocity, body masses $kM,(1-k)M$ and fuel masses $k{m}_0,(1-k)m_0$, the rocket equation gives the displayed final second-stage speed after burning the first-stage fuel, discarding its body without impulse, and burning the second-stage fuel. It assumes $M>0$ and $0\le k<1$. The speed exceeds the single-stage value when $0<k<1$ and fuel is positive. Its limit as $k\to 1^{-}$ concerns a surviving second stage of vanishing mass; at $k=1$ no such stage exists.

If a vehicle moving at speed $v$ ejects mass forward at relative speed $u>0$, thenAs $dM<0$, the vehicle slows down.

The angular momentum of a particle about the origin is $L=r\times p$.

The total angular momentum of an isolated system is constant. More generally, its time derivative equals the external torque.

The rate at which angular momentum crosses a surface, including advection and stresses. With positive inward mass accretion rate, the net inward flux in a steady accretion disk is $\dot{M}h-\mathcal{G}$.

For unit mass in a central force with potential $\Phi (r)$ and conserved angular momentum $L$, a radial turning point $r_{\ast }$ has zero radial speed and therefore satisfies

For unit mass, incoming speed $v>0$, impact parameter $b>0$, and potential energy $\mp \kappa /r$, put $q=\kappa /v^2>0$. At closest approach, conservation of energy and conservation of angular momentum give the displayed radii and tangential speeds $u_{\mp }=bv/p_{\mp }$. Consequently $p_{-}{p}_{+}=b^2$ and $u_{-}{u}_{+}=v^2$. These relations compare equal force magnitudes. The attractive radial case $b=0$ is a singular collision, so the speed-product statement does not extend to that endpoint.

Angular momentum transport moves angular momentum between locations through matter, stresses, waves, or fields. An accretion disk must transport angular momentum outward for most of its mass to move inward.

Torque is the moment of force, $\tau =r\times F$, and equals the rate of change of angular momentum.

Moment balance equates the rate of change of angular momentum to applied torques. A static slender rod obeys ${\partial }_s\mathbf{M}+\mathbf{t}\times \mathbf{F}+\mathbf{m}=0$.

The areal velocity is the rate at which the radius vector sweeps area. For specific angular momentum $h$, it is constant and equals