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Momentum

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From Wikipedia, the free encyclopedia
Property of a mass in motion
This article is about linear momentum. Not to be confused withangular momentum ormoment (physics).
This article is about momentum in physics. For other uses, seeMomentum (disambiguation).

Momentum
A pool break-off shot
Momentum of apool cue ball is transferred to the racked balls after collision.
Common symbols
p,p
SI unitkg⋅m⋅s−1
Other units
slugft/s
Conserved?Yes
DimensionMLT1{\displaystyle {\mathsf {M}}{\mathsf {L}}{\mathsf {T}}^{-1}}
Part of a series on
Classical mechanics
F=dpdt{\displaystyle {\textbf {F}}={\frac {d\mathbf {p} }{dt}}}

InNewtonian mechanics,momentum (pl.:momenta ormomentums; more specificallylinear momentum ortranslational momentum) is theproduct of themass andvelocity of an object. It is avector quantity, possessing a magnitude and a direction. Ifm is an object's mass andv is its velocity (also a vector quantity), then the object's momentump (from Latinpellere "push, drive") is:p=mv.{\displaystyle \mathbf {p} =m\mathbf {v} .}In theInternational System of Units (SI), theunit of measurement of momentum is thekilogrammetre per second (kg⋅m/s), which isdimensionally equivalent to thenewton-second.

Newton's second law of motion states that the rate of change of a body's momentum is equal to the net force acting on it. Momentum depends on theframe of reference, but in anyinertial frame of reference, it is aconserved quantity, meaning that if aclosed system is not affected by external forces, its total momentum does not change. Momentum is also conserved inspecial relativity (with a modified formula) and, in a modified form, inelectrodynamics,quantum mechanics,quantum field theory, andgeneral relativity. It is an expression of one of the fundamental symmetries of space and time:translational symmetry.

Advanced formulations of classical mechanics,Lagrangian andHamiltonian mechanics, allow one to choose coordinate systems that incorporate symmetries and constraints. In these systems the conserved quantity isgeneralized momentum, and in general this is different from thekinetic momentum defined above. The concept of generalized momentum is carried over into quantum mechanics, where it becomes an operator on awave function. The momentum and position operators are related by theHeisenberg uncertainty principle.

In continuous systems such aselectromagnetic fields,fluid dynamics anddeformable bodies, amomentum density can be defined as momentum per volume (avolume-specific quantity). Acontinuum version of the conservation of momentum leads to equations such as theNavier–Stokes equations for fluids or theCauchy momentum equation for deformable solids or fluids.

Classical

Momentum is avector quantity: it has both magnitude and direction. Since momentum has a direction, it can be used to predict the resulting direction and speed of motion of objects after they collide. Below, the basic properties of momentum are described in one dimension. The vector equations are almost identical to the scalar equations (seemultiple dimensions).

Single particle

The momentum of a particle is conventionally represented by the letterp. It is the product of two quantities, the particle'smass (represented by the letterm) and itsvelocity (v):[1]p=mv.{\displaystyle p=mv.}

The unit of momentum is the product of the units of mass and velocity. InSI units, if the mass is in kilograms and the velocity is in meters per second then the momentum is in kilogram meters per second (kg⋅m/s). Incgs units, if the mass is in grams and the velocity in centimeters per second, then the momentum is in gram centimeters per second (g⋅cm/s).

Being a vector, momentum has magnitude and direction. For example, a 1 kg model airplane, traveling due north at 1 m/s in straight and level flight, has a momentum of 1 kg⋅m/s due north measured with reference to the ground.

Many particles

The momentum of a system of particles is the vector sum of their momenta. If two particles have respective massesm1 andm2, and velocitiesv1 andv2, the total momentum isp=p1+p2=m1v1+m2v2.{\displaystyle {\begin{aligned}p&=p_{1}+p_{2}\\&=m_{1}v_{1}+m_{2}v_{2}\,.\end{aligned}}}The momenta of more than two particles can be added more generally with the following:p=imivi.{\displaystyle p=\sum _{i}m_{i}v_{i}.}

A system of particles has acenter of mass, a point determined by the weighted sum of their positions:rcm=m1r1+m2r2+m1+m2+=imiriimi.{\displaystyle r_{\text{cm}}={\frac {m_{1}r_{1}+m_{2}r_{2}+\cdots }{m_{1}+m_{2}+\cdots }}={\frac {\sum _{i}m_{i}r_{i}}{\sum _{i}m_{i}}}.}

If one or more of the particles is moving, the center of mass of the system will generally be moving as well (unless the system is in pure rotation around it). If the total mass of the particles ism{\displaystyle m}, and the center of mass is moving at velocityvcm, the momentum of the system is:

p=mvcm.{\displaystyle p=mv_{\text{cm}}.}

This is known asEuler's first law.[2][3]

Relation to force

If the net forceF applied to a particle is constant, and is applied for a time intervalΔt, the momentum of the particle changes by an amountΔp=FΔt.{\displaystyle \Delta p=F\Delta t\,.}

In differential form, this isNewton's second law; the rate of change of the momentum of a particle is equal to the instantaneous forceF acting on it,[1]F=dpdt.{\displaystyle F={\frac {{\text{d}}p}{{\text{d}}t}}.}

If the net force experienced by a particle changes as a function of time,F(t), the change in momentum (orimpulseJ) between timest1 andt2 isΔp=J=t1t2F(t)dt.{\displaystyle \Delta p=J=\int _{t_{1}}^{t_{2}}F(t)\,{\text{d}}t\,.}

Impulse is measured in thederived units of thenewton second (1 N⋅s = 1 kg⋅m/s) ordyne second (1 dyne⋅s = 1 g⋅cm/s)

Under the assumption of constant massm, it is equivalent to write

F=d(mv)dt=mdvdt=ma,{\displaystyle F={\frac {{\text{d}}(mv)}{{\text{d}}t}}=m{\frac {{\text{d}}v}{{\text{d}}t}}=ma,}

hence the net force is equal to the mass of the particle times itsacceleration.[1]

Example: A model airplane of mass 1 kg accelerates from rest to a velocity of 6 m/s due north in 2 s. The net force required to produce this acceleration is 3 newtons due north. The change in momentum is 6 kg⋅m/s due north. The rate of change of momentum is 3 (kg⋅m/s)/s due north which is numerically equivalent to 3 newtons.

Conservation

In aclosed system (one that does not exchange any matter with its surroundings and is not acted on by external forces) the total momentum remains constant. This fact, known as thelaw of conservation of momentum, is implied byNewton's laws of motion.[4][5] Suppose, for example, that two particles interact. As explained by the third law, the forces between them are equal in magnitude but opposite in direction. If the particles are numbered 1 and 2, the second law states thatF1 =dp1/dt andF2 =dp2/dt. Therefore,

dp1dt=dp2dt,{\displaystyle {\frac {{\text{d}}p_{1}}{{\text{d}}t}}=-{\frac {{\text{d}}p_{2}}{{\text{d}}t}},}with the negative sign indicating that the forces oppose. Equivalently,

ddt(p1+p2)=0.{\displaystyle {\frac {\text{d}}{{\text{d}}t}}\left(p_{1}+p_{2}\right)=0.}

If the velocities of the particles arevA1 andvB1 before the interaction, and afterwards they arevA2 andvB2, then

mAvA1+mBvB1=mAvA2+mBvB2.{\displaystyle m_{A}v_{A1}+m_{B}v_{B1}=m_{A}v_{A2}+m_{B}v_{B2}.}

This law holds no matter how complicated the force is between particles. Similarly, if there are several particles, the momentum exchanged between each pair of particles adds to zero, so the total change in momentum is zero. The conservation of the total momentum of a number of interacting particles can be expressed as[4]mAvA+mBvB+mCvC+=constant.{\displaystyle m_{A}v_{A}+m_{B}v_{B}+m_{C}v_{C}+\ldots ={\text{constant}}.}

This conservation law applies to all interactions, includingcollisions (bothelastic andinelastic) and separations caused by explosive forces.[4] It can also be generalized to situations where Newton's laws do not hold, for example in thetheory of relativity and inelectrodynamics.[6]

Dependence on reference frame

Momentum is a measurable quantity, and the measurement depends on theframe of reference. For example: if an aircraft of mass 1000 kg is flying through the air at a speed of 50 m/s its momentum can be calculated to be 50,000 kg.m/s. If the aircraft is flying into a headwind of 5 m/s its speed relative to the surface of the Earth is only 45 m/s and its momentum can be calculated to be 45,000 kg.m/s. Both calculations are equally correct. In both frames of reference, any change in momentum will be found to be consistent with the relevant laws of physics.

Supposex is a position in an inertial frame of reference. From the point of view of another frame of reference, moving at a constant speedu relative to the other, the position (represented by a primed coordinate) changes with time as

x=xut.{\displaystyle x'=x-ut\,.}

This is called aGalilean transformation.

If a particle is moving at speeddx/dt =v in the first frame of reference, in the second, it is moving at speed

v=dxdt=vu.{\displaystyle v'={\frac {{\text{d}}x'}{{\text{d}}t}}=v-u\,.}

Sinceu does not change, the second reference frame is also an inertial frame and the accelerations are the same:

a=dvdt=a.{\displaystyle a'={\frac {{\text{d}}v'}{{\text{d}}t}}=a\,.}

Thus, momentum is conserved in both reference frames. Moreover, as long as the force has the same form, in both frames, Newton's second law is unchanged. Forces such as Newtonian gravity, which depend only on the scalar distance between objects, satisfy this criterion. This independence of reference frame is called Newtonian relativity orGalilean invariance.[7]

A change of reference frame can often simplify calculations of motion. For example, in a collision of two particles, a reference frame can be chosen where one particle begins at rest. Another commonly used reference frame is thecenter of mass frame – one that is moving with the center of mass. In this frame, the total momentum is zero.

Application to collisions

If two particles, each of known momentum, collide and coalesce, the law of conservation of momentum can be used to determine the momentum of the coalesced body. If the outcome of the collision is that the two particles separate, the law is not sufficient to determine the momentum of each particle. If the momentum of one particle after the collision is known, the law can be used to determine the momentum of the other particle. Alternatively if the combinedkinetic energy after the collision is known, the law can be used to determine the momentum of each particle after the collision.[8] Kinetic energy is usually not conserved. If it is conserved, the collision is called anelastic collision; if not, it is aninelastic collision.

Elastic collisions

Main article:Elastic collision
Elastic collision of equal masses
Elastic collision of unequal masses

An elastic collision is one in which nokinetic energy is transformed into heat or some other form of energy. Perfectly elastic collisions can occur when the objects do not touch each other, as for example in atomic or nuclear scattering where electric repulsion keeps the objects apart. Aslingshot maneuver of a satellite around a planet can also be viewed as a perfectly elastic collision. A collision between twopool balls is a good example of analmost totally elastic collision, due to their highrigidity, but when bodies come in contact there is always somedissipation.[9]

A head-on elastic collision between two bodies can be represented by velocities in one dimension, along a line passing through the bodies. If the velocities arevA1 andvB1 before the collision andvA2 andvB2 after, the equations expressing conservation of momentum and kinetic energy are:

mAvA1+mBvB1=mAvA2+mBvB212mAvA12+12mBvB12=12mAvA22+12mBvB22.{\displaystyle {\begin{aligned}m_{A}v_{A1}+m_{B}v_{B1}&=m_{A}v_{A2}+m_{B}v_{B2}\\{\tfrac {1}{2}}m_{A}v_{A1}^{2}+{\tfrac {1}{2}}m_{B}v_{B1}^{2}&={\tfrac {1}{2}}m_{A}v_{A2}^{2}+{\tfrac {1}{2}}m_{B}v_{B2}^{2}\,.\end{aligned}}}

A change of reference frame can simplify analysis of a collision. For example, suppose there are two bodies of equal massm, one stationary and one approaching the other at a speedv (as in the figure). The center of mass is moving at speedv/2 and both bodies are moving towards it at speedv/2. Because of the symmetry, after the collision both must be moving away from the center of mass at the same speed. Adding the speed of the center of mass to both, we find that the body that was moving is now stopped and the other is moving away at speedv. The bodies have exchanged their velocities. Regardless of the velocities of the bodies, a switch to the center of mass frame leads us to the same conclusion. Therefore, the final velocities are given by[4]

vA2=vB1vB2=vA1.{\displaystyle {\begin{aligned}v_{A2}&=v_{B1}\\v_{B2}&=v_{A1}\,.\end{aligned}}}

In general, when the initial velocities are known, the final velocities are given by[10]

vA2=(mAmBmA+mB)vA1+(2mBmA+mB)vB1vB2=(mBmAmA+mB)vB1+(2mAmA+mB)vA1.{\displaystyle {\begin{aligned}v_{A2}&=\left({\frac {m_{A}-m_{B}}{m_{A}+m_{B}}}\right)v_{A1}+\left({\frac {2m_{B}}{m_{A}+m_{B}}}\right)v_{B1}\\v_{B2}&=\left({\frac {m_{B}-m_{A}}{m_{A}+m_{B}}}\right)v_{B1}+\left({\frac {2m_{A}}{m_{A}+m_{B}}}\right)v_{A1}\,.\end{aligned}}}

If one body has much greater mass than the other, its velocity will be little affected by a collision while the other body will experience a large change.

Inelastic collisions

Main article:Inelastic collision
a perfectly inelastic collision between equal masses

In an inelastic collision, some of the kinetic energy of the colliding bodies is converted into other forms of energy (such asheat orsound). Examples includetraffic collisions,[11] in which the effect of loss of kinetic energy can be seen in the damage to the vehicles; electrons losing some of their energy to atoms (as in theFranck–Hertz experiment);[12] andparticle accelerators in which the kinetic energy is converted into mass in the form of new particles.

In a perfectly inelastic collision (such as a bug hitting a windshield), both bodies have the same motion afterwards. A head-on inelastic collision between two bodies can be represented by velocities in one dimension, along a line passing through the bodies. If the velocities arevA1 andvB1 before the collision then in a perfectly inelastic collision both bodies will be travelling with velocityv2 after the collision. The equation expressing conservation of momentum is:

mAvA1+mBvB1=(mA+mB)v2.{\displaystyle {\begin{aligned}m_{A}v_{A1}+m_{B}v_{B1}&=\left(m_{A}+m_{B}\right)v_{2}\,.\end{aligned}}}

If one body is motionless to begin with (e.g.u2=0{\displaystyle u_{2}=0}), the equation for conservation of momentum is

mAvA1=(mA+mB)v2,{\displaystyle m_{A}v_{A1}=\left(m_{A}+m_{B}\right)v_{2}\,,}

so

v2=mAmA+mBvA1.{\displaystyle v_{2}={\frac {m_{A}}{m_{A}+m_{B}}}v_{A1}\,.}

In a different situation, if the frame of reference is moving at the final velocity such thatv2=0{\displaystyle v_{2}=0}, the objects would be brought to rest by a perfectly inelastic collision and 100% of the kinetic energy is converted to other forms of energy. In this instance the initial velocities of the bodies would be non-zero, or the bodies would have to be massless.

One measure of the inelasticity of the collision is thecoefficient of restitutionCR, defined as the ratio of relative velocity of separation to relative velocity of approach. In applying this measure to a ball bouncing from a solid surface, this can be easily measured using the following formula:[13]

CR=bounce heightdrop height.{\displaystyle C_{\text{R}}={\sqrt {\frac {\text{bounce height}}{\text{drop height}}}}\,.}

The momentum and energy equations also apply to the motions of objects that begin together and then move apart. For example, anexplosion is the result of a chain reaction that transforms potential energy stored in chemical, mechanical, or nuclear form into kinetic energy, acoustic energy, and electromagnetic radiation.Rockets also make use of conservation of momentum: propellant is thrust outward, gaining momentum, and an equal and opposite momentum is imparted to the rocket.[14]

Multiple dimensions

Two-dimensional elastic collision. There is no motion perpendicular to the image, so only two components are needed to represent the velocities and momenta. The two blue vectors represent velocities after the collision and add vectorially to get the initial (red) velocity.

Real motion has both direction and velocity and must be represented by avector. In a coordinate system withx,y,z axes, velocity has componentsvx in thex-direction,vy in they-direction,vz in thez-direction. The vector is represented by a boldface symbol:[15]

v=(vx,vy,vz).{\displaystyle \mathbf {v} =\left(v_{x},v_{y},v_{z}\right).}

Similarly, the momentum is a vector quantity and is represented by a boldface symbol:

p=(px,py,pz).{\displaystyle \mathbf {p} =\left(p_{x},p_{y},p_{z}\right).}

The equations in the previous sections, work in vector form if the scalarsp andv are replaced by vectorsp andv. Each vector equation represents three scalar equations. For example,

p=mv{\displaystyle \mathbf {p} =m\mathbf {v} }

represents three equations:[15]

px=mvxpy=mvypz=mvz.{\displaystyle {\begin{aligned}p_{x}&=mv_{x}\\p_{y}&=mv_{y}\\p_{z}&=mv_{z}.\end{aligned}}}

The kinetic energy equations are exceptions to the above replacement rule. The equations are still one-dimensional, but each scalar represents themagnitude of the vector, for example,

v2=vx2+vy2+vz2.{\displaystyle v^{2}=v_{x}^{2}+v_{y}^{2}+v_{z}^{2}\,.}

Each vector equation represents three scalar equations. Often coordinates can be chosen so that only two components are needed, as in the figure. Each component can be obtained separately and the results combined to produce a vector result.[15]

A simple construction involving the center of mass frame can be used to show that if a stationary elastic sphere is struck by a moving sphere, the two will head off at right angles after the collision (as in the figure).[16]

Objects of variable mass

See also:Variable-mass system

The concept of momentum plays a fundamental role in explaining the behavior of variable-mass objects such as arocket ejecting fuel or astaraccreting gas. In analyzing such an object, one treats the object's mass as a function that varies with time:m(t). The momentum of the object at timet is thereforep(t) =m(t)v(t). One might then try to invoke Newton's second law of motion by saying that the external forceF on the object is related to its momentump(t) byF =dp/dt, but this is incorrect, as is the related expression found by applying the product rule tod(mv)/dt:[17]

F=m(t)dvdt+v(t)dmdt.(incorrect){\displaystyle F=m(t){\frac {{\text{d}}v}{{\text{d}}t}}+v(t){\frac {{\text{d}}m}{{\text{d}}t}}.{\text{(incorrect)}}}

This equation does not correctly describe the motion of variable-mass objects. The correct equation is

F=m(t)dvdtudmdt,{\displaystyle F=m(t){\frac {{\text{d}}v}{{\text{d}}t}}-u{\frac {{\text{d}}m}{{\text{d}}t}},}

whereu is the velocity of the ejected/accreted massas seen in the object's rest frame.[17] This is distinct fromv, which is the velocity of the object itself as seen in an inertial frame.

This equation is derived by keeping track of both the momentum of the object as well as the momentum of the ejected/accreted mass (dm). When considered together, the object and the mass (dm) constitute a closed system in which total momentum is conserved.

P(t+dt)=(mdm)(v+dv)+dm(vu)=mv+mdvudm=P(t)+mdvudm{\displaystyle P(t+{\text{d}}t)=(m-{\text{d}}m)(v+{\text{d}}v)+{\text{d}}m(v-u)=mv+m{\text{d}}v-u{\text{d}}m=P(t)+m{\text{d}}v-u{\text{d}}m}

Generalized

See also:Analytical mechanics

Newton's laws can be difficult to apply to many kinds of motion because the motion is limited byconstraints. For example, a bead on an abacus is constrained to move along its wire and a pendulum bob is constrained to swing at a fixed distance from the pivot. Many such constraints can be incorporated by changing the normalCartesian coordinates to a set ofgeneralized coordinates that may be fewer in number.[18] Refined mathematical methods have been developed for solving mechanics problems in generalized coordinates. They introduce ageneralized momentum, also known as thecanonical momentum orconjugate momentum, that extends the concepts of both linear momentum andangular momentum. To distinguish it from generalized momentum, the product of mass and velocity is also referred to asmechanical momentum,kinetic momentum orkinematic momentum.[6][19][20] The two main methods are described below.

Lagrangian mechanics

InLagrangian mechanics, a Lagrangian is defined as the difference between the kinetic energyT and thepotential energyV:

L=TV.{\displaystyle {\mathcal {L}}=T-V\,.}

If the generalized coordinates are represented as a vectorq = (q1,q2, ... ,qN) and time differentiation is represented by a dot over the variable, then the equations of motion (known as the Lagrange orEuler–Lagrange equations) are a set ofN equations:[21]

ddt(Lq˙j)Lqj=0.{\displaystyle {\frac {\text{d}}{{\text{d}}t}}\left({\frac {\partial {\mathcal {L}}}{\partial {\dot {q}}_{j}}}\right)-{\frac {\partial {\mathcal {L}}}{\partial q_{j}}}=0\,.}

If a coordinateqi is not a Cartesian coordinate, the associated generalized momentum componentpi does not necessarily have the dimensions of linear momentum. Even ifqi is a Cartesian coordinate,pi will not be the same as the mechanical momentum if the potential depends on velocity.[6] Some sources represent the kinematic momentum by the symbolΠ.[22]

In this mathematical framework, a generalized momentum is associated with the generalized coordinates. Its components are defined as

pj=Lq˙j.{\displaystyle p_{j}={\frac {\partial {\mathcal {L}}}{\partial {\dot {q}}_{j}}}\,.}

Each componentpj is said to be theconjugate momentum for the coordinateqj.

Now if a given coordinateqi does not appear in the Lagrangian (although its time derivative might appear), thenpj is constant. This is the generalization of the conservation of momentum.[6]

Even if the generalized coordinates are just the ordinary spatial coordinates, the conjugate momenta are not necessarily the ordinary momentum coordinates. An example is found in the section on electromagnetism.

Hamiltonian mechanics

InHamiltonian mechanics, the Lagrangian (a function of generalized coordinates and their derivatives) is replaced by a Hamiltonian that is a function of generalized coordinates and momentum. The Hamiltonian is defined as

H(q,p,t)=pq˙L(q,q˙,t),{\displaystyle {\mathcal {H}}\left(\mathbf {q} ,\mathbf {p} ,t\right)=\mathbf {p} \cdot {\dot {\mathbf {q} }}-{\mathcal {L}}\left(\mathbf {q} ,{\dot {\mathbf {q} }},t\right)\,,}

where the momentum is obtained by differentiating the Lagrangian as above. The Hamiltonian equations of motion are[23]

q˙i=Hpip˙i=HqiLt=dHdt.{\displaystyle {\begin{aligned}{\dot {q}}_{i}&={\frac {\partial {\mathcal {H}}}{\partial p_{i}}}\\-{\dot {p}}_{i}&={\frac {\partial {\mathcal {H}}}{\partial q_{i}}}\\-{\frac {\partial {\mathcal {L}}}{\partial t}}&={\frac {{\text{d}}{\mathcal {H}}}{{\text{d}}t}}\,.\end{aligned}}}

As in Lagrangian mechanics, if a generalized coordinate does not appear in the Hamiltonian, its conjugate momentum component is conserved.[24]

Symmetry and conservation

Conservation of momentum is a mathematical consequence of thehomogeneity (shiftsymmetry) of space (position in space is thecanonical conjugate quantity to momentum). That is, conservation of momentum is a consequence of the fact that the laws of physics do not depend on position; this is a special case ofNoether's theorem.[25] For systems that do not have this symmetry, it may not be possible to define conservation of momentum. Examples where conservation of momentum does not apply includecurved spacetimes ingeneral relativity[26] ortime crystals incondensed matter physics.[27][28][29][30]

Momentum density

See also:Mass flux

In deformable bodies and fluids

Conservation in a continuum

Main article:Cauchy momentum equation
Motion of a material body

In fields such asfluid dynamics andsolid mechanics, it is not feasible to follow the motion of individual atoms or molecules. Instead, the materials must be approximated by acontinuum in which, at each point, there is a particle orfluid parcel that is assigned the average of the properties of atoms in a small region nearby. In particular, it has a densityρ and velocityv that depend on timet and positionr. The momentum per unit volume isρv.[31]

Consider a column of water inhydrostatic equilibrium. All the forces on the water are in balance and the water is motionless. On any given drop of water, two forces are balanced. The first is gravity, which acts directly on each atom and molecule inside. The gravitational force per unit volume isρg, whereg is thegravitational acceleration. The second force is the sum of all the forces exerted on its surface by the surrounding water. The force from below is greater than the force from above by just the amount needed to balance gravity. The normal force per unit area is thepressurep. The average force per unit volume inside the droplet is the gradient of the pressure, so the force balance equation is[32]

p+ρg=0.{\displaystyle -\nabla p+\rho \mathbf {g} =0\,.}

If the forces are not balanced, the droplet accelerates. This acceleration is not simply the partial derivativev/t because the fluid in a given volume changes with time. Instead, thematerial derivative is needed:[33]

DDtt+v.{\displaystyle {\frac {D}{Dt}}\equiv {\frac {\partial }{\partial t}}+\mathbf {v} \cdot {\boldsymbol {\nabla }}\,.}

Applied to any physical quantity, the material derivative includes the rate of change at a point and the changes due toadvection as fluid is carried past the point. Per unit volume, the rate of change in momentum is equal toρDv/Dt. This is equal to the net force on the droplet.

Forces that can change the momentum of a droplet include the gradient of the pressure and gravity, as above. In addition, surface forces can deform the droplet. In the simplest case, ashear stressτ, exerted by a force parallel to the surface of the droplet, is proportional to the rate of deformation orstrain rate. Such a shear stress occurs if the fluid has a velocity gradient because the fluid is moving faster on one side than another. If the speed in thex direction varies withz, the tangential force in directionx per unit area normal to thez direction is

σzx=μvxz,{\displaystyle \sigma _{zx}=-\mu {\frac {\partial v_{x}}{\partial z}}\,,}

whereμ is theviscosity. This is also aflux, or flow per unit area, ofx-momentum through the surface.[34]

Including the effect of viscosity, the momentum balance equations for theincompressible flow of aNewtonian fluid are

ρDvDt=p+μ2v+ρg.{\displaystyle \rho {\frac {D\mathbf {v} }{Dt}}=-{\boldsymbol {\nabla }}p+\mu \nabla ^{2}\mathbf {v} +\rho \mathbf {g} .\,}

These are known as theNavier–Stokes equations.[35]

The momentum balance equations can be extended to more general materials, including solids. For each surface with normal in directioni and force in directionj, there is a stress componentσij. The nine components make up theCauchy stress tensorσ, which includes both pressure and shear. The local conservation of momentum is expressed by theCauchy momentum equation:

ρDvDt=σ+f,{\displaystyle \rho {\frac {D\mathbf {v} }{Dt}}={\boldsymbol {\nabla }}\cdot {\boldsymbol {\sigma }}+\mathbf {f} \,,}

wheref is thebody force.[36]

The Cauchy momentum equation is broadly applicable todeformations of solids and liquids. The relationship between the stresses and the strain rate depends on the properties of the material (seeTypes of viscosity).

Acoustic waves

A disturbance in a medium gives rise to oscillations, orwaves, that propagate away from their source. In a fluid, small changes in pressurep can often be described by theacoustic wave equation:

2pt2=c22p,{\displaystyle {\frac {\partial ^{2}p}{\partial t^{2}}}=c^{2}\nabla ^{2}p\,,}

wherec is thespeed of sound. In a solid, similar equations can be obtained for propagation of pressure (P-waves) and shear (S-waves).[37]

The flux, or transport per unit area, of a momentum componentρvj by a velocityvi is equal toρvjvj.[dubiousdiscuss] In the linear approximation that leads to the above acoustic equation, the time average of this flux is zero. However, nonlinear effects can give rise to a nonzero average.[38] It is possible for momentum flux to occur even though the wave itself does not have a mean momentum.[39]

In electromagnetics

Particle in a field

InMaxwell's equations, the forces between particles are mediated by electric and magnetic fields. The electromagnetic force (Lorentz force) on a particle with chargeq due to a combination ofelectric fieldE andmagnetic fieldB is

F=q(E+v×B).{\displaystyle \mathbf {F} =q(\mathbf {E} +\mathbf {v} \times \mathbf {B} ).}

(inSI units).[40]: 2 It has anelectric potentialφ(r,t) andmagnetic vector potentialA(r,t).[22]In the non-relativistic regime, its generalized momentum is

P=mv+qA,{\displaystyle \mathbf {P} =m\mathbf {\mathbf {v} } +q\mathbf {A} ,}

while in relativistic mechanics this becomes

P=γmv+qA.{\displaystyle \mathbf {P} =\gamma m\mathbf {\mathbf {v} } +q\mathbf {A} .}

The quantityV =qA is sometimes called thepotential momentum.[41][42][43] It is the momentum due to the interaction of the particle with the electromagnetic fields. The name is an analogy with the potential energyU =qφ, which is the energy due to the interaction of the particle with the electromagnetic fields. These quantities form a four-vector, so the analogy is consistent; besides, the concept of potential momentum is important in explaining the so-calledhidden momentum of the electromagnetic fields.[44]

Conservation

In Newtonian mechanics, the law of conservation of momentum can be derived from thelaw of action and reaction, which states that every force has a reciprocating equal and opposite force. Under some circumstances, moving charged particles can exert forces on each other in non-opposite directions.[45] Nevertheless, the combined momentum of the particles and the electromagnetic field is conserved.

Vacuum

The Lorentz force imparts a momentum to the particle, so by Newton's second law the particle must impart a momentum to the electromagnetic fields.[46]

In a vacuum, the momentum per unit volume is

g=1μ0c2E×B,{\displaystyle \mathbf {g} ={\frac {1}{\mu _{0}c^{2}}}\mathbf {E} \times \mathbf {B} \,,}

whereμ0 is thevacuum permeability andc is thespeed of light. The momentum density is proportional to thePoynting vectorS which gives the directional rate of energy transfer per unit area:[46][47]

g=Sc2.{\displaystyle \mathbf {g} ={\frac {\mathbf {S} }{c^{2}}}\,.}

If momentum is to be conserved over the volumeV over a regionQ, changes in the momentum of matter through the Lorentz force must be balanced by changes in the momentum of the electromagnetic field and outflow of momentum. IfPmech is the momentum of all the particles inQ, and the particles are treated as a continuum, then Newton's second law gives

dPmechdt=Q(ρE+J×B)dV.{\displaystyle {\frac {{\text{d}}\mathbf {P} _{\text{mech}}}{{\text{d}}t}}=\iiint \limits _{Q}\left(\rho \mathbf {E} +\mathbf {J} \times \mathbf {B} \right){\text{d}}V\,.}

The electromagnetic momentum is

Pfield=1μ0c2QE×BdV,{\displaystyle \mathbf {P} _{\text{field}}={\frac {1}{\mu _{0}c^{2}}}\iiint \limits _{Q}\mathbf {E} \times \mathbf {B} \,dV\,,}

and the equation for conservation of each componenti of the momentum is

ddt(Pmech+Pfield)i=σ(jTijnj)dΣ.{\displaystyle {\frac {\text{d}}{{\text{d}}t}}\left(\mathbf {P} _{\text{mech}}+\mathbf {P} _{\text{field}}\right)_{i}=\iint \limits _{\sigma }\left(\sum \limits _{j}T_{ij}n_{j}\right){\text{d}}\Sigma \,.}

The term on the right is an integral over the surface areaΣ of the surfaceσ representing momentum flow into and out of the volume, andnj is a component of the surface normal ofS. The quantityTij is called theMaxwell stress tensor, defined as[46]

Tijϵ0(EiEj12δijE2)+1μ0(BiBj12δijB2).{\displaystyle T_{ij}\equiv \epsilon _{0}\left(E_{i}E_{j}-{\frac {1}{2}}\delta _{ij}E^{2}\right)+{\frac {1}{\mu _{0}}}\left(B_{i}B_{j}-{\frac {1}{2}}\delta _{ij}B^{2}\right)\,.}

Media

The above results are for themicroscopic Maxwell equations, applicable to electromagnetic forces in a vacuum (or on a very small scale in media). It is more difficult to define momentum density in media because the division into electromagnetic and mechanical is arbitrary. The definition of electromagnetic momentum density is modified to

g=1c2E×H=Sc2,{\displaystyle \mathbf {g} ={\frac {1}{c^{2}}}\mathbf {E} \times \mathbf {H} ={\frac {\mathbf {S} }{c^{2}}}\,,}

where the H-fieldH is related to the B-field and themagnetizationM by

B=μ0(H+M).{\displaystyle \mathbf {B} =\mu _{0}\left(\mathbf {H} +\mathbf {M} \right)\,.}

The electromagnetic stress tensor depends on the properties of the media.[46]

Non-classical

Quantum mechanical

Further information:Momentum operator

Inquantum mechanics, momentum is defined as aself-adjoint operator on thewave function. TheHeisenberguncertainty principle defines limits on how accurately the momentum and position of a single observable system can be known at once. In quantum mechanics, position and momentum areconjugate variables.

For a single particle described in the position basis the momentum operator can be written as

p=i=i,{\displaystyle \mathbf {p} ={\hbar \over i}\nabla =-i\hbar \nabla \,,}

where is thegradient operator,ħ is thereduced Planck constant, andi is theimaginary unit. This is a commonly encountered form of the momentum operator, though the momentum operator in other bases can take other forms. For example, inmomentum space the momentum operator is represented by theeigenvalue equation

pψ(p)=pψ(p),{\displaystyle \mathbf {p} \psi (p)=p\psi (p)\,,}

where the operatorp acting on a wave eigenfunctionψ(p) yields that wave function multiplied by the eigenvaluep, in an analogous fashion to the way that the position operator acting on a wave functionψ(x) yields that wave function multiplied by the eigenvaluex.

For both massive and massless objects, relativistic momentum is related to thephase constantβ by[48]

p=β{\displaystyle p=\hbar \beta }

Electromagnetic radiation (includingvisible light,ultraviolet light, andradio waves) is carried byphotons. Even though photons (the particle aspect of light) have no mass, they still carry momentum. This leads to applications such as thesolar sail. The calculation of the momentum of light withindielectric media is somewhat controversial (seeAbraham–Minkowski controversy).[49][50]

Relativistic

See also:Mass in special relativity andTests of relativistic energy and momentum

Lorentz invariance

Newtonian physics assumes thatabsolute time and space exist outside of any observer; this gives rise toGalilean invariance. It also results in a prediction that thespeed of light can vary from one reference frame to another. This is contrary to what has been observed. In thespecial theory of relativity, Einstein keeps the postulate that the equations of motion do not depend on the reference frame, but assumes that the speed of lightc is invariant. As a result, position and time in two reference frames are related by theLorentz transformation instead of theGalilean transformation.[51]

Consider, for example, one reference frame moving relative to another at velocityv in thex direction. The Galilean transformation gives the coordinates of the moving frame as

t=tx=xvt{\displaystyle {\begin{aligned}t'&=t\\x'&=x-vt\end{aligned}}}

while the Lorentz transformation gives[52]

t=γ(tvxc2)x=γ(xvt){\displaystyle {\begin{aligned}t'&=\gamma \left(t-{\frac {vx}{c^{2}}}\right)\\x'&=\gamma \left(x-vt\right)\,\end{aligned}}}

whereγ is theLorentz factor:

γ=11v2/c2.{\displaystyle \gamma ={\frac {1}{\sqrt {1-v^{2}/c^{2}}}}.}

Newton's second law, with mass fixed, is not invariant under a Lorentz transformation. However, it can be made invariant by making theinertial massm of an object a function of velocity:

m=γm0;{\displaystyle m=\gamma m_{0}\,;}

m0 is the object'sinvariant mass.[53]

The modified momentum,

p=γm0v,{\displaystyle \mathbf {p} =\gamma m_{0}\mathbf {v} \,,}

obeys Newton's second law:

F=dpdt.{\displaystyle \mathbf {F} ={\frac {d\mathbf {p} }{dt}}\,.}

Within the domain of classical mechanics, relativistic momentum closely approximates Newtonian momentum: at low velocity,γm0v is approximately equal tom0v, the Newtonian expression for momentum.

Four-vector formulation

Main article:Four-momentum

In the theory of special relativity, physical quantities are expressed in terms offour-vectors that include time as a fourth coordinate along with the three space coordinates. These vectors are generally represented by capital letters, for exampleR for position. The expression for thefour-momentum depends on how the coordinates are expressed. Time may be given in its normal units or multiplied by the speed of light so that all the components of the four-vector have dimensions of length. If the latter scaling is used, an interval ofproper time,τ, defined by[54]

c2dτ2=c2dt2dx2dy2dz2,{\displaystyle c^{2}{\text{d}}\tau ^{2}=c^{2}{\text{d}}t^{2}-{\text{d}}x^{2}-{\text{d}}y^{2}-{\text{d}}z^{2}\,,}

isinvariant under Lorentz transformations (in this expression and in what follows the(+ − − −)metric signature has been used, different authors use different conventions). Mathematically this invariance can be ensured in one of two ways: by treating the four-vectors asEuclidean vectors and multiplying time by−1; or by keeping time a real quantity and embedding the vectors in aMinkowski space.[55] In a Minkowski space, thescalar product of two four-vectorsU = (U0,U1,U2,U3) andV = (V0,V1,V2,V3) is defined as

UV=U0V0U1V1U2V2U3V3.{\displaystyle \mathbf {U} \cdot \mathbf {V} =U_{0}V_{0}-U_{1}V_{1}-U_{2}V_{2}-U_{3}V_{3}\,.}

In all the coordinate systems, the (contravariant) relativistic four-velocity is defined by

UdRdτ=γdRdt,{\displaystyle \mathbf {U} \equiv {\frac {{\text{d}}\mathbf {R} }{{\text{d}}\tau }}=\gamma {\frac {{\text{d}}\mathbf {R} }{{\text{d}}t}}\,,}

and the (contravariant)four-momentum is

P=m0U,{\displaystyle \mathbf {P} =m_{0}\mathbf {U} \,,}

wherem0 is the invariant mass. IfR = (ct,x,y,z) (in Minkowski space), then

P=γm0(c,v)=(mc,p).{\displaystyle \mathbf {P} =\gamma m_{0}\left(c,\mathbf {v} \right)=(mc,\mathbf {p} )\,.}

Using Einstein'smass–energy equivalence,E =mc2, this can be rewritten as

P=(Ec,p).{\displaystyle \mathbf {P} =\left({\frac {E}{c}},\mathbf {p} \right)\,.}

Thus, conservation of four-momentum is Lorentz-invariant and implies conservation of both mass and energy.

The magnitude of the momentum four-vector is equal tom0c:

P2=PP=γ2m02(c2v2)=(m0c)2,{\displaystyle \|\mathbf {P} \|^{2}=\mathbf {P} \cdot \mathbf {P} =\gamma ^{2}m_{0}^{2}\left(c^{2}-v^{2}\right)=(m_{0}c)^{2}\,,}

and is invariant across all reference frames.

The relativistic energy–momentum relationship holds even for massless particles such as photons; by settingm0 = 0 it follows that

E=pc.{\displaystyle E=pc\,.}

In a game of relativistic "billiards", if a stationary particle is hit by a moving particle in an elastic collision, the paths formed by the two afterwards will form an acute angle. This is unlike the non-relativistic case where they travel at right angles.[56]

The four-momentum of a planar wave can be related to a wave four-vector[57]

P=(Ec,p)=K=(ωc,k){\displaystyle \mathbf {P} =\left({\frac {E}{c}},{\vec {\mathbf {p} }}\right)=\hbar \mathbf {K} =\hbar \left({\frac {\omega }{c}},{\vec {\mathbf {k} }}\right)}

For a particle, the relationship between temporal components,E =ħω, is thePlanck–Einstein relation, and the relation between spatial components,p =ħk, describes ade Brogliematter wave.

History of the concept

Impetus

Main article:Theory of impetus

John Philoponus

In about 530 AD,John Philoponus developed a concept of momentum inOn Physics, a commentary toAristotle'sPhysics. Aristotle claimed that everything that is moving must be kept moving by something. For example, a thrown ball must be kept moving by motions of the air. Philoponus pointed out the absurdity in Aristotle's claim that motion of an object is promoted by the same air that is resisting its passage. He proposed instead that an impetus was imparted to the object in the act of throwing it.[58]

Ibn Sīnā

Engraving of Ibn Sīnā
Ibn Sīnā
(980–1037)

In 1020,Ibn Sīnā (also known by hisLatinized name Avicenna) read Philoponus and published his own theory of motion inThe Book of Healing. He agreed that an impetus is imparted to a projectile by the thrower; but unlike Philoponus, who believed that it was a temporary virtue that would decline even in a vacuum, he viewed it as a persistent, requiring external forces such asair resistance to dissipate it.[59][60][61]

Peter Olivi, Jean Buridan

In the 13th and 14th century,Peter Olivi andJean Buridan read and refined the work of Philoponus, and possibly that of Ibn Sīnā.[61] Buridan, who in about 1350 was made rector of the University of Paris, referred toimpetus being proportional to the weight times the speed. Moreover, Buridan's theory was different from his predecessor's in that he did not consider impetus to be self-dissipating, asserting that a body would be arrested by the forces of air resistance and gravity which might be opposing its impetus.[62][63]

Quantity of motion

René Descartes

InPrinciples of Philosophy (Principia Philosophiae) from 1644, the French philosopherRené Descartes defined "quantity of motion" (Latin: quantitas motus) as the product of size and speed,[64] and claimed that the total quantity of motion in the universe is conserved.[64][65]

Portrait of René Descartes
René Descartes
(1596–1650)

If x is twice the size of y, and is moving half as fast, then there's the same amount of motion in each.

[God] created matter, along with its motion ... merely by letting things run their course, he preserves the same amount of motion ... as he put there in the beginning.

This should not be read as a statement of the modern law ofconservation of momentum, since Descartes had no concept of mass as distinct from weight and size. (The concept of mass, as distinct from weight, was introduced by Newton in 1686.)[66] More important, he believed that it is speed rather than velocity that is conserved. So for Descartes, if a moving object were to bounce off a surface, changing its direction but not its speed, there would be no change in its quantity of motion.[67][68][69]Galileo, in hisTwo New Sciences (published in 1638), used theItalian wordimpeto to similarly describe Descartes's quantity of motion.

Christiaan Huygens

Portrait of Christiaan Huygens
Christiaan Huygens
(1629–1695)

In the 1600s,Christiaan Huygens concluded quite early thatDescartes's laws for the elastic collision of two bodies must be wrong, and he formulated the correct laws.[70] An important step was his recognition of theGalilean invariance of the problems.[71] His views then took many years to be circulated. He passed them on in person toWilliam Brouncker andChristopher Wren in London, in 1661.[72] What Spinoza wrote toHenry Oldenburg about them, in 1666 during theSecond Anglo-Dutch War, was guarded.[73] Huygens had actually worked them out in a manuscriptDe motu corporum ex percussione in the period 1652–1656. The war ended in 1667, and Huygens announced his results to the Royal Society in 1668. He published them in theJournal des sçavans in 1669.[74]

Momentum

John Wallis

In 1670,John Wallis, inMechanica sive De Motu, Tractatus Geometricus, stated the law of conservation of momentum: "the initial state of the body, either of rest or of motion, will persist" and "If the force is greater than the resistance, motion will result".[75] Wallis usedmomentum for quantity of motion, andvis for force.

Gottfried Leibniz

In 1686,Gottfried Wilhelm Leibniz, inDiscourse on Metaphysics, gave an argument against Descartes' construction of the conservation of the "quantity of motion" using an example of dropping blocks of different sizes different distances. He points out that force is conserved but quantity of motion, construed as the product of size and speed of an object, is not conserved.[76]

Isaac Newton

Portrait of Isaac Newton by James Thronill, after Sir Godfrey Kneller
Isaac Newton
(1642–1727)

In 1687,Isaac Newton, inPhilosophiæ Naturalis Principia Mathematica, just like Wallis, showed a similar casting around for words to use for the mathematical momentum. His Definition II definesquantitas motus, "quantity of motion", as "arising from the velocity and quantity of matter conjointly", which identifies it as momentum.[77] Thus when in Law II he refers tomutatio motus, "change of motion", being proportional to the force impressed, he is generally taken to mean momentum and not motion.[78]

John Jennings

In 1721,John Jennings publishedMiscellanea, where the momentum in its current mathematical sense is attested, five years before the final edition of Newton'sPrincipia Mathematica.MomentumM or "quantity of motion" was being defined for students as "a rectangle", the product ofQ andV, whereQ is "quantity of material" andV is "velocity",s/t.[79]

In 1728, theCyclopedia states:

TheMomentum,Impetus, or Quantity of Motion of any Body, is theFactum [i.e., product] of its Velocity, (or the Space it moves in a given Time, seeMotion) multiplied into its Mass.

See also

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  • Landau, L.D.; Lifshitz, E.M. (2000).The classical theory of fields. English edition, reprinted with corrections; translated from the Russian by Morton Hamermesh (4th ed.). Oxford: Butterworth Heinemann.ISBN 978-0-7506-2768-9.
  • Rindler, Wolfgang (1986).Essential Relativity: Special, general and cosmological (2nd ed.). New York: Springer.ISBN 978-0-387-10090-6.
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