Momentum

The momentum of a particle is conventionally represented by the letter p. It is the product of two quantities, the particle's mass (represented by the letter m) and its velocity (v):[1]

${\displaystyle p=mv.}$

The unit of momentum is the product of the units of mass and velocity. In SI 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). In cgs 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.

The momentum of a system of particles is the vector sum of their momenta. If two particles have respective masses m₁ and m₂, and velocities v₁ and v₂, the total momentum is

${\displaystyle {\begin{aligned}p&=p_{1}+p_{2}\\&=m_{1}v_{1}+m_{2}v_{2}\,.\end{aligned}}}$

The momenta of more than two particles can be added more generally with the following:

${\displaystyle p=\sum _{i}m_{i}v_{i}.}$

A system of particles has a center of mass, a point determined by the weighted sum of their positions:

${\displaystyle r_{\text{cm}}={\frac {m_{1}r_{1}+m_{2}r_{2}+\cdots }{m_{1}+m_{2}+\cdots }}={\frac {\sum \limits _{i}m_{i}r_{i}}{\sum \limits _{i}m_{i}}}.}$

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 is ${\displaystyle m}$, and the center of mass is moving at velocity v_cm, the momentum of the system is:

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

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

If the net force F applied to a particle is constant, and is applied for a time interval Δt, the momentum of the particle changes by an amount

${\displaystyle \Delta p=F\Delta t\,.}$

In differential form, this is Newton's second law; the rate of change of the momentum of a particle is equal to the instantaneous force F acting on it,[1]

${\displaystyle F={\frac {dp}{dt}}.}$

If the net force experienced by a particle changes as a function of time, F(t), the change in momentum (or impulse J ) between times t₁ and t₂ is

${\displaystyle \Delta p=J=\int _{t_{1}}^{t_{2}}F(t)\,dt\,.}$

Impulse is measured in the derived units of the newton second (1 N⋅s = 1 kg⋅m/s) or dyne second (1 dyne⋅s = 1 g⋅cm/s)

Under the assumption of constant mass m, it is equivalent to write

${\displaystyle F={\frac {d(mv)}{dt}}=m{\frac {dv}{dt}}=ma,}$

hence the net force is equal to the mass of the particle times its acceleration.[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.

In a closed system (one that does not exchange any matter with its surroundings and is not acted on by external forces) the total momentum is constant. This fact, known as the law of conservation of momentum, is implied by Newton's laws of motion.[4][5] Suppose, for example, that two particles interact. Because of the third law, the forces between them are equal and opposite. If the particles are numbered 1 and 2, the second law states that F₁ = dp₁/dt and F₂ = dp₂/dt. Therefore,

${\displaystyle {\frac {dp_{1}}{dt}}=-{\frac {dp_{2}}{dt}},}$

with the negative sign indicating that the forces oppose. Equivalently,

${\displaystyle {\frac {d}{dt}}\left(p_{1}+p_{2}\right)=0.}$

If the velocities of the particles are u₁ and u₂ before the interaction, and afterwards they are v₁ and v₂, then

${\displaystyle m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}.}$

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 up to zero, so the total change in momentum is zero. This conservation law applies to all interactions, including collisions and separations caused by explosive forces.[4] It can also be generalized to situations where Newton's laws do not hold, for example in the theory of relativity and in electrodynamics.[6]

Newton's apple in Einstein's elevator. In person A's frame of reference, the apple has non-zero velocity and momentum. In the elevator's and person B's frames of reference, it has zero velocity and momentum.

Momentum is a measurable quantity, and the measurement depends on the motion of the observer. For example: if an apple is sitting in a glass elevator that is descending, an outside observer, looking into the elevator, sees the apple moving, so, to that observer, the apple has a non-zero momentum. To someone inside the elevator, the apple does not move, so, it has zero momentum. The two observers each have a frame of reference, in which, they observe motions, and, if the elevator is descending steadily, they will see behavior that is consistent with those same physical laws.

Suppose a particle has position x in a stationary frame of reference. From the point of view of another frame of reference, moving at a uniform speed u, the position (represented by a primed coordinate) changes with time as

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

This is called a Galilean transformation. If the particle is moving at speed dx/dt = v in the first frame of reference, in the second, it is moving at speed

${\displaystyle v'={\frac {dx'}{dt}}=v-u\,.}$

Since u does not change, the accelerations are the same:

${\displaystyle a'={\frac {dv'}{dt}}=a\,.}$

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 or Galilean 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 the center of mass frame – one that is moving with the center of mass. In this frame, the total momentum is zero.

By itself, the law of conservation of momentum is not enough to determine the motion of particles after a collision. Another property of the motion, kinetic energy, must be known. This is not necessarily conserved. If it is conserved, the collision is called an elastic collision; if not, it is an inelastic collision.

Elastic collision of equal masses

Elastic collision of unequal masses

An elastic collision is one in which no kinetic energy is absorbed in the collision. 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 them apart. A slingshot maneuver of a satellite around a planet can also be viewed as a perfectly elastic collision. A collision between two pool balls is a good example of an almost totally elastic collision, due to their high rigidity, but when bodies come in contact there is always some dissipation.[8]

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 are u₁ and u₂ before the collision and v₁ and v₂ after, the equations expressing conservation of momentum and kinetic energy are:

${\displaystyle {\begin{aligned}m_{1}u_{1}+m_{2}u_{2}&=m_{1}v_{1}+m_{2}v_{2}\\{\tfrac {1}{2}}m_{1}u_{1}^{2}+{\tfrac {1}{2}}m_{2}u_{2}^{2}&={\tfrac {1}{2}}m_{1}v_{1}^{2}+{\tfrac {1}{2}}m_{2}v_{2}^{2}\,.\end{aligned}}}$

A change of reference frame can simplify analysis of a collision. For example, suppose there are two bodies of equal mass m, one stationary and one approaching the other at a speed v (as in the figure). The center of mass is moving at speed v/2 and both bodies are moving towards it at speed v/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 speed v. 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]

${\displaystyle {\begin{aligned}v_{1}&=u_{2}\\v_{2}&=u_{1}\,.\end{aligned}}}$

In general, when the initial velocities are known, the final velocities are given by[9]

${\displaystyle v_{1}=\left({\frac {m_{1}-m_{2}}{m_{1}+m_{2}}}\right)u_{1}+\left({\frac {2m_{2}}{m_{1}+m_{2}}}\right)u_{2}\,}$

${\displaystyle v_{2}=\left({\frac {m_{2}-m_{1}}{m_{1}+m_{2}}}\right)u_{2}+\left({\frac {2m_{1}}{m_{1}+m_{2}}}\right)u_{1}\,.}$

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.

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 as heat or sound). Examples include traffic collisions,[10] 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 the Franck–Hertz experiment);[11] and particle 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. If one body is motionless to begin with, the equation for conservation of momentum is

${\displaystyle m_{1}u_{1}=\left(m_{1}+m_{2}\right)v\,,}$

so

${\displaystyle v={\frac {m_{1}}{m_{1}+m_{2}}}u_{1}\,.}$

In a frame of reference moving at the speed v), the objects are brought to rest by the collision and 100% of the kinetic energy is converted to other forms of energy.

One measure of the inelasticity of the collision is the coefficient of restitution C_R, 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:[12]

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

The momentum and energy equations also apply to the motions of objects that begin together and then move apart. For example, an explosion 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.[13]

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 a vector. In a coordinate system with x, y, z axes, velocity has components v_x in the x-direction, v_y in the y-direction, v_z in the z-direction. The vector is represented by a boldface symbol:[14]

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

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

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

The equations in the previous sections, work in vector form if the scalars p and v are replaced by vectors p and v. Each vector equation represents three scalar equations. For example,

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

represents three equations:[14]

${\displaystyle {\begin{aligned}p_{x}&=mv_{x}\\p_{y}&=mv_{y}\\p_{z}&=mv_{z}.\end{aligned}}}$

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

${\displaystyle v^{2}=v_{x}^{2}+v_{y}^{2}+v_{z}^{2}\,.}$

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.[14]

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).[15]

The concept of momentum plays a fundamental role in explaining the behavior of variable-mass objects such as a rocket ejecting fuel or a star accreting 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 time t is therefore p(t) = m(t)v(t). One might then try to invoke Newton's second law of motion by saying that the external force F on the object is related to its momentum p(t) by F = dp/dt, but this is incorrect, as is the related expression found by applying the product rule to d(mv)/dt:[16]

${\displaystyle F=m(t){\frac {dv}{dt}}+v(t){\frac {dm}{dt}}.}$ (incorrect)

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

${\displaystyle F=m(t){\frac {dv}{dt}}-u{\frac {dm}{dt}},}$

where u is the velocity of the ejected/accreted mass as seen in the object's rest frame.[16] This is distinct from v, 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.

${\displaystyle P(t+dt)=(m-dm)(v+dv)+dm(v-u)=mv+mdv-udm=P(t)+mdv-udm}$

Newtonian physics assumes that absolute time and space exist outside of any observer; this gives rise to Galilean invariance. It also results in a prediction that the speed of light can vary from one reference frame to another. This is contrary to observation. In the special 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 light c is invariant. As a result, position and time in two reference frames are related by the Lorentz transformation instead of the Galilean transformation.[17]

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

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

while the Lorentz transformation gives[18]

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

where γ is the Lorentz factor:

${\displaystyle \gamma ={\frac {1}{\sqrt {1-v^{2}/c^{2}}}}.}$

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

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

m₀ is the object's invariant mass.[19]

The modified momentum,

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

obeys Newton's second law:

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

Within the domain of classical mechanics, relativistic momentum closely approximates Newtonian momentum: at low velocity, γm₀v is approximately equal to m₀v, the Newtonian expression for momentum.

In the theory of special relativity, physical quantities are expressed in terms of four-vectors that include time as a fourth coordinate along with the three space coordinates. These vectors are generally represented by capital letters, for example R for position. The expression for the four-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 of proper time, τ, defined by[20]

${\displaystyle c^{2}d\tau ^{2}=c^{2}dt^{2}-dx^{2}-dy^{2}-dz^{2}\,,}$

is invariant 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 as Euclidean vectors and multiplying time by √−1; or by keeping time a real quantity and embedding the vectors in a Minkowski space.[21] In a Minkowski space, the scalar product of two four-vectors U = (U₀,U₁,U₂,U₃) and V = (V₀,V₁,V₂,V₃) is defined as

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

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

${\displaystyle \mathbf {U} \equiv {\frac {d\mathbf {R} }{d\tau }}=\gamma {\frac {d\mathbf {R} }{dt}}\,,}$

and the (contravariant) four-momentum is

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

where m₀ is the invariant mass. If R = (ct,x,y,z) (in Minkowski space), then

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

Using Einstein's mass-energy equivalence, E = mc², this can be rewritten as

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

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 to m₀c:

${\displaystyle ||\mathbf {P} ||^{2}=\mathbf {P} \cdot \mathbf {P} =\gamma ^{2}m_{0}^{2}(c^{2}-v^{2})=(m_{0}c)^{2}\,,}$

and is invariant across all reference frames.

The relativistic energy–momentum relationship holds even for massless particles such as photons; by setting m₀ = 0 it follows that

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

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.[22]

The four-momentum of a planar wave can be related to a wave four-vector[23]

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

For a particle, the relationship between temporal components, E = ħ ω, is the Planck–Einstein relation, and the relation between spatial components, p= ħ k, describes a de Broglie matter wave.

In Lagrangian mechanics, a Lagrangian is defined as the difference between the kinetic energy T and the potential energy V:

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

If the generalized coordinates are represented as a vector q = (q₁, q₂, ... , q_N) and time differentiation is represented by a dot over the variable, then the equations of motion (known as the Lagrange or Euler–Lagrange equations) are a set of N equations:[27]

${\displaystyle {\frac {d}{dt}}\left({\frac {\partial {\mathcal {L}}}{\partial {\dot {q}}_{j}}}\right)-{\frac {\partial {\mathcal {L}}}{\partial q_{j}}}=0\,.}$

If a coordinate q_i is not a Cartesian coordinate, the associated generalized momentum component p_i does not necessarily have the dimensions of linear momentum. Even if q_i is a Cartesian coordinate, p_i will not be the same as the mechanical momentum if the potential depends on velocity.[6] Some sources represent the kinematic momentum by the symbol Π.[28]

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

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

Each component p_j is said to be the conjugate momentum for the coordinate q_j.

Now if a given coordinate q_i does not appear in the Lagrangian (although its time derivative might appear), then

${\displaystyle p_{j}={\text{constant}}\,.}$

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.

In Hamiltonian 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

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

where the momentum is obtained by differentiating the Lagrangian as above. The Hamiltonian equations of motion are[29]

${\displaystyle {\begin{aligned}{\dot {q}}_{i}&={\frac {\partial {\mathcal {H}}}{\partial p_{i}}}\\-{\dot {p}}_{i}&={\frac {\partial {\mathcal {H}}}{\partial q_{i}}}\\-{\frac {\partial {\mathcal {L}}}{\partial t}}&={\frac {d{\mathcal {H}}}{dt}}\,.\end{aligned}}}$

As in Lagrangian mechanics, if a generalized coordinate does not appear in the Hamiltonian, its conjugate momentum component is conserved.[30]

Conservation of momentum is a mathematical consequence of the homogeneity (shift symmetry) of space (position in space is the canonical 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 of Noether's theorem.[31]

In Maxwell's equations, the forces between particles are mediated by electric and magnetic fields. The electromagnetic force (Lorentz force) on a particle with charge q due to a combination of electric field E and magnetic field B is

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

(in SI units).[32]^:2 It has an electric potential φ(r, t) and magnetic vector potential A(r, t).[28] In the non-relativistic regime, its generalized momentum is

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

while in relativistic mechanics this becomes

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

In Newtonian mechanics, the law of conservation of momentum can be derived from the law 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.[33] Nevertheless, the combined momentum of the particles and the electromagnetic field is conserved.

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.[34]

In a vacuum, the momentum per unit volume is

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

where μ₀ is the vacuum permeability and c is the speed of light. The momentum density is proportional to the Poynting vector S which gives the directional rate of energy transfer per unit area:[34][35]

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

If momentum is to be conserved over the volume V over a region Q, 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. If P_mech is the momentum of all the particles in Q, and the particles are treated as a continuum, then Newton's second law gives

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

The electromagnetic momentum is

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

and the equation for conservation of each component i of the momentum is

${\displaystyle {\frac {d}{dt}}\left(\mathbf {P} _{\text{mech}}+\mathbf {P} _{\text{field}}\right)_{i}=\iint \limits _{\sigma }\left(\sum \limits _{j}T_{ij}n_{j}\right)d\Sigma \,.}$

The term on the right is an integral over the surface area Σ of the surface σ representing momentum flow into and out of the volume, and n_j is a component of the surface normal of S. The quantity T_ij is called the Maxwell stress tensor, defined as

${\displaystyle T_{ij}\equiv \epsilon _{0}\left(E_{i}E_{j}-{\frac {1}{2}}\delta _{ij}E^{2}\right)+{\frac {1}{\mu _{0}}}\left(B_{i}B_{j}-{\frac {1}{2}}\delta _{ij}B^{2}\right)\,.}$[34]

The above results are for the microscopic 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

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

where the H-field H is related to the B-field and the magnetization M by

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

The electromagnetic stress tensor depends on the properties of the media.[34]

Motion of a material body

In fields such as fluid dynamics and solid mechanics, it is not feasible to follow the motion of individual atoms or molecules. Instead, the materials must be approximated by a continuum in which there is a particle or fluid parcel at each point that is assigned the average of the properties of atoms in a small region nearby. In particular, it has a density ρ and velocity v that depend on time t and position r. The momentum per unit volume is ρv.[39]

Consider a column of water in hydrostatic 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 is ρg, where g is the gravitational 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 the pressure p. The average force per unit volume inside the droplet is the gradient of the pressure, so the force balance equation is[40]

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

If the forces are not balanced, the droplet accelerates. This acceleration is not simply the partial derivative ∂v/∂t because the fluid in a given volume changes with time. Instead, the material derivative is needed:[41]

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

Applied to any physical quantity, the material derivative includes the rate of change at a point and the changes due to advection as fluid is carried past the point. Per unit volume, the rate of change in momentum is equal to ρDv/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, a shear stress τ, exerted by a force parallel to the surface of the droplet, is proportional to the rate of deformation or strain 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 the x direction varies with z, the tangential force in direction x per unit area normal to the z direction is

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

where μ is the viscosity. This is also a flux, or flow per unit area, of x-momentum through the surface.[42]

Including the effect of viscosity, the momentum balance equations for the incompressible flow of a Newtonian fluid are

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

These are known as the Navier–Stokes equations.[43]

The momentum balance equations can be extended to more general materials, including solids. For each surface with normal in direction i and force in direction j, there is a stress component σ_ij. The nine components make up the Cauchy stress tensor σ, which includes both pressure and shear. The local conservation of momentum is expressed by the Cauchy momentum equation:

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

where f is the body force.[44]

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

A disturbance in a medium gives rise to oscillations, or waves, that propagate away from their source. In a fluid, small changes in pressure p can often be described by the acoustic wave equation:

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

where c is the speed of sound. In a solid, similar equations can be obtained for propagation of pressure (P-waves) and shear (S-waves).[45]

The flux, or transport per unit area, of a momentum component ρv_j by a velocity v_i is equal to ρ v_jv_j. 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.[46] It is possible for momentum flux to occur even though the wave itself does not have a mean momentum.[47]