Momentum

 ( Conservation Laws ) 

In Newtonian mechanics, momentum (: momenta or momentums; more specifically linear momentum or translational momentum) is the Multiplication of the mass and velocity of an object. It is a Euclidean vector quantity, possessing a magnitude and a direction. If is an object's mass and is its velocity (also a vector quantity), then the object's momentum (from Latin "push, drive") is: $\backslash mathbf\{p\}\; =\; m\; \backslash mathbf\{v\}.$ In the International System of Units (SI), the unit of measurement of momentum is the kilogram metre per second (kg⋅m/s), which is dimensionally equivalent to the newton-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 the frame of reference, but in any inertial frame of reference, it is a conserved quantity, meaning that if a closed system is not affected by external forces, its total momentum does not change. Momentum is also conserved in special relativity (with a modified formula) and, in a modified form, in electrodynamics, quantum mechanics, quantum field theory, and general relativity. It is an expression of one of the fundamental symmetries of space and time: translational symmetry.

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

In continuous systems such as electromagnetic fields, fluid dynamics and deformable bodies, a momentum density can be defined as momentum per volume (a volume-specific quantity). A continuum version of the conservation of momentum leads to equations such as the Navier–Stokes equations for fluids or the Cauchy momentum equation for deformable solids or fluids.

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.

A system of particles has a center of mass, a point determined by the weighted sum of their positions: $$r\_\backslash text\{cm\}\; =\; \backslash frac\{m\_1\; r\_1\; +\; m\_2\; r\_2\; +\; \backslash cdots\}\{m\_1\; +\; m\_2\; +\; \backslash cdots\}\; =\; \backslash frac\{\backslash sum\_\{i\}m\_ir\_i\}\{\backslash sum\_\{i\}m\_i\}.$$

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 is $m$, and the center of mass is moving at velocity , the momentum of the system is:

$$p=\; mv\_\backslash text\{cm\}.$$

This is known as Euler's first law.

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 acting on it, $$F\; =\; \backslash frac\{\backslash text\{d\}p\}\{\backslash text\{d\}t\}.$$

If the net force experienced by a particle changes as a function of time, , the change in momentum (or impulse ) between times and is $$\backslash Delta\; p\; =\; J\; =\; \backslash int\_\{t\_1\}^\{t\_2\}\; F(t)\backslash ,\; \backslash text\{d\}t\backslash ,.$$

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 , it is equivalent to write

$$F\; =\; \backslash frac\{\backslash text\{d\}(mv)\}\{\backslash text\{d\}t\}\; =\; m\backslash frac\{\backslash text\{d\}v\}\{\backslash text\{d\}t\}\; =\; m\; a,$$

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

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.

$$\backslash frac\{\backslash text\{d\}p\_1\}\{\backslash text\{d\}t\}\; =\; -\; \backslash frac\{\backslash text\{d\}p\_2\}\{\backslash text\{d\}t\},$$ with the negative sign indicating that the forces oppose. Equivalently,

$$\backslash frac\{\backslash text\{d\}\}\{\backslash text\{d\}\; t\}\; \backslash left(p\_1\; +\; p\_2\backslash right)=\; 0.$$

If the velocities of the particles are and before the interaction, and afterwards they are and , then

$$m\_A\; v\_\{A1\}\; +\; m\_B\; v\_\{B1\}\; =\; m\_\{A\}\; v\_\{A2\}\; +\; m\_B\; v\_\{B2\}.$$

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 $$m\_Av\_A+m\_Bv\_B+m\_Cv\_C\; +\; \backslash ldots\; =\; \backslash text\{constant\}.$$

This conservation law applies to all interactions, including (both elastic and inelastic) and separations caused by explosive forces. It can also be generalized to situations where Newton's laws do not hold, for example in the theory of relativity and in electrodynamics.

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

$$x\text{'}\; =\; x\; -\; ut\backslash ,.$$

This is called a Galilean transformation.

If a particle is moving at speed in the first frame of reference, in the second, it is moving at speed

$$v\text{'}\; =\; \backslash frac\{\backslash text\{d\}x\text{'}\}\{\backslash text\{d\}t\}\; =\; v-u\backslash ,.$$

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

$$a\text{'}\; =\; \backslash frac\{\backslash text\; \{d\}v\text{'}\}\{\backslash text\{d\}t\}\; =\; a\backslash ,.$$

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.

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.

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 and before the collision and and after, the equations expressing conservation of momentum and kinetic energy are:

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

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

$$\backslash begin\{align\}$$

v_{A2} &= \left( \frac{m_{A} - m_{B}}{m_{A} + m_{B}} \right) v_{A1} + \left( \frac{2 m_{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{2 m_{A}}{m_{A} + m_{B}} \right) v_{A1}\,.
     

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.

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 are and before the collision then in a perfectly inelastic collision both bodies will be travelling with velocity after the collision. The equation expressing conservation of momentum is:

If one body is motionless to begin with (e.g. $u\_2\; =\; 0$), the equation for conservation of momentum is

$$m\_A\; v\_\{A1\}\; =\; \backslash left(\; m\_A\; +\; m\_B\; \backslash right)\; v\_2\backslash ,,$$

so

$$v\_2\; =\; \backslash frac\{m\_\{A\}\}\{m\_\{A\}+m\_\{B\}\}\; v\_\{A1\}\backslash ,.$$

In a different situation, if the frame of reference is moving at the final velocity such that $v\_2\; =\; 0$, 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 the coefficient of restitution , 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:

$$C\_\backslash text\{R\}\; =\; \backslash sqrt\{\backslash frac\{\backslash text\{bounce\; height\}\}\{\backslash text\{drop\; height\}\}\}\backslash ,.$$

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. also make use of conservation of momentum: propellant is thrust outward, gaining momentum, and an equal and opposite momentum is imparted to the rocket.

$$\backslash mathbf\{v\}\; =\; \backslash left(v\_x,v\_y,v\_z\; \backslash right).$$

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

$$\backslash mathbf\{p\}\; =\; \backslash left(p\_x,p\_y,p\_z\; \backslash right).$$

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

$$\backslash mathbf\{p\}=\; m\; \backslash mathbf\{v\}$$

represents three equations:

$$\backslash begin\{align\}$$

p_x &= m v_x\\
p_y &= m v_y \\
p_z &= m v_z.
     

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,

$$v^2\; =\; v\_x^2+v\_y^2+v\_z^2\backslash ,.$$

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.

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).

$$F\; =\; m(t)\; \backslash frac\{\backslash text\{d\}v\}\{\backslash text\{d\}t\}\; +\; v(t)\; \backslash frac\{\backslash text\{d\}m\}\{\backslash text\{d\}t\}.\; \backslash text\{(incorrect)\}$$

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

$$F\; =\; m(t)\; \backslash frac\{\backslash text\{d\}v\}\{\backslash text\{d\}t\}\; -\; u\; \backslash frac\{\backslash text\{d\}m\}\{\backslash text\{d\}t\},$$

where is the velocity of the ejected/accreted mass as seen in the object's rest frame. This is distinct from , 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 (). When considered together, the object and the mass () constitute a closed system in which total momentum is conserved.

$$P(t+\backslash text\{d\}t)\; =\; (\; m\; -\; \backslash text\{d\}m\; )\; (\; v\; +\; \backslash text\{d\}v\; )\; +\; \backslash text\{d\}m\; (\; v\; -\; u\; )\; =\; mv+m\; \backslash text\{d\}v\; -\; u\; \backslash text\{d\}m\; =\; P(t)\; +m\; \backslash text\{d\}v\; -\; u\; \backslash text\{d\}m$$

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

If the generalized coordinates are represented as a vector 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 equations:

$$\backslash frac\{\backslash text\{d\}\}\{\backslash text\{d\}t\}\backslash left(\backslash frac\{\backslash partial\; \backslash mathcal\{L\}\; \}\{\backslash partial\backslash dot\{q\}\_j\}\backslash right)\; -\; \backslash frac\{\backslash partial\; \backslash mathcal\{L\}\}\{\backslash partial\; q\_j\}\; =\; 0\backslash ,.$$

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

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

$$p\_j\; =\; \backslash frac\{\backslash partial\; \backslash mathcal\{L\}\; \}\{\backslash partial\; \backslash dot\{q\}\_j\}\backslash ,.$$

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

Now if a given coordinate does not appear in the Lagrangian (although its time derivative might appear), then is constant. This is the generalization of the conservation of momentum.

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.

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

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

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

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 , where 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 . The average force per unit volume inside the droplet is the gradient of the pressure, so the force balance equation is The Feynman Lectures on Physics Vol. II Ch. 40: The Flow of Dry Water

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

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

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

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 . 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 , 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 direction varies with , the tangential force in direction per unit area normal to the direction is

$$\backslash sigma\_\{zx\}\; =\; -\backslash mu\backslash frac\{\backslash partial\; v\_x\}\{\backslash partial\; z\}\backslash ,,$$

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

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

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

These are known as the Navier–Stokes equations.

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

$$\backslash rho\; \backslash frac\{D\; \backslash mathbf\{v\}\}\{D\; t\}\; =\; \backslash boldsymbol\{\backslash nabla\}\; \backslash cdot\; \backslash boldsymbol\{\backslash sigma\}\; +\; \backslash mathbf\{f\}\backslash ,,$$

where is the body force.

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).

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

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

The flux, or transport per unit area, of a momentum component by a velocity is equal to . 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.

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

(in SI units). It has an electric potential and magnetic vector potential . In the non-relativistic regime, its generalized momentum is

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

while in relativistic mechanics this becomes

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

The quantity is sometimes called the potential momentum.

 t' &= t  \\
 x' &= x - v t
     

 t' &= \gamma \left( t - \frac{v x}{c^2} \right)  \\
 x' &= \gamma \left( x - v t \right)\,
     

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

$$\backslash mathbf\{U\}\; \backslash cdot\; \backslash mathbf\{V\}\; =\; U\_0\; V\_0\; -\; U\_1\; V\_1\; -\; U\_2\; V\_2\; -\; U\_3\; V\_3\backslash ,.$$

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

$$\backslash mathbf\{U\}\; \backslash equiv\; \backslash frac\{\backslash text\{d\}\backslash mathbf\{R\}\}\{\backslash text\{d\}\backslash tau\}\; =\; \backslash gamma\; \backslash frac\{\backslash text\{d\}\backslash mathbf\{R\}\}\{\backslash text\{d\}t\}\backslash ,,$$

and the (contravariant) four-momentum is

$$\backslash mathbf\{P\}\; =\; m\_0\backslash mathbf\{U\}\backslash ,,$$

where is the invariant mass. If (in Minkowski space), then

$$\backslash mathbf\{P\}\; =\; \backslash gamma\; m\_0\; \backslash left(c,\backslash mathbf\{v\}\backslash right)\; =\; (m\; c,\; \backslash mathbf\{p\})\backslash ,.$$

Using Einstein's mass–energy equivalence, , this can be rewritten as

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

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 :

$$\backslash |\backslash mathbf\{P\}\backslash |^2\; =\; \backslash mathbf\{P\}\; \backslash cdot\; \backslash mathbf\{P\}\; =\; \backslash gamma^2\; m\_0^2\; \backslash left(c^2\; -\; v^2\backslash right)\; =\; (m\_0c)^2\backslash ,,$$

and is invariant across all reference frames.

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

$$E\; =\; pc\backslash ,.$$

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.

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

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

For a particle, the relationship between temporal components, , is the Planck–Einstein relation, and the relation between spatial components, , describes a de Broglie matter wave.