Editing Momentum (section)

===Relativistic===
{{See also|Mass in special relativity|Tests of relativistic energy and momentum}}

====Lorentz invariance====
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 what has been observed. 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 {{mvar|c}} is invariant. As a result, position and time in two reference frames are related by the [[Lorentz transformation]] instead of the [[Galilean transformation]].<ref name=RindlerCh2>{{harvnb|Rindler|1986|loc=Chapter 2}}</ref>

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

<math display="block">\begin{align}
  t' &= t  \\
  x' &= x - v t
\end{align}</math>

while the Lorentz transformation gives<ref name=FeynmanCh15>[https://feynmanlectures.caltech.edu/I_15.html#Ch15-S2 ''The Feynman Lectures on Physics''] Vol. I Ch. 15-2: The Lorentz transformation</ref>

<math display="block">\begin{align}
  t' &= \gamma \left( t - \frac{v x}{c^2} \right)  \\
  x' &= \gamma \left( x - v t \right)\,
\end{align}</math>

where {{mvar|γ}} is the [[Lorentz factor]]:

<math qid=Q599404 display="block">\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}. </math>

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'' {{mvar|m}} of an object a function of velocity:

<math display="block">m = \gamma m_0\,;</math>

{{math|{{var|m}}{{sub|0}}}} is the object's [[invariant mass]].<ref name=Rindler>{{harvnb|Rindler|1986|pp=77–81}}</ref>

The modified momentum,

<math display="block"> \mathbf{p} = \gamma m_0 \mathbf{v}\,,</math>

obeys Newton's second law:

<math display="block"> \mathbf{F} = \frac{d \mathbf{p}}{dt}\,.</math>

Within the domain of classical mechanics, relativistic momentum closely approximates Newtonian momentum: at low velocity, {{math|{{var|γ}}{{var|m}}{{sub|0}}'''v'''}} is approximately equal to {{math|{{var|m}}{{sub|0}}'''v'''}}, the Newtonian expression for momentum.

====Four-vector formulation====
{{Main|Four-momentum}}
In the theory of special relativity, physical quantities are expressed in terms of [[four-vector]]s that include time as a fourth coordinate along with the three space coordinates. These vectors are generally represented by capital letters, for example {{math|'''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]], {{mvar|τ}}, defined by<ref>{{harvnb|Rindler|1986|p=66}}</ref>

<math display="block">c^2\text{d}\tau^2 = c^2\text{d}t^2-\text{d}x^2-\text{d}y^2-\text{d}z^2\,,</math>

is [[Invariant (physics)|invariant]] under Lorentz transformations (in this expression and in what follows the {{nowrap|(+ − − −)}} [[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 vector]]s and multiplying time by {{math|[[Imaginary unit|{{sqrt|−1}}]]}}; or by keeping time a real quantity and embedding the vectors in a [[Minkowski space]].<ref>{{cite book|last1=Misner|first1=Charles W.|first2=Kip S. |last2=Thorne |first3=John Archibald |last3=Wheeler  |title=Gravitation|date=1973|publisher=W. H. Freeman|location=New York|isbn=978-0-7167-0344-0|page=51|others=24th printing.}}</ref> In a Minkowski space, the [[scalar product]] of two four-vectors {{math|1='''U''' = ({{var|U}}{{sub|0}}, {{var|U}}{{sub|1}}, {{var|U}}{{sub|2}}, {{var|U}}{{sub|3}})}} and {{math|1='''V''' = ({{var|V}}{{sub|0}}, {{var|V}}{{sub|1}}, {{var|V}}{{sub|2}}, {{var|V}}{{sub|3}})}} is defined as

<math display="block"> \mathbf{U} \cdot \mathbf{V} =  U_0 V_0 - U_1 V_1 - U_2 V_2 - U_3 V_3\,. </math>

In all the coordinate systems, the ([[Covariance and contravariance of vectors|contravariant]]) relativistic four-velocity is defined by

<math display="block"> \mathbf{U} \equiv \frac{\text{d}\mathbf{R}}{\text{d}\tau} = \gamma \frac{\text{d}\mathbf{R}}{\text{d}t}\,,</math>

and the (contravariant) [[four-momentum]] is

<math qid=Q1068463 display="block">\mathbf{P} = m_0\mathbf{U}\,,</math>

where {{math|{{var|m}}{{sub|0}}}} is the invariant mass. If {{math|1='''R''' = ({{var|c}}{{var|t}}, {{var|x}}, {{var|y}}, {{var|z}})}} (in Minkowski space), then

<math display="block">\mathbf{P} = \gamma m_0 \left(c,\mathbf{v}\right) = (m c, \mathbf{p})\,.</math>

Using Einstein's [[mass–energy equivalence]], {{math|1={{var|E}} = {{var|m}}{{var|c}}{{sup|2}}}}, this can be rewritten as

<math display="block">\mathbf{P} = \left(\frac{E}{c}, \mathbf{p}\right)\,.</math>

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 {{math|{{var|m}}{{sub|0}}{{var|c}}}}:

<math display="block">\|\mathbf{P}\|^2 = \mathbf{P} \cdot \mathbf{P} = \gamma^2 m_0^2 \left(c^2 - v^2\right) = (m_0c)^2\,,</math>

and is invariant across all reference frames.

The relativistic energy–momentum relationship holds even for massless particles such as photons; by setting {{math|1={{var|m}}{{sub|0}} = 0}} it follows that

<math display="block">E = pc\,.</math>

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.<ref>{{harvnb|Rindler|1986|pp=86–87}}</ref>

The four-momentum of a planar wave can be related to a wave four-vector<ref>{{cite book
 |title=Introduction to Special Relativity
 |edition=2nd
 |first1=Wolfgang
 |last1=Rindler
 |publisher=Oxford Science Publications
 |year=1991
 |isbn=978-0-19-853952-0
 |pages=[https://archive.org/details/introductiontosp0000rind/page/82 82–84]
 |url=https://archive.org/details/introductiontosp0000rind/page/82
 }}</ref>

<math display="block">\mathbf{P} = \left(\frac{E}{c},\vec{\mathbf{p}}\right) = \hbar \mathbf{K} = \hbar \left(\frac{\omega}{c},\vec{\mathbf{k}}\right)</math>

For a particle, the relationship between temporal components,  {{math|1={{var|E}} = {{var|ħ}}{{var|ω}}}}, is the [[Planck–Einstein relation]], and the relation between spatial components, {{math|1='''p''' = {{var|ħ}}'''k'''}}, describes a [[de Broglie]] [[matter wave]].