Editing Mass in special relativity (section)

==Relativistic energy–momentum equation==
{{More citations needed|date=February 2016}}
[[File:Invariant and additive masses.svg|thumb|right|Dependency between the rest mass and ''E'', given in 4-momentum {{math|(''p''<sub>0</sub>, ''p''<sub>1</sub>)}} coordinates, where {{math|1=''p''<sub>0</sub>''c'' = ''E''}}|427x427px]]
The relativistic expressions for {{mvar|E}} and {{mvar|p}} obey the relativistic [[energy–momentum relation]]:<ref name=taylor />
<math display="block">E^2 - (pc)^2 = \left(mc^2\right)^2</math>
where the <var>m</var> is the rest mass, or the invariant mass for systems, and {{mvar|E}} is the total energy.

The equation is also valid for photons, which have {{math|1=<var>m</var> = 0}}:
<math display="block">E^2 - (pc)^2 = 0</math>
and therefore
<math display="block">E = pc</math>

A photon's momentum is a function of its energy, but it is not proportional to the velocity, which is always {{math|''c''}}.

For an object at rest, the momentum {{mvar|<var>p</var>}} is zero, therefore
<math display="block">E = mc^2.</math> Note that the formula is true only for particles or systems with zero momentum.

The rest mass is only proportional to the total energy in the rest frame of the object.

When the object is moving, the total energy is given by
<math display="block">E = \sqrt{\left(mc^2\right)^2 + (pc)^2}</math>

To find the form of the momentum and energy as a function of velocity, it can be noted that the four-velocity, which is proportional to <math>\left(c, \vec{v}\right)</math>, is the only four-vector associated with the particle's motion, so that if there is a conserved four-momentum <math>\left(E, \vec{p}c\right)</math>, it must be proportional to this vector. This allows expressing the ratio of energy to momentum as
<math display="block">pc = E \frac{v}{c} ,</math>
resulting in a relation between {{mvar|<var>E</var>}} and {{mvar|<var>v</var>}}:
<math display="block">E^2 = \left(mc^2\right)^2 + E^2 \frac{v^2}{c^2},</math>

This results in 
<math display="block">E = \frac{mc^2}{\sqrt{1 - \dfrac{v^2}{c^2}}}</math>
and
<math display="block">p = \frac{mv}{\sqrt{1 - \dfrac{v^2}{c^2}}}.</math>

these expressions can be written as
<math display="block">\begin{align}
E_0 &= mc^2 , \\
E &= \gamma mc^2 , \\
p &= mv \gamma ,
\end{align}</math>
where the factor <math display="inline">\gamma = {1}/{\sqrt{1-\frac{v^2}{c^2}}}.</math>

When working in [[system of units|units]] where {{math|1=''c'' = 1}}, known as the [[natural unit system]], all the relativistic equations are simplified and the quantities [[energy]], [[momentum]], and [[mass]] have the same natural dimension:<ref name="QFT">{{cite book |title=Quantum Field Theory |edition=2nd |first1=Franz |last1=Mandl |first2=Graham |last2=Shaw |publisher=John Wiley & Sons |year=2013 |isbn=978-1-118-71665-6 |page=70 |url=https://books.google.com/books?id=jTBzQfctvHAC}} [https://books.google.com/books?id=jTBzQfctvHAC&pg=PT70 Extract of page 70]</ref>

<math display="block">m^2 = E^2 - p^2.</math>

The equation is often written this way because the difference <math>E^2 - p^2</math> is the relativistic length of the energy [[4-momentum|momentum four-vector]], a length which is associated with rest mass or invariant mass in systems. Where {{math|''m'' > 0}} and {{math|1=''p'' = 0}}, this equation again expresses the mass–energy equivalence {{math|1=''E'' = ''m''}}.