Editing Kinetic energy (section)

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

If a body's speed is a significant fraction of the [[speed of light]], it is necessary to use relativistic mechanics to calculate its kinetic energy. In relativity, the total energy is given by the [[energy-momentum relation]]:

<math display=block>E^2 = (p \textrm c)^2 + \left(m_0 \textrm c^2\right)^2\,</math>

Here we use the relativistic expression for linear momentum: <math>p = m\gamma v</math>, where <math display="inline">\gamma = 1/\sqrt{1 - v^2/c^2}</math>.
with <math>m</math> being an object's [[invariant mass|(rest) mass]], <math>v</math> speed, and ''c'' the speed of light in vacuum.
Then kinetic energy is the [[Energy–momentum_relation#Connection_to_E_=_mc2 | total relativistic energy minus the rest energy]]:
<math display=block>E_K = E - m_{0}c^2 = \sqrt{(p \textrm c)^2 + \left(m_0 \textrm c^2\right)^2}- m_{0}c^2</math>

At low speeds, the square root can be expanded and the rest energy drops out, giving the Newtonian kinetic energy. 

[[Image:KEvsMOMENTUM.png|thumb|450px|center| Log of relativistic kinetic energy versus log relativistic momentum, for many objects of vastly different scales. The intersections of the object lines with the bottom axis approaches the rest energy. At low kinetic energy the slope of the object lines reflect Newtonian mechanics. As the lines approach <math>c</math> the slope bends at the lightspeed barrier.]]

=== Derivation ===
Start with the expression for linear momentum <math>\mathbf{p} = m\gamma \mathbf{v}</math>, where <math display="inline">\gamma = 1/\sqrt{1 - v^2/c^2}</math>. [[Integration by parts|Integrating by parts]] yields:<math display="block">E_\text{k} =
  \int \mathbf{v} \cdot d \mathbf{p} =
  \int \mathbf{v} \cdot d (m \gamma \mathbf{v}) =
  m \gamma \mathbf{v} \cdot \mathbf{v} - \int m \gamma \mathbf{v} \cdot d \mathbf{v} =
  m \gamma v^2 - \frac{m}{2} \int \gamma d \left(v^2\right)
</math>Since <math>\gamma = \left(1 - v^2/c^2\right)^{-\frac{1}{2}}</math>,<math display="block">\begin{align}
  E_\text{k} &= m \gamma v^2 - \frac{- m c^2}{2} \int \gamma d \left(1 - \frac{v^2}{c^2}\right) \\
             &= m \gamma v^2 + m c^2 \left(1 - \frac{v^2}{c^2}\right)^\frac{1}{2} - E_0
\end{align}</math>Where <math>E_0</math> is a [[constant of integration]] for the [[indefinite integral]]. Rearranging to combine the factor <math>m\gamma</math> gives
<math display="block">\begin{align}
  E_\text{k} &= m \gamma \left(v^2 + c^2 \left(1 - \frac{v^2}{c^2}\right)\right) - E_0 \\
             &= m \gamma \left(v^2 + c^2 - v^2\right) - E_0 \\
             &= m \gamma c^2 - E_0
\end{align}</math><math>E_0</math> is found by observing that when <math>\mathbf{v} = 0,\ \gamma = 1</math> and <math> E_\text{k} = 0</math>, the result is the "rest energy":

<math display="block">E_0 = m c^2 </math>

and resulting in the formula:

<math display="block">E_\text{k} = m \gamma c^2 - m c^2 = \frac{m c^2}{\sqrt{1 - \frac{v^2}{c^2}}} - m c^2 = (\gamma - 1) m c^2</math>
This formula shows that the work expended accelerating an object from rest approaches infinity as the velocity approaches the speed of light. Thus it is impossible to accelerate an object across this boundary.

=== Low speed limit ===

The mathematical by-product of this calculation is the [[mass–energy equivalence]] formula, that mass and energy are essentially the same thing:<ref name=Einstein1922>{{cite book |last1=Einstein |first1=Albert |title=The Meaning of Relativity: Four Lectures Delivered at Princeton University, May, 1921 |date=1922 |publisher=Methuen & Company Limited |pages=51–52 |url=https://books.google.com/books?id=0nIxAQAAMAAJ&q=%22Mass%20and%20energy%20are%20therefore%20essentially%20alike%22}}</ref>{{rp|51}}<ref>{{Cite book |last=Resnick |first=Robert |url=https://archive.org/details/robertresnickintroductiontospecialrelativitywiley1968/page/n131/mode/2up?q=%22kinetic+energy%22 |title=Introduction to special relativity |date=1968 |publisher=Wiley |isbn=978-0-471-71725-6 |location=New York |access-date=2024-11-17}}</ref>{{rp|121}}

<math display="block">E_\text{rest} = E_0 = m c^2 </math>

At a low speed (''v'' ≪ ''c''), the relativistic kinetic energy is approximated well by the classical kinetic energy. To see this, apply the [[binomial approximation]] or take the first two terms of the [[Taylor expansion]] in powers of <math>v^2</math> for the reciprocal square root:<ref name=Einstein1922/>{{rp|51}}

<math display="block">E_\text{k} \approx m c^2 \left(1 + \frac{1}{2} \frac{v^2}{c^2}\right) - m c^2 = \frac{1}{2} m v^2</math>

So, the total energy <math>E_k</math> can be partitioned into the rest mass energy plus the non-relativistic kinetic energy at low speeds.

When objects move at a speed much slower than light (e.g. in everyday phenomena on Earth), the first two terms of the series predominate. The next term in the Taylor series approximation

<math display="block">E_\text{k} \approx m c^2 \left(1 + \frac{1}{2} \frac{v^2}{c^2} + \frac{3}{8} \frac{v^4}{c^4}\right) - m c^2 = \frac{1}{2} m v^2 + \frac{3}{8} m \frac{v^4}{c^2}</math>

is small for low speeds. For example, for a speed of {{convert|10|km/s|mph|abbr=on}} the correction to the non-relativistic kinetic energy is 0.0417&nbsp;J/kg (on a non-relativistic kinetic energy of 50&nbsp;MJ/kg) and for a speed of 100&nbsp;km/s it is 417&nbsp;J/kg (on a non-relativistic kinetic energy of 5&nbsp;GJ/kg).

The relativistic relation between kinetic energy and momentum is given by

<math display="block">E_\text{k} = \sqrt{p^2 c^2 + m^2 c^4} - m c^2</math>

This can also be expanded as a [[Taylor series]], the first term of which is the simple expression from Newtonian mechanics:<ref>{{cite web
 |url=http://farside.ph.utexas.edu/teaching/qmech/Quantum/node107.html
 |title=Fine Structure of Hydrogen
 |first=Richard
 |last=Fitzpatrick
 |date=20 July 2010
 |work=Quantum Mechanics
 |access-date=20 August 2016
 |archive-date=25 August 2016
 |archive-url=https://web.archive.org/web/20160825061339/http://farside.ph.utexas.edu/teaching/qmech/Quantum/node107.html
 |url-status=live
 }}</ref>
:<math>E_\text{k} \approx \frac{p^2}{2 m} - \frac{p^4}{8 m^3 c^2}.</math>

This suggests that the formulae for energy and momentum are not special and axiomatic, but concepts emerging from the equivalence of mass and energy and the principles of relativity.

===General relativity===
{{See also|Schwarzschild geodesics}}
Using the convention that

<math display="block">g_{\alpha\beta} \, u^{\alpha} \, u^{\beta} \, = \, - c^2</math>

where the [[four-velocity]] of a particle is

<math display="block">u^{\alpha} \, = \, \frac{d x^{\alpha}}{d \tau}</math>

and <math>\tau </math> is the [[proper time]] of the particle, there is also an expression for the kinetic energy of the particle in [[general relativity]].

If the particle has momentum

<math display="block">p_{\beta} \, = \, m \, g_{\beta\alpha} \, u^{\alpha}</math>

as it passes by an observer with four-velocity ''u''<sub>obs</sub>, then the expression for total energy of the particle as observed (measured in a local inertial frame) is

<math display="block">E \, = \, - \, p_{\beta} \, u_{\text{obs}}^{\beta}</math>

and the kinetic energy can be expressed as the total energy minus the rest energy:

<math display="block">E_{k} \, = \, - \, p_{\beta} \, u_{\text{obs}}^{\beta} \, - \, m \, c^2 \,.</math>

Consider the case of a metric that is diagonal and spatially isotropic (''g''<sub>''tt''</sub>, ''g''<sub>''ss''</sub>, ''g''<sub>''ss''</sub>, ''g''<sub>''ss''</sub>). Since

<math display="block">u^{\alpha} = \frac{d x^{\alpha}}{d t} \frac{d t}{d \tau} = v^{\alpha} u^{t} </math>

where ''v''<sup>α</sup> is the ordinary velocity measured w.r.t. the coordinate system, we get

<math display="block">-c^2 = g_{\alpha\beta} u^{\alpha} u^{\beta} = g_{tt} \left(u^{t}\right)^2 + g_{ss} v^2 \left(u^{t}\right)^2 \,.</math>

Solving for ''u''<sup>t</sup> gives

<math display="block">u^{t} = c \sqrt{\frac{-1}{g_{tt} + g_{ss} v^2}} \,.</math>

Thus for a stationary observer (''v'' = 0)

<math display="block">u_{\text{obs}}^{t} = c \sqrt{\frac{-1}{g_{tt}}} </math>

and thus the kinetic energy takes the form

<math display="block">E_\text{k} = -m g_{tt} u^t u_{\text{obs}}^t - m c^2 = m c^2 \sqrt{\frac{g_{tt}}{g_{tt} + g_{ss} v^2}} - m c^2\,.</math>

Factoring out the rest energy gives:

<math display="block">E_\text{k} = m c^2 \left( \sqrt{\frac{g_{tt}}{g_{tt} + g_{ss} v^2}} - 1 \right) \,.</math>

This expression reduces to the special relativistic case for the flat-space metric where

<math display="block">\begin{align}
  g_{tt} &= -c^2 \\
  g_{ss} &=  1 \,.
\end{align}</math>

In the Newtonian approximation to general relativity

<math display="block">\begin{align}
  g_{tt} &= -\left( c^2 + 2\Phi \right) \\
  g_{ss} &= 1 - \frac{2\Phi}{c^2}
\end{align}</math>

where Φ is the Newtonian [[gravitational potential]]. This means clocks run slower and measuring rods are shorter near massive bodies.