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Line 63: − The energy of a particle in [[relativity]] is:+ − + + + + + + + + + + + + + + − where <math>\gamma</math> is the Lorentz factor, <math>m_0</math> is the [[rest mass]] and c is the [[speed of light]].+ − + − + + + + + + +
Added derivation
== Kinetic Energy in Relativity ==
== Kinetic Energy in Relativity ==
In [[relativity]], kinetic energy can be expressed as:
<math> E = \gamma m_0 c^2 </math>
<math> K = (\gamma - 1) m_0 c^2 </math>
where
:<math>\gamma</math> is the [[Lorentz factor]]
:<math>m_0</math> is the [[rest mass]]
:<math>c</math> is the [[speed of light]]
===Derivation===
The kinetic energy is the [[work|work done]] accelerating a particle from rest to some speed <math>v</math>. Suppose the particle is at rest at <math>x_1</math> and speed <math>v</math> at position <math>x_2</math>. Hence:
<math>K = \int^{x_2}_{x_1} \vec{F} \dot d \vec{x}</math>
Since <math>\vec{F} = \vec{F}_{\perp} + \vec{F}_{\parallel}</math>, and a perpendicular force does no work, we can ignore the perpendicular component and write:
<math>K = \int^{x_2}_{x_1} F_{\parallel} dx = \int^{x_2}_{x_1} \gamma^3 m_0 a dx</math>
Since this includes the mass energy of the particle, we must subtract a factor of <math> m_0 c^2 </math> to get the energy due to the particle's motion, the kinetic energy, as:
Since <math>a \, dx = \frac{dv_{\parallel}}{dt} dx = \frac{dx}{dt} dv</math>, we find the integral can be rewritten as:
<math> K = (\gamma - 1) m_0 c^2 </math>
<math>K = \int^{v_2}_{v_1} \frac{mv}{(1- \frac{v^2}{c^2})} dv</math>
where <math>v_1</math> is the initial speed and hence 0 by definition and <math>v_2</math> is the final speed.
Performing the integration reveals the kinetic energy as:
<math>K=(\gamma -1)m_0 c^2</math>
==References==
==References==
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