Editing Bremsstrahlung (section)

== Classical description ==
{{main|Larmor formula}}
[[File:Bremsstrahlung.gif|thumb|Field lines and modulus of the electric field generated by a (negative) charge first moving at a constant speed and then stopping quickly to show the generated Bremsstrahlung radiation.]]

If [[quantum mechanics|quantum]] effects are negligible, an accelerating charged particle radiates power as described by the [[Larmor formula]] and its relativistic generalization.

=== Total radiated power ===
The total radiated power is<ref>''A Plasma Formulary for Physics, Technology, and Astrophysics'', D. Diver, pp. 46–48.</ref>
<math display="block">P = \frac{2 \bar q^2 \gamma^4}{3 c} 
    \left( \dot{\beta}^2 + \frac{\left(\boldsymbol{\beta} \cdot \dot{\boldsymbol{\beta}}\right)^2}{1 - \beta^2}\right),</math>
where <math display="inline">\boldsymbol\beta = \frac{\mathbf v}{c}</math> (the velocity of the particle divided by the speed of light), <math display="inline">\gamma = {1}/{\sqrt{1-\beta^2}}</math> is the [[Lorentz factor]], <math>\varepsilon_0</math> is the [[vacuum permittivity]], <math>\dot{\boldsymbol\beta}</math> signifies a time derivative of {{nowrap|<math>\boldsymbol\beta</math>,}} and {{math|''q''}} is the charge of the particle.
In the case where velocity is parallel to acceleration (i.e., linear motion), the expression reduces to<ref>{{cite book | title = Introduction to Electrodynamics | first = D. J. | last = Griffiths | pages = 463–465 }}</ref>
<math display="block">P_{a \parallel v} = \frac{2 \bar q^2 a^2 \gamma^6}{3 c^3},</math>
where <math>a \equiv \dot{v} = \dot{\beta}c</math> is the acceleration. For the case of acceleration perpendicular to the velocity (<math>\boldsymbol{\beta} \cdot \dot{\boldsymbol{\beta}} = 0</math>), for example in [[synchrotron]]s, the total power is
<math display="block">P_{a \perp v} = \frac{2 \bar q^2 a^2 \gamma^4 }{3c^3}.</math>

Power radiated in the two limiting cases is proportional to <math>\gamma^4</math> <math>\left(a \perp v\right)</math> or <math>\gamma^6</math> <math>\left(a \parallel v\right)</math>. Since <math>E = \gamma m c^2</math>, we see that for particles with the same energy <math>E</math> the total radiated power goes as <math>m^{-4}</math> or <math>m^{-6}</math>, which accounts for why electrons lose energy to bremsstrahlung radiation much more rapidly than heavier charged particles (e.g., muons, protons, alpha particles). This is the reason a TeV energy electron-positron collider (such as the proposed [[International Linear Collider]]) cannot use a circular tunnel (requiring constant acceleration), while a proton-proton collider (such as the [[Large Hadron Collider]]) can utilize a circular tunnel. The electrons lose energy due to bremsstrahlung at a rate <math>(m_\text{p}/m_\text{e})^4 \approx 10^{13}</math> times higher than protons do.

=== Angular distribution ===
The most general formula for radiated power as a function of angle is:<ref name=Jackson>{{cite book | last = Jackson | title = Classical Electrodynamics | at = §14.2–3 }}</ref>
<math display="block">\frac{d P}{d\Omega} = \frac{\bar q^2}{4\pi c} \frac{\left|\hat{\mathbf n} \times \left(\left(\hat{\mathbf n} - \boldsymbol{\beta}\right) \times \dot{\boldsymbol{\beta}}\right)\right|^2}{\left(1 - \hat{\mathbf n}\cdot\boldsymbol{\beta}\right)^5}</math>
where <math>\hat{\mathbf n}</math> is a unit vector pointing from the particle towards the observer, and <math>d\Omega</math> is an infinitesimal solid angle.

In the case where velocity is parallel to acceleration (for example, linear motion), this simplifies to<ref name=Jackson/>
<math display="block">\frac{dP_{a \parallel v}}{d\Omega} = \frac{\bar q^2a^2}{4\pi c^3}\frac{\sin^2 \theta}{(1 - \beta \cos\theta)^5}</math>
where <math>\theta</math> is the angle between <math>\boldsymbol{\beta}</math> and the direction of observation <math> \hat{\mathbf n}</math>.