Widdekind Posted December 17, 2012 Posted December 17, 2012 There is an observed relation, between the brightness & power of relativistic jets, common to Quasars & GRBs: http://phys.org/news/2012-12-common-physics-black-holes.html Could the following calculations help explain the same? [math]P = \frac{e^2}{6 \pi \epsilon_0 c} \left( \gamma^3 \dot{\beta} \right)^2[/math] [math] = \frac{d}{dt} \left( \gamma m_e c^2 \right) = m_e c^2 \left( \gamma^3 \beta \dot{\beta} \right) [/math] So: [math]\gamma^3 \dot{\beta} = \frac{3 c}{2} \frac{m_e c^2}{\left( \frac{e^2}{4 \pi \epsilon_0} \right) } = \frac{3 c}{2 r_e} \beta [/math] [math]\frac{\gamma^3}{\beta} d\beta = \frac{3 c}{2 r_e} dt[/math] [math]\frac{\gamma^3}{\beta} \frac{d\gamma}{\frac{d\gamma}{d\beta}} = \frac{3 c}{2 r_e} dt[/math] [math]\frac{\gamma^3}{\beta} \frac{d\gamma}{\gamma^3 \beta} = \frac{3 c}{2 r_e} dt[/math] [math]\frac{\gamma^2}{\gamma^2-1} d\gamma = \frac{3 c}{2 r_e} dt[/math] By partial fractions: [math]\frac{\gamma^2}{\gamma^2-1} = 1 + \frac{1/2}{\gamma-1} - \frac{1/2}{\gamma+1}[/math] So the integration yields: [math]\Delta \left( \gamma + ln \left( \sqrt{\frac{\gamma - 1}{\gamma + 1}} \right) \right) = \frac{3 c}{2 r_e} \Delta t[/math] For relativistic jets [math]\gamma \gg 1[/math], and most of the power is radiated early on, at high [math]\gamma[/math]. So: [math]\Delta \gamma \approx \frac{3 c}{2 r_e} \Delta t[/math] [math]\boxed{ t_{cool} \approx \frac{2 r_e}{3 c} \gamma_0 }[/math] But the initial electron energy was also proportional to [math]\gamma_0[/math]. So, the average power of emissions is (calculated to be) constant at all energy scales: [math]\bar{P} \equiv \frac{E_0}{t_{cool}} \approx \frac{\gamma_0 m_e c^2}{ \frac{2 r_e}{3 c} \gamma_0 } = \frac{3 c}{2 r_e} m_e c^2 \approx 10^{10} W[/math] Perhaps some similar sort of scale invariance, whereby the power emitted by decelerating electrons is quasi-constant, could account, for the observed Jet brightness / power relation.
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