
for energy in heavier particles (baryons), which may range anything from 1 to 2000 or so. It should be noted that if the  
plasma comprises both electrons and positrons (a pair plasma), 
then k=0, as $N_0$ includes both electrons and positrons whose contributions are indistinguishable in incoherent synchrotron cases 
(Singal 2012). The expression for energy density is similar to as in the
standard case of a single spectral index (Pacholczyk 1970; Moffet 1972), except that the co-efficient $a$ is  somewhat different. 

After minimizing the total energy with $B$, we get
\begin{eqnarray}
\label{eq:64o}
U_{\rm min} = 3.0 \times 10^{-10} \left[ \frac {(1+k)(1+z)^{3.5} \: S_{0}\:  \nu_{0} ^{0.5}\: }  { \phi \: \psi_{x}  \psi_{y}\: s\: \sin^{1.5}\!\theta} 
\left\{\frac {(\nu_1/\nu_0)^{-\alpha_1+0.5} -1 }{\alpha_1-0.5}
-\frac {(\nu_2/\nu_0)^{-\alpha_2+0.5} -1}{\alpha_2-0.5}  \right\}\right] 
^{4/7} {\rm erg\: cm}^{-3}. 
\end{eqnarray}
Here $z$ is the redshift, $S_{0}$ (Jy) is the flux density at the break frequency 
$\nu_{0}$ (GHz), $\alpha_1$ and $\alpha_2$ are the spectral indices (defined as S$\propto\nu^{-\alpha}$) before and after the break, with $\nu_{1}$ 
and $\nu_{2}$ (GHz) as the lower and upper cut off
frequencies presumably for the observed radio spectrum, $ \psi_{x}$ and $ \psi_{y}$ (arcsec) represent the size of the source component along its major 
and minor radio axes, 
and $s$ is the path length through the component along the line of sight in kpc. 
It should be noted that the expression within curly brackets on the right hand side in Equation~(\ref{eq:64o}) is independent 
of redshift if $\nu_{1}, \nu_0, \nu_{2}$ are all in observer's frame, 
determined, for instance, from observations. 

The magnetic field value corresponding to the minimum energy is given by
\begin{eqnarray}
\label{eq:64q1}
B_{\rm min} =\left[\frac {24 \pi\: U_{\rm min}}{7}\right]^{0.5}=3.3\: U_{\rm min}^{0.5}\; G.
\end{eqnarray}
It should be noted that $B_{\rm min}$ is not the minimum value of the magnetic field in the source under consideration. In fact it yields nearly the 
equipartition value of the magnetic field. 

For a single spectral index case, where $S_{0}$  is the flux density at any frequency 
$\nu_{0}$ with $\nu_{1} < \nu_0 < \nu_{2}$, we put $\alpha_1=\alpha_2=\alpha$ in Equation~(\ref{eq:64o}) to get,
\begin{eqnarray}
\label{eq:64p}
U_{\rm min} = 3.0 \times 10^{-10} \left[ \frac {(1+k)(1+z)^{3.5} \: S_{0}\:  \nu_{0} ^{\alpha}\: }  {\phi\: \psi_{x} \psi_{y}\: s\: \sin^{1.5}\!\theta } 
\left\{\frac {\nu_1^{-\alpha+0.5} - \nu_2^{-\alpha+0.5}}{\alpha-0.5} \right\}\right] 
^{4/7} {\rm erg\: cm}^{-3}. 
\end{eqnarray}
If $\nu_{1}$ and $\nu_{2}$ represent the cut-off frequency values in the rest frame of the source (specified based on some other 
arguments), then the factor $(1+z)^{3.5}$ in Equation~(\ref{eq:64p}) gets replaced with  $(1+z)^{3+\alpha}$ to yield the same expression as available in the 
literature (Miley 1980; Singal et al. 2004).
\section{THE PITCH ANGLE PARADOX AND RADIATIVE LIFE TIMES IN A SYNCHROTRON SOURCE}
In synchrotron radiation there is a paradox whether or not the pitch angle of a radiating charge varies. 
The conventional wisdom is that the pitch angle does not change during the radiation process. 
Since the radiation is confined to a narrow cone of angle $1/\gamma$ 
around the instantaneous direction of motion of the charge, (Jackson 1975; Pacholczyk 1970; Rybicky \& Lightman 1979), 
any radiation reaction on the charge will be in a direction just opposite to its instantaneous velocity vector.
This means that the direction of motion of the charge will remain unaffected, implying no change in 
the pitch angle of the charge (Kardashev 1962;  Pacholczyk 1970). The accordingly derived formulas for 
energy losses of synchrotron electrons in radio galaxies are the standard text-book 
material for the last 50 years. However, using the momentum transformation laws from special relativity, we shall show that the pitch angle of a radiating 
charge varies. While the velocity 
component parallel to the magnetic field remains unaffected, the perpendicular component does 
reduce in magnitude due to radiative losses, implying a change in the pitch angle.

With the pitch angle as a constant of motion, half life-times of radiating electrons have been calculated 
(Kardashev 1962; Pacholczyk 1970; Rybicki \& Lightman 1989), using a radiation power loss formula 
\begin{equation}
\label{eq:30.1}
\frac{{\rm d}{\gamma}}{{\rm d}t} = -{\eta}\sin^2 \theta\;\gamma^2.
\end{equation}
Here ${\eta}=2e^4 B^2/(3m_{\rm o}^3c^5)=1.94 \times 10^{-9}B^2\; {\rm s}^{-1}$.
If $\gamma_{\rm 0}$ represents the initial energy  at $t=0$, then from Equation~(\ref{eq:30.1}) the energy of the radiating charge at $t=\tau$ becomes 
\begin{equation}
\label{eq:30.4}
\gamma= \frac{\gamma_{\rm 0}}{1+\eta\sin^2 \theta\;  \tau\; \gamma_{\rm 0}}\;.
\end{equation}
It follows that energy of the electron reduces to half its value in  time $\tau_{1/2} = 1/(\eta\sin^2 \theta \;\gamma_{\rm o})$ 
(Kardashev 1962; Kellermann 1964; van der Laan \& Perola 1969; Pacholczyk 1970; Miley 1980).  

However from Equation~(\ref{eq:30.4}) one can easily see that while it might be true  
for a highly relativistic charge ($\gamma\gg1$), it cannot be a general result. 
This is because from Equation~(\ref{eq:30.4}) $\gamma\rightarrow 0$ when $\tau\rightarrow\infty$, while we know that $\gamma\ge1$ always. Actually an 
approximation $\beta \approx 1$ has been used in Equation~(\ref{eq:30.1}); 
the exact equation for the energy loss rate (Melrose 1971; Jackson 1975; Longair 2011) is
\begin{equation}
\label{eq:30.2}
\frac{{\rm d}{\gamma}}{{\rm d}t} = -{\eta}\sin^2 \theta\;\beta ^2\gamma^2=-{\eta} \sin^2 \theta \;(\gamma^2-1).
\end{equation}
Now, consider a charge particle in its gyrocenter (GC) frame ${\cal K}'$, where 
the charge has no component of velocity parallel to the magnetic field and has 
only a circular motion in a plane perpendicular to the magnetic field (with a pitch angle $\theta=\pi/2$). 
In this frame, due to radiative losses by the charge, there will be a decrease in the velocity which is solely  
in a plane perpendicular to the magnetic field, and the charge will never ever attain a velocity component parallel to the magnetic field because of radiation 
losses.

Now we look at this particle from the lab frame ${\cal K}$, in which the charge has (at least to begin with) a motion, 
$\beta_{\parallel}$ along 
the magnetic field. Since in the inertial frame ${\cal K}'$, the charge never gets a velocity component parallel 
to the magnetic field  and the two inertial frames (${\cal K}$ and ${\cal K}'$)