**Minimum Energy and
Equipartition**

The existence of a synchrotron source implies the presence of relativistic electrons with some energy density $U_{\rm e}$ and a magnetic field whose energy density is $U_{\rm B} = B^2 / (8 \pi)$. What is the minimum energy in relativistic particles and magnetic fields required to produce a synchrotron source of a given luminosity?

To estimate $U_{\rm e}$, we assume a
power-law electron energy
distribution

$$ N(E) \approx K E^{-\delta} $$ spanning the energy
range $E_{\rm min}$ to $E_{\rm max}$ that is needed to produce
synchrotron
radiation over any frequency range $\nu_{\rm min}$ to
$\nu_{\rm max}$. Then $$U_{\rm e} = \int_{E_{\rm
min}}^{E_{\rm max}} E N(E)\, d E$$ For a given synchrotron luminosity

$$L = \int_{\nu_{\rm min}}^{\nu_{\rm max}} L_\nu d \nu ,$$ $$
{U_{\rm e} \over L} \propto { \int_{E_{\rm min}}^{E_{\rm max}} E N(E)\,
dE \over -\int_{E_{\rm min}}^{E_{\rm max}} (dE/dt) N(E)\, dE}$$
Substituting $N(E) = K E^{-\delta}$ and the synchrotron power emitted
per electron
$(-dE/dt) \propto B^2 E^2$ gives

$${U_{\rm e} \over L} \propto { K
\int_{E_{\rm min}}^{E_{\rm max}} E^{1 - \delta} \, dE \over K B^2
\int_{E_{\rm min}}^{E_{\rm max}} E^{2-\delta} \, dE}$$ $${U_{\rm e}
\over L} \propto {E^{2 - \delta} \vert_{E_{\min}}^{E_{\rm max}} \over
B^2 E^{3 - \delta} \vert_{E_{\rm min}}^{E_{\rm max}} }$$

Since electrons with energy $E$ emit
most of the radiation seen at
frequency $\nu \propto E^2 B$, the electron energy needed to produce
radiation at frequency $\nu$ scales as

$$E \propto B^{-1/2}$$ If we
consider the energy content of only those electrons that emit in a
fixed frequency range (e.g., from $\nu_{\rm min} \sim 10^7$ Hz to
$\nu_{\rm max} \sim 10^{11}$ Hz), then the energy limits $E_{\rm min}$
and $E_{\rm max}$ are both proportional to $B^{-1/2}$ and

$${U_{\rm e}
\over L } \propto { (B^{-1/2})^{2 - \delta} \over B^2 (B^{-1/2})^{3 -
\delta} } = {B^{-1 + \delta/2} \over B^2 B^{-3/2 + \delta/2}} =
B^{-3/2}$$ We conclude that

$$\bbox[border:3px blue solid,7pt]{U_{\rm e} \propto
B^{-3/2}}\rlap{\quad \rm {(5D1)}} $$

and we already
know that $$U_{\rm B} \propto B^2~.$$

The "invisible" cosmic-ray protons
and heavier ions emit negligible synchrotron power but they still
contribute
to the total cosmic-ray particle energy. If we call the ion/electron
energy ratio
$\eta$, then the total energy density in cosmic rays is $(1 + \eta)
U_{\rm e}$. The total energy density $U$ of both cosmic rays and
magnetic
fields is

$$\bbox[border:3px blue solid,7pt]{U = (1 + \eta) U_{\rm e} + U_{\rm
B}}\rlap{\quad \rm {(5D2)}}$$

We cannot measure
$\eta$ directly in distant radio sources, but cosmic rays collected
near the Earth have $\eta \approx 40$.

The greatly differing
dependences of $U_{\rm e}$ and $U_{\rm B}$ on $B$ means that the total
(cosmic ray plus magnetic) energy density $U(B)$ has
a fairly sharp minimum near the point at which $(1 + \eta) U_{\rm e}
\approx U_{\rm B}$.