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Tuesday, May 21, 2019

Few concepts in astrophysics radiation

Luminosity:
Power that the source emits in radiation
$[L]=erg/s=10^{-7}W$
$E = \int Ldt$

Flux:
Power received per unit area $\phi = L/(4\pi d^2)$
$[\phi] = W/{m^2}$
$E=\int_{all area} dtdA$

Flux density/ specific density (specific refers to $Hz^{-1}$)
Power received per unit area and unit frequency
Flux is the integrate of  flux density $\phi = \int F_{\nu}d\nu$
Astronomers often say flux when they mean flux density
$[F_{\nu}]=W/m^2/Hz$
$Jy=10^{-26}W/m^2/Hz$
$E=\int F_{\nu}d\nu dAdt$

Surface luminosity/ specific intensity
Power received or emitted per unit are per unit frequency per unit solid angle
$dF_{\nu}=I_{\nu} \cos(\theta)d\Omega$
Specific intensity is independent on the distance ?
mean specific intensity $J_{\nu}=\frac{1}{4\pi}\int I_{\nu}d\Omega$
$[I_{\nu}= W/m^2/sr/Hz]$
The perceived measurement area orthogonal to the incident flux is significantly reduced at oblique angles, causing energy to be spread out over a wider area than it would if it was falling perpendicular to the surface.

Brightness temperature
It is the temperature that a black-body source would have in order to be the same brightness
Black-body for Rayleigh-Jeans region where $h\nu << kT$ gives $I_{\nu}=2kT/{\lambda^2}$
$T_B(v) = I_{\nu}\lambda^2/(2K)$

Radiation energy density
$u_v(\Omega)$ is the specific energy density. It is the energy in the radiation field per unit frequency per unit volume per unit solid angle. The volume of field incident on the target is $dV=c\cos{\theta} dA dt$.
Therefore $dE= u_v(\Omega) c\cos{\theta} dA dtd\nu d\Omega $.
$dE= I_{\nu}\cos{\theta}d\nu dAdt$
So, $u_v(\Omega)= \frac{I_{\nu}}{c}$.
$u_{\nu}$ is $u_{\nu}(Ω)$ integrated over all angles; it has units $Jm^{-3}/Hz$
 $U$ is $u_{\nu}$ integrated over all frequencies i.e. the total energy density in EM fields; it has units $Jm^{-3}$.

Radiation pressure
For photons of momentum $p=E/c$. Since $p_{\perp}=p\cos{\theta}$, then the momentum per unit area per unit time per unit frequency of a radiation field is
$p_{\nu}=\frac{1}{c}\int I_{\nu}\cos{\theta}^2d\Omega$.
$p_{\perp}=p\cos{\theta}=E/c\cos{\theta}$
The definition of $p_{\nu}$ is $dp_{\perp}/dA/dt/d\nu=dE/c/dA/dt/d\nu \cos{\theta}=\frac{1}{c}F_{\nu}\cos{\theta}$, and $F_{\nu}=\int I_{\nu}\cos\theta d\Omega$.
$[p_{\nu}]=\frac{s}{cm}erg/cm^2/s/sr/Hz *sr=erg/(cm/s)/(cm^2)=momentum/s/area/Hz$.

For isotropic radiation perfectly reflected by a wall (so the total momentum change is twice the incident momentum), and $P=dF/dA$ and $F=dp/dt$
$F = \Delta p/(unit \,time)=2p/(unit\, time)$
$P=\int_{\nu}\int_{2\pi} \frac{2}{c}I_{\nu}\cos{\theta}^2d\Omega d\nu=\frac{1}{3}U$
$[P]=Force/{cm}^2$

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Note that the number of photons per unit area hitting the film is proportional to $\cos\theta$ if the normal to the film is tilted by an angle $\theta$ from the ray direction. This is just the same projection effect that reduces the amount of water collected by a tilted rain gauge by $\cos\theta$.
If energy $dE_{\nu}$ flows through $dA$ in time $dt$ in the frequency range $\nu$ to
$\nu+d\nu$ within the solid angle $d\Omega$ on a ray which points an angle $\theta$ away the surface normal,
$dE_{\nu}=(dAdtd\nu d\Omega I_{\nu}) \cos\theta$,
$I_{\nu} = dE_{\nu}/dA/dt/d\nu/d\Omega/cos\theta$.

If a source is discrete, meaning that it subtends a well-defined solid angle, the spectral power received by a detector of unit projected area is called the source flux density $F_{\nu}$.
$F_{\nu}=dE_{\nu}/dt/dA/d\nu=\int_{source} I_{\nu}\cos\theta d\Omega$
Flux density is dependent of source distance $d$, since $\int_{source}\propto \frac{1}{d^2}$.

The flux density of the Sun.
$\theta=arcsin(R_s/d)$
$F_{\nu}=\int_{sun}I_{\nu}\cos\theta d\Omega = 2\pi \int_0^{\theta}I_{\nu}\sin\theta \cos\theta d\theta $
$F_{\nu}=\int_{sun}I_{\nu}\cos\theta d\Omega = 2\pi \int_{sun}I_{\nu}\sin\theta d\sin\theta $
$F_{\nu}=2\pi I_{\nu} \frac{1}{2}\sin^2\theta=\pi I_{\nu} \theta^2$.
flux density = $(\pi\theta^2=\Omega)$ specific intensity


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