IR derivation

From Tauwiki

1 Energy input
2 Temperature calculation
3 Energy output
4 Detection of Surface brightness
5 Extra points taken into consideration

[edit] Energy input

Energy input in a volume dV at any wavelength $(λ)$ is

$E_\lambda =(1-a_\lambda)n_\lambda \frac {hc}{\lambda}$

where $a λ$ is the albedo at a given wavelength and $n λ$ is the number of photons absorbed.

$E_T =\int d\lambda (1-a_\lambda)n_\lambda hc/\lambda$

$n λ$ comes from the simulation and it is not possible to run the simulatio for all $λ$ , hence we make an approximation that $n λ$ will scale as the single scattering flux.

$n_\lambda = \frac{s_\lambda \sigma_\lambda}{s_0 \sigma_0} n_0$

where, $s λ$ is the stellar flux and $σ λ$ is the cross section. Also, $s 0,σ 0, n 0$ are at the wavelength of simulation.

Because the simulation only includes $107$ photons, there is a scale factor of $S 0 / 10 7$ to be applied before calculating the energy.

Hence we have

$E_T = \frac {S_0}{10^7}\int d\lambda (1-a_\lambda)\frac{s_\lambda \sigma_\lambda}{s_0 \sigma_0} n_0 hc/\lambda$

which is now the total energy input into the volume dV.

The energy input in the volume dV is $E T$ which is spread over N grains so the average energy per grain is $E T / N$ .

[edit] Temperature calculation

Hence the temperature of each grain can be defined by

$E T / N = 4π < Q > σ T 4$

[edit] Energy output

We need the spectral energy distribution of the output energy and assuming blackbody radiation, energy from each grain is

$\int d\lambda B_{\lambda} (T) 4\pi r_g^2 = E_T/N$

Let's say the grain temperature is 15 K. And $r g = 0.1μ = 10 - 5 c m$ . The energy sent out in one of the IRAS bands will be

$dV \int f(\lambda) B_\lambda (T) d\lambda 4\pi r_g^2 n = F$

where,

$n = N / d V$

is the number density.

For number density n =300 dust grains per cm^3,

in CGS Units,

$BBflux(100 \mu ,T_{g})= \frac{2hc^2}{\lambda^5}\frac{1}{exp(\frac{hc}{k\lambda T_{g}})-1}= 8.13796 \times10^{-8} ergs/s/Sr/cm^2/A$

taking $d λ = 5μ = 50000 A$ ,

$F = dV \int f(\lambda) B_\lambda (T) d\lambda 4\pi r_g^2 n = dV \times 1.15835\times10^{-8}$

If we consider $d V = W 3 = (0.0545 p c) 3$ as described in the section below,

then $F = 1.15835 \times 10^{-8} \times (0.0545)^3\times (3.08 \times 10^{18})^3 = 5.47873 \times 10^{43} ergs/s$

[edit] Detection of Surface brightness

At the detector, the detected energy

$= \frac{F}{4\pi r^2}d A_d$

as always, we want the surface brightness, so we want

$\frac{F dA_d}{4\pi r^2}\frac{1}{dA_d d\Omega \Delta\lambda} = \frac{F }{4\pi r^2}\frac{r^2}{dA_c \Delta \lambda}= \frac {dz}{\Delta \lambda} \int d\lambda f(\lambda) B_\lambda (T) 4\pi r_g^2 n$

We can define

$\Delta \lambda = \int f(\lambda) d\lambda = 332650.$

Using above values surface brightness is

$= \frac {0.0545 \times 3\times 10^{18}}{332650} \times 1.15835 \times 10^{-8}= 5845.18$

This will give brightness in ergs/cm^2/s/A/Sr. Using

$1 \frac{erg}{ cm^2 \times s\times sr\times A }= 3.336 \times 10^6 \times (\lambda(\mu m)^2)MJy / Sr$

we get brightness in MJy/Sr.

i.e. the expected output is

$= 5845.18 \times 3.336 \times 10^6 \times (100)^2 = 1.94995 \times 10^{14} MJy/Sr$

Maximum of IRAS output is at $1.5 \times 10^4 MJy/Sr$ . Since we have to combine 32 pixels of IRAS in one pixel of simulation, maximum output from simulation can be taken as $4.8 \times 10^5 MJy/Sr$