IR Check

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1 Procedure for Tally with published results
2 Calculation of Grain temperatures
3 Calculation of IR Brightness per Hydrogen Column density
4 Comparison with IRAS observations

[edit] Procedure for Tally with published results

Calculation of Temperature: Calculating temperatures for Draine's grain model with Mathis ISRF, and then substituting our stellar radiation in place of Mathis ISRF.
Calculation of 100 micron Brightness per Hydrogen column density for temperatures given by Draine, with his grain size distribution. Later replacing with the grain temperatures calculated for Coalsack region.
Comparison with data: deriving the Brightness per H in Jy cm^2/Sr units from Low et al data, using BH units and comparing with values quoted by Draine. Calculating Brightness per H in Jy cm^2/Sr units from IRAS data. Comparing with Brightness calculated in previous step.
Preparing the FITS file for IR simulation results, using best fit dust file.

[edit] Calculation of Grain temperatures

Mezger, Mathis, Panagia (1982) give the average interstellar radiation mean intensity, i.e.

$F λ = 4π J λ e r g / c m 2 / s / μ m$

as follows:

$F_{\lambda}=38.57 \lambda_{\mu m}^{3.4172} ,0.0912<\lambda<0.110 \mu m$

$F_{\lambda}=2.045\times10^{-2} ,0.110<\lambda<0.134 \mu m$

$F_{\lambda}=7.115\times10^{-4} \lambda_{\mu m}^{-1.6678} ,0.134<\lambda<0.246 \mu m$

$F_{\lambda}= 4\pi\Sigma_i B_{\lambda}(T_i) W_i, 0.246\mu m< \lambda < \infty$

Where, $W=[1,10^{-14},10^{-13},4\times 10^{-13}]$ and $T = [2.9,7500,4000,3000]$ .

Using this ISRF and the grain model described in next section, Draine and Lee(1984) compute the Temperatures of Graphite and Silicate grains as listed in the table of next section.

The Energy balance equation used is:

$\int u_{\lambda} c Q_{\lambda} d \lambda = 4 <Q(a,T_{gr})>\sigma T_{gr}^4$

Here the left hand side contains the absorption cross sections, and the Planck Averaged emissivities are defined as :

$<Q(a,T_{gr})> = 15 (\frac{hc}{\pi k T})^4 \int Q_{abs}(a,\lambda ) \lambda^{-5}[exp\frac{hc}{\lambda k T }-1]^{-1} d \lambda$

We have,

$u λ c = F λ$ .

Using these equations, following temperatures for Graphite are obtained:

Description of Grains used in Model
Grainsize	Graphite		Silicate
microns	Mine	Draine	Mine	Draine
0.01	20.53	20.0	16.33	17.9
0.02	20.16	19.7	15.62	17.3
0.03	20.10	19.6	15.26	17.0
0.05	20.04	19.3	15.09	16.5
0.1	19.49	18.8	14.91	15.4
0.2	17.58	17.3	14.36	15.4
0.3	16.50	16.2	14.30	14.7
0.5	14.95	14.8	13.75	14.4
1.0	12.88	12.7	13.07	13.4

My values are somewhat higher for Graphite and lower for Silicates, than Draine's values.

[edit] Equations used for calculation of temperature:

$E(dust_{index})=intens \times (1-albedo) \times \frac{hc}{\lambda} \times f_{spec} \times \frac{BBflux}{N_{phot}}$

$f_{spec}=\frac {\int B(\lambda, T_{star})d\lambda}{B(\lambda_0,T_{star})}$

$E(dust_{index})=E(dust_{index}) / hits(dust_{index})\times \frac {dN}{\int dN}$

$E = 4 <Q> a \sigma T_{gr}^4.$

[edit] Deriving relevant stellar flux factors for Simulation:

This section will be corrected soon.

Let's consider a star at Temperature $T s t a r$ . It will emit the Kurucz model spectrum, for simplicity we can consider right now the Planck Spectrum. We run the simulation at single wavelength, say $λ 0$ . The Flux at this wavelength will be:

$B(\lambda_0,T_{star})= \frac{2hc^2}{\lambda_0^5}\frac{1}{exp(\frac{hc}{\lambda k T_{star}})-1} erg/s/cm^2/A/Sr$

This means there are following number of photons available at $λ 0$ :

$\frac{B(\lambda_0,T_{star})}{hc/\lambda_0}photons/s/cm^2/A/Sr$ .

Out of these, we run the simulation for $N p h o t$ photons. These $N p h o t$ photons travel in all random directions and get scattered after a randomly chosen optical depth

$\tau(\lambda)=\int n(r)C_{abs}(\lambda)dr$ .

Each of these rays travelling in different directions, represents

$\frac{B(\lambda_0,T_{star})}{hc/\lambda_0}\times \frac{1}{N_{phot}} photons/s/cm^2/A/Sr$

Since its a Monte Carlo Simulation, we assume that all these photons, represented by the ray, survive the randomly generated optical depth $τ$ . After having travelled this optical depth, let's consider scattering and absorption of these rays by a volume of 1cm^3. The energy given to the grains in this cm^3 is

$E_0=\frac{B(\lambda_0,T_{star})}{hc/\lambda{0}}\times\frac{1}{N_{phot}} \times(1-a)\times\frac{hc}{\lambda_0} erg/s/cm^2/A/Sr$

If the simulation was carried over all wavelengths, the energy given to grains will be

$E_T=\int \frac{B(\lambda,T_{star})}{N_{phot}(\lambda)}(1-a (\lambda))d\lambda erg/s/cm^2/Sr$

Assuming constant albedo and number of photons in the simulation, we get,

$f_{spec}=\frac{E_T}{E_0}=\frac{\int B(\lambda,T_{star})d\lambda}{B(\lambda_0,T_{star})}=\frac{\sigma T_{star}^4}{\pi B(\lambda_0,T_{star})} A$

Where,

$\sigma = \frac{2\pi^5k^4}{15c^2h^3} = 5.67\times 10^{-5} erg/s/cm^2/K^4$

In the simulation, we use the variable intens set to 1 for the first scattering. As the multiple scatterings happen, this value diminishes upto a certain minimum value,or the photons leave the dust boundary. Since this variable gives the strength of the ray, viz. number of photons it represents, we need to multiply by the actual number of photons represented, to the variable intens. Thus the energy given to 1cm^3 volume will be

$E_T = intens \times \frac{B(\lambda_0,T_{star})}{N_{phot}\times hc/\lambda_0} \times f_{spec}\times \frac{hc}{\lambda_0}(1-a) erg/s/cm^2/Sr$

Since the rays are randomly distributed in all steradians, we get

$E_T = 4\pi \times intens \times \frac{B(\lambda_0,T_{star})}{N_{phot}\times hc/\lambda_0}\times f_{spec}\times \frac{hc}{\lambda_0}(1-a) erg/s/cm^2$

$= 4\pi \times intens \times \frac{B(\lambda_0,T_{star})}{N_{phot} }\times f_{spec}\times (1-a) erg/s/cm^2$

$= 4\pi \times intens \times \frac{1}{N_{phot} }\times \frac{\sigma T_{star}^4}{\pi}\times (1-a) erg/s/cm^2$

$= 4 \times intens\times \frac{\sigma T_{star}^4}{N_{phot} }\times (1-a) erg/s/cm^2$

This energy will get stored in the variable $E T$ and will give total energy absorbed in erg/s/cm^2. This is distributed in different grainsizes. Each cm^3 has $n_H\times \int dN$ grains. A grains of radius a will get the energy

$E_T(a)=E_T\times\frac{dN(a)}{\int dN}\times n_H erg/s/cm^2$

A single grain of radius a then gets the energy of the amount

$E_T(a)=E_T\times\frac{1}{\int dN}\times n_H erg/s/cm^2$

In our dust model,

$d N = d N g r a + d N s i l$

This needs to be equated to the RHS of energy balance equation to get temperature:

$E_T\times\frac{1}{\int dN}erg/s/cm^2=4<Q(a,T_{gr})>\sigma T_{gr}^4$

[edit] Volume distribution of Photon hits in Simulation

The energy $E T$ is assumed to be absorbed by a single $c m 3$ of dust grains. However, in the simulation, the $E T$ is stored in same location for $1 p c 3$ binsize. In the simulation, the energy could have been absorbed in any of the various $1 c m 3$ boxes inside this $1 p c 3$ bin. The Average energy in a $1 c m 3$ box is found by noting the Hits of photons in the 1pc³ bin, and dividing the total $E T$ by this number of hits, before using the energy balance equation.

[edit] Calculation of IR Brightness per Hydrogen Column density

Following Draine and Lee (1984),(henceforth DL) we assume graine sizes and temperatures as given in following table. MRN distribution is assumed: $dN_i = A_i \times n_H \times a^{-3.5} da$

$A g r a = 10 - 25.16 c m 2.5 / H$ and $A s i l = 10 - 25.11 c m 2.5 / H$

To get number density per Hydrogen, we take

$dn_i = A_i \times a^{-3.5} da$

Description of Grains used in Model
Grainsize	Graphite	Silicate	da	dn_gra	dn_sil	Relative
microns	K	K	microns	per n_H	per n_H	Ratio %
0.01	20.0	17.9	0.01	6.91831e-11	7.76246e-11	86.7538
0.02	19.7	17.3	0.01	6.11498e-12	6.86111e-12	7.66803
0.03	19.6	17.0	0.02	2.95873e-12	3.31975e-12	3.71017
0.05	19.3	16.5	0.05	1.23759e-12	1.38859e-12	1.55190
0.1	18.8	15.4	0.1	2.18776e-13	2.45471e-13	0.274340
0.2	17.3	15.4	0.1	1.93373e-14	2.16967e-14	0.0242484
0.3	16.2	14.7	0.2	9.35633e-15	1.04980e-14	0.0117326
0.5	14.8	14.4	0.5	3.91359e-15	4.39111e-15	4.90754e-03
1.0	12.7	13.4	1	6.91831e-16	7.76246e-16	8.67538e-05

As we can see, most of the emission will come from 0.01 micron grains, as they have larger temperature and number density. These sizes were chosen as the graphs for absorption efficiencies of these grain sizes are given in DL:

Emission efficiency is taken as equal to absorption efficiency.

A blackbody at temperature T will emit the Planck Spectrum:

$B(\lambda,T)=\frac{2hc^2}{\lambda^5}\frac{1}{exp(\frac{hc}{\lambda kT})-1} erg/s/cm^2/A/Sr$

The emission cross section is $Q a b s π a 2$ , and the effective emission per steredian for our grains will be $B(\lambda , T)\times Q_{abs}(\lambda)\times \pi a^2, erg/s/A/Sr$ . We need to sum this over all grain sizes for both graphite and silicate,including their per Hydrogen number density, and integrate it over the wavelength. The part of this emission that will actually get observed by IRAS is the fraction of this multiplied by IRAS spectral response, so the integration with respect to wavelength must contain $f I R A S (λ)$ .

So we get,

$I_{em}=\Sigma\int B(\lambda,T_{gra})Q_{gra}(\lambda,a_i)\pi a_i^2\times dn_{gra}(a_i)\times f_{IRAS}(\lambda)d \lambda$ $+ \Sigma\int B(\lambda,T_{sil})Q_{sil}(\lambda,a_i)\pi a_i^2\times dn_{sil}(a_i)\times f_{IRAS}(\lambda)d \lambda , erg/s/Sr \times n_H$

IRAS Bandwidth is given by $\Delta\lambda=\int f_{IRAS} d\lambda , \mu m$ . To get the bandwidth per Hertz, we convert this to the frequency range , $\Delta\nu=\frac{c}{\lambda^2}\Delta\lambda$ . With $λ$ in $μ m$ and $c = 3 * 10 14 μ m / s$ ,this will give $Δν$ in Hz.