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Last active June 9, 2020 20:07
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{
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{
"cell_type": "markdown",
"metadata": {},
"source": [
"A blackbody has the form:\n",
"\n",
"$$B_\\nu = \\frac{2 h \\nu^3}{c^2} (1-e^{h \\nu / k_B T})^{-1}$$\n",
"\n",
"with units $[B_\\nu] = \\mathrm{Jy}$\n",
"\n",
"The spectral flux density from a blackbody $F_{\\nu} = \\pi B_{\\nu}$ from integrating over a hemisphere (see eqn 2.107-2.110 from https://www.cv.nrao.edu/~sransom/web/Ch2.html)."
]
},
{
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"metadata": {},
"source": [
"The modified blackbody for dust is only valid in the Rayleigh-Jeans regime ($\\nu << \\nu_{peak}$, where $\\nu_{peak}$ is the frequency of the peak of the blackbody at a given temperature) and is given by:\n",
"\n",
"$$F_\\nu = \\pi \\frac{2 h \\nu^3}{c^2} (1-e^{h \\nu / k_B T})^{-1} (1-e^{-\\kappa_\\nu \\Sigma})$$\n",
"\n",
"where $\\kappa_\\nu = \\kappa_{\\nu_0} \\left(\\frac{\\nu}{\\nu_0}\\right)^{\\beta}$ is the dust opacity index with units $[\\kappa_\\nu] = \\frac{\\mathrm{cm}^2}{\\mathrm{g}}$. \n",
"\n",
"$\\Sigma$ is the surface density in $[\\Sigma] = \\frac{\\mathrm{g}}{\\mathrm{cm^2}}$. \n",
"\n",
"The optical depth $\\tau = \\kappa_nu \\Sigma$\n",
"\n",
"This approximates, in the $\\tau<<1$ regime, to:\n",
"$$F_\\nu = \\pi \\frac{2 h \\nu^3}{c^2} (1-e^{h \\nu / k_B T})^{-1} \\kappa_\\nu \\Sigma$$\n",
"\n",
"The mass of an object $M = \\Sigma A$, where $A$ is the object's area. We usually assume spheres (or, projected circles), such that $M = \\Sigma \\pi r^2$, where $r$ is the object's radius.\n",
"\n",
"$$F_\\nu = \\pi \\frac{2 h \\nu^3}{c^2} (1-e^{h \\nu / k_B T})^{-1} \\kappa_\\nu \\frac{M}{\\pi r^2}$$\n",
"\n",
"\n",
"\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"A blackbody of size $r$ at distance $d$ will produce an observed spectral flux density\n",
"\n",
"$$S_\\nu = B_\\nu \\frac{\\pi r^2}{d^2}$$\n",
"\n",
"\n",
"(section 1.4, eqn 1.6 of Tools of Radio Astronomy by Rohlfs & Wilson, 2009)\n",
"\n",
"Note that $\\Omega = \\frac{\\pi r^2}{d^2}$ is the angle a circular (or spherical) object subtends on the sky."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Putting these together, in the $\\tau << 1$ limit, the received flux from a blackbody is:\n",
"\n",
"$$S_\\nu = \\frac{2 h \\nu^3}{c^2} (1-e^{h \\nu / k_B T})^{-1} \\kappa_\\nu \\frac{M}{\\pi r^2} \\frac{\\pi r^2}{ d^2}$$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"The source areas cancel:\n",
"\n",
"$$S_\\nu = \\frac{2 h \\nu^3}{c^2} (1-e^{h \\nu / k_B T})^{-1} \\kappa_\\nu M \\frac{1}{ d^2}$$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Some rearrangements yield:\n",
"\n",
"$$S_\\nu = \\frac{2 h \\nu^3}{c^2} (1-e^{h \\nu / k_B T})^{-1} \\frac{\\kappa_\\nu M}{d^2}$$\n",
"\n",
"$$S_\\nu = B_{\\nu} \\frac{\\kappa_\\nu M}{d^2}$$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Solving for $M$ yields:\n",
"$$M = \\frac{S_{\\nu} d^2}{\\kappa_\\nu B_\\nu}$$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"We can replace $B_\\nu$ with the Rayleigh-Jeans version\n",
"$$B_\\nu = \\frac{2\\nu^2 k_B T}{c^2}$$\n",
"to obtain\n",
"$$M = \\frac{S_\\nu d^2 c^2}{2\\kappa_\\nu \\nu^2 k_B T}$$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"and\n",
"$$S_\\nu = \\frac{2 M \\kappa_\\nu \\nu^2 k_B T}{d^2 c^2}$$"
]
},
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