# Specific Intensity

### Units

Here are some terms pertaining to telescope observations:

aperture area ($\Delta A$ ), solid angle on sky ($\Delta \Omega$ ), exposure time ($\Delta t$ ), collects energy ($\Delta E$ ), over waveband ($\Delta \nu$ ), but $\Delta \lambda \neq {c \over \Delta \nu }$ .

${\Delta E \over \Delta t\Delta \nu \Delta A\Delta \Omega }\iff {dE \over dtd\nu dAd\Omega }\equiv I_{\nu }\,\!$ $I_{\nu }$ is the specific intensity per unit frequency.

Flux density is power per unit frequency passing through a differential area whose normal is ${\hat {n}}$ . Thus, flux density is:

${F_{\nu }\equiv \int I_{\nu }\cos \theta d\Omega }\,\!$ ### Proof that Specific Intensity is conserved along a ray

The power received by the telescope is:

$P_{rec}=I_{\nu }d\Omega dA\,\!$ where $I_{\nu }(\alpha ,\delta )$ is the intensity as a function of right-ascension ($\alpha$ ) and declination ($\delta$ ). Say that $\Sigma _{\nu }(\alpha ,\delta )$ is the surface luminosity of a patch of sky (that is, the emitted intensity). Then power emitted by patch of sky is:

$P_{emit}=\Sigma _{\nu }{dA \over r^{2}}d{\tilde {A}}\,\!$ Recognizing that $d{\tilde {A}}=d\Omega r^{2}$ :

$\Sigma _{\nu }=I_{\nu }\,\!$ This derivation assumes that we are in a vacuum and that the frequencies of photons are constant. If frequencies change, then though specific intensity $I_{\nu }$ is not conserved, ${I_{\nu } \over \nu ^{3}}$ is. Also, for redshift $z$ ,

$I_{\nu }\propto \nu ^{3}\propto {\frac {1}{(1+z)^{3}}}\,\!$ so intensity decreases with redshift. Finally:

${I_{\nu } \over \eta ^{2}}\,\!$ is conserved along a ray, where $\eta$ is the index of refraction.

### The Blackbody

A blackbody is the simplest source: it absorbs and reemits radiation with 100% efficiency. The frequency content of blackbody radiation is given by the Planck Function:

$B_{\nu }={h\nu \over \lambda ^{2}}{2 \over (e^{h\nu \over kT}-1)}\,\!$ ${B_{\nu }={2h\nu ^{3} \over c^{2}(e^{h\nu \over kT}-1)}\neq B_{\lambda }}\,\!$ (The Planck Function for Black Body Radiation)

Derivation:

The # density of photons having frequency between $\nu$ and $\nu +d\nu$ has to equal the # density of phase-space cells in that region, multiplied by the occupation # per cell. Thus:

$n_{\nu }d\nu ={4\pi \nu ^{2}d\nu \over c^{3}}{2 \over e^{h\nu \over kT}-1}\,\!$ However,

$h\nu {n_{\nu }c \over 4\pi }=I_{\nu }=B_{\nu }\,\!$ so we have it. In the limit that $h\nu \gg kT$ :

$B_{\nu }\approx {2h\nu ^{3} \over c^{2}}e^{-{h\nu \over kT}}\,\!$ Wein tail

If $h\nu \ll kT$ :

$B_{\nu }\approx {2kT \over \lambda ^{2}}\,\!$ Rayleigh-Jeans tail Note that this tail peaks at $\sim {3kT \over h}$ . Also,

$\nu I_{\nu }=\lambda I_{\lambda }\,\!$ ### Reiteration: Conservation of specific intensity

Conservation of specific intensity told us the intensity collected by your telescope is absolutely equal to the surface intensity emitted into the angle leftrightarrowonding to your pixel on the sky. Specific intensity is distance independent. $1m^{2}$ of sky emitted into one square degree specific intensity measured by a $1m^{2}$ telescope pointed at a one square degree pixel of sky.

### Units

We have flux, flux density, surface brightness, and spectral energy distribution to describe luminosity. When in doubt, look at the units. Surface brightness, like specific intensity, is distance independent. It is simply specific intensity integrated over frequency.

### Lenses

If you look at something through a lens, the specific intensity of a point is conserved. In the lens, the specific intensity may be greater, because $I_{\nu } \over \eta ^{2}$ is conserved (remember that $\eta$ is the index of refraction of the medium), but once the light goes out again, even if the object looks bigger, the specific intensity is the same. This does not violate conservation of energy because the light is being focused to a smaller area.

Now if something is inside the lens (say, a bowl of water), then the lens bends light rays closer to each other, increasing the specific intensity. Thus, the object is emitting/scattering higher density photons, and appears more luminous, and the object also appears larger. This means that there must be some places where the object, omitting the lens, would have been visible, but now isn’t.