# Difference between revisions of "Specific Intensity"

### Related Topics

In order to motivate the useful properties of specific intensity, it’s helpful to list the variety of ways we have of describing the energy transfer coming from electromagnetic waves (or photons) striking a telescope. Below are a few that are commonly used:

### 1.1 Voltage

Voltage (with units of Volts, ${\displaystyle V}$) gives the most direct view of the shape of the electromagnetic waves that are striking a (usually radio) telescope. However, astronomical signals are usually noise, so the random fluctuations in voltage aren’t usually terribly illuminating, and you can’t average them directly. This makes voltage a poor unit for learning about the sky.

### 1.2 Power

Power (with units of ergs/s, or watts, or dBm), which is generally proportional to voltage squared, is a much more useful quantity. For example, it can be averaged over time. However, it encodes no information about what frequency interval (bandwidth) that the measurement was made over, and the power of a measurement is generally proportional to the bandwidth of the signals you let in.

The unit dBm may not be familiar to most astronomers. It refers to decibels relative to a milliwatt:

${\displaystyle P_{\rm {dBm}}=10\log _{10}\left({\frac {P}{1~{\rm {mW}}}}\right)\,\!}$

### 1.3 Power Density

Power density (with units of ergs/s/Hz, or dBm/Hz), divides out by the bandwidth ${\displaystyle B}$ that the measurement is made over. However, it contains no information about how large an area this signal was collected over.

### 1.4 Flux

Flux (with units of ergs/s/cm${\displaystyle ^{2}}$) divides power received by the area the signal was collected over, but it does not divide by the bandwidth.

### 1.5 Flux Density

Flux density (with units of ergs/(s cm${\displaystyle ^{2}}$ Hz), or Jy) combines power density and flux to get a measurement that divides out bandwidth and collecting area. Most astronomers can agree on the flux density of a source, but if the beam of your telescope is smaller than the source on the sky, you can get a different answer because you are not be getting all the photons from that source.

Radio astronomers use Janskies (which are units of flux density) commonly. A Jansky is defined as:

${\displaystyle 1~{\rm {Jy}}\equiv 1e-23{\frac {\rm {erg}}{{\rm {s}}\cdot {\rm {cm}}^{2}\cdot {\rm {Hz}}}}\,\!}$

### 1.6 Specific Intensity

Specific intensity (with units of ergs/(s cm${\displaystyle ^{2}}$ Hz sr), Jy/beam) divides flux density by the angular area of the measurement (or of the source), and is intrinsic to source. As - conserved along a ray

### 1.7 Brightness Temperature

Brightness temperature (with units of K) uses the Rayleigh-Jeans tail of a blackbody spectrum to define an equivalent temperature corresponding to a specific intensity. It is often used as a proxy for specific intensity. (See below and Black-Body Radiation.)

${\displaystyle I_{\nu }={\frac {2kT_{b}}{\lambda ^{2}}}\,\!}$

## 2 Specific Intensity, Specifically

Here are some terms pertaining to telescope observations:

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

${\displaystyle {\Delta E \over \Delta t\Delta \nu \Delta A\Delta \Omega }\iff {dE \over dtd\nu dAd\Omega }\equiv I_{\nu }\,\!}$

${\displaystyle I_{\nu }}$ is the specific intensity per unit frequency.

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

${\displaystyle {F_{\nu }\equiv \int I_{\nu }\cos \theta d\Omega }\,\!}$

### Proof that Specific Intensity is conserved along a ray

The power received by the telescope is:

${\displaystyle P_{rec}=I_{\nu }d\Omega dA\,\!}$

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

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

Recognizing that ${\displaystyle d{\tilde {A}}=d\Omega r^{2}}$:

${\displaystyle \Sigma _{\nu }=I_{\nu }\,\!}$

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

${\displaystyle I_{\nu }\propto \nu ^{3}\propto {\frac {1}{(1+z)^{3}}}\,\!}$

so intensity decreases with redshift. Finally:

${\displaystyle {I_{\nu } \over \eta ^{2}}\,\!}$

is conserved along a ray, where ${\displaystyle \eta }$ 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:

${\displaystyle B_{\nu }={h\nu \over \lambda ^{2}}{2 \over (e^{h\nu \over kT}-1)}\,\!}$
${\displaystyle {B_{\nu }={2h\nu ^{3} \over c^{2}(e^{h\nu \over kT}-1)}\neq B_{\lambda }}\,\!}$

(The Planck Function for Black Body Radiation)

Derivation:

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

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

However,

${\displaystyle h\nu {n_{\nu }c \over 4\pi }=I_{\nu }=B_{\nu }\,\!}$

so we have it. In the limit that ${\displaystyle h\nu \gg kT}$:

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

Wein tail

If ${\displaystyle h\nu \ll kT}$:

${\displaystyle B_{\nu }\approx {2kT \over \lambda ^{2}}\,\!}$

Rayleigh-Jeans tail Note that this tail peaks at ${\displaystyle \sim {3kT \over h}}$. Also,

${\displaystyle \nu I_{\nu }=\lambda I_{\lambda }\,\!}$

### 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. ${\displaystyle 1m^{2}}$ of sky emitted into one square degree specific intensity measured by a ${\displaystyle 1m^{2}}$ 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 ${\displaystyle I_{\nu } \over \eta ^{2}}$ is conserved (remember that ${\displaystyle \eta }$ 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.