# Why is intensity not an observable quantity?

I found that in astronomy we usually use flux $$F$$ instead of intensity $$I$$ in observations. My teacher tells me it is because intensity is not a observable quantity. So why can we only measure flux, not intensity?

$$F = \frac{\mathrm{d}E}{\mathrm{d}A \mathrm{d}t}$$

$$I = \frac{\mathrm{d}E}{\mathrm{d}A \mathrm{d}t \mathrm{d}\Omega}$$

• astro.vaporia.com/start/intensity.html HI :) there is one interesting line here: "Astronomically, it can only be measured for extended sources (e.g., for galaxies), often termed brightness. This particular measure has the property that it does not decline with distance, defined in such a way that it remains unaffected by the spread of EMR." Could your teacher clarify: are we talking about the intensity of radiation in general from all observable objects?
– dtn
Jan 8 at 5:04
• Re "except for": Do you mean "instead of"? Jan 8 at 15:30

You note the definition of both, flux and intensity in your question: The difference is that intensity is flux integrated over all angles or put differently: flux is the amount of energy which is sent in an angle such that it arrives at the detector (but not taking into account that most energy goes elsewhere but not into your detector).

Usually it's a quite good approximation in astronomy to treat sources as point sources, and more important: the angle the radiation is received from the source is hard to determine, not least due to the uncertainty of the distance - but you can still measure the flux regardless of whether you know the distance and or the angle of the source you receive your radiation from.

• I was wondering, what if there are many such point sources and they are all located at a short distance from each other? In this case, we need to be able to separate one flow from another, and prevent their “mixing and mutual influence", right?
– dtn
Jan 8 at 12:24
• @dtn Happens all the time. One way to handle it is to assume no contamination, estimate the flux of each individual image, and then estimate the cross contamination from that. Correct your flux estimates by the contamination yielding a better set of flux estimates. Repeat until the process converges. You can use more sophisticated math in some cases, and real data has other sorts of contamination, but this is basically the approach. Jan 8 at 15:08

I like to make an addition because radiometric quantities are confusing. I also experienced that different communities are using different terminologies although there are precise definitions (see this table on Wikipedia).

It's always a good idea to check units. Energy per time $$d E / d t$$ is just power $$P$$ in Watts (W), which in Radiometry is also called flux and often denoted as $$\Phi$$. So the first quantity $$F$$ is in units $$W/m^2$$ and this is called irradiance. It's an observable in the sense that photodetectors or powermeters provide this value. However, often photodetectors just give a value in W, which is wrong. Every detector has an area (e.g. a photodiode) and measures the power falling on that area, so in case your powermeter gives $$W$$, you need to be aware that it is actually $$W/m^2$$ (you can measure the photodetector's area often yourself). However, good instruments directly report $$W/m^2$$. Another very common source of confusion: Irradiance is often called intensity. That's pretty common, I have to admit that I do it myself although hardly working on it, so it's ok because everyone knows what you mean, but strictly speaking it's wrong (intensity is W/s, see again here).

The second quantity $$I$$ is power per area per solid angle, in units $$W/m^2/sr$$, which is called radiance. For example the radiance of a laser is its emitted power divided by the emitting area (small in case of the laser) divided by the solid angle in which the laser emits (also small), meaning its radiance is large (division by two small numbers). In case of the sun's radiance observed from Earth it can be calculated from the total sun's power, which is $$P = 3.846 \cdot 10^{26} \ W$$. Half of the sun we don't see, so we go with $$P/2$$. For the radiance, $$P$$ (resp. $$P/2$$) has to be divided by the area perpendicular to the vector from sun to Earth (also called projected area). This is a disk with radius $$R_{sun} = 6.96340 \cdot 10^8 \ m$$, so $$A_{proj, sun} = \pi \cdot R_{sun}^2$$. Light from this (projected) area is emitted in the entire half-sphere, so the solid angle is $$\Omega = 2\pi$$. The radiance of the sun (often called $$L$$ instead of $$I$$) is then $$L = \frac{P/2}{A_{proj, sun} \cdot 2 \cdot \pi} = 20091245.3 \ W/m^2/sr$$. (To increase confusion, we could have used the real surface area of the sun as well, which is not a disk but a half-sphere (it's a sphere, but the side facing us is a half-sphere); however, the solid angle is then only $$\pi$$ as the projected angle needs to be used, which was driving me crazy some time ago).

The first cool thing about the radiance is that it is the same whether observed from the emitter (sun) or the receiver (Earth). So on Earth we receive the same radiance $$L$$ that the sun emits, which (in German) is called "Photometrisches Grundgesetz" (found nothing corresponding on the English wiki page, but equations and images in that link are self explaining). From the Earth, the sun is (as mentioned above) a disk of area $$A_{proj, sun}$$ and the solid angle subtended by this area is $$\Omega = A_{proj, sun} / r_{AU}^2 = 6.8 \cdot 10^{-5} \ sr$$ where $$r_{AU}$$ is one astronomical unit ($$149.6 \cdot 10^{9} \ m$$). The irradiance in $$W/m^2$$ measured by a powermeter on top of the atmosphere can be derived from that by $$F = L \cdot \Omega \approx 1367 \ W/m^2$$, which is indeed the solar constant!

The second cool thing about the radiance is that it is a preserved quantity, so it doesn't change with distance from the source. This is surprising and in contrast to the irradiance, which falls with $$1/r^2$$. (The reason why the radiance stays constant is that it is divided by the solid angle, which also falls with $$1/r^2$$, so maybe not that surprising...). For example, the distance between Mars and sun is approximately $$r_m = 1.524$$ AU. Calculating $$\Omega_m = A_{proj, sun} / r_m^2$$ and multiplying with $$L$$ gives $$F_m = L \cdot \Omega_m = 588.7 \ W/m^2$$, which is very close to the solar constant for Mars.

Anyways, I just wanted to make the point that clear definitions of radiometric quantities exist, but every community use their own terminologies for historical reasons. And I wanted to refer to these cool facts about the radiance. But as you see, you always need solid angles and calculate with projected areas, whereas the irradiance in $$W/m^2$$ is just provided by measurements performed with a powermeter. I think this is what your teacher meant.

People receive information about celestial bodies using electromagnetic radiation, making observations in the optical, radio, UV, X-ray and gamma- ranges of the electromagnetic spectrum. The concept of “radiation intensity” (sometimes called “surface intensity”) is used in the theory of macroscopic EMR transfer for a quantitative understanding of the processes occurring in astrophysical sources. Let's imagine a small observation platform $$dA$$ in space filled with radiation from different sources. The orientation of the observation platform is characterized by the normal vector $$\overrightarrow{n}$$ to its surface. The radiation intensity in a given direction is the power of light energy passing through an area of a unit cross-section, calculated through the elementary solid angle $$d\Omega$$, the frequency interval $$d\nu$$ or the elementary wavelength $$d\lambda$$.

It turns out that if the angle between the observation platform and the selected observation direction is $$\theta$$, then $$I_\nu=\frac{dE}{\cos(\theta) dA dt d\nu d\Omega}$$ or $$I_\lambda=\frac{dE}{\cos(\theta) dA dt d\lambda d\Omega}$$.

Using these formulas, one important property of radiation intensity can be revealed: it does not depend on the distance from the source to which the observation site is located. This is due to the fact that the increase in distance $$D$$ is compensated by a decrease in the solid angle at which the source is visible, according to the same law.

It is further assumed that the radiation from the source is isotropic (although, as far as I understand, for real sources this is a rather rough approximation). For a source with isotropic radiation, the radiation flux is $$F_{\nu,\lambda}=\pi I_{\nu,\lambda}$$.

It is the “pointness” of the source and the isotropy of its radiation that is used to demonstrate that astronomers (using telescopes or other instruments) can only record the radiation flux (and not the intensity).

Let's imagine a spherically symmetric source ("star") with a radius $$R$$ and a distance $$D$$. Due to isotropy, the star will be visible as a disk of uniform brightness. The directly measured radiation flux from this star is by definition equal to $$F_{detect}=I_{\nu,detect} d\Omega$$, where $$I_{\nu,detect}$$ is the radiation intensity at the detector point, $$d\Omega= \pi \frac{R^2}{D^2}$$ is the solid angle at which the source is visible. Taking into account the flux per unit surface of the source for isotropic intensity $$F_{\nu,isosource}=\pi I_{\nu,isosource }$$ and then neglecting absorption (i.e. $$I_{\nu,detect}=I_{\nu,isosource}$$), we find for the measured quantity $$F_{\nu,detect}=(\frac{R}{D})^2F_{\nu,isosource}$$.

For a point source, the factor in brackets is $$<<1$$ and is a priori unknown. The transition from a directly measured quantity $$F_{\nu,detect}$$ to intensity $$I_{\nu,detect}$$ would be possible only if the angular size $$\frac{R}{D}$$ of the source was known, i.e. if it is not perceived as a point.

Thus, as far as I understand, such assumptions (about isotropy, pointness, etc.) allow us to simultaneously demonstrate what you asked about. But I don’t know how things would be taking into account all sorts of effects (anisotropy, absorption).