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## Surface brightness, the Tolman effect and school-boy errors

Last week I blogged about the most distant galaxy so far found, a galaxy at a spectroscopically measured redshift of z=11.1. When I posted this news item on Twitter (and before I had blogged about it), I made the comment that the surface brightness of a galaxy is constant with distance. As was pointed out to me by numerous colleagues, this is only true for nearby galaxies; something I knew but had forgotten! (I don’t do research in distant galaxies, so that is my excuse for having forgotten this important detail 😉 )

So, let me explain why surface brightness is constant for nearby galaxies but not for more distant ones. First of all, however, I will explain what surface brightness actually is.

## The surface brightness of an extended object

When I started my PhD in Cardiff, the research group which I joined was probably best known for work on something called surface brightness selection effects. I won’t go into the details of what this means, but I will explain what the concept of surface brightness means in astronomy.

Apart from stars, pretty much everything else in astronomy is an extended object, which means that it covers an area on the sky. This is particularly true of nebulae, which include emission and reflection nebulae, but also galaxies (which used to be called ‘nebulae’ before it was realised that they were outside of the Milky Way, see my blog here for more about that).

So, for example, the Andromeda galaxy has an apparent visual (V-band) magnitude of 3.44, which should render it easily visible. But it is not, as this magnitude is spread over an area which is about 6 times larger than the Moon. It is the fact that the light is spread out over such a large area which renders the Andromeda galaxy so difficult to see to the naked eye.

How bright (luminous) an object appears to be decreases with distance; it goes as the square of the distance, so an object three times further away appears nine times less luminous. But, the luminosity per unit area, which we call the surface brightness, is constant with distance. Let me explain why.

As an object moves further away an area of e.g. 1 arc-second squared on the object will encompass more of the object. Let us suppose that some galaxy is initially at 10 Mega parsecs, and 1 square arc-second on the object encompasses an area where there are 10 million stars. The surface brightness we measure will be due to the light from these 10 million stars, as seen from our distance of 10 Mega parsecs away.

If we now move the same galaxy to twice the distance, 20 Mega parsecs, and we assume that the distribution of stars is uniform across the galaxy (a simplification which is not generally true), then the light from each star is reduced by a factor of four. But, now our 1 square arc-second encompasses a larger area of the galaxy. In fact, whereas before we said there were 10 million stars in our one square arc-second, when we move it to twice as far away there will now be 40 million stars in our 1 square arc-second, because the area on the galaxy goes up to four times as large as previously. So, even though the brightness of each star is reduced by a factor of four, we have four times as many stars in our 1 square arc-second, and so the two effects compensate and we have a constant surface brightness with distance.

As I have mentioned before (for example here), astronomers use the strange units of magnitudes to express the brightness/luminosity of objects. In these units, we can express the surface brightness $S$ of an extended object (in units of magnitude per square arc-second) as

$S = m + 2.5 log_{10} A \text{ (1) }$

where $m$ is the integrated magnitude of the object over an area of $A$ square arc-seconds.

## The Tolman effect

Richard Tolman was a mathematical physicist who worked at Caltech, and he is mentioned in my book The Cosmic Microwave Background (click here for more information) for being the first person to show that a blackbody spectrum retains its characteristic shape in an expanding Universe. He also devised a number of tests to determine whether the Universe was expanding or not. One of these (which he suggested in 1930) predicted that the surface brightness of a galaxy would not be constant with distance if the Universe were expanding. I will explain why below.

In 1934, in a book entitled Relativity, Thermodynamics, and Cosmology, Richard Tolman showed that a blackbody spectrum in an expanding Universe would cool due to the expansion of space, but crucially would retain its characteristic shape (this is a screen capture from my book The Cosmic Microwave Background)

Richard Tolman with Albert Einstein at Caltech in 1932

There are three effects which lead to the surface brightness not being constant with distance in an expanding Universe. These are

• that photons from a distant galaxy arrive at a different rate compared to in a static Universe
• the wavelength of each photon from a distant galaxy is different compared to in a static Universe
• the angular size of the galaxy when the light was emitted was larger, because the galaxy was closer, than its current angular size

Let me explain each of these in more detail.

## The rate of arrival of photons

Let us assume that, in a static Universe, a galaxy is emitting a certain number of photons every second in some astronomical passband, such as the B-band filter (see my blog here about astronomical filters). Our detector (a camera) detects, let us say, a million of these photons each second. Of course the galaxy is emitting a lot more photons than this, as we are only intercepting a tiny fraction of the emitted photons with our telescope, and no telescope or detector is 100% efficient. But, at our detector, we record 1 million photons each second through our filter in this static Universe.

Now, let us consider exactly the same galaxy but in an expanding Universe. Because of the expansion, each second which lapses means that the galaxy we are observing is slightly further away than it was the second before. The photons have a further distance to travel as time passes, due to the expansion of the Universe. As a consequence, the number of photons we detect is not going to stay constant, it is going to go down.

## The wavelength of each photon

Not only will the number of photons arriving each second go down due to the expansion, but the wavelength of each photon arriving will be longer, due to the cosmological redshift. As space expands each photon’s wavelength gets stretched, this is why we refer to the redshift of distant objects, their light gets shifted to the red end of the spectrum.

This also leads to the light from stars being shifted into a longer wavelength pass-band. This would not matter if the light from galaxies were spread equally across wavelengths, but it is not. So, for example, if a particular galaxy’s spectrum peaks in the blue part of the spectrum in its rest frame, this peak will shift to e.g. the V-band or R-band due to the stretching of the wavelength of each photon, so the light seen in the B-band will be reduced, and hence the surface brightness measured in the B-band will also be reduced.

## The luminosity distance $d_{L}$

The luminosity distance is, as the name implies, a distance determined from the object’s measured (apparent) luminosity. A distance $d_{L}$ can be calculated if its intrinsic luminosity is known. In magnitude units we can write

$m - M = 5 log d_{L} - 5$
where $m$ is the apparent magnitude, $M$ is the absolute magnitude, and $d_{L}$ is the luminosity distance. Re-arranging this, we can write that the luminosity distance $d_{L}$ is given by

$5 log d_{L} = m - M + 5 \rightarrow log d_{L} = \frac{m-M}{5} + 1 \rightarrow \; \boxed{ d_{L} =10^{(m-M)/5 -1} }$.

If you prefer to think in terms of luminosity and flux, we can write

$F = \frac{ L }{ 4 \pi d_{L}^{2} }$
where $F$ is the flux and $L$ is the luminosity. This would then re-arrange to
$d_{L} = \sqrt{ \frac{ L }{ 4 \pi F } }$

In a static, non-expanding Universe, the transverse comoving distance $d_{M}$ is the same as the luminosity distance, $d_{M}=d_{L}$. But, in an expanding Universe this is not the case. In an expanding Universe we can write

$\boxed{ d_{L} = (1 + z) d_{M} } \text{ (2) }$

## The angular diameter distance $d_{A}$

The angular size of an object is simply the angle that the object subtends on the sky. We can measure this. Let us suppose we have a galaxy which is 100 kilo parsecs (100 kpc) across (which is about the size of our Milky Way). If it is at a distance of 10 Mega parsecs then, in a static Universe, its angular size will be measured to be

$\theta \text{ (in radians)} = \frac{D_{A}}{d_{A}} = \frac{ 100 \times 10^{3} }{ 10 \times 10^{6} } = 1 \times 10^{-2} \text{ radians} = 0.57^{\circ}$

where $D_{A}$ is the actual diameter of the galaxy and $d_{A}$ is its distance. In a static Universe this is all very straightforward, but in an expanding Universe, the measured angular size of the galaxy $\theta$ changes. When the photons left the galaxy it was closer that it is when the light is received, so $\theta$ will be measured to be larger than the angular size the galaxy would currently subtend. As a consequence, $\theta$ is an overestimate of its size, this means that the inferred distance $d_{A}$ is too small. We can write that

$\boxed{ d_{A} = \frac{ d_{M} }{ (1 + z) } } \text{ (3) }$

where, again, $d_{M}$ is the transverse comoving distance.

## Doing the equations

We can write that the surface brightness, which in luminosity units we will write as $B$, is given by

$B = \frac{ F}{ d\omega }$
where $F$ is the flux of the object, and $d\omega$ is an element of the solid angle. But, this is the same as luminosity $L$ per unit area, so we can also write
$B = \frac{ L }{ d_{L}^{2} } \cdot \frac{ d_{A}^{2} }{ dl^{2} }$

In a static Universe $d_{M} = d_{L} = d_{A}$, but in an expanding Universe (irrespective of the assumed cosmology)
$d_{A} = \frac{ d_{M} }{ (1+z) } \text{ (Eq. 3 above) } \rightarrow d_{A}^{2} = \left( \frac{d_{M}}{ (1+z) } \right)^{2}$
and
$d_{L} = (1+z) d_{M} \text{ (Eq. 2 above) } \rightarrow d_{L}^{2} = (1+z)^{2} d_{M}^{2}$
This leads to the our being able to write that
$B = \frac{ L }{ dl^{2} } \cdot \frac{ d_{M}^{2} }{ (1+z)^{2} } \cdot \frac{ 1 }{ d_{M}^{2}(1+z)^{2} }$
$\boxed{ B = \frac{ L }{ dl^{2} } \frac{ 1 }{ (1+z)^{4} } }$
So, as we can see, the surface brightness depends on redshift in the sense that it is proportional to $(1+z)^{-4}$, a very strong dependence on redshift. For small redshift, $z << 1$, this reduces to what we stated at first, that surface brightness is independent of distance.