It is customary in gravitational lensing problems, to project both the background source and the deflecting mass (e.g. a background quasar, and a foreground galaxy acting as a lens) in a plane.

Then, the lensing problem can be regarded as a mapping between the unlensed source plane, and the lensed image plane. In such transformations, the Jacobian evaluated at a point of the source plane, expresses how an infinitesimal area located around that point increases.

Lens mass and mass distribution, relative positions and distances involved give rise to different scenarios. The special case in which the distortions are too small to be resolved by telescopes, is called microlensing regime. Typically, a dark, unseen object like a floating planet, happens to cross transversally in front of a background star.

The image of the background star suffers amplification and distortions that are unresolved, but a change in brightness is detected, with a very typical light curve shape. The measured light curve of a microlensing event can be related to physical parameters of the problem, because the change in brightness of a lensed image can be modelled simply by dividing the area of the lensed image by that of the unlensed source image. If that can be done, it is because the mean surface flux of the image equals that of the source.

That is, gravitational lensing can make a tiny source appear bigger in the sky but in plain terms, every square inch of the image has the same brightness of every square inch of the source. Here comes my question, because that seems to me rather counter-intuitive and, when I try to find a rigorous justification to it, I find the same arcane sentence in each book, in each review, in each paper I have seen:

'Because of Liouville's theorem, gravitational lensing conserves surface brightness'

(... and therefore the magnification is found by dividing the subtended area of the image by that of the source). Every single author I have read, drops that sentence as if it were something very obvious, and quickly goes into other questions.

I have tried to trace-back the origin of the idea, by consulting the bibliography of every book or document in which that thing is stated. Interestingly, I have recognized sort of a fingerprint of obscure points like this one, a patter that is repeated in many of the documents, as if some authors didn't understand and merely copied from each other, developing and personalizing only the parts they understand in between.

I have rigorously developed and resolved every one of the dark points in that pattern, but this one remains unresolved. Is it perhaps something obvious? How is Liouville's theorem applied to photons along null geodesics? I will accept an appropiate link or paper reference as a good answer.

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    $\begingroup$ See also (related but definitely not a duplicate) physics.stackexchange.com/q/31534 - Qmechanic's answer to that question might be helpful to you. $\endgroup$
    – N. Virgo
    Mar 7, 2013 at 8:46
  • $\begingroup$ @Nathaniel, thanks (+1). It seem at first sight related. I'll have a closer look, maybe it puts me on the right track. $\endgroup$ Mar 7, 2013 at 17:07

3 Answers 3


This question is pretty old but I figured other people than OP might be interested in the answer. I had the exact same question, so I asked my advisor and he told me that the answer can be found in two places. First (and unsurprisingly), there's a full derivation in Misner, Thorne and Wheeler, particularly on sections 22.5 and 22.6. There's also an attempt at a more intuitive explanation in Schutz's Gravity from the Ground Up, a very interesting book which attempts to explain GR with only high school math. I'm not actually sure if the explanation given there makes sense, but it has helped me arrive at one that does.

If I had to summarize the whole thing, I'd say that the conservation of surface brightness (or specific intensity, or étendue, or...) is the result of an exchange between position space volume and momentum space volume, or between solid angle and area.

I won't reproduce all the math because it can be found in MTW, but the basic idea is as follows. The first important fact is that you can think of light propagation as a bunch of "classical" photons moving along null geodesics, and the number of photons is conserved.

This photon picture allows you to use kinetic theory, as developed in section 22.6 of MTW, where Liouville's theorem is derived: given a bunch of $N$ particles, the volume $\mathcal{V}$ in phase space occupied by them, given by the product $\mathcal{V} = \mathcal{V}_x \mathcal{V}_p$ of the volume in 3-space and in momentum space as measured by a locally inertial observer, is Lorentz invariant and constant along the world line. Therefore, the number density $\mathcal{N} = N/\mathcal{V}$ is constant as well. I won't go through the proof since it is much easier to just look at the pictures in MTW, but this is the key step which has all the counterintuitive consequences.

The last step is relating the number density to the specific intensity or surface brightness $I_\nu$, the amount of energy per unit time, area of detector, frequency, and solid angle of the photons' momenta. This is a pretty standard derivation, and it makes sense. The result is that $\mathcal{N} = h^{-4} I_\nu/\nu^3$.

The frequency can change in crazy ways along the geodesic, but as long as there's no cosmological redshift between the source and the observer the net change will be zero, so for our purposes $I_\nu$ is constant, which is the result we needed. A given piece of solid angle from the observer's perspective receives the same flux no matter if there's a lens in the way or not.

As you've noticed, this has fairly counterintuitive consequences. From conservation of energy you would (wrongly) expect a bigger image to be dimmer, because a given solid angle of observation covers less of the lensed image than of the unlensed image. But the image also appears closer, so it's also brighter. Or, to put it in more careful terms, given a point on the source, light from a larger solid angle will reach the observer than if there were no lens (hope this sentence makes sense!). Conservation of phase space volume guarantees that these two effects exactly cancel each other out.

If anyone is interested in thinking through this, I have some advice. You must be careful when drawing 2D pictures, because using solid angles is essential: it can happen that images get compressed in one direction and stretched in the other, so in a 2D diagram the image would look smaller while it's actually larger. Also, you have to consider light rays emanating in a given solid angle from a point on the source as well as light rays emanating from an area on the source which converge on a point at the observer with some solid angle. This is the tradeoff between solid angle and area I mentioned earlier.


What they're referring to is a property of light from geometrical optics. The conserved property is "entendue" (see wikipedia article), and the constancy of brightness can be demonstrated in a bunch of ways (Hamiltonian optics i.e. Liouville's theorem, second law of thermodynamics as above, etc.).

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    $\begingroup$ You are pointing in the very right direction (+1), the conservation of éntendue is the non-relativistic analogous to this, but what I am looking for is the derivation in the context of gravitational lensing with general relativity, geodesics and so on, where you cannot for instance happily say "energy is conserved across here and there...". I know this issue is waiting for me, I will have to stop sooner or later and spend the necessary time to try to understand and derive it by myself. But I would like to see the original derivation all authors are ignoring, because now I am short on time. $\endgroup$ Sep 3, 2013 at 23:00
  • $\begingroup$ Yeah, I figured you were looking for the real derivation. I wish I could help but my classical-mechanics-fu is not so strong. $\endgroup$
    – Nanite
    Sep 4, 2013 at 6:29
  • $\begingroup$ (I would guess that the original non-relativistic proof (if you can find it), if put into strictly hamiltonian terms, should work well enough because the hamiltonian mechanics can be used also to model relativistic motion, with suitable choice of coordinates.) $\endgroup$
    – Nanite
    Sep 4, 2013 at 6:41
  • $\begingroup$ By the way, ouch, the correct spell is "étendue". It is a frech word meaning sort of "extended" or "extensive". I also spelled it wrong in my comment above, which cannot be edited any more (unlike your answer, that can be edited easily) $\endgroup$ Sep 4, 2013 at 12:18

This is just the second law of thermodynamics.

Suppose you have some big source black body and some little target black body. You build a whole bunch of geometric optics to focus light from the source to the target (e.g. mirrors behind the target and lenses in front of it). Eventually an observer standing on the target looks up and sees light from the source at every point in the sky.

All this incoming light heats the target blackbody up until it's radiating light away as fast as it's coming in. That means the target comes up to the same temperature as the source. If you could increase the flux per unit solid angle, the target would get hotter than the source, breaking the second law.

Is this enough, or did you want a technical analysis of the lensing equation?

  • $\begingroup$ I am afraid it is not enough. (i) I don't grasp why the observer standing at the target cannot simply receive less light from, say, the mirrors standing behind, but more light from the front lens, thus having parts of the image brighter at the cost of having other parts of the image darker. (ii) In real problems, the observer receives only part of the flux, the rest goes to another directions. I don't see why a spatially extended image cannot be dimmer (that would be my guess). So the technical analysis of the lensing equation would be fantastic, if you feel like writing it... $\endgroup$ Mar 7, 2013 at 5:19
  • $\begingroup$ I had read too about the violation of the 2nd principle if you could concentrate light in too small a point with a magnifying glass, but I don't see how that translates to this question. That is why I want to find the mathematical details, that will surely give rise to the correct understanding. $\endgroup$ Mar 7, 2013 at 5:44
  • $\begingroup$ The second-law argument does work. For example if there is less light from the mirror behind the black body, just paint the back of it perfect white. Re: Liouville's theorem in geometric optics; I might have some time for that later on but I can't just do it in a couple minutes off the top of my head like this above argument. $\endgroup$ Mar 7, 2013 at 18:15

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