I've learnt that the universe is expanding, and that red-shifts light in the direction it's travelling, but since Intensity = Power / Area, and expansion would mean the particles of light in a square wavefront would surely be moving away from each other?

if so, What equation can be made to describe the Intensity purely in terms of expansion of the size of the square due to this effect w.r.t time [ignoring the effect of cosmological red-shifting in the direction of motion]?


This question is regarding a SQUARE WAVEFRONT moving through space, I only want to understand if the expansion of the universe drags out photons in the plane perpendicular to it's motion i.e. away from each other in the plane of the wavefront.


4 Answers 4


The answer is relatively simple, if you realize that in fact to explore distance in an expanding universe it is useful often to use the redshift, z. It is an unambiguous way to get the distance. z is the redshift due to the expansion of the universe (considering us as the observer, the universe expanding away from us, and slightly adjusting for the peculiar velocity of the earth, or Galaxy, and our cluster/supercluster relative to the average flow of the universe, the so called Hubble flow).

See a description of distances in cosmology, and z, at https://en.m.wikipedia.org/wiki/Distance_measures_(cosmology)

One of those distances described is basically what you are asking. You are asking how does a star's or galaxy's luminosity (the total energy per unit area per unit time from it photons) change in the expanding universe as its light moves away from it - it is like asking what does it's luminosity look like as function of distance from it. If there was no expansion it would decrease as 1/$d^2$, with d the distance. You can also use time in the equation because if no expansion, in a flat universe, d= ct, c the speed of light, and t the time since it was emitted.

But those equations change in an expanding universe. And one simple way of writing those are using the redshift z of where that light was emitted. If we use physical distance, D, (also called proper distance, and for us it is also the comoving distance), the apparent luminosity distance $D_l$ is related to it as

$D_l$ = (1+z)D

With z the redshift, which is 0 or higher.

That is the luminosity distance is bigger than the physical distance. Objects appear to be further than they are, because they are more faint. Those galaxies or stars appear fainter to us because their photons spread out more, traveled further, because the universe expanded under them.

An example is in the wiki article referenced above. See the figure, the lower one, also shown below. At a redshift z of about 1 the physical or comoving distance is about 10 Gly, 10 billion light years. But the distance we see, the luminosity distance, is about 20 billion light years. With the equation above,

$D_l$ = (1+z)D = (1+1)D = 2 times 10 billion light years = 20 billion light years. Just like the graphic ((hard to get exact numbers in the graphic).

So it look fainter, we think it's twice as far, because it looses luminosity to the universe expansion.

It is a significant, and not a trivial effect, and astronomers adjust for it when measuring distances and luminosities, and other physical entities. So far, they have been consistently valid.

It should be noted that some of the numbers in some of the other answers are wrong. At a z of approximately 1.5 the speed of expansion is already the speed of light, c. That's at about a proper/physical/comoving distance of about 14 billion light years. At the particle horizon at a distance of about 46-47 billion light years the universe is expanding at roughly 3c. We can't see anything further, that is the horizon of our observable universe. There is no good known reason it stops there, and a billion years from now it will have expanded.

See the figure for the red shift and distances here at https://en.m.wikipedia.org/wiki/Distance_measures_(cosmology)#/media/File%3ACosmoDistanceMeasures_z_to_1e4.png

Hope this helps get some thing un-confused.

  • $\begingroup$ It seems like most people either aren't reading my question or are ignoring it. I asked regarding a square wavefront, exclusively because I don't care about the inverse square law here, and I'm asking about expansion in the plane of the wavefront, bringing the points of the wave further out from each other in that plane. $\endgroup$
    – user95137
    Commented Feb 7, 2017 at 23:51
  • $\begingroup$ I am glad you edited and added what you wanted. The 'square' didn't seem a key part of the question. It still may be somewhat misleading. But if you want to know whether the expansion pulls in the plane of the square, sure, the expansion is radially out and so anything off its radial line to the plane will be pulled out and outwards to that on the plane. If no expansion you can have a plane wave solution in GR (infinite of course, like classically). But I'm not sure what you are really getting at. ----- continued $\endgroup$
    – Bob Bee
    Commented Feb 8, 2017 at 2:32
  • $\begingroup$ The expansion of null geodesic bundles can be calculated in any GR field, Sachs did a lot of the optics equations in GR, and the Newman Penrose parameters relate to it. You can get expansion, shear etc. But for the expanding universe with no perturbations you won't get shear, and you should get probably the intuitive picture (but I've not calculated it, and there may be something less obvious). This might be what you are looking for, but not certain. If not try to ask it differently $\endgroup$
    – Bob Bee
    Commented Feb 8, 2017 at 2:41

From my somewhat limited understanding of your premises, there seem to be a few things that need clearing up.

1) Red shifting is light apparently changing to a longer wavelength due to velocity or to expansion of space as seen by some observer. Some galaxies are actually apparently blue shifted to us because their relative velocities towards us are greater than the expansion of space.

Your question seems based on an incomplete understanding of what red shift is. I could be wrong, but that's what I'm getting from your statements.

2) It appears you are grasping towards describing the Cosmic Microwave Background. Is it the case that you are seeking an equation to calculate the energy density of the CMB? That would be something like: εrad = Enγ ∝ (1 + z)4

  • $\begingroup$ I was referring to Cosmological Redshift? $\endgroup$
    – user95137
    Commented Feb 6, 2017 at 8:43
  • $\begingroup$ If all points in the universe are expanding away from each other due to this Hubble expansion, then surely a square wavefront of a light wave [i.e. one that in the absence of expansion and loss of energy would be constant intensity w.r.t distance from source or time.] would actually expand outwards, in the plane of the wavefront, and therefore the intensity would decrease w.r.t the time it was travelling surely? $\endgroup$
    – user95137
    Commented Feb 6, 2017 at 8:45
  • $\begingroup$ Please explain why this isn't true $\endgroup$
    – user95137
    Commented Feb 6, 2017 at 8:47
  • $\begingroup$ @Phase It does decrease, so if you use luminosity to measure distance you will think it's further away because it is fainter. See my answer below. The expansion has a big effect. $\endgroup$
    – Bob Bee
    Commented Feb 7, 2017 at 6:33
Surface brightness in Euclidian space

Yes, as photons leave a light source, they also move away from each other. Since the photons are thus "spread out" on a larger area with distance, and since the area of a spherical surface is proportional to its radius squared, this makes the flux of photons from a light source fall off with distance $d$ from the light source as $1/d^2$. This is the "inverse-square law".

If you observe an extended source such as a uniformly bright screen, the surface brightness (SB) of some region of the source, spanned by some solid angle $\Omega$, is constant. This is because although the flux decreases, the actual area that $\Omega$ covers increases by the same factor.

Surface brightness in an expanding Universe

In an expanding Universe, this is more complicated, because of several factors.

As you say, the expansion of the Universe makes photons be carried away from each other at a higher rate. However, if we consider the motion of the wavefront in comoving coordinates — i.e. coordinates that expand along with the Universe — this effect disappears.

Still, in physical coordinates, the effect is of course there, in addition to other factors:


First, the photons redshift such that photons emitted at a time when the Universe was a size $a$ relative to now, today are observed redshifted by a factor $1+z=1/a$. For instance, photons emitted when the Universe was 1/4 the size of today are redshifted by a factor 4 (and are said to "lie at a redshift of $z=3$").

Angular diameter

Second, although more distant objects appear smaller, when the light we see today was emitted they were nearer and hence larger. In fact, their surface area increase by a factor $(1+z)^2$ wrt. the comoving coordinates. This has the peculiar effect of making nearby galaxies look smaller and smaller with distance (as expected), but after a certain distance (roughly 15 billion lightyears, corresponding to a redshift of $z\simeq1.5$) they actually appear larger and larger. This is another way to describe what your question is referring to: That photons are "carried farther apart by expansion". Note that what I just said is true only in a "flat" universe. In open or closed universes, the "angular diameter distance" — which is what describes this effect — is different. The relevant equations are found here.

Time dilation

Finally, the recession velocity causes a "normal", special relativistic time dilation. That means that the atoms (or whatever) emitting the photons do so at a rate that we observe to be slower by a factor $1+z$, and thus the observed flux is decreased further by the same factor. That is, time in a distant galaxy runs slower.

All in all, the SB of galaxies fall off with distance by a factor $(1+z)^{-4}$, making observations of distant galaxies excdeedingly difficult. For instance, a galaxy lying at $z=3$ has a SB which is $\sim250$ times smaller than local galaxies. And the current record holder for most distant galaxies, GN-z11 at $z=11.1$ has a SB 20,000 times smaller.


There is an intensity reduction in the light due to the expansion of the universe - it is not the only effect to consider though ;never-the-less just for this effect and therefore only in a classical treatment have a look at the figures...

The hubble constant is about $H_0$ = 67.15 ± 1.2 (km/s)/Mpc So if the light came frome 1million parsecs away the rate of expansion is 67km/s. (A parsec is about 3.26 light year.)

If we took light from Alpha Centuri at about 4(.37) light years, then by the time it arrives the distance travelled will be very roughly 1 parsec. Over this distance the expansion rate will be $0.07$m/s so after the 4 years flight time the universe will have expanded very roughly by 2000km

This causes a reduction in intensity which is inversely proportional to the square of the radius (from the surface area of a sphere), that is $$\bigg(\frac{4.37ly+2000km}{4.37ly}\bigg)^2 \approx 1 + (4\times 10^{-10})$$ With a light year being $9.4\times 10^{15}$m. You can see this is a small effect.

The effect should be larger for more distant objects, though. OK so let's imagine light from an object on the edge of the observable universe, 14Gpc away. Over this distance the rate of expansion is 67000(km/s)/Gpc for 14Gpc = $9.38\times 10^8$, about 3c. Light takes 45Gyears to get to us, during which time the expansion will be $1.4\times 10^{27}$m, which is about $1.4\times 10^{11}$ly. Thus the intensity reduction effect for distant objects is aprox

$$\bigg(\frac{45Gly+150Gly}{45Gly}\bigg)^2 \approx 16$$

  • $\begingroup$ Sorry @JMLCarter, your answer is wrong on the effect on luminosity and recession velocities. You're probably using the small z limit of the equations, need to use the GR equations,. See my answer below. At the particle horizon the expansion speed is about 3c. $\endgroup$
    – Bob Bee
    Commented Feb 7, 2017 at 6:30
  • $\begingroup$ I found an arithmetic error and have corrected it. $\endgroup$
    – JMLCarter
    Commented Feb 9, 2017 at 21:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.