# Wavelengths increasing with space?

1. Feb 9, 2010

### earlofwessex

please could somebody please explain how and why the wavelengths of photons increase with the expansion of space?

thanks.

i thought i understood the whole expansion acording to the hubble law and the scale factor thing, and the surface of last scattering, and the redshift, but this concept is messing my head up.

unless it is the redshift?

thanks V. Much

2. Feb 9, 2010

### nutgeb

Please read my post "The cosmological redshift". It describes how wavelengths and strings of photons can stretch without any need for space itself to stretch. However, this does not rule out the (rather suspect) possibility that space itself stretches photon wavelengths and intervals.

3. Feb 10, 2010

### Chalnoth

There are a large number of different ways of looking at this. One somewhat intuitive way would be to consider what happens if you have a gas of photons that is expanding.

That is, consider that we have a gas of photons in a box with sides of length a. Now, those photons have a pressure equal to one third their energy density. If we increase the size of the box to 2a on a side, then that pressure does work on each side. And since the photons do work, they lose energy. They have to do work in each direction, so they pick up a factor of three due to the three dimensions of space that cancels the one third in front of the pressure. This makes it so that an increase of the volume of the box by a factor of two decreases the total energy in the box by a factor of two: the photons have all doubled in wavelength.

There are many other ways of looking at this exact same phenomenon, this is but one.

4. Feb 10, 2010

### Chronos

I fail to see how expansion is a suspect explanation for redshift. So far as I can see, it is the only logical, observationally supported explanation.

5. Feb 10, 2010

### earlofwessex

thanks for the replies.

I've read the cosmological redshift post, but its going to take time for me to understand.

doesn't that imply that the photons are causeing the expansion though? if thats true does it still hold in the vacuum energy dominated expansion? (is the ratio of 1:1 still valid?)

-----

just for clarification, is this the same observed effect as the redshift?

6. Feb 10, 2010

### Chalnoth

No, not at all. The interaction between all forms of matter (including photons) and gravity is what causes the expansion to change with time. If you want to talk about what started it all off, well, that's a whole 'nother ball of wax. But the fact here is that when you have a universe dominated by photons, the expansion actually slows down more rapidly than if you just have a universe dominated by normal matter.

I'm not sure what ratio you're talking about.

7. Feb 10, 2010

### BillSaltLake

I think that the theory here is that the photons redshift as they climb "out" of the gravity well. As a result they lose energy. Massive particles can also lose kinetic energy but their mass remains constant. (All quantities are as measured by a local, comoving observer).

8. Feb 10, 2010

### Chalnoth

Well, that's true, but how does that apply to photons traversing an expanding space-time?

9. Feb 10, 2010

### mikeph

Photons climbing out of a gravity well is another phenomenon (or I guess the same phenomenon but due to a different form of metric). I'm also not sure what to make of post #2.

My understanding is that the expansion of space stretches everything, but photons are unique in that they have no internal binding energy and therefore cannot resist the expansion. An atom, for example, if expanded, will return to equilibrium position due to the internal forces, ie. it has a binding energy which means the "unstreched" state is energetically favourable to the "streched" state. Same applies for solar systems, galaxies, protons, etc. But not to photons.

10. Feb 10, 2010

### Chalnoth

It's not really quite that simple, though, because if you, for instance, take a closed box here on Earth and attempt to see if photons within the box stretch, you won't find any such stretching going on.

The basic reasoning here is that the stretching of photons as space expands is a gravitational phenomenon. But, when you look at a bound system like a galaxy, that galaxy is bound by the force of gravity, and is actually quite stable against expansion. The same gravity that causes photons traversing intergalactic space to stretch makes it so that photons traveling within a galaxy don't (except for going through local gravitational potentials).

11. Feb 11, 2010

### Chronos

12. Feb 11, 2010

### earlofwessex

wow, now i'm really confused!

i replied to this shortly after you posted, but it seems to have vanished. oh well. the ratio i was talking about was only valid if what i said was true. which it wasn't.

a thought experiment i would really appreciate some help with:

so, if i take an closed and bounded space (3D), filled with only one photon, then would it expand? why?

i think theres only two types of energy here, vacume energy and photon energy, so if it expands, energy conservation says the wavelength will increase, proportional to a-3 right?

now take two massive point particles, and place them in the expanding space. as they retract from each other, take one to be in an inertial frame.
then the other gains both potential and kinetic energy. do these cancel each other? does the vacume energy change? (assuming the space is large enough that the presence of the point particles do not effect the vacume significantly, and that the photon is far enough away as to have no significant effect)

13. Feb 11, 2010

### Ich

Inertial frames in cosmology - sorry, can't resist.
No, the other falls down a potential well, gaining kinetic energy. If you send a photon after him, it will get blueshifted in the potential, but the acceleration of the other wins, so there is a net redshift. If the photon reaches the other at the horizon, the redshift is z=1, just as it is for someone falling in a black hole.
No, it is static and independent of the reference frame.

This is just a different description of cosmological redshift. It does not explain why 1+z ~ a, but it provides a "mechanism", if that's what you're looking for.

If you're looking for a reason why 1+z ~ a, maybe a symmetry argument in cosmological coordinates is better. Just let me know.

14. Feb 11, 2010

### Chalnoth

Well, first we would have to know the initial conditions. Also, if there's some vacuum energy, that's going to effect things quite dramatically as well.

As before, it depends quite a lot upon the initial conditions. Basically, gravity effects how things change in time, so if you don't know how the system started out, and all you have is a law saying how it changes, then you can't say very much about what's going on.

15. Feb 11, 2010

### earlofwessex

thanks both of you
i've not come across z, so i can't really follow this. you seem to be saying that vacuum energy is just a different way of looking at redshift due to expansion. hit me with the symmetry argument, if you dont mind, it might help to clear things up. in the mean time i'll go and look up z.

is vacuum energy density not a constant? what other conditions need to be specified?
sorry if thats a stupid question

can i say,
take a space big enough to contain a horizon, and completely empty, which obeys all the laws that our universe does. the space is neither expanding nor contracting at t=0
does one of those laws dictate that it expands? or is that a meaningless question, as theres nothing there to move?

is it enough to say that they are moving away from each other faster than the escape velocity, and are of equal mass? actually, lets leave this alone untill I understand the single photon case.

its amazing how many wrong ideas i've got, and how they almost fit together perfectly.

16. Feb 12, 2010

### Chalnoth

Z is the variable used to describe the redshift:

$$\lambda' = (1 + z)\lambda$$

Here $$\lambda'$$ is the wavelength after the redshift is applied, with $$\lambda$$ being the wavelength when the photon was emitted.

Well, we don't yet know where the vacuum energy comes from, so we can't say. It's basically constant within our own visible universe, but this doesn't mean it isn't different elsewhere. If we're going to take a hypothetical scenario, there's no a priori reason why we should assume that the cosmological constant should be the same as what we measure in our observable region.

Anyway, that said, in order to get the behavior of a region of the universe, you have to specify the total energy density for each component of the universe (e.g. radiation, normal matter, dark matter, vacuum energy). Then you have to specify an expansion rate at a particular time (such as the current rate of expansion). Once you've done this, the universe follows the first Friedmann equation:

$$H^2(a) = \frac{8 \pi G}{3} \rho - \frac{k c^2}{a^2}$$

Here $$\rho$$ is the total energy density of the universe, including any vacuum energy, and $$k$$ is the spatial curvature. If we know the expansion rate at a given time, we can calculate the curvature. Basically, if you have a universe with low energy density and a fast expansion rate, then you have a large negative curvature (open universe) that will tend to expand forever. If, by contrast, you have a high energy density and a low expansion rate, then you have a large positive curvature (closed universe) that tends to recollapse back on itself under its own gravity. If you have a lot of vacuum energy (positive or negative), this can change things, but I think that this gives a nice intuitive picture of how it works.

Now, once you have this equation, you need to know how the various components of the universe scale in energy density with expansion. That is, the parameter that defines the expansion is $$a$$. If this parameter doubles (meaning things will be, on average, twice as far apart), how does the energy density change?

With normal matter, this is easy: increase the distance by a factor of $$a$$, and you increase the volume by a factor of $$a^3$$. The energy per bit of normal matter stays about the same, but now occupies a much larger volume, meaning the energy density is reduced by a factor of $$1/a^3$$.

Radiation is very similar: increase the scale factor, and you have the same number of photons in a larger region of space. So the number of photons also drops by $$1/a^3$$. However, the photons are also redshifted by the expansion, such that the energy per photon is reduced by a factor of $$1/a$$, and so the photon energy density scales as $$1/a^4$$.

Vacuum energy keeps the same density with time, so that remains unchanged.

Did this help?

Well, yeah, it would be pretty much meaningless if there's nothing there.

17. Feb 12, 2010

### Ich

Hmm,

right now I'm not entirely sure what exactly you want to know.
The increase in wavelength is redshift, and if you want a cause for it, it's mainly a doppler shift, but in a changing environment, one can't pin it down exactly.

If you want an argument to show that redshift is proportional to the scale factor a, let me try.

Our cosmology model simplifies things a bit. There's a (possibly infinite) number of "comoving" observers, say galaxies or clusters of galaxies. If any two such observers have now distance x, they will have distance a*x at some time in the future, where a is the scale factor at that time.
The model is based on the "cosmological principle". That says that for any such observer, the universe looks and works essentially the same, and that there's no preferred direction.
For example, if you send out a flash of light now, and it is at a distance y after a certain time, this works also for any other observer, e.g. the one who is now at a distance x from you. A flash of light that he sends now in the same direction is then also at a distance y from him.
So now the flashes are at position 0 and x, respectively (difference: x).
Then, they are at y and a*x+y, respectively (difference: a*x). You know, the other one is now a*x away, and the light y more.
Now, these flashes might be prepared exactly to mark two subsequent wave crests of some wave you send out, such that the wavelength now (the distance between the flashes) is x right now.
The flashes go at the same speed as the wave crests, so they will mark them also then. So the wavelength then is x*a.