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Effect of expansion of space on CMB

+ 1 like - 0 dislike
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Is it true that the expansion of space time cause the CMB to become microwaves from a shorter wavelength. If it is has the amplitude been increased?

Seeing as the amplitude has decreased; why hasn't it increased (/"stretched") in the same way the wavelength has?

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user Jonathan.
asked Mar 11, 2011 in Theoretical Physics by Jonathan. (10 points) [ no revision ]
I tried to answer the first part of your question, but I can't do a good job at answering the second. I don't know why you would have expected the amplitude to increase, so I can't give a very satisfactory answer to the question "why not".

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user Ted Bunn
I would have expected it to increase because it "stretches" similar to the wavelength, but I see now from the answers that the amplitude is to do with the energy (which I knew already but didn't completely make the connection)

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user Jonathan.
Hi Jonathan, imagine that the Universe is a surface of a ball - a sphere. A photon is moving along the equator and has $N$ maxima of the wave on it. Clearly, $N$ will be conserved during expansion. It follows that the wavelength will grow proportionally to the radius of the Universe, and because the photon's energy is inversely proportional to wavelength, the energy of the photon will go down inversely proportionally. You shouldn't imagine the "waves' amplitudes" to be geometric quantities similar to length because they (e.g. the electric field) don't even have the units of length.

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user Luboš Motl

3 Answers

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Just to be clear: by "amplitude" you mean the amplitude of a classical electromagnetic wave -- that is, the peak value of the electric field -- right? In that case, the answer is that the amplitude goes down.

For definiteness, let's consider a wave packet of electromagnetic radiation with some fairly well-defined wavelength. At some early time, it has a wavelength $\lambda_1$ and energy $U_1$. (I'm not calling it $E$ because I want to reserve that for the electric field.) After the Universe has expanded for a while, it has a longer wavelength $\lambda_2$ and a smaller energy $U_2$. (Fine print: wavelengths and energies are measured by a comoving observer -- that is, one who's at rest in the natural coordinates to use.) In fact, the ratios are both just the factor by which the Universe has expanded: $$ {\lambda_2\over\lambda_1}={U_1\over U_2}={a_2\over a_1}\equiv 1+z, $$ where $a$ is the "scale factor" of the Universe. $1+z$ is the standard notation for this ratio, where $z$ is the redshift.

The physical extent of the wave packet is also stretched by the same factor. So the energy density in the wave packet goes down by a factor $(1+z)^2$.

What does that mean about the amplitude of the wave? The energy density in the wave packet is proportional to the electric field amplitude squared. So if the energy density has gone down by $(1+z)^2$, the electric field amplitude must have gone down by $(1+z)$.

Specifically, if the Universe doubles in size, the wavelength of any given wave packet doubles, and the amplitude (peak value of ${\bf E}$) is cut in half.

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user Ted Bunn
answered Mar 11, 2011 by Ted Bunn (140 points) [ no revision ]
If E and frequency are halved simultanously, then Energy of the wave paket should be quartered, right?

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user Georg
Nope. Energy density in an EM wave goes like $E^2$ (independent of frequency). So total energy goes like $E^2L$, where $L$ is the physical extent of the wave packet. $E\propto (1+z)^{-1}$, $L\propto (1+z)$, so $E^2L\propto (1+z)^{-1}$ as desired.

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user Ted Bunn
+ 1 like - 0 dislike

Yes to the first part of your question. This phenomenon is known as cosmological redshift. Also due to the increase in the volume due to cosmlogical expansion the same amount of photons, as were present during the CMB, now have to occupy a much larger region. Consequently the average temperature ("amplitude" in your words) of such a gas drops from a high of $T \sim 150,000$ deg. Kelvin (corresponding to $T=E/k_B$ with $E$ being the ionization energy of hydrogen (13.6 ev) and $k_B$ is Boltzmann's constant $k_B = 8.6 \times 10^{-5}$ eV/Kelvin) to the currently observed $T'=2.7$ deg. Kelvin.

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user user346
answered Mar 11, 2011 by Deepak Vaid (1,975 points) [ no revision ]
If the wavelength of the radiation was increased then why has the amplitude of the waves decreased?

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user Jonathan.
I agree with every word of this answer except the all-important first one! Don't you mean "No"? The question asks if the amplitude increases, and the answer is that it decreases. (Also, I wouldn't have guessed that "amplitude" meant "temperature," but maybe you're right that that was the intent.)

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user Ted Bunn
@Ted good point. That is liable to create a big misunderstanding.

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user user346
@Jonathan I'm guessing that your reasoning (regarding the amplitude) is based upon some notion of energy conservation applied to the CMB photons. Keep in mind that this energy is not conserved. It is lost in overcoming the (effective) gravitational potential well the photon experiences on its journey from the time of reionization uptil now.

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user user346
Excellent point, Deepak.

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user Ted Bunn
Thanks @Ted ....

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user user346
+ 0 like - 0 dislike

It is easier to think in terms of a gas of photons. The photons (in thermal equilibrium) have their wavelength distributed according with the Bose-Einstein (BE) distribution $$n(p) = \frac{2}{\exp(pc/k_BT) - 1}.$$

From the null geodesics in a Friedmann metric we have that the photon momentum decreases with $a^{-1}$ (the wavelength is stretched with the scale factor $a$) and applying the Vlasov equation in a Friedmann metric we have that the temperature also decreases with $a^{-1}$.

Using the BE distribution and the results above one obtain that the total number of photons ($\propto T^3V$) in a volume $V$ is conserved, therefore, the density of photons decreases with $a^{-3}$. This shows that the energy density decreases with $a^{-4}$ ($a^{-1}$ from the stretch of the wavelength and $a^{-3}$ from the decrement of the number density). Other way to obtain that is to use the continuity equation for a gas of relativistic particles in Friedmann i.e, $$\dot{\rho} + 3\frac{\dot{a}}{a}(\rho+p) = 0,$$ and since $p=\rho/3$ we obtain $\rho \propto a^{-4}$.

Finally, since the energy density is proportional to the electric field squared $E^2$, then, the electric field decreases with $E \propto a^{-2}$.

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user Sandro Vitenti
answered Sep 18, 2012 by Sandro Vitenti (0 points) [ no revision ]

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