# Is there radiation colder than the CMBR?

1. Aug 6, 2014

### Lino

Hi, As I understand it, the CMBR presents with a temperature equivalent to approx 2.5k. If a WMAP-type experiment was run to mapa radiation signature at a colder temperature, say 2, or 1.5, or 1 degree, would it find anything?

Noel.

2. Aug 6, 2014

Staff Emeritus
I don't understand the question.

Let's go back to the basics. I measure the temperature of a pot of boiling water to be 100 degrees. I won't measure it to be any other number, right? So what is different about the CMBR?

3. Aug 6, 2014

### Lino

Vanadium, I'm not suggesting a different temperature for the CMBR, but wondering if there is "other radiation " with a lower temperature. The logic behind the question is simply - the CMBR represents the radiation from as far back as we can see (& it has been "stretched " as much as it can be), so am I right in thinking that nothing has been stretched more or can / does present with a colder temperature?

Regards,

Noel.

4. Aug 6, 2014

### marcus

they found a kind of lop-sided bell curve covering a range of photon energies. The temperature corresponds to the peak of the curve by a formula named after herr doktor Wien (pronounced "veen")

There were plenty of photons below the peak.
I basically am of same mind as Vanadium.

If they kept building and launching larger colder more sensitive antennas and receivers looking for lower energy longer wavelength photons, I expect they would keep on finding radiation but it would keep on fitting the SAME LOPSIDED BELL CURVE, that we already have, with the same Wien peak and bespeaking the same temperature.

That is just the way black body radiation is. Generic radiation from something at a given temperature, it is a kind of bellish mix. You can't "run a map at a lower temperature". You have to run a map at lower frequencies and see what temperature Nature gives you, what shape curve.

There could be a surprise though. The surprise would be if there were an unexpected little BUMP on the curve, which they discovered as they were exploring lower frequencies and longer wavelengths. That would certainly represent new physics and new cosmology. Nobody expects that. There is no reason AFAIK there should be anything in the Background of Ancient Light besides the redshifted glow of the hot gas circa year 380,000---a glow which fits the Planck black body curve with remarkable precision. It would be a shock t discover a second little peak in a range frequencies lower than have been explored yet.

5. Aug 6, 2014

### Lino

Thanks Marcus.

6. Aug 6, 2014

### Chronos

The CMB was quite cool compared say to a star - only around 3K kelvin. It is certainly conceivable cooler objects at lesser redshifts could be detected by radio telescopes. But, even at z~100 you would be talking an object with an intrinsic temperature of only a few hundred kelvin. It would be a pretty faint signal.

7. Aug 6, 2014

### Lino

Thanks Chronos.

8. Aug 6, 2014

### Lino

Marcus, Is this curve the "power spectrum" representation that is frequently shown?

Regards,

Noel.

9. Aug 6, 2014

### marcus

Yes, black body power spectrum. The curve can be be plotted either with frequency on the x-axis or with wavelength on the x-axis. They are reciprocals so naturally the shape is different. the curve looks the same shape for every temperature only it is skootched over to the right or left so the peak comes at a different place.

the study COBE discovered that the cmb has an amazingly accurate fit to the black body power spectrum. Everybody was astounded, at the meeting where they first showed the slide, the audience immediately rose and gave them a standing ovation. Now we take that fit for granted but in the 1990s it was astonishing.

Max Planck discovered the formula for that curve right around 1900.

95-some years later the satellite COBE (cosmic background explorer) found that the most ancient light in the universe has that exact shape

10. Aug 7, 2014

### Lino

Thanks again Marcus. Just so that I'm clear / can confirm; at very very small variations (in frequency /wavelength, which are representations of temperature variations) from the CMB there's also radiation, and the only quantity of radiation at these small variations is that needed to "complete" the power spectrum curve. Is that correct? (As I write that, my words sound very doubting, and that's not the intention, I fully accept and appreciate that the calculated curve represents a very detailed / specific prediction, which has been completely born out by observation. I'm just trying to understand the observation at the CMBR temperature versus the observation that would be made at 0.5k higher than the CMBR or 0.5k lower than the CMBR.)

11. Aug 7, 2014

### Drakkith

Staff Emeritus
I think you misunderstand what the power spectrum curve means. The CMB is not a single frequency, but a spectrum of frequencies. The power curve simply tells us how much power we receive at each frequency. See the graph below.

Note how the curve peaks around 100-200 GHz, yet we still see radiation at frequencies higher and lower than this. It is this specific shape of the curve that corresponds to an object at 2.73 kelvin. An object that is hotter or colder would have a different shape and would emit different amounts of radiation at each frequency, but would still emit all across the EM spectrum.

Play with the calculator here: http://lamp.tu-graz.ac.at/~hadley/ss1/emfield/blackbody.php

12. Aug 7, 2014

### phsopher

While the above replies are correct, I guess it's also possible that there is a "dark" radiation component which interacts very weakly with ordinary matter. If it decoupled early, at least before electron-positron annihilation, then it would have a lower temperature than the CMB. Sort of like neturinos would have been if they had been massless. Of course we wouldn't be able to measure it with conventional instruments.

13. Aug 7, 2014

### Drakkith

Staff Emeritus
Not as far as we know. If this is personal speculation, please be aware that PF rules don't allow it.

14. Aug 7, 2014

### Lino

Thanks Darkkith & phsopher.

Darkkith, If I use the calculator that you linked to and plot the graph for 2k, 2.3k, and 3k, I can see that (for example) the frequency of 2000μm appears for each, however, the power obtained for that frequency varies for each plot. So, is an experiment like CODE measuring the amount of radiation with a specific frequency and a specific "power" to obtain power spectrum?

And (assuming the last statement was correct), maybe I could ask a followup ... I thought that power was determined by frequency (i.e. that a specific frequency always has a specific power) - is it energy, not power, that is determined by frequency (and can you recommend any reading on the difference between energy and power)?

Noel.

15. Aug 7, 2014

### Drakkith

Staff Emeritus
You mean COBE? It measures the amount of power from each frequency. It doesn't look for a specific amount of power, it simply measures whatever is coming in.

I think you're referring to photons here. Each frequency will interact with matter with photons of a specific amount of energy. The higher the frequency, the more energy each photon has. Power is energy over time. For example, if we have 1 joule of energy being deposited into the detector every second, then we have 1 watt of power. Two joules of energy every second would be two watts of power.

16. Aug 7, 2014

### Lino

That all makes sense. Thanks again Drakkith.

17. Aug 7, 2014

### phsopher

"Not as far as we know" as in it's not possible or as in there does't exist a dark radiation component? I was not aware that the forum rules don't allow discussion of unproven things. I was under the impression that they merely prohibit things that challenge the mainstream. To my knowledge an additional relativistic component is not such and there are dozens of published papers exploring this idea. Some here. Furthermore, not too long ago there were hints, in the SPT and ACT data I believe, that $N_{\mathrm{eff}}$ was around 4 suggesting an additional relativistic component, though I believe the value has since come down and also Planck is consistent with 3 (admittedly I haven't followed the issue very closely). Even more furthermore, it was my impression that there only exist bounds on the differences between neutrino masses, in which case there is still the possibility that one neutrino species is massless or extremely light and thus would constitute a radiation component with lower temperature than the CMB.

However, if I'm mistaken and the discussion of these issues is not allowed then I would like to apologize and retract my comment.

18. Aug 8, 2014

### cristo

Staff Emeritus
The forum rules limit discussion to material which appears in published, mainstream articles. You're right that there have been papers considering a dark, relativistic species, so you are free to discuss that here.

19. Aug 8, 2014

### Chronos

Yes, the 'dark' sector has generated significant interest in mainstream literature. I haven't studied it much yet, there is still too much of the 'light' sector left to fathom.

20. Aug 8, 2014

### Drakkith

Staff Emeritus
Interesting. I'd never even heard of "dark radiation" before this thread, hence why I thought it was personal speculation at first.

21. Aug 8, 2014

### Lino

All, Maybe I could ask a basic question about power spectrums generally. I think that my problems come from a basic lack of understanding of power spectrums, and if you could recommend a "starter" book or article, that would be very much appreciated.

In the meantime ... if I may, I would have thought that if I set up an experiment to measure energy for a specific unit of time, and graphed the result (power on the y-axis and the frequency at which the power was measured on the x-axis) then I would have a power spectrum. But (where I think that I am going wrong and) based on my understanding of the previous comments, I assume that I need to "split" the power reading into various spectrums so that I have the right result for the spectrum that I am looking for (the example that I used previously was: plot the graph for 2k, 2.3k, and 3k, I can see that (for example) the frequency of 2000μm appears for each). Or is it that the measured result for 2000μm (and each of the other frequencies) is that value that is required for 2.3k black body curve and no splitting of the measured result is required. (I am hoping that it is the later, as this makes more sense at a basic level ... but even then I would not expect the measured result for 3500μm (for example) to give an un-contaminated "low" result, that only fits the 2.3k black body curve.)

Noel.

22. Aug 8, 2014

### Drakkith

Staff Emeritus
No, when you do the measurement you will only get one spectrum (a spectrum is a plot of light intensity or power as a function of frequency or wavelength). So you measure the incoming radiation, plot it, and you will come up with a spectrum. In the case of the CMB, the spectrum is practically identical to one that is emitted by a perfect black body at 2.7k.

23. Aug 9, 2014

### Lino

Thanks Drakkith, and there in lies my problem of understanding (and I appreciate that it is "my problem of understanding"). I would completely accept that if (using the calculator and graph provided earlier - http://lamp.tu-graz.ac.at/~hadley/ss1/emfield/blackbody.php and graph reproduced for easy reference below) for frequency of less than 1GHz, or greater than 999GHz, the power (or incoming radiation) was always less than 10^-22Iv, but for example, I know that the power at levels much greater than 99GHz is far in excess of this (i.e solar and nearby stellar radiation). I just assumed that there must be some (logical / mathematical) way of limiting the range or excluding certain aspects of the incoming radiation!

24. Aug 9, 2014

### Drakkith

Staff Emeritus
Okay, I think I understand. When I say that there's only one spectrum, I mean that when you measure the radiation from an object (and only that object) you get one spectrum. If I want to measure the spectrum of the Sun, I just point my telescope or antenna at the Sun so that only the radiation from the Sun is hitting the detectors. For the CMB this is a little more difficult, as we have background sources to contend with. Luckily, most of the radiation at the wavelengths associated with the CMB are from the CMB itself. Other stray sources can be filtered out before or after measurement using various techniques.

25. Aug 9, 2014

### ChrisVer

2 nice papers on the topic:
http://arxiv.org/pdf/1303.0049.pdf

http://arxiv.org/pdf/1111.5715.pdf
I guess the first is closer to the "mass" question.