# Can my TV set pick up the CMBR?

1. May 28, 2005

### Mariko

Nice Topic and info Nick! I heard that when you turn your T.V. on and you get that snow or in otherwords no reception that, the T.V. is picking up is acctualy some sort of energy from the Big Bang.. what do you think?

2. May 28, 2005

### Nereid

Staff Emeritus
May I add my support and congrats too - this is excellent!
Would you like to do a little digging on your own Mariko?

Do you know, when a TV set is tuned to station x, what the frequency is of the radio waves it's picking up? You've probably heard of VHF and UHF - do you what they are?

You've no doubt read that the 'cosmic microwave background radiation' (CMBR) pervades the universe, and was 'created' when the universe was only a few hundred thousand years young. You may also have read that it has a 'temperature' of 2.7K. The 'microwave' part gives you a clue as to the frequency , and the 'temperature' to the distribution of the frequency (it's a 'black body').

Putting two and two together ...

3. May 28, 2005

### Mariko

Yes I would!

The radio wave frequency of station X = 8-12 GHz

VHF and UHF are one of the bands that make up the electromagnetic spectrum. Here are the diffrent bands in order starting with the highest frequency:
Gamma rays, X-rays, ultraviolet, Visible, infrared, microwaves, UHF VHF,Short wave radio and long wave radio

4. Jun 2, 2005

### Nereid

Staff Emeritus
5. Jun 2, 2005

### Nereid

Staff Emeritus
Here is a quick summary of the names we humans have agreed to use for various parts of the radio/microwave spectrum, and http://map.gsfc.nasa.gov/DIMES/dimes_spectrum.html [Broken] is a nice summary of the cosmic microwave background spectrum, including its intensity and where various observations have been made (note that, confusingly, the axes go in opposite directions).

Last edited by a moderator: May 2, 2017
6. Jun 2, 2005

### Berislav

I remember attending a popular science lecture a few years back where the speaker said that a small part of static (i.e, "snow") is due to the CMBR.

7. Jun 2, 2005

### Nereid

Staff Emeritus
As it was a popsci lecture, the speaker may not have given any details ... do you think we could work this out for ourselves?

I've moved Mariko's idea forward a bit ... would you like to continue Berislav?

8. Jun 2, 2005

### Berislav

Okay. :)

It is notoriously difficult (if not impossible) to detect the CMBR on Earth. This is because it interferes with other microwave radiation. But that only makes it difficult to determine the exact origin of the signal, not to detect the signal itself, which interfered, of course.

Cosmic noise is a well known phenomena in signal noise research:
Cosmic Noise

As mentioned it the article most of it arises from radiation emmited by the sun and distant stars. This because the Sun's radiation has a longer wave-lenght than the CMBR:
http://en.wikipedia.org/wiki/Cmbr

The CMBR has the characteristics of black-body with temparature 2.725 K, while the Sun has the characteristics of a black-body with temparature 6000K (if I recall the temparature correctly).
So a small part of the noise one detects with their television will be from the CMBR.

9. Jun 2, 2005

### SpaceTiger

Staff Emeritus
Theoretically, yes, but the difficulty would be in figuring out all of the other sources of noise. Determining the level of the CMB in that frequency range is simply a matter of integrating the blackbody spectrum, but the important thing for answering the question is how it compares to the rest of the static.

10. Jun 2, 2005

### SpaceTiger

Staff Emeritus
Most of the sun's radiation is of a much shorter wavelength than the CMBR, but it's so close and bright that it contributes a sizable amount of noise even in the microwave regime.

The temperatures you quoted are right, but that's not enough to determine that the CMB should contribute a small amount relative to the sun (or other sources). You need also to consider proximity, surface area, and directionality (for example, the sun won't contribute at night).

11. Jun 2, 2005

### Berislav

Yes, of course.

One can see that by observing Wien's law and the temparatures of the two:
$$\lambda=\frac{C}{T}$$
where C is a constant.