How do you find the temperature of the boomerang nebula?

[melodyman888]

So i have a physics presentation to do on tuesday and i have no idea how to find the temperature of a of objects in space, specifically the boomeran nebula. I did some research and found stuff involving black body radiation, wiens law, and stefan Boltzmann law but i don't really understand it too well.

Some of the things i don't understand:

wiens law states that wavelength(in meters) is equal to 0.0029/temperature(K). so is finding the temperature of the boomerang nebula really as easy as simply searching for the wavelength? is there a mathematical way to find the wavelength?

how can something so cold be so bright?

does the black body radiation curve have anything to do with this?

If none of this is the right approach to finding the temperature, how did they do it?

Any help will be greatly appreciated

thanks, melodyman888

 

[davenn]

hi there

welcome to PF

never heard of the boomerang nebula, what is its NGC or IC number ?

how can something so cold be so bright?

it may be a reflection nebula of which a large majority are

BTW DONT post the same question in multiple forum sections
it leads to lots of confusion!


cheers
Dave

 

[Simon Bridge]

Basically the color of a hot object depends on it's temperature... so, yes, it is a matter of measuring the wavelengths of the light the gas gives off. The overall spectrum is usually needed though.

The brightness of the light depends on how big the object is, and how far away it is, as well as it's temperature. You'll find the nebula is exremely big.

 

[Drakkith]

Note that the Boomerang Nebula is visible because it reflects and scatters light from the central star, not because it is glowing.

 

[Drakkith]

As I said in your other thread, the nebula is a reflection nebula, so it's visible because of the light it reflects scatters, not because it emits light. I'm afraid I don't know how they found the temperature though.

 

[glappkaeft]

This paper describes how they did it.

 

[Nanosuit]

Luminosity, which is a measure of the total output power of a star(unit: Watts) can be used to find the temperature of the surrounding stars. The luminosity of a star is proportional to its radius and temperature raised to the power 4.That is, L=σ x 4πr²x T⁴

Wien's law, on the other hand is a direct measure of the surface temperature of the star from the peak wavelength λ_mof the radiation spectrum emitted by the star.That is,
λ_m x T = 2.90x10^-3 m K

 

[melodyman888]

Ok this helps guys, still not completely sure how they did it but i will keep trying to figure it out

Thanks, melodyman888

 

[Simon Bridge]

Did you want to know how it was actually done, historically, or the methods by which it is done today?

You have been told two things above:
1. the principles by which is possible to discover the temperature of distant objects;
2. The exact method that was used to discover the temperature of this nebula (the link from post #6).

The trouble with the second is that the description assumes you already have a detailed knowledge of the first.
The relationship between the color an object glows and it's temperature is not a simple one and a lot of work had to go into figuring these things out first.

I think, though, that the descriptions of (1) missed something ... you can also tell the temperature of something that does not glow by how dark it is ... and, specifically, in what way it is dark.

When light from a star shines through a nebula - different frequencies get blocked by different amounts. The amounts depend on what the nebula is made of and the nebulas temperature.

If we know what the light is supposed to look like with nothing in the way, then we can figure out the temperature and composition of the bits of the nebula in the way.

In this specific case:
http://www.sci-news.com/astronomy/science-boomerang-nebula-01493.html
... astronomers used the radiation left over from the "big bang" to do the comparison.
This radiation is much the same from all directions so we know what it is supposed to be.
But where it passes through a cold nebula to get here, it is different.