Redshift

[Smigatron (Ben)]

Something has been bugging me all day.

My understanding of redshift is that as space expands between two objects, the wavelength of the light wave is stretched causing it to appear redder, because it is getting closer to the redder end of the visible spectrum.

Assuming that this expansion affects all wavelengths, I can understand that all visible light from a star/galaxy will appear redder by the time it gets to us. But doesn't this also mean that waves in the ultra-violet spectrum would also be redshifted into visible light?

And if this was the case, wouldn't that mean the object would not appear redder at all?

Please don't hesitate to embarrass me if I am missing something extremely simple here.

[KenGee]

Hi Ben,
No such thing as a silly question just silly answers.
When they talk about redshifting they are not saying the light will look red. What they are saying is that the emissions lines are shifted relative to their normal position.

[NereidT]

As KenGee said, there are no silly questions.

On this: "But doesn't this also mean that waves in the ultra-violet spectrum would also be redshifted into visible light?" you are 100% correct.

If we call our Sun 'yellow', then as it is redshifted more and more it will appear 'orange', then 'red', then a very dim red, as the peak of its spectrum (around 480 nm) moves further and further to the red. And as more and more of its ultraviolet spectrum comes into view (so to speak). As the Sun a very dim, comparatively speaking, in the ultraviolet, its colour will look more and more red.

There are stars - very hot ones - whose spectral peak is in the ultraviolet; we usually call these 'blue'. As they are redshifted, they simply get brighter; then, as the peak comes into the visual range, they start to look 'white', then 'yellow', etc, just like the Sun (but you need a much higher redshift for these to become 'red').

There are some AGN (active galactic nuclei) whose spectral peak lies so far into the ultraviolet that it's almost into the x-ray part of the spectrum. These objects are often seen as quasars, and sometimes at very high redshift (>~7). They look red, but not because their spectral peak has moved that far, but because all light with wavelengths shorter than 91.2 nm has been absorbed!

[Smigatron (Ben)]

Originally Posted by NereidT

As KenGee said, there are no silly questions.

On this: "But doesn't this also mean that waves in the ultra-violet spectrum would also be redshifted into visible light?" you are 100% correct.

If we call our Sun 'yellow', then as it is redshifted more and more it will appear 'orange', then 'red', then a very dim red, as the peak of its spectrum (around 480 nm) moves further and further to the red. And as more and more of its ultraviolet spectrum comes into view (so to speak). As the Sun a very dim, comparatively speaking, in the ultraviolet, its colour will look more and more red.

There are stars - very hot ones - whose spectral peak is in the ultraviolet; we usually call these 'blue'. As they are redshifted, they simply get brighter; then, as the peak comes into the visual range, they start to look 'white', then 'yellow', etc, just like the Sun (but you need a much higher redshift for these to become 'red').

There are some AGN (active galactic nuclei) whose spectral peak lies so far into the ultraviolet that it's almost into the x-ray part of the spectrum. These objects are often seen as quasars, and sometimes at very high redshift (>~7). They look red, but not because their spectral peak has moved that far, but because all light with wavelengths shorter than 91.2 nm has been absorbed!

Awesome answer, just what I was looking for.

I think the mistake I was making was assuming that the object was pumping out radiation in even amounts across the spectrum.

I didn't really know about the 'spectral peak' which makes the whole thing make more sense.

Thanks to both of you.

[NereidT]

I'm glad you found it helpful.

Have you heard of the black-body curve? The Planck law?

A perfect 'black body' (i.e. one which does not reflect any light at all) will give off electromagnetic radiation. If you plot the intensity of that radiation against wavelength (or frequency), you get a very distinct curve. Around the end of the 19th century the shape of this curve was pretty much known, from experiments. And both the long wavelength tail and the fact that the peak wavelength decreased as the temperature increased could be explained, by separate equations - the Rayleigh-Jeans Law and Wien's Displacement Law, respectively - but there seemed no way to get (derive) the actual curve from the then theory of electromagnetism (Maxwell's equations, in a nutshell). This was maddening, because that theory was so spectacularly successful otherwise.

Along came Max Planck, who introduced the concept of the photon, a quantum of electromagnetic energy; with this, he could produce a curve that matched the observational data very well. But he thought of the photon as a computational aid only, a sort of trick; in 1905 Einstein took the idea and ran with it, and showed that it explained the experimental data on the photoelectric effect, if you assume photons are real.

It turns out that lots of astronomical objects radiate with a spectrum close to that of a black body, the Sun is one such (HII regions and planetary nebulae are exceptions).

Here is a website which explains this in more detail.

[Smigatron (Ben)]

Haha you are the man. Very cool stuff and I appreciate the information. I'm going to have to read into it a lot more. I do remember reading a bit about Planck back in my high school days but there has been a lot of alcohol consumed since then.