• WANTED: Happy members who like to discuss audio and other topics related to our interest. Desire to learn and share knowledge of science required. There are many reviews of audio hardware and expert members to help answer your questions. Click here to have your audio equipment measured for free!

Tomorrow Dec 25th 2021, is a very big day! James Webb Scope is headed out.

Blumlein 88

Grand Contributor
Forum Donor
Joined
Feb 23, 2016
Messages
20,524
Likes
37,057
Yeah - you're not gonna get (infra) redshift with the speed of a car though. :cool:
Actually I'd think you have to on principle. It might be hard to measure, but there is no question the wavelengths have to be shifted longer by the speed of a car departing.

And there is always this:

1663956063473.png
 

MAB

Major Contributor
Joined
Nov 15, 2021
Messages
2,137
Likes
4,782
Location
Portland, OR, USA
Thanks :)

Whenever I see nice pictures of something on the universe I tend to think that someone has taken some liberties to translate the original image to the published one. This Neptune one looks more credible than the first ones to me. I don't mean the other are fake, but it's hard to believe the isn't a human touch on them.
There totally is human touch;). So, yeah the images are a bit subject to that. By design, JWST takes images that are primarily outside of our eye's frequency response so they gotta do something to the raw data so we can see what the mad-scientists have been up to... And lots of the images are stitched together, even composites from different instruments. Mapping the IR to human-detectable wavelengths is just math, applied to the raw data. And you get an astoundingly pretty picture out of it! While the pictures are great, it's the data that's exiting. I don't think there is anything more or less credible between images since the raw data and the spectra of all the bits and pieces of each image are what is important.
 

Doodski

Grand Contributor
Forum Donor
Joined
Dec 9, 2019
Messages
20,751
Likes
20,766
Location
Canada
There totally is human touch;). So, yeah the images are a bit subject to that. By design, JWST takes images that are primarily outside of our eye's frequency response so they gotta do something to the raw data so we can see what the mad-scientists have been up to... And lots of the images are stitched together, even composites from different instruments. Mapping the IR to human-detectable wavelengths is just math, applied to the raw data. And you get an astoundingly pretty picture out of it! While the pictures are great, it's the data that's exiting. I don't think there is anything more or less credible between images since the raw data and the spectra of all the bits and pieces of each image are what is important.
I imagine with what the commoner people are seeing now that budgets may increase and funding for this sort of stuff will increase. The people that I know are astonished.
 
  • Like
Reactions: MAB

antcollinet

Master Contributor
Joined
Sep 4, 2021
Messages
7,409
Likes
12,294
Location
UK/Cheshire
Actually I'd think you have to on principle. It might be hard to measure, but there is no question the wavelengths have to be shifted longer by the speed of a car departing.

And there is always this:

View attachment 232820
Sure - you get doppler shift. But for redshift you have to change the frequency enough to change visible spectrum to predominantly red. I mean, I've tried - but I've never been able to see a car moving away from me change its colour. :cool:
 
Last edited:

xaviescacs

Major Contributor
Forum Donor
Joined
Mar 23, 2021
Messages
1,494
Likes
1,971
Location
La Garriga, Barcelona
Sure - you get doppler shift. But for redshift you have to change the frequency enough to change visible spectrum to predominantly red. I mean, I've tried - but I've never been able to see a car moving away from me change its colour. :cool:
Redshift or blueshift means decrease or increase of frequency, that is, going towards the red (less energy) or going towards the blue (more energy). Is has nothing to do with being red or blue. A redshifted very energetic light emitter doesn't look red to us, just a bit more red :)

I guess what you meant by your first comment on this matter that at the moment I didn't get is that there are mainly two kinds of redshifts on the universe: Doppler and cosmological. The Doppler redshift (or blueshift sometimes) is due to the relative velocity of the body, like what the traffic radar does. This happens with close objects. Cosmological redshift, which is the most famous (Hubble constant etc etc) is due to the expansion of universe, a property of spacetime that is expanding and that make photons to loose energy as they travel. The farthest they are, the more energy they loose until they arrive at us, or the more redshifted we observe them.

Doppler effect on space needs a relativistic treatment indeed, which complicates a bit the formulas, but not too much.

Actually I'd think you have to on principle. It might be hard to measure, but there is no question the wavelengths have to be shifted longer by the speed of a car departing.

Now, with respect to Doppler effect here on earth, if the speed of the wave is much greater than the speed of the moving body, and the direction of the body is the same as the direction of the wave, the formula that relates the initial and the received frequency (f) or wavelength (l) is quite simple (I make use of c = f*l):

f' = (1 + v/c)*fº l' = (1 + v/c)^-1*lº = lº/(1 + v/c)

Where c is the speed of the wave (EM wave in this case, so ~ 3e8 m/s) and v is the negative of the relative velocity and hence the coefficient (1 + v/c) is greater than 1 if the moving body is approaching the emitter (v is positive) and less than 1 if it's moving away from it, which implies that in the first case the frequency increases and in the second case decreases, and the opposite with the wavelength. Which is what we know intuitively.

At a given velocity of the moving body, the ratio change of frequency or wavelength is the same, independently of the frequency itself, meaning that a microwave or visible light will be subjected to the same change, proportionally. If the relative velocity is 30 m/s, then 1 + v/c = 1 + 3e1/3e8 = 1 + 1e-7 = 1.0000001, which is is the same coeficient in both cases, so the change is tiny in both cases in relative terms. That explains why @tonycollinet doesn't appreciate a change in color when looking at a moving red car, no matter how hard he tries. However, with eyes sensible to microwaves, I don't think he could do any better either. :)

In absolute terms, just for curiosity, we can calculate what is called the doppler frequency (fd) or the difference in frequency:

fd = 2v*fº/(c - v) ~ 2v*fº/c

And if we use a microwave of lets say 10 GHz = 1e10 Hz and a visible light of 400 THz = 4e14 Hz, that is, one 4e4 greater than the other, the difference in frequency in both cases is:

fd_microwave = 2*30*1e10/3e8 = 6e11/3e8 = 2e3 = 2000 Hz

fd
_visible = 2*30*4e14/3e8 = 2.4e16/3e8 = 8e7 = 80000000 Hz

exactly, 4e4 times greater than the other xD

So in absolute terms of course the difference in frequency will be quite different.

Please correct me if I'm missing something or (specially) if not getting it at all.
 

fredoamigo

Addicted to Fun and Learning
Forum Donor
Joined
Mar 11, 2018
Messages
604
Likes
998
Location
South East France
OWI4TVC25RFUNPBPUDMD6ISZHA.jpg


The "Pillars of Creation" are located 6,500 light years from Earth, in our galaxy, the Milky Way. More precisely, they are located in the Eagle Nebula. They were made famous by the Hubble Space Telescope, which took a first picture of them in 1995, revisited in 2014(left photo). But thanks to its infrared capabilities, the James Webb telescope, launched into space less than a year ago, can pierce the opacity of the pillars, revealing many new stars in formation -- bright red balls.
 

pseudoid

Master Contributor
Forum Donor
Joined
Mar 23, 2021
Messages
5,122
Likes
3,420
Location
33.58 -117.88
Redshift or blueshift means decrease or increase of frequency, that is, going towards the red (less energy) or going towards the blue (more energy).
:oops: The following are not 'fighting words' (or questioning your knowledge) but are made because this statement created "red" flags inside my cranial.
I think that there maybe more energy (under the curve, rms) in the 'redshift' than the 'blueshift', at least from my vantage as I think of the 'red' part of the spectrum to be similar to a (say) 20Hz signal; whereas, the 'blue' part to be equivalent of (say) a 20kHz signal. Please educate me!
Thank you!
 

xaviescacs

Major Contributor
Forum Donor
Joined
Mar 23, 2021
Messages
1,494
Likes
1,971
Location
La Garriga, Barcelona
:oops: The following are not 'fighting words' (or questioning your knowledge) but are made because this statement created "red" flags inside my cranial.
I think that there maybe more energy (under the curve, rms) in the 'redshift' than the 'blueshift', at least from my vantage as I think of the 'red' part of the spectrum to be similar to a (say) 20Hz signal; whereas, the 'blue' part to be equivalent of (say) a 20kHz signal. Please educate me!
Thank you!
Energy of a photon is E = h*v where h is Planck constant and v is frequency. More frequency, more energy. We know what's the temperature of a star, or any other process involving photon emission, by its color. Photons in the red spectrum are emitted by quantum leaps less energetic that photons in the blue region, so we know whatever particle, usually electrons, are loosing energy, they are decaying from a more energetic state, so we know the electrons on that systems reach higher energetic states, meaning there is more energy available. Another good example is a flame. A blue flame is hotter than a red flame. From inner to outer, from blue to yellow and red.

Edit: After a second read, there is something worth clarifying. The energy level from which the electron decays is not what matters, but the difference from it to the state where it lands, namely the difference between the excited temporary state and the "home" state. When more energy is available, electrons coming from certain natural state (damn, what is the word for such states? I'm sure there is one but I've forgotten :rolleyes:) absorb more energy and reach more energetic states. That is, an hydrogen atom will produce blue light in a blue dwarf star and red light in a giant red star, because the former is warmer, more energy density, and the latter is cooler, less energy density.
 
Last edited:

symphara

Addicted to Fun and Learning
Joined
Jan 24, 2021
Messages
632
Likes
592
Energy of a photon is E = h*v where h is plank constant and v is frequency. More frequency, more energy. We know what's the temperature of a star, or any other process involving photon emission, by its color. Photons in the red spectrum are emitted by quantum leaps less energetic that photons in the blue region, so we know what ever particle, usually electrons, are loosing energy, they are decaying from a more energetic state, so we know the electrons on that systems reach higher energetic states, meaning there is more energy available. Another good example is a flame. A blue flame is hotter than a red flame. From inner to outer, from blue to yellow and red.
In fairness, Planck not plank. He deserves that.
 

pseudoid

Master Contributor
Forum Donor
Joined
Mar 23, 2021
Messages
5,122
Likes
3,420
Location
33.58 -117.88
202401_JWST-CosmicCliffs.jpg
202401_JWST-PillarsOfCreation.jpg


The U.S. Postal Service issued [two] Priority Mail Express stamps Jan. 22, 2024, highlighting [two images] from NASA’s James Webb Space Telescope.
Greg Breeding, an art director for the U.S. Postal Service, designed the stamps with images provided by NASA, ESA, CSA, and the Space Telescope Science Institute.

From <https://science.nasa.gov/missions/w...rvice-stamps-feature-iconic-nasa-webb-images/>

It's up to you to determine what those numerical designations mean, at lower left of each stamp.:facepalm:
 
Top Bottom