54 replies to this topic

[Freezout]

One question came in my mind during a debate about the use of electronic devices for astronomy, and the usual quarrel "pure" visual vs electronic astronomy.

Disclaimer: I do not want this topic to be a way to take any side or advocate one of these practices against the other, it's a pure physics question but that will still, I think, be interesting for our hobby. 

 

We know one of the usual arguments for visual use is "we receive the true photons from the stars".

Now, how much are these photons modified when they come to our eye? (we know that afterwards they are just transformed into electric impulse to our brain "forming" the image).

 

The wavelength/photon can lose energy, be absorbed, scattered, go through fluorescence, transmitted, bounce etc depending the environment/matter it goes through. 

A photon re-emitted by an electron that absorbed it cannot be considered as the original photon.

I have been reading that no particle of light emitted by the object is unmodified when it reaches our eye. It seems strange to me, as it would imply that no single photon makes it with pure transmission without having been absorbed/re-emitted. But I'm happy to be proven the opposite. I don't consider loss of energy as a real modification (the wavelength still comes directly from the original object).

 

As far as I know stars or nebula and all objects we observe emit some visual spectrum (excepted dark nebula). How much of the photons reaching our eye are those really emitted at the surface of the star or nebula we "see"?

[daveco2]

Everything we see naturally in daily life, that is, without emitters of their own - from road signs to mountains - result from reflected light.  The original photon from the Sun experienced many bounces and transmissions through various media in its travels.  

Of course, photons from very distant galaxies also experienced cosmological red shift that photons from stars in our galaxy did not.

[Phil Cowell]

It takes up to a million years for the gamma ray to work its way from the suns core to its “escape” during that time it has lost energy and its frequency altered to possibly the visual light range a human can see. 
The human eye is a poor sensor at night with over half its detectors never hitting their detection threshold. But we have moved on from that with better detectors.

Everything we see naturally in daily life, that is, without emitters of their own - from road signs to mountains - result from reflected light.  The original photon from the Sun experienced many bounces and transmissions through various media in its travels.  

Of course, photons from very distant galaxies also experienced cosmological red shift that photons from stars in our galaxy did not.

[skysurfer]

And what about the story that light 'takes millions of years' to travel from the stellar core to the surface of the star ? No star has a size of millions of light years. In the case of the Sun it is slightly more than two seconds (R=696000km == 2.3 light seconds) and not 'millions of years'.

Or does light travel vvvveeerrrrrrrrryyyyyyy slowly inside the body of a star ?

[Freezout]

Hi Phil, point taken but I am not asking about the rays from the core of the star, I am inquiring about the ones escaping, those giving its color to the star. For the sake of my question let’s consider the beginning of the birth photon like that moment, the one it escapes into space void.
The stars at that point emit in visual range also. We still see (even if poorly) stars in the sky.

[artliddle]

And what about the story that light 'takes millions of years' to travel from the stellar core to the surface of the star ? No star has a size of millions of light years. In the case of the Sun it is slightly more than two seconds (R=696000km == 2.3 light seconds) and not 'millions of years'.

Or does light travel vvvveeerrrrrrrrryyyyyyy slowly inside the body of a star ?

Yes, you could say that light travels "vvvveeerrrrrrrrryyyyyyy slowly inside the body of a star."

What is actually happening? A photon created at the center of the sun immediately bumps into another atom. This atom absorbs the energy and then reemits it. Although this process is nearly instantaneous, it does take some amount of time. And of course the newly emitted photon immediately bumps into another atom--and the process repeats. But even this does not explain why it takes so long to reach the sun's surface. The reemitted photon does not follow the path of the original photon. It may go off in any direction, including back towards the center. So the path taken by a photon originating at the center of the sun is extremely convoluted.
 

[Sebastian_Sajaroff]

The original photons that left stars were modified lots of times in the middle.
They’re altered everytime hit an interstellar gas molecule or dust particle.
The last kilometers through our atmosphere add a lot
of modifications, and the last few centimeters across our eyeballs even more.
So, no, no original photons hit our retinas.

[Phil Cowell]

The particle collides with other particles on the way out. 
There you go you learnt something. https://futurism.com...rney-center-sun
 

 

And what about the story that light 'takes millions of years' to travel from the stellar core to the surface of the star ? No star has a size of millions of light years. In the case of the Sun it is slightly more than two seconds (R=696000km == 2.3 light seconds) and not 'millions of years'.

Or does light travel vvvveeerrrrrrrrryyyyyyy slowly inside the body of a star ?

[Phil Cowell]

Do you want X-Ray’s? IR? UV? They all have already been modified. 
So you mean the modified particle that falls into the visual range.

 

Hi Phil, point taken but I am not asking about the rays from the core of the star, I am inquiring about the ones escaping, those giving its color to the star. For the sake of my question let’s consider the beginning of the birth photon like that moment, the one it escapes into space void.
The stars at that point emit in visual range also. We still see (even if poorly) stars in the sky.

[AstroVPK]

One question came in my mind during a debate about the use of electronic devices for astronomy, and the usual quarrel "pure" visual vs electronic astronomy.

Disclaimer: I do not want this topic to be a way to take any side or advocate one of these practices against the other, it's a pure physics question but that will still, I think, be interesting for our hobby. 

 

We know one of the usual arguments for visual use is "we receive the true photons from the stars".

Now, how much are these photons modified when they come to our eye? (we know that afterwards they are just transformed into electric impulse to our brain "forming" the image).

 

The wavelength/photon can lose energy, be absorbed, scattered, go through fluorescence, transmitted, bounce etc depending the environment/matter it goes through. 

A photon re-emitted by an electron that absorbed it cannot be considered as the original photon.

I have been reading that no particle of light emitted by the object is unmodified when it reaches our eye. It seems strange to me, as it would imply that no single photon makes it with pure transmission without having been absorbed/re-emitted. But I'm happy to be proven the opposite. I don't consider loss of energy as a real modification (the wavelength still comes directly from the original object).

 

As far as I know stars or nebula and all objects we observe emit some visual spectrum (excepted dark nebula). How much of the photons reaching our eye are those really emitted at the surface of the star or nebula we "see"?

 

From a physics perspective, no photons leaving the surface of a star (whatever that means) or nebula (again, there's no such thing as the 'edge' of the nebula) that reach our eyes are original. On the way from those sources, the photons end up going through multiple interactions (in the quantum mechanical sense). From a quantum mechanics perspective, the outgoing photon from each interaction is a brand new particle, completely distinct from the photon incoming into the interaction. Any time the energy-momentum of the photon changes (due to an interaction), it is a new photon. Even the expansion of the Universe which stretches the wavelength of the photon (i.e. redshifts it) is an interaction that changes the photon into a new photon.

[12BH7]

Hi Phil, point taken but I am not asking about the rays from the core of the star, I am inquiring about the ones escaping, those giving its color to the star. For the sake of my question let’s consider the beginning of the birth photon like that moment, the one it escapes into space void.
The stars at that point emit in visual range also. We still see (even if poorly) stars in the sky.

I'm a little confused about  exactly what you're asking. The birth of a photon is not the same photon that started the photon you're seeing. 

 

Here's a very basic model of what happens. Photon 1 hits atom 2.  Atom 2 absorbs that energy and then emits photon 3. This goes on for a few trillion trillion times and eventually it runs out of atoms to run into - and that photon hits your eye. 

 

But if we want to get picky about it, would you rather look at photons reflecting off the Grand Canyon or reflecting off a picture of the Grand Canyon?

[Freezout]

Thanks a lot gentlemen for your patient explanations. I understood now.

The argument of the "true photon" as used by many loses its value.

 

12BH7, I prefer to look at the photons from galaxies going through space, intergalactic dust, atmosphere, meniscus glass, mirror (twice), prism, glass of eyepieces than additionally re-emitted via a screen. There is here a difference of nature in the medium that has some influence on me. Pure personal preference. 

[12BH7]

Pure personal preference. 

And that is what it's all about. Anything that gets you out under the stars is OK with me.

[KBHornblower]

The details of what happens to the light while in transit from the emitters (stars, etc.) to the receptor (eye or camera) are irrelevant to comparisons of the direct and electronic experiences.  

 

I would make an analogy to listening to music.  Observing directly is analogous to listening to an orchestra in the concert hall, while looking at a photographically or electronically generated image is analogous to listening on the radio or playing back a recording.

[Neptunus Rex]

Interesting discussion. A couple of points from me.
If photons go through so may interactions, why do we still see point sources in the night sky?

Another is the interior of stars. I think I read somewhere in one of my many books years ago that the gas in a stars deep interior is opaque to visable light due to temperature, pressure and ionization.

Edited by Neptunus Rex, 21 February 2024 - 09:28 PM.

[Phil Cowell]

Not a good analogy. Move the frequency of your music into the visual frequency range is a better analogy (yes the frequency is going up not down).

 

The details of what happens to the light while in transit from the emitters (stars, etc.) to the receptor (eye or camera) are irrelevant to comparisons of the direct and electronic experiences.  

 

I would make an analogy to listening to music.  Observing directly is analogous to listening to an orchestra in the concert hall, while looking at a photographically or electronically generated image is analogous to listening on the radio or playing back a recording.

Edited by Phil Cowell, 21 February 2024 - 11:45 PM.

[viewer]

Basically collecting a lot of photons and concentrating them with a telescope is 'bending the rules' too. One threshold could be if we see the effects of the photons in real time, say within half a second after the photons hits. But that threshold is also a bit arbitrary. An image contains the work of photons too, just collected a bit longer, and at times with some relatively small interaction by the human hand.

 

I think in the end the most important is what you do yourself. Are you relatively much in charge of how the photons hit?

Edited by viewer, 22 February 2024 - 02:02 AM.

[Freezout]

Interesting discussion. A couple of points from me.
If photons go through so may interactions, why do we still see point sources in the night sky?
 

This is the point that I find the most complex, how would the objects keep their color especially. When we send a telescope in outer space, for the visible spectrum, it sees the stars in same color than those on Earth under atmosphere. I was thinking that most of the photons, due to their speed, where purely transmitted.

Apparently when hitting another particle this one emits still enough photons of same wavelength in same direction?

 

I know it cannot be true for X-ray, because I worked for a company making X-ray fluorescence and diffraction systems, and the fact that photons, when hitting the target, are bouncing back or trigger the emission of another one having another angle or wavelength than the initial one, is at the core of the functioning of the system.  

Edited by Freezout, 22 February 2024 - 01:18 AM.

[viewer]

We see the stars as points because they are utterly far away, making them look small. So the photons come from practically one direction only, forming a point.

Edited by viewer, 22 February 2024 - 01:49 AM.

[Tom Barnacle]

If none of the photons reaching the detector are original what would be the point of spectroscopy?

[skysurfer]

Interesting discussion. A couple of points from me.
If photons go through so may interactions, why do we still see point sources in the night sky?

Another is the interior of stars. I think I read somewhere in one of my many books years ago that the gas in a stars deep interior is opaque to visible light due to temperature, pressure and ionization.

Interesting. So how does light escape from a star at all ? I know that stars like the Sun have an opaque photosphere, although very large and tenuous red giants (Antares, Betelgeuse, VY CMa)  might not have a hard photosphere.

[Tom Barnacle]

Interesting. So how does light escape from a star at all ? I know that stars like the Sun have an opaque photosphere, although very large and tenuous red giants (Antares, Betelgeuse, VY CMa)  might not have a hard photosphere.

Not an expert here but I guess photons escape from a star once their 'free mean path' - that is the distance they travel before the interact with another particle becomes effectively unbound in interstellar space. I guess many then interact along the way to Earth and get scattered or re-radiated in another direction, but many would reach the Earth without interacting or scattering - because otherwise, as I said above spectroscopy would be pointless?

[viewer]

Interesting. So how does light escape from a star at all ? I know that stars like the Sun have an opaque photosphere, although very large and tenuous red giants (Antares, Betelgeuse, VY CMa)  might not have a hard photosphere.

The light escapes when the star gets sufficiently transparent, which happens somewhere before the space vacuum, with the free path. If energy wasn't released, it would just heat up some outer layer, which would release it.

[Phil Cowell]

The first answer here might help.

https://physics.stac...gular-distances

 

We see the stars as points because they are utterly far away, making them look small. So the photons come from practically one direction only, forming a point.

[Phil Cowell]

Your spectroscopy answer is here, you just have to read and digest.

https://ocw.mit.edu/..._Background.pdf

 

Not an expert here but I guess photons escape from a star once their 'free mean path' - that is the distance they travel before the interact with another particle becomes effectively unbound in interstellar space. I guess many then interact along the way to Earth and get scattered or re-radiated in another direction, but many would reach the Earth without interacting or scattering - because otherwise, as I said above spectroscopy would be pointless?