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22 minutes ago, andrew s said:

it seek patterns where none exists, it fills in were data is missing.

One of my own "experiments" to see if what I'm seeing is this filled in data or noise or... was my pursuit of IFN. Which I do see, for sure. However my view was tainted by the fact I looked at Bartels sketches first- strike one against. However I do not see the IFN excatly as he does + one for the observation.  I first reported 6 years ago seeing a bunch of patches of "liquid grey" nebulosity, very very faint. As I have observed more and more some of these patches have names attached to them for example around IC1318. 

The Pleiades is another example of seeing some really faint things- it can be stark in appearance, the so called "Bubble" and is beautiful. I don't see it quite like Bartels but a colleague of his confirmed my description of it.

One of my main goals is to understand the nature of light, photons included as pertains to visual astronomy. If I wreck my eyes I do see noise in the sky even after partially adapted and in my case it results in the inability to see very faint objects.

Now for aperture...

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Due to the random nature of photons, to measure the brightness (flux) from the object to say 1% you need 100^2 = 10^4 photons To resolve the object into say 100x100 spacial points you then need t

I have opposite opinion. I think they are elements of reality - regardless of the fact that we can't measure wave functions themselves.  

Here is some starter info: http://math.ucr.edu/home/baez/physics/Quantum/see_a_photon.html Maybe most important paragraph is the first one: As for daytime exhaustion of Rhodopsin - ta

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18 minutes ago, vlaiv said:

How is this for starting to define what light is?

Thanks so much Vlaiv, Andrew and yourself are helping immensely and it is much appreciated.

I will go over this info repeatedly.

To fastrack to my aperture question which you and @andrew s can answer-  will an aperture of say 8" contain the same information an aperture of say 24"? I might think it would if the photon level in the smaller is enough to form the same image?

Are more photons or light packets in the larger aperture? can and will this be seen by the eye?

Many believe in "light grasp", I'm undecided as traditionally most serious observers think in terms of magnification at a certain exit pupil level, which aperture is a component of.

My 24" might be changing my mind a bit...

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@vlaiv nice explanation. Just two points I might have put differently.

I don't think you need to use a quantum wave function you can use the classical EM field but with discrete additions of energy ( creation, adding a photo and absorption, annihilation of a photon).

There are no instantaneous changes to the whole EM field in QED as far as I know. QED is comparable with special relativity. (Keeping away from entanglement for now.)

Regards Andrew 

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1 hour ago, andrew s said:

I don't think you need to use a quantum wave function you can use the classical EM field but with discrete additions of energy ( creation, adding a photo and absorption, annihilation of a photon).

Classical EM field approach prevents us from understanding of some of the phenomena and people will get the sense that wave in the field is the thing that is carrying momentum and energy.

1 hour ago, andrew s said:

There are no instantaneous changes to the whole EM field in QED as far as I know. QED is comparable with special relativity. (Keeping away from entanglement for now.)

Well - that was all about entanglement or rather decoherence.  How else are we supposed to explain case with "single" photon and two detectors - wave reaches both detectors but once there is detection on one of them, energy is removed from whole wave function - and that wave function indeed changes. It happens with the speed of entanglement.

In fact - let's do that thought experiment like this - there is hydrogen atom, and electron jumps down to ground state and photon is emitted. We have our rock in a pond - it is spherical shell that starts propagating from that atom. Two detectors on opposite sides of this atom far away - one closer by few miles to the atom.

Wave function "arrives" to the first detector and photon is detected - it is removed from quantum field - does wave function reach other detector in the same state? It can't. As we know it represents probability of detection of a photon - but there is no probability (I know that one photon example is flawed as we assume one and only one photon) for second detector to fire - hence wave function must have evolved so.

It indeed evolves so by mechanism of decoherence and entanglement - as wave function is entangled with first detector it changes and we know (or rather think) that entanglement is instantaneous (but it can't carry information so all is good with GR).

I made this example because without it - people keep getting back to little bullet photon traveling thru the space. With this example and above explanation they can see that it is wave function that is traveling - thus it is traveling along every possible path at the same time and interference is clear.

It also explains that photons do really exist only in two instants - once when energy is given off into field - "ripple stars" and when photon is absorbed - decoherence happens and "ripple ends". In all other cases they are "smeared" or "interwoven" into this thing we call wave function.

I know that notion of wave function gets us back to plain old QM - but I'm more talking about ripples in EM field than anything else.

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@vlaivI know its very difficult  to get the right balance of concepts in these discussions.

For me the real thing is the EM field. It interacts with the "particles and compound particles" of the other quantum fields e.g. atoms.

If you start with one excitation of the EM field at a time, as in the slit experiment, then all you can do is calculate the probability of a detection at one point or another. When you see a detection you instantly know the probability  of detection elsewhere is zero. However this is an information update local to the observer.

To talk about a EM spherical wave spreading out from a point source you need a very large number of excitations/photons and then you can use the classical Maxwell equations to good effect.

I know Mott did a famous paper on how particles localise along a track in QM.  I will look it up.

You may well have the better approach  in explaining it than I.

Regards Andrew 

 

 

 

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1 minute ago, andrew s said:

If you start with one excitation of the EM field at a time, as in the slit experiment, then all you can do is calculate the probability of a detection at one point or another. When you see a detection you instantly know the probability  of detection elsewhere is zero. However this is an information update local to the observer.

To talk about a EM spherical wave spreading out from a point source you need a very large number of excitations/photons and then you can use the classical Maxwell equations to good effect.

Let's take that dual slit experiment then - how do you calculate interference pattern on the screen?

Feynman-Mirror-1.jpg

We know what this diagram means. You take a little clock with hand going at some "speed" around the clock and you set it on a certain path and when it hits something - you again let it fly all over the place - there are some rules like when you hit something clock hand flips and so on ...

Now instead of tracking all possible paths one by one and adding probability - track them at the same time, take each path and when hand is at 12 o'clock - mark that position. Join all those points. Know what you get? A wave front of a wave function propagating in space.

Wave that I'm describing is not classical wave - and you should not look at it as EM wave that you mentioned - a bunch of photons - it is wave in above sense - all path integrals represented as single entity rather than bunch of separate paths.

 

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@vlaiv that's the problem of confusing how you do the calculation and the physical reality.

The path integral method is a way of doing the calculation. It is based on an expansion of the full QED equations. It leads you down the path of thinking virtual particles are travelling all possible routes to interfering at the end to give to a real particle (photon) to detect.

Virtual particles  don't exist they are calculation tools just as the terms in an expansion of sin(x) are.

Personally I don think wave functions are real . No one has ever observed one to the best of my knowledge. 

Yes, it's remarkable it gives the right answer but that is true of Lagrange multipliers, Taylor expansions and the like.

Regards  Andrew 

PS I should have said the physical reality is that over time the probabilities predicted by QED correlate well with the results of experiment.  But not individual events, they remain random.

Edited by andrew s
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9 minutes ago, andrew s said:

Personally I don think wave functions are real . No one has ever observed one to the best of my knowledge. 

I have opposite opinion. I think they are elements of reality - regardless of the fact that we can't measure wave functions themselves.

 

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11 minutes ago, vlaiv said:

I have opposite opinion. I think they are elements of reality - regardless of the fact that we can't measure wave functions themselves.

That's fine. We have different philosophical positions.  

9 minutes ago, vlaiv said:

Btw - when I say wave function I actually mean quantum state.

Now that's different some quantum states are observable some are not.

What fun. I will freely admit that my knowledge is limited and given that leading scientists can't agree on these things I don't feel too bad about it.

I have tried to purge the worst of my understanding of pop science views but that still leaves a vast area for discussion.

Regards Andrew 

 

Edited by andrew s
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3 hours ago, jetstream said:

Thanks so much Vlaiv, Andrew and yourself are helping immensely and it is much appreciated.

I will go over this info repeatedly.

To fastrack to my aperture question which you and @andrew s can answer-  will an aperture of say 8" contain the same information an aperture of say 24"? I might think it would if the photon level in the smaller is enough to form the same image?

Are more photons or light packets in the larger aperture? can and will this be seen by the eye?

Many believe in "light grasp", I'm undecided as traditionally most serious observers think in terms of magnification at a certain exit pupil level, which aperture is a component of.

My 24" might be changing my mind a bit...

Assuming you are seeing limited, both will have the same effective resolution.  The larger aperture will for the same magnification of extended objects give a brighter image allowing you to see fainter detail.

There is a reason astronomer seek larger telescopes.

Regards Andrew 

Edited by andrew s
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1 hour ago, andrew s said:

Assuming you are seeing limited, both will have the same effective resolution.  The larger aperture will for the same magnification of extended objects give a brighter image allowing you to see fainter detail.

There is a reason astronomer seek larger telescopes.

Regards Andrew 

Sounds good and I hold this view but- do you think its possible that the larger aperture can give "brighter" images due to the aperture itself? Bartels believes in throughput or "entendue". More waves and resulting photons or whatever? Sorry if this is a redundant question.

In my scope jumps I went for 1 stellar magnitude increase- I had to use something as a gauge- but my 24" f4.1 seems to blow the doors off the 15" in unexpected ways, M1's detail for example or the Veils. Is this all due to increased magnification at the same(ish) exit pupils? I never thought there would be such a difference Andrew.

Btw, back to the brain filling in info- an optics acquaintance said there no way I can see "real" red or blue in M42- that it is an illusion from exposure to the bright green I do see and that the brain fills in the other colours some how. Oddly enough where I see red or blue is in the right places from images I've seen.

No clue but the colours are there for me now but were not at first.

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2 minutes ago, jetstream said:

Sounds good and I hold this view but- do you think its possible that the larger aperture can give "brighter" images due to the aperture itself? 

Yes that where the brighter image cones from it captures move photons.

In low light it's your rods (black and white) that work the cones (colour) need much more light to work.

Regards Andrew 

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1 minute ago, andrew s said:

Yes that where the brighter image cones from it captures move photons.

In low light it's your rods (black and white) that work the cones (colour) need much more light to work.

Regards Andrew 

Thank you sir!

I have been wanting (needing) a confirmation of what Ive noticed and what some other various and serious astronomers have told me. Packing more light into the exit pupil or however we want to describe it. This idea seems to send some into fits and they always refer to mag at exit pupil as the only factors.

I have to say that fast f ratios and large apertures under dark skies give absolutely remarkable views. I cannot believe what the 24" shows, maybe around this aperture crosses the line or something light wise, view wise, eye brain wise etc etc.

 I truly appreciate you helping me along again Andrew, @vlaiv too. I hope you guys continue on in the thread- I might be a few posts back trying to absorb but I'll be here.

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3 hours ago, jetstream said:

 do you think its possible that the larger aperture can give "brighter" images due to the aperture itself?

Yes. The larger the aperture the more photons you collect in a given time. (You might decide to spread them over the detector to a different extent eg by choosing the focal ratio but that is independent of aperture).

The more photons you collect the more you know about the object as I demonstrated up the thread so the larger the aperture and the longer you "stare" at it, the more information about the object you acquire.

In a perfect  system (no noise, no distortion of the signal between the object and you) :-

The first photon tells you there is an object emitting light

the second photon (or more accurately, the time between them and the difference in direction) tells you a little bit more about the object (A hint towards deciding if  it is a point or an extended source and very, very roughly its brightness) 

The third and subsequent photons increases the precision of the brightness estimate and if it is an extended object adds information about the size, structure and  brightness variation across it.

Each subsequent photon adds more detail in the way I described.

Different equipment may have different abilities to distinguish which direction they are coming from (spacial resolution) but at the end of the day the more photons, the more accurate description of the object you get

Robin

Edited by robin_astro
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4 hours ago, robin_astro said:

Yes. The larger the aperture the more photons you collect in a given time

 Yes, thank you Robin. Its great to hear this from you- most say aperture cannot give brighter images etc visually- actually most start off believing it until "corrected" by others. Excellent info about the photons and how the image is formed, I did not realize the image is built up this way in our brain. The 24" gives so much more detail than my 15" it really puzzled me until now. Both the scope builder and optician said it would surprise me. I also have a 200mm f3.8 for comparison.

Now for the exit pupil... I know there is a sweet spot for nebula, my favourite objects with and without filters. So if I maintain a "proper" exit pupil range of say 4.5mm to 5.5mm could it be true then that the smaller exit pupil has more photons available for my eye? ie packing more in the disk?

Edit: just thought about it...the photon level must be dictated by primary mirror aperture so the same number must pass through the eyepiece(s). The smaller exit pupil must have more concentrated photon levels?

Edited by jetstream
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