A revolution in thought

[Spacemonkey]

Originally Posted by thedoc

So the photons that are at the object and allow us to see the object are the same photons that are at a persons eye and the same photons that interact with the film in a camera. Presumably if there were several people looking at an object and several taking pictures of that object, the photons at the object allowing everyone to see and photograph the object, would also instantly be at everyones eye and the film in several cameras at once?

PS. could someone bump this just incase Peacegirl really uses 'ignore'.

Bump.

[peacegirl]

Originally Posted by Spacemonkey

If you think I'm back to talking about stationary photons, I'm not. I'm just saying that a camera works the same way the eyes work up to the film. The only difference is that a camera develops a photograph from the light that is instantly at the film (as long as the object is in the field of view), and the eyes are able to see the actual object from the light that is instantly at the retina (as long as the object is within visual range).

Bump.

Of course. Why wouldn't the light be at everyone's film in several cameras at once? It's not like there is a shortage of light if the many cameras are in the field of view of the object. For example, at a wedding there are often more than one photographer taking pictures. Just because one takes a picture doesn't mean there's not enough light at the film for another photographer to take a picture? Or what if everyone is looking at the same landscape. The light is at everyone's retina at once. There's not a shortage of light, but distance does matter. If one person is closer to something that he's looking at than someone else, it will show up differently on his retina than someone who is further away. How is this any different than what optics tells us?

[LadyShea]

Originally Posted by peacegirl

I explained why a camera would get an instantaneous photograph because the lens of a camera and eye work similarly. A camera cannot see an object because it doesn't have a brain, but the light that is at the film is the same light that is at the retina, because both work in the same way.

Lessans didn't say light has to be "at the retina", that's what he was trying to demonstrate with his "sun turned on at noon" hypothetical. He very clearly stated there need not be any photons "at the retina" in order to see the sun when he stated the photons would not arrive on Earth for 8.5 minites. You are trying to make his ideas fit reality, but he really didn't think photons had to be interacting physically with/at the eyes at all. He never said one word about photons touching retina....he said nothing about instant mirror images allowing photons to be in two places at once, breaking the laws of physics. He thought the brain looked out and saw what was there to be seen. He clearly was not referring to lenses or cameras.

And, given his hypothetical of the sun turned on at noon, meaning there are no photons on Earth, you could not take a photograph at noon without photons teleporting from the sun to the surface of the film 93 million miles away.

[thedoc]

Originally Posted by peacegirl

Of course. Why wouldn't the light be at everyone's film in several cameras at once? It's not like there is a shortage of light if the many cameras are in the field of view of the object. For example, at a wedding there are often more than one photographer taking pictures. Just because one takes a picture doesn't mean there's not enough light at the film for another photographer to take a picture? Or what if everyone is looking at the same landscape. The light is at everyone's retina at once. There's not a shortage of light, but distance does matter. If one person is closer to something that he's looking at than someone else, it will show up differently on his retina than someone who is further away. How is this any different than what optics tells us?

Lets keep it simple they are all standing together so different distance is not a factor.

So you are saying that the exact same photons that are at the object are the same photons that are at everyones eyes and the at the film of several cameras at once if everyone is looking and the others just happen to snap the photo at the same instant, these individual photons were at the object are now at several peoples eyes and several cameras, each individual photon is at all those places at once?

PS. could someone bump this just incase Peacegirl really uses 'ignore'.

[LadyShea]

Bump

Originally Posted by thedoc

Lets keep it simple they are all standing together so different distance is not a factor.

So you are saying that the exact same photons that are at the object are the same photons that are at everyones eyes and the at the film of several cameras at once if everyone is looking and the others just happen to snap the photo at the same instant, these individual photons were at the object are now at several peoples eyes and several cameras, each individual photon is at all those places at once?

PS. could someone bump this just incase Peacegirl really uses 'ignore'.

[Spacemonkey]

Originally Posted by Spacemonkey

You've just told me the blue-wavelength light/photons are in two places at once. You just said they are both instantly at the film and that they are in the light bouncing off the object and travelling away from it.

The same light/photons cannot be in two places at once. So where are they? If they are instantly at the film, then they are not also contained in the white light bouncing off the object. And if they are in the white light bouncing off the object, then they cannot also be instantly at the film.

That's not what I asked. You haven't answered my question.

There are blue-wavelength photons in the collection of photons (white light) hitting the ball. Either they bounce off the ball or they don't. If they do, then there are two ways this might happen. They may be the only photons bouncing off, in which case the light bouncing off the ball will be blue light only. Or they may be bouncing off along with all of the other photons which were hitting the ball. In that case the totality of light bouncing off will not be blue light, but it will still contain this blue light (just as it did before hitting the ball). On both of these options the blue-wavelength light is bouncing off the ball. So here's the question again for you to answer:

When sunlight (including light of all wavelengths, including blue) hits a blue object, what happens to the blue-wavelength light as it hits that object? At one moment it is traveling towards the object along with all the light of other wavelengths. Then it hits the surface of the object. Then what?

Does it bounce off the surface to travel away from it? [Y/N?]

It is true by definition. If something is sucked in and used up, it can't still be there bouncing off the object. If it is absorbed it is gone. If it bounces off then it is still there. Nothing can be both absorbed and bouncing off.

If these non-blue wavelength photons are being sucked in and used up, then they cannot also be still there in the light bouncing off the ball.

It must do, because what you are saying about it is simply contradictory according to its normal meaning.

Then it can't be contained within the white light bouncing off the object, because all of that light is bouncing off something.

Any light that is anywhere has either travelled there, teleported there, or come into existence there. There are no other options. If the light which is instantly at the film consists of the same blue-wavelength photons which were immediately previously hitting the ball, then (i) they have teleported; and (ii) they can't still be in the white light bouncing off the object.

You have yet to explain the difference between (P)reflection and teleportation (as there is no difference between getting somewhere instantly and instantly relocating there), and you still haven't told me where my blue-wavelength photons will be immediately after hitting the ball or told me whether or not they are bouncing off it.

Bump.

[LadyShea]

Originally Posted by Spacemonkey

Any light that is anywhere has either traveled there, teleported there, or come into existence there.

This

[LadyShea]

OMG the dupes are trying to kill me

[Spacemonkey]

Originally Posted by peacegirl

Of course. Why wouldn't the light be at everyone's film in several cameras at once? It's not like there is a shortage of light if the many cameras are in the field of view of the object.

The problem isn't that you're positing light at multiple places at once. It's that you're positing the same light - i.e. the exact same photons - at multiple places at once.

[peacegirl]

Originally Posted by peacegirl

The blue light is (P) reflected which means that we can only utilize that blue wavelength light when we're looking directly at the object in real time (which makes light a condition of sight in the efferent model, not a cause), or when the lens of a camera is focused on the object, for then the blue light is instantly at the film. The white light is constantly bouncing off of the object with the blue light in it.

Originally Posted by Spacemonkey

You've just told me the blue-wavelength light/photons are in two places at once. You just said they are both instantly at the film and that they are in the light bouncing off the object and travelling away from it.

Anytime an object absorbs light, the non-absorbed light is at the film/retina if that object is in our field of view. This light is not (N) reflected. That non-absorbed blue light is what allows us to see the object when we're looking at it. As the photons get dispersed as the object gets further and further away from the lens, the (P) light is no longer showing blue; it is showing white on the film/retina. White light is all that shows up due to the inverse square law. White light is what continues on through space and time, not the blue wavelength light.

There's a third option. The white light is traveling but as it passes over (or bounces off of) the object, the non-blue light gets absorbed. This means that when we're looking at the object, we see it due to its absorptive property. If the object is out of our field of view, we don't see it because the blue wavelength light is limited to how far the object is from our retina (the inverse square law). This is why the object must be present, not just the light. We don't resolve light alone and get an image, which is what the afferent version of sight states.

No. You're getting confused because you're not understanding how efferent vision works.

That's true, the blue wavelength light is allowing us to see the object due to the object absorbing the non-blue wavelength light, but as soon as that blue wavelength light fades out (due to dispersion), we get white light. The blue wavelength light doesn't bounce so it doesn't travel. It is there because the eyes see it due to the way the brain works. The only way that this will be a viable model is to do more testing, otherwise the tendency will be to resist such a huge error on the part of science.

You're right, only the blue wavelength light is seen while the white light continues to replace the old photons. The object continues to suck in the new photons while the (P) reflection is there when we are looking. Once again, as soon as the object gets too tiny for the lens to resolve due to the dispersion of photons as distance becomes a factor, all we get is white light on the lens. We do not resolve blue when the blue object is no longer large enough to be seen.

Maybe there's just a misunderstanding because my definition is the same as yours.

White light is bouncing off and as it does the object is continually absorbing certain photons, which is why blue light is present at the film/retina when we're looking directly at the object. But that blue light does not travel at all; it becomes a condition of sight; it does not cause sight which only means it does not travel through space and time and eventually reach the film/retina without the object being within visual range.

No Spacemonkey, they have not teleported. Remember, the white light is traveling, and it is not the same light when we look at the object a moment later. These are new photons, but we see blue because the object is still absorbing that non-blue light. So when the lens is focused on the object, that blue light is instantly at the film/retina as long as the object is within the field of view.

Bump.

I told you that they are being renewed by new white photons but the object is absorbing those photons so when we look at the object efferently, the light is instantly at the film/retina. The inverse square law allows that (P) light to instantly be at the film/retina when the lens is focused on the OBJECT.

[thedoc]

Originally Posted by peacegirl

The inverse square law allows that (P) light to instantly be at the film/retina when the lens is focused on the OBJECT.

How does the 'Inverse Square Law' allow the light to be instantly at the film/retnia when the lens is focused on the object?

[davidm]

LOL.

Hey, peacegirl, if real time seeing is true, why does NASA depend on delayed time seeing, as measured by the speed of light, to send space probes to mars and other celestial bodies? Can't answer that, can you, oh my!

[thedoc]

And now for something completely different,

P. I. Tchaikovsky - Violin Concerto in D major,...

[thedoc]

And now for something completely different,

http://www.youtube.com/watch?v=GQAWJHITdhg

[Spacemonkey]

Originally Posted by peacegirl

Anytime an object absorbs light, the non-absorbed light is at the film/retina if that object is in our field of view. This light is not (N) reflected. That non-absorbed blue light is what allows us to see the object when we're looking at it. As the photons get dispersed as the object gets further and further away from the lens, the (P) light is no longer showing blue; it is showing white on the film/retina. White light is all that shows up due to the inverse square law. White light is what continues on through space and time, not the blue wavelength light.

The white light hitting the ball simply consists of the blue light and the non-blue light that it contains. Nothing else. That's all white light is. If the non-blue part is sucked in and used up by the ball, and the blue part is instantly at the film, then there is nothing left to be bouncing off the object. Not only can there be no white light bouncing off, but there cannot be any light at all bouncing off. White light cannot be continuing on through space and time, and there is nothing for dispersion or the inverse square law to apply to. Light that is absorbed cannot still be bouncing off, and light that turns up at the distant film cannot still be bouncing off. And the white light hitting the ball does not contain anything else but the absorbed non-blue light and the (P)reflected blue light.

Originally Posted by peacegirl

There's a third option. The white light is traveling but as it passes over (or bounces off of) the object, the non-blue light gets absorbed. This means that when we're looking at the object, we see it due to its absorptive property. If the object is out of our field of view, we don't see it because the blue wavelength light is limited to how far the object is from our retina (the inverse square law). This is why the object must be present, not just the light. We don't resolve light alone and get an image, which is what the afferent version of sight states.

That's not a third option for the blue-wavelength photons bouncing off the object. In fact this doesn't even mention those photons, or in any way address the question. Either those photons bounce off the ball or they don't. And if they do, then they either bounce off alone, or they do not. There are no other options here.

Originally Posted by peacegirl

No. You're getting confused because you're not understanding how efferent vision works.

I'm not the one contradicting myself. If those photons are not bouncing off the ball, then where are they immediately after hitting it? If they are at the distant film then they have instantaneously relocated themselves, i.e. teleported.

Originally Posted by peacegirl

That's true, the blue wavelength light is allowing us to see the object due to the object absorbing the non-blue wavelength light, but as soon as that blue wavelength light fades out (due to dispersion), we get white light. The blue wavelength light doesn't bounce so it doesn't travel. It is there because the eyes see it due to the way the brain works. The only way that this will be a viable model is to do more testing, otherwise the tendency will be to resist such a huge error on the part of science.

You can't do any testing until you first have a logically consistent and coherent model, and you still don't. If the non-blue parts of the white light hitting the ball are sucked in and used up, then they can't still be there in the light bouncing off the ball. If that light bouncing off is lacking its non-blue parts then it is no longer white light. White light minus the non-blue part is blue light. Unless of course the blue parts are also disappearing to relocate themselves to nearby films and retinas, in which case there won't be any light bouncing off the object at all.

Originally Posted by peacegirl

White light is bouncing off and as it does the object is continually absorbing certain photons, which is why blue light is present at the film/retina when we're looking directly at the object. But that blue light does not travel at all; it becomes a condition of sight; it does not cause sight which only means it does not travel through space and time and eventually reach the film/retina without the object being within visual range.

If the light bouncing off the object no longer contains the blue photons then it is no longer white light. If it no longer contains the non-blue photons then it is no longer white light. If it no longer contains either the blue photons or the non-blue photons then there is nothing left, and no light can be bouncing off the object at all.

Originally Posted by peacegirl

No Spacemonkey, they have not teleported. Remember, the white light is traveling, and it is not the same light when we look at the object a moment later. These are new photons, but we see blue because the object is still absorbing that non-blue light. So when the lens is focused on the object, that blue light is instantly at the film/retina as long as the object is within the field of view.

If those blue-wavelength photons which at one moment were just hitting the surface of the ball are, at the very next moment, present at the distant camera film, then they have instantaneously relocated - i.e. they teleported there. That's just what teleportation means.

Originally Posted by peacegirl

I told you that they are being renewed by new white photons but the object is absorbing those photons so when we look at the object efferently, the light is instantly at the film/retina. The inverse square law allows that (P) light to instantly be at the film/retina when the lens is focused on the OBJECT.

If the blue-wavelength photons that were hitting the object are instantly at the distant film, then they have teleported. If they are instantly at the film then they are not also in the light bouncing off the object. If they are not in that light bouncing off the object then the light bouncing off no longer contains light of all wavelengths and cannot be white light. And there are no such things as white photons.

[Angakuk]

Originally Posted by peacegirl

The light is at everyone's retina at once. There's not a shortage of light, but distance does matter. If one person is closer to something that he's looking at than someone else, it will show up differently on his retina than someone who is further away.

Why should distance matter. If the light which allows us to see an object is instantly at the film/retina without regard to how far away the object is from the film/retina why would someone closer to the object have different visual experience than someone who is further away?

Originally Posted by peacegirl

That's true, the blue wavelength light is allowing us to see the object due to the object absorbing the non-blue wavelength light, but as soon as that blue wavelength light fades out (due to dispersion), we get white light. The blue wavelength light doesn't bounce so it doesn't travel.

Dispersion is a function of light that is traveling. If the blue wavelength light does not travel, why and how does it get dispersed?

You keep claiming that this blue wavelength light does not travel yet you persist in describing its behavior by using terms that only apply to light that is traveling. This is very strange.

[Angakuk]

Originally Posted by Spacemonkey

And there are no such things as white photons.

You sir, are a racist.

[peacegirl]

Originally Posted by Kael

No, it's not, for the simple reason that the inverse-square law only applies to a physical quantity that travels from its source to whatever is detecting it. You have repeatedly stated that the light does not need to travel between an object and the eye, it is the very core of Lessans' ridiculous notions. If that were true, the inverse-square law would not apply, because distance simply wouldn't make a difference in the amount of light being detected (which is what the inverse-square law is about).

You're wrong right there. I also said that (P) light is constantly being replaced by new white light. The only difference is that there is no time element involved when the lens is focused on the object, or when a camera snaps a picture due to the efferent model, which means there is no time element involved. But the distance from the object (the inverse square law) is still in effect.

Originally Posted by Kael

If there is no time element then there can be no travel, if there is no travel then distance is irrelevant, and the inverse-square law will not apply.

But it does apply. The blue wavelength light, in Spacemonkey's example, is there because we see it, not because it travels to us, but the light is constantly being replaced by new photons. The (P) reflection therefore has everything to do with the distance of the object from the observer. The farther away the object is, the smaller it will be on the retina when the lens is focused on the object.

Originally Posted by Kael

If the inverse-square law does not apply, an object would have to appear equally bright at any distance. It does not, so you will have to find another explanation for why that is the case. The inverse-square law does not fit Lessans' suppositions. It really is that simple.

You're wrong. The inverse-square law does apply. There is very little that changes except for one thing: Images (or the non-absorbed wavelength light) does not bounce off and travel through space and time. This light is present when we're looking at the object, but, to repeat, it does not travel through space and time as the full visible spectrum does.

The only difference is the the lens must be focused on the object. It does not focus the light ONLY.

Originally Posted by Kael

Lenses do not focus on things. They bend light that passes through them, creating a point past the lens where that light comes together, or focuses. That's all they do, and it only does any good if A) there is light physically traveling through the lens, and B) there is a detector at the focal point of the light, so that a clear image can be produced from that information.

None of this is inconsistent with what I'm saying. The only difference is that the object must be present for the light to come together or to be focused. There is no image possible without the object being present if the efferent model is correct.

Originally Posted by Kael

Yes, if the efferent model (aka, "viola, we see!") is correct then the object must be present. Since we know definitively that an object does not, in fact, have to be present the instant the picture is taken (it must only be present the instant the light we are capturing interacted with the object, not the instant it interacts with the eye/film/detector. We are also increasingly capable of simply making incoming like behave as though it had interacted with a given object, rather than needing any actual object.), then it would follow that the efferent model (aka, "viola, we see!") is not correct.

You're simply repeating what is being disputed. It is believed that an object does not have to, in fact, be present, but so far that is an established theory, nothing more. I'd like to see how incoming light can behave as though it had interacted with a given object rather than needing any actual object. Can you show me an instance of this?

Originally Posted by Kael

The fact that you have deliberately refused to understand or even acknowledge most of the evidence of this we have presented does not have any bearing on this. Reality is inconsistent with your/Lessans' claims. It really is that simple.

I'm all about reality Kael. I'm not arguing for argument's sake.

Originally Posted by Kael

Ignorance of optics does not mean you can use it to support your personal notions. We know too much about how it actually works, from too many decades and centuries of repeated testing, for you to be able to get away with bullshitting and buzzword-babbling. I recommend you go back to talking about Free Will and the No-Blame Environment, you've got a better shot there, though that's not saying much.

If you think I should not talk about optics, why in God's name did you stir the conversation up again, and in the wrong forum to boot? And who's talking bullshit?

Originally Posted by Kael

I simply responded in the thread you posted the comment in. And you are the one talking bullshit, in case there was any confusion.

I beg to differ.

[Spacemonkey]

Originally Posted by Spacemonkey

The white light hitting the ball simply consists of the blue light and the non-blue light that it contains. Nothing else. That's all white light is. If the non-blue part is sucked in and used up by the ball, and the blue part is instantly at the film, then there is nothing left to be bouncing off the object. Not only can there be no white light bouncing off, but there cannot be any light at all bouncing off. White light cannot be continuing on through space and time, and there is nothing for dispersion or the inverse square law to apply to. Light that is absorbed cannot still be bouncing off, and light that turns up at the distant film cannot still be bouncing off. And the white light hitting the ball does not contain anything else but the absorbed non-blue light and the (P)reflected blue light.

That's not a third option for the blue-wavelength photons bouncing off the object. In fact this doesn't even mention those photons, or in any way address the question. Either those photons bounce off the ball or they don't. And if they do, then they either bounce off alone, or they do not. There are no other options here.

I'm not the one contradicting myself. If those photons are not bouncing off the ball, then where are they immediately after hitting it? If they are at the distant film then they have instantaneously relocated themselves, i.e. teleported.

You can't do any testing until you first have a logically consistent and coherent model, and you still don't. If the non-blue parts of the white light hitting the ball are sucked in and used up, then they can't still be there in the light bouncing off the ball. If that light bouncing off is lacking its non-blue parts then it is no longer white light. White light minus the non-blue part is blue light. Unless of course the blue parts are also disappearing to relocate themselves to nearby films and retinas, in which case there won't be any light bouncing off the object at all.

If the light bouncing off the object no longer contains the blue photons then it is no longer white light. If it no longer contains the non-blue photons then it is no longer white light. If it no longer contains either the blue photons or the non-blue photons then there is nothing left, and no light can be bouncing off the object at all.

If those blue-wavelength photons which at one moment were just hitting the surface of the ball are, at the very next moment, present at the distant camera film, then they have instantaneously relocated - i.e. they teleported there. That's just what teleportation means.

If the blue-wavelength photons that were hitting the object are instantly at the distant film, then they have teleported. If they are instantly at the film then they are not also in the light bouncing off the object. If they are not in that light bouncing off the object then the light bouncing off no longer contains light of all wavelengths and cannot be white light. And there are no such things as white photons.

Bump.

These are the two key questions you are completely failing to consistently answer:

1) Where are the blue-wavelength photons, contained within the sunlight striking the blue ball, at the point in time immediately after they hit the ball?

2) Where were the blue photons, which are at the film interacting with it to produce a photographic image of the blue ball when the photograph is taken, at the point in time immediately before the photograph is taken.

You also have no consistent story about which parts of the light hitting the object are absorbed, which parts bounce off, and which parts turn up instantly at distant films and retinas. These three options are each mutually exclusive - no individual photon, or collection of photons of a given wavelength or set of wavelengths, can be doing any two of these together.

If a photon is absorbed, then it cannot bounce off, and it cannot turn up at any distant film or retina.

If a photon bounces off to travel away from the object, then it has not been absorbed, and it cannot also be instantly present at any distant film or retina.

If a photon is present at some distant film or retina immediately after striking the object, then it has not been absorbed, and has not bounced off the object to travel away from it.

The sunlight striking the blue ball simply consists of photons of all wavelengths in the visible spectrum. This spectrum is often represented by the acronym ROYGBIV (red, orange, yellow, green, blue, indigo, violet). So let's use a simplified model where there is one photon of each of these colors hitting the blue ball (like little marbles).

These seven differently colored photons are hitting the ball. They comprise the sunlight hitting the ball. I want you to tell me where each one of them is 0.0001sec after this collection of photons hits the ball. Which ones are absorbed (such that they get sucked in and used up, and do not bounce off)? Which ones bounce off and start traveling away from the ball at the speed of light? Which ones instantly appear at distant films or retinas? Which ones, if any, are in more than one place 0.0001sec after hitting the ball?

If you want to speak of blue light, non-blue light, or white sunlight, then the following definitions apply with respect to our seven photons:

Blue light =(def) The blue photon.

Non-blue light =(def) The red, orange, yellow, green, indigo, and violet photons.

White sunlight =(def) All seven photons.

[peacegirl]

Originally Posted by Spacemonkey

The white light hitting the ball simply consists of the blue light and the non-blue light that it contains. Nothing else. That's all white light is. If the non-blue part is sucked in and used up by the ball, and the blue part is instantly at the film, then there is nothing left to be bouncing off the object. Not only can there be no white light bouncing off, but there cannot be any light at all bouncing off.

That's false.

Originally Posted by Spacemonkey

White light cannot be continuing on through space and time, and there is nothing for dispersion or the inverse square law to apply to. Light that is absorbed cannot still be bouncing off,

The light that is absorbed is not bouncing off; that's why the blue wavelength light is present at the film/retina, and why the inverse square law has application properties.

Originally Posted by Spacemonkey

and light that turns up at the distant film cannot still be bouncing off. And the white light hitting the ball does not contain anything else but the absorbed non-blue light and the (P)reflected blue light.

The light that turns up at the film (it's not distant if you think in terms of the object being in the camera's field of view) is not bouncing off of anything. As white light bounces off, it gets absorbed and (P) reflected which means that only when the lens is focused on the object can that blue wavelength light be put to use, so to speak. The inverse square law is still intact because the farther away the object is from the observer, the more dispersed the photons become as we look directly at the object. The blue light is dispersing as the white light continues to bounce off the object and the non-blue photons get absorbed.

Originally Posted by peacegirl

There's a third option. The white light is traveling but as it passes over (or bounces off of) the object, the non-blue light gets absorbed. This means that when we're looking at the object, we see it due to its absorptive property. If the object is out of our field of view, we don't see it because the blue wavelength light is limited to how far the object is from our retina (the inverse square law). This is why the object must be present, not just the light. We don't resolve light alone and get an image, which is what the afferent version of sight states.

Originally Posted by Spacemonkey

That's not a third option for the blue-wavelength photons bouncing off the object. In fact this doesn't even mention those photons, or in any way address the question. Either those photons bounce off the ball or they don't. And if they do, then they either bounce off alone, or they do not. There are no other options here.

I think it does address the question. The blue wavelength light does not bounce off the object with the full spectrum light or without it. It is present as we look at the object. You're missing the whole model.

Originally Posted by peacegirl

No. You're getting confused because you're not understanding how efferent vision works.

Originally Posted by Spacemonkey

I'm not the one contradicting myself. If those photons are not bouncing off the ball, then where are they immediately after hitting it? If they are at the distant film then they have instantaneously relocated themselves, i.e. teleported.

Wrong Spacemonkey. You're not getting it. Those blue photons are present because of the ability of the object to absorb the non-blue photons, but this is a continual process where full spectrum visible light first strikes the object and the non-absorbed and absorbed light gets split up, so to speak, but this light does not bounce and travel. Only white light does this. The difference here is that, if sight is efferent, we are able to get an instant mirror image on our retina or film due to this (P) light and how the eyes work as they look out at the world.

Originally Posted by peacegirl

That's true, the blue wavelength light is allowing us to see the object due to the object absorbing the non-blue wavelength light, but as soon as that blue wavelength light fades out (due to dispersion), we get white light. The blue wavelength light doesn't bounce so it doesn't travel. It is there because the eyes see it due to the way the brain works. The only way that this will be a viable model is to do more testing, otherwise the tendency will be to resist such a huge error on the part of science.

Originally Posted by Spacemonkey

You can't do any testing until you first have a logically consistent and coherent model, and you still don't. If the non-blue parts of the white light hitting the ball are sucked in and used up, then they can't still be there in the light bouncing off the ball. If that light bouncing off is lacking its non-blue parts then it is no longer white light. White light minus the non-blue part is blue light. Unless of course the blue parts are also disappearing to relocate themselves to nearby films and retinas, in which case there won't be any light bouncing off the object at all.

I said that the blue light exists only because the object has absorbed the non-blue light and continues to do so as the full spectrum light bounces off of the object. So what is left is (P) reflected light which continues to be at the film/retina. When the blue photons have dispersed (the inverse square law), the object can no longer be seen, in which case there will be no image, just white light.

Originally Posted by peacegirl

White light is bouncing off and as it does the object is continually absorbing certain photons, which is why blue light is present at the film/retina when we're looking directly at the object. But that blue light does not travel at all; it becomes a condition of sight; it does not cause sight which only means it does not travel through space and time and eventually reach the film/retina without the object being within visual range.

Originally Posted by Spacemonkey

If the light bouncing off the object no longer contains the blue photons then it is no longer white light. If it no longer contains the non-blue photons then it is no longer white light. If it no longer contains either the blue photons or the non-blue photons then there is nothing left, and no light can be bouncing off the object at all.

You're right, it is no longer white light when the non-blue photons are absorbed. This is what allows us to see the object in real time. But this does not mean the blue wavelength light is traveling. It is revealing. I wish you could get a glimpse of how the efferent version of sight works.

Originally Posted by peacegirl

No Spacemonkey, they have not teleported. Remember, the white light is traveling, and it is not the same light when we look at the object a moment later. These are new photons, but we see blue because the object is still absorbing that non-blue light. So when the lens is focused on the object, that blue light is instantly at the film/retina as long as the object is within the field of view.

Originally Posted by Spacemonkey"

If those blue-wavelength photons which at one moment were just hitting the surface of the ball are, at the very next moment, present at the distant camera film, then they have instantaneously relocated - i.e. they teleported there. That's just what teleportation means.

That's because you think the blue wavelength light is traveling, and you're also believing that one moment they are here and the next moment they are there. But there is no there. The photograph that is taken is a mirror image of the object as it is in real time. There is no difference because the object and the light that is (P) reflected are one and the same or the opposite side of the coin.

Originally Posted by peacegirl

I told you that they are being renewed by new white photons but the object is absorbing those photons so when we look at the object efferently, the light is instantly at the film/retina. The inverse square law allows that (P) light to instantly be at the film/retina when the lens is focused on the OBJECT.

Originally Posted by Spacemonkey

If the blue-wavelength photons that were hitting the object are instantly at the distant film, then they have teleported. If they are instantly at the film then they are not also in the light bouncing off the object. If they are not in that light bouncing off the object then the light bouncing off no longer contains light of all wavelengths and cannot be white light. And there are no such things as white photons.

I said the light that is (P) reflected is not white light; it is blue light. It is there because we are looking at the object. It is not there if the object is not there.

Originally Posted by Spacemonkey

These are the two key questions you are completely failing to consistently answer:

1) Where are the blue-wavelength photons, contained within the sunlight striking the blue ball, at the point in time immediately after they hit the ball?

The blue-wavelength photons are (P) reflected until the light fades (due to the inverse square law). When that blue wavelength light is too far away from the object, white light continues traveling.

Originally Posted by Spacemonkey

2) Where were the blue photons, which are at the film interacting with it to produce a photographic image of the blue ball when the photograph is taken, at the point in time immediately before the photograph is taken.

The blue photon is just coming into existence as new photons are constantly being absorbed and (P) reflected by the object.

Originally Posted by Spacemonkey

You also have no consistent story about which parts of the light hitting the object are absorbed, which parts bounce off, and which parts turn up instantly at distant films and retinas. These three options are each mutually exclusive - no individual photon, or collection of photons of a given wavelength or set of wavelengths, can be doing any two of these together.

I have been very consistent. The non-blue wavelength light is being absorbed. The blue wavelength light is being (P) reflected. Depending on the brightness and size of the object will determine the inverse square law. There is nothing turning up instantly at distant films and retinas Spacemonkey. When the lens is focused on the object, the light, being a mirror image, is instantly at the film/retina. You have to remember that the distance between the moon and us, and a streetlight and us, meets the same requirements so it really doesn't matter how far away an object is as long as the object is bright enough and large enough to be seen.

Originally Posted by Spacemonkey

If a photon is absorbed, then it cannot bounce off, and it cannot turn up at any distant film or retina.

I never said a photon that is absorbed can turn up anywhere.

Originally Posted by Spacemonkey

If a photon bounces off to travel away from the object, then it has not been absorbed, and it cannot also be instantly present at any distant film or retina.

You're back to square one. I don't think you've grasped even a little bit of this concept. Sorry.

Originally Posted by Spacemonkey

If a photon is present at some distant film or retina immediately after striking the object, then it has not been absorbed, and has not bounced off the object to travel away from it.

So what are you asking?

Originally Posted by Spacemonkey

The sunlight striking the blue ball simply consists of photons of all wavelengths in the visible spectrum. This spectrum is often represented by the acronym ROYGBIV (red, orange, yellow, green, blue, indigo, violet). So let's use a simplified model where there is one photon of each of these colors hitting the blue ball (like little marbles).

These seven differently colored photons are hitting the ball. They comprise the sunlight hitting the ball. I want you to tell me where each one of them is 0.0001sec after this collection of photons hits the ball. Which ones are absorbed (such that they get sucked in and used up, and do not bounce off)? Which ones bounce off and start traveling away from the ball at the speed of light? Which ones instantly appear at distant films or retinas? Which ones, if any, are in more than one place 0.0001sec after hitting the ball?

The red, orange, green, indigo, and violet photons get absorbed. The blue photon is (P) reflected and appears instantly at the retina because it meets the requirements of efferent vision. Cameras work the same way in that if the object is in the field of view, a mirror image will show up instantly at the film. White light continues to bounce off of the object, getting absorbed and (P) reflected as it bounces off of objects so that the light becomes a condition that allows us to see said object. White light continues to bounce off of objects and travel through space and time. There are no photons at more than one place except for white light.

Originally Posted by Spacemonkey

If you want to speak of blue light, non-blue light, or white sunlight, then the following definitions apply with respect to our seven photons:

Blue light =(def) The blue photon.

Non-blue light =(def) The red, orange, yellow, green, indigo, and violet photons.

White sunlight =(def) All seven photons.

Got it.

[Kael]

Originally Posted by peacegirl

But it does apply.

If there is no traveling then distance is irrelevant. If distance is irrelevant then the inverse-square law does not apply, and is also irrelevant.

The blue wavelength light, in Spacemonkey's example, is there because we see it, not because it travels to us

But how do we "see" it, if it does not travel to us? What is the mechanism? This is why "viola, we see!" is such an accurate summation of these ideas. You don't actually tell us how we see, only that we do. The distinction is largely lost on you though, I am sure.

but the light is constantly being replaced by new photons. The (P) reflection therefore has everything to do with the distance of the object from the observer.

Turquoise bicycle shoe fins, therefore pink radish planter hats!

You're wrong. The inverse-square law does apply. There is very little that changes except for one thing: Images (or the non-absorbed wavelength light) does not bounce off and travel through space and time. This light is present when we're looking at the object, but, to repeat, it does not travel through space and time as the full visible spectrum does.

All light is light, peacegirl. There are no distinctions between reflected light, what you call "the non-absorbed wavelength," and light from some generating source like the Sun, what you call "the full visible spectrum." It is all the same, and it all behaves the same. You are imagining differences of a sort that do not exist, and attempting to pry that imaginary gap wide enough to fit Lessans' ramblings in. When light strikes a blue ball, it does not suddenly become some different, special kind of light that does not need to travel through space and time in order to be detected. It is just light, and if it had to travel to the ball, then it has to travel to the eye.

[thedoc]

Originally Posted by peacegirl

Wrong Spacemonkey. You're not getting it. Those blue photons are present because of the ability of the object to absorb the non-blue photons, but this is a continual process where full spectrum visible light first strikes the object and the non-absorbed and absorbed light gets split up, so to speak, but this light does not bounce and travel. Only white light does this. The difference here is that, if sight is efferent, we are able to get an instant mirror image on our retina or film due to this (P) light and how the eyes work as they look out at the world.

to be cont...

fortunately Spacemonkey does 'get it' there is just light and it behaves just as science has described it for many years now. Lessans was a fool, Peacegirl is an idiot, and efferent vision is nonsense, I don't really care if Peacegirl ever answeres my questions because her answers would be nonsense and I was just asking to trap her in a lie. It's not really as much fun to try to catch her in obvious stupidity any more she really has no grasp of reality.

[Spacemonkey]

Originally Posted by peacegirl

That's false.

Which part of what I said was false, and why do you think it is false?

Originally Posted by peacegirl

The light that is absorbed is not bouncing off...

The light that turns up at the film [...] is not bouncing off of anything...

As white light bounces off...

You've completely missed my entire point. The white sunlight hitting the ball consists of two parts: the blue light and the non-blue light. Nothing else. If the non-blue part of that sunlight is absorbed, then that part doesn't bounce off. If the blue part of it turns up at the distant film, then that part doesn't bounce off either. So if neither part of the sunlight hitting the ball bounces off, then how can white sunlight still be bouncing off the ball?

Originally Posted by peacegirl

I think it does address the question. The blue wavelength light does not bounce off the object with the full spectrum light or without it. It is present as we look at the object. You're missing the whole model.

If the blue part of the spectrum is not bouncing off, then what is bouncing off cannot be the full spectrum. The blue part will be missing.

Originally Posted by peacegirl

Wrong Spacemonkey. You're not getting it. Those blue photons are present because of the ability of the object to absorb the non-blue photons, but this is a continual process where full spectrum visible light first strikes the object and the non-absorbed and absorbed light gets split up, so to speak, but this light does not bounce and travel. Only white light does this. The difference here is that, if sight is efferent, we are able to get an instant mirror image on our retina or film due to this (P) light and how the eyes work as they look out at the world.

This doesn't show that what I said was wrong at all. If the blue-wavelength photons are not bouncing off the ball, but are instead at the distant film immediately after hitting it, then they have instantaneously relocated themselves. That means they have teleported.

Originally Posted by peacegirl

I said that the blue light exists only because the object has absorbed the non-blue light and continues to do so as the full spectrum light bounces off of the object. So what is left is (P) reflected light which continues to be at the film/retina. When the blue photons have dispersed (the inverse square law), the object can no longer be seen, in which case there will be no image, just white light.

The blue light existed before it ever got to the ball. It was a part of the sunlight traveling towards that ball. The full spectrum cannot bounce off the ball if the non-blue part of that spectrum has been absorbed by the ball. What has been absorbed cannot still bounce off, and white light minus the non-blue part does not still equal a full spectrum. And the blue-wavelength photons cannot 'disperse' unless they are traveling. Only traveling light can disperse.

Originally Posted by peacegirl

You're right, it is no longer white light when the non-blue photons are absorbed.

So what bounces off the ball cannot be full spectrum white sunlight, can it?

Originally Posted by peacegirl

The blue-wavelength photons are (P) reflected until the light fades (due to the inverse square law). When that blue wavelength light is too far away from the object, white light continues traveling.

I didn't ask what happens to them. I asked you: Where are they? You haven't answered the question.

Originally Posted by peacegirl

The blue photon is just coming into existence as new photons are constantly being absorbed and (P) reflected by the object.

Coming into existence means that they didn't exist previously, and therefore cannot be the same (P)reflected photons that were previously at the object. And you previously rejected the answer that the photons at the film are newly existing photons. So you still haven't answered the question: Where were these photons just before the photograph was taken? Are they newly existing, magically popping into existence from nowhere at the film? Or were they at the object just immediately before the photograph was taken? Or were they somewhere else?

Originally Posted by peacegirl

[I]I have been very consistent. The non-blue wavelength light is being absorbed. The blue wavelength light is being (P) reflected.

Then nothing is left to bounce off the object, unless you think some light can be both absorbed and still bouncing off, or (P)reflected and still bouncing off?

Originally Posted by peacegirl

There is nothing turning up instantly at distant films and retinas Spacemonkey.

The blue photon is (P) reflected and appears instantly at the retina because it meets the requirements of efferent vision.

Make up your mind and stop contradicting yourself.

Originally Posted by peacegirl

The red, orange, green, indigo, and violet photons get absorbed. The blue photon is (P) reflected and appears instantly at the retina because it meets the requirements of efferent vision.

Then none of them bounce off the object, do they? And the blue photon just teleported from the object to the retina.

Originally Posted by peacegirl

White light continues to bounce off of the object...

Bzzzzzzzzt! The white light just is all seven photons. And you just told me none of them are bouncing off the object. Therefore white light cannot be bouncing off the object.

Originally Posted by peacegirl

Got it.

Obviously you didn't.

[naturalist.atheist]

peavegirl, what happens if my observation is different from Lessans?

[peacegirl]

Originally Posted by Spacemonkey

The problem isn't that you're positing light at multiple places at once. It's that you're positing the same light - i.e. the exact same photons - at multiple places at once.

How can that be if light is in constant motion which I have never debated?