Photons don’t work

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9 minutes ago, Strange said:

They couldn't.

Just chance. Imagine throwing a large (but not infinite) number of grains of rice into a large crowd (like a football stadium full, for example). Some people would get a handful of rice, some would get 1 or 2 grains and some would get none.

The problem seems to be that you are thinking of light as something continuous, more like water, say. If you filled the stadium with water, then everyone would get wet!

i don't think you are being thick. A lot of people struggle with this, I think. The problem is one of scale and our intuitions. We are used to thinking of light as continuous and infinitely divisible, because that is how it seems to behave in everyday situations. And there are such a large number of photons in most light sources that it is pretty close to being continuous. That is why using the photon model isn't very helpful in a lot of cases.

He's possibly thinking of light as wave which is hard to imagine as discrete.

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I get that too, a photon can only react to one eys and  stange's point that it would be determined by "chance" don't seem to carry the definitive logic here.

I want you to consider everything taken to the n'th degree and you don't seem to be doing that, iwant you to consider what I propose as being achievableable and the resultsant out come.

So, going back to the example of a laser being fired from London to Paris.......1. there is only a finite number (lots) of photons created by this event.  2. No extra photons are created during the journey. 3. (Most importantly) viewers who see the beam between london and paris are recieving photons to their retina that have been "scatterred" by dust,water or whatetever.

So my questions remain ?? If there is an observer at every angle at every nano centimetre of the journey, how can that may photons be "scatterred" without detratcting from the light beam.

What if ther was no observer? would the photons still form the same scatter pattern.

If the scatter pattern didn't involve the light being scatterred in 306 degrees it would not be seen by some observers ?? if it did involve scattering to 360 degrees at evry nano cntimteremt point the photons would be dissappattended fron the lase light beam which doesn't happen. Help me ??

Regards

Tim

My first concern with light came through I think it was called hugyens theory of secondary wavelets and I knew as soon as i saw that, many years ago that the theory of light as a particle, wave and gravity are inexsricablley linked but our current ideas of them are incorrect

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but surely this relates to the wave model also. Imagine the laser is fired from London to Paris and we consider the wave model. That is even worse??

An analogoy for showing the wave model  is crap is this:

Think of a perfect sphere,  a pin point of light or a planet.   This emits light. now there are two things to consider, it emits it as a wave  or as as a stream of photons.

So at the surface of the pin/planet the seperationion between the photons /lightwaves will be neglible say o. ten trilloin degrees, but as the waves, phototons journey in a straight line fron the sphere the degree seperating them from, each other will grow, meaninging that as you increase your hight form the pin / planet the degree between you and the neighbouring photon or wave increases, which will result in "holes" between you as your striaght line increases into the miles of space. Convesrsation may cure my inarticulate ability!

Regards

Tim

Early flight must quit, but thanks for your really helpful information, hopefully can resume tomorrow.

Stay Safe

Regards

Tim

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8 hours ago, timharvey027 said:

I want you to consider everything taken to the n'th degree and you don't seem to be doing that, iwant you to consider what I propose as being achievableable and the resultsant out come.

But it is wrong, so why should we consider it.

8 hours ago, timharvey027 said:

If there is an observer at every angle at every nano centimetre of the journey, how can that may photons be "scatterred" without detratcting from the light beam.

They do detract from the light beam. The amount of light in the laser beam itself will be less when it passes through a medium with a lot of particles that scatter the light than if it went through a vacuum. Eventually, all the light would be scattered and there will be no beam.

8 hours ago, timharvey027 said:

What if ther was no observer? would the photons still form the same scatter pattern.

Yes. (Apart from the philosophical "tree in a forest" question of "how could we know"  )

8 hours ago, timharvey027 said:

If the scatter pattern didn't involve the light being scatterred in 306 degrees it would not be seen by some observers ??

Correct. At the extreme, if there were no scattering because the laser was in a vacuum then it would no be visible from the side.

8 hours ago, timharvey027 said:

if it did involve scattering to 360 degrees at evry nano cntimteremt point the photons would be dissappattended fron the lase light beam which doesn't happen.

I'm not sure what you think doesn't happen. Photons (light) gets scattered from the laser, that is why you can see it in a medium with a lot of particles present.

7 hours ago, timharvey027 said:

but surely this relates to the wave model also. Imagine the laser is fired from London to Paris and we consider the wave model. That is even worse??

Not sure why.

Some of the light waves are scattered from particles and, as Huygens showed about 400 years ago, each particle can then be thought of as a new source of light. The total amount of light remains the same (apart from what is absorbed) but now you have light waves going in all directions.

7 hours ago, timharvey027 said:

Think of a perfect sphere,  a pin point of light or a planet.   This emits light. now there are two things to consider, it emits it as a wave  or as as a stream of photons.

So at the surface of the pin/planet the seperationion between the photons /lightwaves will be neglible say o. ten trilloin degrees, but as the waves, phototons journey in a straight line fron the sphere the degree seperating them from, each other will grow, meaninging that as you increase your hight form the pin / planet the degree between you and the neighbouring photon or wave increases, which will result in "holes" between you as your striaght line increases into the miles of space.

In the wave model, the wavefront gets spread out over a large and larger sphere. The surface area of that sphere increases as the square of the distance, therefore the brightness falls off as the inverse of the square of the distance. Confirming the well-known inverse square law.

In the case of photons, then they get more widely separated as the spread out from the source (again, by an inverse square law).

If we consider that there was a single flash of light, then eventually there will come a point where there are gaps and not all observers will receive photons and will not see the flash.

If it is a steady source of light (a distant star or galaxy, for example) then at any moment some of these distant observers might see a photon and others not. But a moment later the one who didn't see a photon might get one. And maybe some of those who got a photon before won't get one now. So, on average, they all get smaller and smaller numbers of photons with distance.

As they get further away, the average number of photons they get over time will fall in an inverse square law. They will never stop seeing photons. There might be photons occasionally passing to the left or right that they miss, but eventually they will detect one.

And note that this is not just a random example based on theory. This is what is actually observed to happen. Either in very low light conditions (where, for example, night vision scopes use photomultipliers(1) to go from an undetectably small number of photons to a large enough number to allow the eye to see them) or for light from distant galaxies (where we receive single photons from the source).

(1) I don't think photomultipliers based on the photoelectric effect are used anymore

7 hours ago, timharvey027 said:

An analogoy for showing the wave model  is crap is this:

So you don't like the photon model and you don't like the wave model (despite that fact that they both work). What do you like?

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Posted (edited)

Tim, I didn't read EVERYTHING here, but I think we need to take some facts into account, your thought experiment has certain assumptions that (if true) do make for some very weird interactions.

Let's only think of the light as discrete photons and don't add in wave-duality.
Photons go in straight lines.
Only photons that reach our eyes can be seen.

Photons won't randomly turn, but when they go through something such as air (many particles), some photons will interact with the air and bounce in other directions.
That means, if there are no particles in between point A and point B, the photons keep going straight and if the laser isn't pointing into your eye, you won't see it.
I am adding some numbers that I think I saw in the thread, their accuracy isn't relevant (a few zero's extra or more won't change much).

Let's say our beam of photons is 1019 photons/second. That is 1016 every millisecond or 1/1000th of a second.  The air is constantly moving, and there are many particles in the air for the light to scatter onto, so it's not too hard to imagine it scattering in 360 degrees.
Even if only 1% of the light scatters over said distance, that is 1012 every 1/1000th of a second.

These photons ARE detracting from the amount of photons in the beam. But the amount is so minimal that for us observes it seems like there is no difference (when talking about air, and distances that aren't too far).
When you shine light through thick fog, the light that 'comes out' on the other side, is less strong than without the fog right?
The same happens in air, but so little of the total light is scattered that it may not seem like any has scattered.

If I shine a laser through 100 meters of vacuum, or through 100 meters of air and measure how many photons reach the other end per second, then the laser through air will have lost some photons to scattering.
The scattering itself is independent of whether there are observes (people, we aren't talking about observers in the sense of the observer effect. Each gas particle that the photon beam interacts with, would be considered an 'observer' in the quantum sense, but I don't think this is relevant) I hope I use the terminology correctly for quantum observers in relation to the air particles, if other members could correct me on this, that would be great!

Now the thing that you say about these 'holes' that will form where no photons will go (as more and more scatter), is (I think, but I could be wrongtrue, however since the scattering is random, and people don't see 1 static picture but observe over time, those gaps will sometimes be filled (and other gaps will form). Additionally, over small distances such as the scales of our planet, those 'gaps/holes' where for brief moments there are no photons, will be so small (much smaller than our eyes), so that it doesn't really matter on these scales.

But with a finite amount of photons, and great enough distances, eventually only 1 photon would reach you. It is just that in the thought experiment you have constructed, the distances are way too small for the amounts of photons that there are.

Hope this helps and hopefully I didn't say anything wrong/incorrect

-Dagl

Edited by Dagl1
edited a sentence to make it make sense

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12 hours ago, timharvey027 said:

Hi Swansont,

So tell me, if photons scatter of of something e.g dust, do they scatter in 360 degrees.

Depends on the kind of scattering. Mie scattering has a forward bias. Rayleigh scattering does not (though it has a wavelength dependence)

But isotropic scatter is a reasonable approximation for this discussion.

Quote

So when everyone on say Hong Kong sees the laser beam projected into the sky, each person..lets suppose there are observers at every angle at every nanometre will all see the laser beam. If we can work out the number of photons hwich is not infinite, what happens when that number of photons is reached, or "runs out" will observers simply not see the beam

OK, well, let's say you had a 1 W laser at 620 nm - that's 2 eV per photon. You would be emitting a little over 3 x 10^18 photons per second. Even if you had a billion people watching, that's the capacity to deliver 3 billion photons per second to every observer. We get far fewer photons into the eye when we observe

If the number of photons falls below what the eye can detect, you don't see anything. Dim stars and/or ones far away can't be seen with the naked eye. You need to use a telescope, which has more light-gathering ability, in order to do so. And if you take a picture, which integrates the light gathering over time, you can see objects you can't see in real-time.

edit: I now see Strange gave a similar example already.

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Posted (edited)
On 3/11/2020 at 7:42 PM, StringJunky said:

He's possibly thinking of light as wave which is hard to imagine as discrete.

When he said 360 I pictured a single pebble tossed into a puddle that produces a ring where potentially an infinite number of observers in line with the ring produced by the single pebble. The next step up in thought would be spherical where again that word infinite seems to neatly fit in since the potential for observation points is infinite.

This possibly being a misunderstood perception of particle-wave duality. The infinite number of potential observation points would imply the need for an infinite number of of photons, but though there can be an apparent infinite number of observation points, well not even stars shine forever and the laser emitter generally comes with an off switch and a limited power supply, so an infinite number laser photons being emitted isn't likely to occur. Bringing to the forefront one of the problems with the word infinite. Often it is used wrongly. Of course my reasoning almost always tends to have plenty of weaknesses as opposed to an infinite number of weaknesses.

Actually, I used to think that the reason we see things was because as they absorbed a photon they would then emit a photon of the proper wavelength thus explaining color, so believe me Tim I am not attempting to be critical in any way, I have enjoyed the thread. Attempting to figure out where you are coming from is as much a learning experience as the answers you have been given 🙂.

Edited by jajrussel

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Posted (edited)
On 3/11/2020 at 10:51 PM, timharvey027 said:

what happens when that number of photons is reached, or "runs out" will observers simply not see the beam

I haven't followed the follow-ups very closely, as first answers were very satisfactory IMO. If you're going to spend any length of time thinking about photons, be careful, though. I just want to point out that #(photons) is not a conserved quantity. Photons are very 'dangerous' to think about in terms of little bundles of 'something.'

Edited by joigus
minor stylistic correction

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