Light transmission over large distances

[John0117]

Recently while posting a theory about how the universe works on another site, I came across a problem when dealing with waves and gravity.

The basis of the problem goes back to one of the ancient greeks and his thoughts on infinity.

If you take a curved surface, in this case a star, and draw as many straight lines as possible radiating out from its surface then no matter how many lines you draw if you draw a larger circle around the original there will be gaps between the lines, the greater the radius of the second circle the bigger the gaps.

Light travels, basically, in straight lines. EM waves oscillate, 2 waves at right angles, simulations are easy to find on the internet.

If looked at end on they would, looking at the simulations, appear as a cross.

Increasing amplitude apparently increases brightness, the only way for a cross shape to spread out is to increase in amplitude or lengthen the arms.

The question then is when looking at distant objects, in terms of millions of light years (the radius of the second circle above), how does light spread out to fill the gaps between the lines without increasing in amplitude. Brightness apparently drops off according to inverse square over distance thus reducing amplitude.

In other words if light travels in a straight line then we are infinitly more likely to see a gap than to see a light wave.

Also if it spreads out as a circle there would be overlaps and gaps thus making the distant object seem to brighten and disappear over time.

There would of course be no problem if the 2 distant objects were flat surfaces opposite each other, although this would cause a similar problem if looked at from any angle.

Hope the above makes sense.

I do have a possible solution but would like to hear from experts before making any final conclusions or causing any arguments.

 

 

 

 

 

 

 

 

[swansont]

I think you may be missing the sheer number of photons involved. 1 watt of visible light is around a few x 10¹⁸ photons/second. Even if you drop to a nanowatt, it's still several billion photons per second. Detectors have finite sizes and integration times. Under most circumstances, there are no gaps.

 

If you were far enough away and the source were weak enough, a single-photon detector would stop detecting photons continually — it would trigger as each photon hit, and be dark the rest of the time.

 

I suspect from your description of an EM wave you are thinking that the waves are changing size as the amplitude changes, but that's not what is being depicted. The electric and magnetic field amplitudes are shown, i.e. how strong the field is, rather than their spatial extent.

[John0117]

It is not the number of photons per second I am concerned about it is how they spread out across space.

That is imagine photons per second like a line of photons one behind the other, it is how the photon at the front spreads out sideways to that line.the ones behind will naturally follow.

Or put another way how does a wave-front of single photons side by side spread out sideways as they are moving forwards. The surface area of something as big as a star is still finite and so presumably only a finite number of photons will sit side by side over that area, these are the end points of the radiating straight lines in post 1.

You are I suspect thinking of frequency which in this case is not applicable unless the back ones are moving faster than the front ones in order to fill the gaps.

[Sensei]

It is not the number of photons per second I am concerned about it is how they spread out across space.

See animation that I made for other thread

http://www.scienceforums.net/topic/82391-how-on-earth-can-light-rays-reflect-a-person-from-a-mirror/#entry798307

 

(it's too uniform I know, add randomness in your imagination)

Edited May 6, 2014 by Sensei

[Delta1212]

You're missing what he said. There will be gaps. But you have to get really, really far away from the light source before the gaps will be wide enough that you will only occasionally get hit by a photon that traveled on the exact right path to reach you instead of detecting them continuously.

[swansont]

You're missing what he said. There will be gaps. But you have to get really, really far away from the light source before the gaps will be wide enough that you will only occasionally get hit by a photon that traveled on the exact right path to reach you instead of detecting them continuously.

 

Yes. To clarify, a billion photons in a square centimeter doesn't really have gaps, even if the sensor is a few orders of magnitude smaller and has a fast response time. For a 10x10 micron sensor, that's still 1000 photons a second, or a photon every millisecond on average. That will look pretty continuous in both time and space

 

When it stops looking continuous, it will look discrete.

[imatfaal]

Additionally - In three dimensions the "photons per square metre" varies in line with the surface area of an imaginary sphere with a radius of the distance of the light source to the observer - same as your example but this time in 3d.

 

If the total power of a star is P [Watts] - the intensity I [Watts/m^2] at a distance r [metres] is:

 

[latex]

I = \frac{total\ power}{surface\ area}= \frac{P}{4 \pi r^2}

[/latex]

 

P stays constant so the intensity varies inversely with the radius squared - this is one of the many instances of the famous inverse square law

[swansont]

To add to my previous response — from a practical point if you were detecting a dim signal you would simply integrate longer if you could, to try and boost your signal. The fluctuations in the photons count would make this noisy (shot noise), so you would notice a dimming and brightening of a detector owing to that noise, but for an integrating signal (like a photograph) it's just a matter of keeping the shutter open longer and figuring out how to not swamp the signal from other sources..

[John0117]

Yes thanks for the animation it shows what I mean exactly. If the dots represent photons and they are touching when in the middle then as they spread out there are gaps where there are no photons and so no light.

Post 6

A billion photons per square centimeter does not sound a lot to me but if that represents part of the surface area of a star like the sun, or a sphere with the radius of the sun then increase that radius to the maximum distance that the sun would be observable at then what area would that square cm represent and what density the photons (there would have to be the same number that we started off with) Remember we can see these objects from wherever we are along the earths orbit so there have to be enough photons available at all points along that orbit to make the source visible.

I assume the diameter of the instrument being used to do the observing would also have to be taken into account or not ?

[John Cuthber]

The simple answer is that there are gaps.

But (because there are so many photons) the gaps are very small and they don't last long, so you don't notice them.

[swansont]

 

A billion photons per square centimeter does not sound a lot to me

 

That was for a nanowatt of visible light. We get about a 100mW/cm^2 of solar radiation on the earth, which means 100 million times more than that billion. 100 million billion. Sound better? And, of course, that has dropped off with the square of the distance from the sun — it's bigger if you get closer.

[John0117]

As you may realise I dont know how many photons it takes to make something visible so I am trying to keep things simple by using density per sq cm which can then be multiplied up by someone who knows how much light is required to make something visible.

Doing some rough calculations taking the suns radius as approx 2.5 light seconds and 1 sq cm equaling 1 sq cm at that point taking photon density as 1.5 billion per squ cm. if area increases at the rate of r2 then at 1 light year the radius has increased by a factor of 12,600,000 if we square that and apply it to our original sq cm it has increased to approx 1.5 bil sq cm or a density of 1 photon per sq cm. So at 10 light years we get a density of 1 photon per 100 sqcm and this is not far in universal terms and I initially thought it would only show up over intergalactic distances. Of course if we look at volume we would start with 1 cubic cm (this would give us a starting point of around 1.8 billion billion (sqrt 1.5 cubed) or somewhat more than 100 million billion mentioned) which would still give us a density of 1.8 bil bil photons divided by 1.5 bil cc at 1 light year and at 10 light years that number divided by 100 per cc. I am only increasing sideways assuming depth remains the same. I am of course willing to accept my maths may be incorrect. At what point or density would something cease to become visible.

Edited May 7, 2014 by John0117

[swansont]

Visible to the naked eye or visible to instruments? I think the eye triggers at a few tens or hundreds of photons per second, over the entire area (with the iris opening up because it's dark). Lab instruments can detect single photons.

 

So if you have a dim source that's a nanowatt/cm² at 1 cm (this more than a million times dimmer than a 100W light bulb.), and we assume that's giving you a billion photons into your eye, at 100 meters that's 10 photons/sec. So at several tens of meters is where you probably couldn't see it anymore, if it were otherwise pitch black.