Expanding Light: Wave's Radius Increase - Why Not Dim?

[vorcil]

here it goes

you know how from the origin of a light wave in space
i.e a star

when the waves of light travel outwards, the radius of that wave increases right?

dosen't that mean that the wave loses energy at those infinite points along the wave as the circumference of that wave increases?

i'll try draw a quick picture in paint,
http://img408.imageshack.us/img408/3512/sunpic.jpg

because the way i see it,
take a photon on the outer edge of the wave, it expands right? to get the radius to increase? so dosen't it lose energy?
so wouldn't that mean that the star light we see is actually more dim?

i think this is the same for all waves,
would someone please explain to me why I'm wrong?

I'm not really sure how to explain it, but I've had this idea in my head ever since i finished college and started Uni

 

[vorcil]

If so, why does the wave increase it's radius in the first place?

shouldn't it go in a straight line? leaving blank spaces between the waves as it gets MUCH further away
http://img408.imageshack.us/img408/2271/sunpicq.jpg
and i know that is false because we see light from millions of lightyears away
so the waves must radiate outwards, bringing me back to my original idea, dosen't it lose energy as it expands radially outwards?

why do waves travel out in a circle?

 

[Tuomaaca]

dosen't that mean that the wave loses energy at those infinite points along the wave as the circumference of that wave increases?

The wave doesn't lose energy, but intensity (energy is proportional to wavelength, intensity to the number of photons).

because the way i see it,
take a photon on the outer edge of the wave, it expands right? to get the radius to increase? so dosen't it lose energy?
so wouldn't that mean that the star light we see is actually more dim?

The wave is spread out over a larger and larger surface (as it travels outward). As the radius increases, the photons are spread further and further apart, and the intensity (# of photons) decreases. So lower intensity means less photons, which means fainter light. (Ignoring redshift/blueshift, these photons travel at the same wavelength and carry the same energy as when they were emitted).

Why does a wave travel outward from its source? I believe that answer is in the wave equation.

 

[DaveC426913]

The wave doesn't lose energy, but intensity (energy is proportional to wavelength, intensity to the number of photons).

The photon model of light had an obvious answer (fewer photons), but I withheld because the OP was asking about the wave model of light...

But you're right. Intensity is what drops. Intensity is represented in the wave model by wave amplitude. So, as the wave spreads, its amplitude decreases.

 

[vorcil]

The photon model of light had an obvious answer (fewer photons), but I withheld because the OP was asking about the wave model of light...

But you're right. Intensity is what drops. Intensity is represented in the wave model by wave amplitude. So, as the wave spreads, its amplitude decreases.

how does it decrease though? what would happens to the photon packets as they get infinitely small? like a photon on the edge of the universe, what's happening to it! the amplitude can't get smaller and smaller forever!

 

[vorcil]

how does it decrease though? what would happens to the photon packets as they get infinitely small? like a photon on the edge of the universe, what's happening to it! the amplitude can't get smaller and smaller forever!

as the radius doubles, the amplitude thus intensity halves?

what would happen to the amplitude at an infinite radius?
would the photon packet just disappear?

 

[chroot]

Yes, the intensity of any light source seen from an infinite distance would be zero.

- Warren

 

[Tuomaaca]

as the radius doubles, the amplitude thus intensity halves?

for a 2-D wave, the amplitude would be divided by the circumference 2*pi*r.
In 3-D, it would decrease by the surface area of the sphere 4*pi*r^2.

 

[russ_watters]

The picture in the OP is wrong: you can demonstrate how waves propagate by dropping a rock in a pond: the wavelength does not increase as they propagate, only the amplitude decreases (as others have said)/

 

[vorcil]

Yes, the intensity of any light source seen from an infinite distance would be zero.

- Warren

wouldn't that mean you have destroyed energy??!
i'm not talking about light seen, I'm talking about the actual photon of light itself, that has traveled that infinite distance, with it's intensity getting smaller and smaller

are you saying that it will reach 0?
what happens when it reaches 0?

 

[russ_watters]

It will reach zero when it reaches infinity...

...it won't reach infinity. Infinity is not a location.

 

[Janus]

wouldn't that mean you have destroyed energy??!
i'm not talking about light seen, I'm talking about the actual photon of light itself, that has traveled that infinite distance, with it's intensity getting smaller and smaller

are you saying that it will reach 0?
what happens when it reaches 0?

It would be more proper to say that the intensity of light approaches zero as the distance approaches infinty. It will never actually become zero because you can never actually be an infinite distance away.

The energy of the photon itself does not decrease, only the number of photons that reach a given point at any time as the photons emitted are spread out over the surface of the expanding wave front.
At an infinite distance, you would have a finite number of photons spread over an infinite surface, so the chaces of any photon being at any point of the wavefront would be 0. But again, in reality you can never have an infinite separation, so this situation can never occur.

 

[DaveC426913]

wouldn't that mean you have destroyed energy??!
i'm not talking about light seen, I'm talking about the actual photon of light itself, that has traveled that infinite distance, with it's intensity getting smaller and smaller

are you saying that it will reach 0?
what happens when it reaches 0?

In the photon model, intensity is modeled by the number of photons. "At infinity", the photons is so highly-dispersed that the number of photons that reach you is simply zero. i.e. you could wait forever for one photon to be detected.

 

[vorcil]

In the photon model, intensity is modeled by the number of photons. "At infinity", the photons is so highly-dispersed that the number of photons that reach you is simply zero. i.e. you could wait forever for one photon to be detected.

This is what I'm talking about, wouldn't their be blank spots between the waves?
and shouldn't we think of the distance from another star to us being very large,

http://img408.imageshack.us/img408/2271/sunpicq.jpg

 

[russ_watters]

This is what I'm talking about, wouldn't their be blank spots between the waves?

Essentially, yes. There are sensors that are capable of detecting a photon at a time or a very small number of photons to register a signal. Telescopes can gather individual photons separated by some distance and even time and focus them on a detector to develop an image.

and shouldn't we think of the distance from another star to us being very large,

Not sure what you mean. Yes, the distance between us and the stars is large compared to what we are typically used to.

 

[vorcil]

Essentially, yes. There are sensors that are capable of detecting a photon at a time or a very small number of photons to register a signal. Telescopes can gather individual photons separated by some distance and even time and focus them on a detector to develop an image.
Not sure what you mean. Yes, the distance between us and the stars is large compared to what we are typically used to.

So potentially we are missing billions of stars because their photons are spread apart and not reaching us?

 

[DaveC426913]

So potentially we are missing billions of stars because their photons are spread apart and not reaching us?

Yes. They are dim.

This is why we make telescopes that are more and more powerful to gather very dim light.

But note, we are not at infinity from any stars, therefore the number of photons coming from them is not zero, just very, very few.