Question on colours

[gogetawb]

Ok so before you start yelling at me because I spell colour with a "u".

Heres my question. I understand that colours are differenciated by the wavelength of their wave and that different surfaces reflect light with different wavelengths. What I dont get is what makes an atom of say gold, reflect light at a different wavelength than say silver. Geez I hope I explained myself enough.

[Jay-qu]

you explianed it fine, and i wont fry you for spelling colour with a 'u' because thats how we do it in Australia :) Atoms of different elements are able to absorb different wavelengths because of the discreet energy levels at which the electons in the atom can exsist. They can only absorb photons of the exact energy that is required to move between these energy levels.

 

These energy levels arise from the electrons only been able to orbit in regions of space where they can set up a standing wave. The radius of these orbits can be calculated by:

 

r = (n*lambda)/(2*pi) , where n = 1,2,3... lambda = de Broglie wavelength of the electron

[infamous]

Ok so before you start yelling at me because I spell colour with a "u".

Heres my question. I understand that colours are differenciated by the wavelength of their wave and that different surfaces reflect light with different wavelengths. What I dont get is what makes an atom of say gold, reflect light at a different wavelength than say silver. Geez I hope I explained myself enough.

Welcome to Hypography gogetawb, and BTW, your allowed to spell color..colour either way according to Websters. At any rate, tell us a little about yourself and show your true colours, wer're pleased to have you on board.................enjoy.

[gogetawb]

Hey. Well two comments:

1-Amazing answer thanks. But what would happen in the case of a translucent, coloured material. ie a material which lets light with a certain wavelength through (stained glass or something of the sort). How come light with a certain wavelength is "let through" and reflected at the same time?.

2-Hmm lets see. Showing my true colours, quick brief about me (which sould have been done in Introduction forum but im too tired for that.) Im your standard mexican adolecent/struggling musician/physicist/chemist/biologist/philosopher/martial artist/writer. Basically I joined the forum so I can suck all information from your brains, learn it, re-phrase it and claim it is my own genius thinking :) (hope you dont mind)

[Jay-qu]

1-Amazing answer thanks. But what would happen in the case of a translucent, coloured material. ie a material which lets light with a certain wavelength through (stained glass or something of the sort). How come light with a certain wavelength is "let through" and reflected at the same time?.

 

I dont know :)

 

I asked my teacher once and he gave me the same answer...

[Bo]

the easiest explaination for that is to simply see a material as a bunch of dots (the atoms), and the light as another dot, trying to pass through. now if the light doesn't hit an atom, it isn't refrected.

(This is a oversimplification of a process that is actually quantum mechanical in nature; the problem lies in the fact that 'light hitting an atom' is a process that can't be described by classical physics and that atoms aren't dots)

 

Bo

[gogetawb]

Staying with the dot analogy. How can (in translucent materials) the light dots hit the atom, be reflected as red dots AND go through the material as red dots. My logic says that if light passed a translucent material which is red, red light would be reflected and ANYTHING BUT red light would go through.

Ive heard a theory that says that when photons hit an atom their energy is "modified" to equal the individual properties of that atom. The electrons would then change shells like crazy which would cause the photons to be produced (these photons would have been emited with a certain amount of energy/wavelength/colour). This implies that the photon that is originally hitting the atom isnt the same photon that "leaves" it (although they are produced from the same energy). This theory would be the one I have encountered which best could explain light acquiring a wavelength WITHOUT it having to be reflected. The only problem is that a light ray leaving a translucent material isnt equally distributed, it is still a ray, (which means it couldnt have been emanated by anything inside the material, which frustrates me). I could be talking out of pure ignorance so please feel extremely free to correct anything ive said. And dont even feel extremely free to answer my question...feel obliged :)

[Torson]

Hello gogetawb,

 

Describing light encountering a transparent and non-transparent material is complicated as you need to look at what is happening at the particle level. I will try to explain it in a “light” version rather than the “heavy”, which requires a lengthy physical outline.

 

Before explaining that, a quick look at vibrations. In general we can say that everything vibrates in one way or another, down to the smallest particle. We may not sense all these vibrations as they are out of a frequency range or amplitude which we cannot sense.

 

Light is observed as a wave (EM), carried by the photon, but you and other masses will experience the wave as vibrations because they are coming towards you standing still in comparison with the light-speed, just as if you were holding the end of a rope and someone were shaking the rope at the other end.

 

If you were backing off from a light-source with the speed of light, you would not see the light even if it had hit you, you would only see one amplitude of the wave.

 

As you are a musician you know about resonance frequencies, a piano-wire is vibrating at its resonance frequency and if you change the wire’s parameters by stretching it or making it longer/shorter, the frequency will go up or down. If you put some chewing gum on the string, you will change the mass of the wire and no matter how hard you hit the key, it gives no sound (maybe a “bump”). The chewing gum is absorbing the energy you put in.

 

Phonon is the mode of vibration in the lattice of atoms in a solid, i.e. a “resonance” frequency of the combinations of atoms and how they are arranged within the solid. There are many phonons in a solid. You can have same atoms in a solid but different arrangement, like hydrocarbons in a diamond or in graphite, one is transparent and the other not, the phonon is not the same in the two solids.

 

A transparent solid is a solid which has no phonons (modes of vibration) corresponding to the wave (or frequency) of light. If light hits a solid which has the phonons of light, the energy will be absorbed and amplifies the existing phonos (ultimately results as heat). If light hits a solid which has no corresponding phonos, the energy can not be absorbed and therefore just temporarily absorbed, then re-transmitted in the direction it came in (looking away from refraction).

 

If the phonon is very close to the lights frequency, some of the light’s energy will be absorbed and the balance of the energy is re-transmitted but with the frequency of the phonon, which may change the colour of the light and reduce intensity. (hope this is answer to your question)

 

At the surface of the transparent solid there is a slight disturbance in the phonon compared to inside the solid, which makes the light reflect (bounced back to the media it came from)

[gogetawb]

Thanks Torson. Thats an amazing piece of physics. No wonder I hadnt figured it out.

[Jay-qu]

If you were backing off from a light-source with the speed of light, you would not see the light even if it had hit you, you would only see one amplitude of the wave.

 

What happened to relativity :) ...light always traveling at c in all intertial frames of reference of the observer?

[Torson]

Relativity is in good shape! The observer in this case is moving with the light at speed c, i.e. the light does not move relative to the observer.

[Torson]

Double posting

[Jay-qu]

hmm... i dont know, I was pretty sure that light goes at the same speed relative to all observers

[Erasmus00]

Relativity is in good shape! The observer in this case is moving with the light at speed c, i.e. the light does not move relative to the observer.

 

Light always moves at c relative to any observer. Thats how relativity works. Not matter how fast you move, light never stops moving.

-Will

[Torson]

Light always moves at c relative to any observer. Thats how relativity works. Not matter how fast you move, light never stops moving.

-Will

 

Maybe you misunderstand a little.

Light has constant speed c except when penetrating transparent solids, gases, fighting gravity etc.

As with anything moving, light is subject to relativity when measuring its speed from an observer.

Example: You are walking down the street with a speed of 10km/hr and I am standing still measuring your speed relative to me, which then will be 10km/hr.

If I am walking with a speed of 5km/hr in the same direction as you, the relative speed between the two of us is 5km/hr.

If I am walking with a speed of 10km/hr in the same direction as you, the relative speed between the two of us is 0km/hr, but relative to a fixed mark on the street we are both walking 10km/hr.

This would be the same with a photon (light), if it is travelling 300 000km/hr and I am travelling at 200 000km/hr beside it in the same direction, the relative speed between us is 100 000 km/hr, and if I am travelling at 300 000km/hr (of course this is not possible, no mass can, but that’s not the point) the relative speed between me and the photon is 0km/hr.

What I think you are confusing the issue with is this: If I am travelling at 100 000km/hr and send out a beam of light in the direction I am travelling, the speed of light will NOT be c + 100 000km/hr, it will be 300 000km/hr. If I am travelling at c and try to send out a beam of light, I would not be able to see the light.

[Jay-qu]

This would be the same with a photon (light), if it is travelling 300 000km/hr and I am travelling at 200 000km/hr beside it in the same direction, the relative speed between us is 100 000 km/hr, and if I am travelling at 300 000km/hr (of course this is not possible, no mass can, but that’s not the point) the relative speed between me and the photon is 0km/hr.

 

This is where you are wrong :) , no matter how fast you go you will always measure light to be going at c! This is a direct concequence of special relativity

[Erasmus00]

This would be the same with a photon (light), if it is travelling 300 000km/hr and I am travelling at 200 000km/hr beside it in the same direction, the relative speed between us is 100 000 km/hr, and if I am travelling at 300 000km/hr (of course this is not possible, no mass can, but that’s not the point) the relative speed between me and the photon is 0km/hr.

What I think you are confusing the issue with is this: If I am travelling at 100 000km/hr and send out a beam of light in the direction I am travelling, the speed of light will NOT be c + 100 000km/hr, it will be 300 000km/hr. If I am travelling at c and try to send out a beam of light, I would not be able to see the light.

 

I believe it is you who has confused the issue. The corner stone of relativity is that the speed of light is always c relative to an observer, regardless of motion. If you travel at .9999c and send out a beam of light, you'll see it move at c. Here is the wiki article on the famous Michelson Morley experiment (which helped prove this counterintuititve effect) http://en.wikipedia.org/wiki/Michelson-Morley_experiment

Follow the special relatitivty link at the bottom for more info.

-Will