Nick Martin said:For all we know it may go really slowly once it gets out of our solar system, or indeed much faster.
I've never thought of that (i.e. the light wavelength compared to the distance of the atoms in a solid medium). That's a good point. However, is it correct to assume that the light-wave doesn't travel at speed c if its full wavelength is not "completed" through its travel? (It's like to say that the sound doesn't travel at the speed of sound if the the distance between the transmitter (speaker) and the receiver (microphone) is less than a full wavelength, which is not correct.)sophiecentaur said:I'm going to throw something else in, here. You are talking about "light" but any model you come up with should also be applicable to 1500m long waves and the shortest of gamma waves.
In a 'condensed' substance, the spacing between atoms / molecules will probably be less than one wavelength of visible light. So it is not realistic to talk in terms of launching a light wave (or photon, if you must) and then it arriving at another atom. (That would be like what happens in a low density gas, where the effect is random scattering from individual atoms.) The fields around each atom (and all the other nearby atoms) will all be involved in setting up the energy levels and the transitions. The model to use is much more classical - coupled oscillators or a transmission line with masses on springs. The 'space' in between (where you could say that c applies) is only part of it; it's the interaction between the charge systems along the line that counts. The contribution to the delay along each step the journey is, of course, affected by the 'c' delay of the fields but each atom is also contributing a significant delay as it reacts with the fields around it and the fields of its neighbouring atoms. This 'loads' the line.
RF model:
If you take a line of parallel dipoles (say each one is 1m long) and you feed the first one with a 150MHz RF signal, that signal will propagate along the line. Some of the power will couple with the each dipole- they each have an equivalent cross section- but slower than c because each dipole introduces a phase shift as the currents flowing will cause a re-radiated signal that lags behind the incoming signal. The signal that emerges from the other end will be the net sum of the incident signal and each of the re-radiated signals. But there's something else here. Each element will be interacting with its neighbours (depending on the spacing). Appropriate spacing can produce a Null in the emerging RF wave in the direction of the line.
In a solid, the system is three dimensional but the same principle is at work and the majority of the power will travel in a straight line.
How do you know that the wave equations are valid inside the atoms?George K said:Every medium consists of 99,99999... % vacuum. So, what happens when the light travels inside these "vast areas" of vacuum (as it travels inside this material)? Is its speed slower than c? (And if yes then how is this possible?)
That's a reasonable question but the answer is that you don't need a 'whole wavelength' to pass in order to be affected by the speed of the wave. Any change in the E field is delayed by a time x/c where x is the distance. A 200kHz signal (λ = 1500m) is still delayed by 3ns over a distance in space of 1m. A minuscule amount of phase shift (delay) but it's there.George K said:However, is it correct to assume that the light-wave doesn't travel at speed c if its full wavelength is not "completed" through its travel?
Except that the 're-emission' is not from individual atoms but the whole region of 'charge systems' that the wave is passing through. That ensures that a ray will pass through without being dispersed so much that it is not recognisable (which would happen if single atoms were involved.)George K said:the medium's atoms are stimulated by the incident light and they re-transmit light themselves.
That's exactly what I meant. E.g. the propagation velocity of an AC signal on a wire is (somewhat less than) c. It doesn't matter if the signal's wavelength is very large and the wire's length is very small (i.e. much smaller than the signal's wavelength). The propagation velocity is always the same (≈c). That's why I think that your specific citation (i.e. the light's wavelength compared to the atom's distance, as mentioned in your previous post) was rather pointless.sophiecentaur said:That's a reasonable question but the answer is that you don't need a 'whole wavelength' to pass in order to be affected by the speed of the wave. Any change in the E field is delayed by a time x/c where x is the distance. A 200kHz signal (λ = 1500m) is still delayed by 3ns over a distance in space of 1m. A minuscule amount of phase shift (delay) but it's there.
Yes, of course, the incident light is interacting with a whole region of the lattice. (The interaction with the single atoms is a rather wrong and misleading explanation.)sophiecentaur said:Except that the 're-emission' is not from individual atoms but the whole region of 'charge systems' that the wave is passing through. That ensures that a ray will pass through without being dispersed so much that it is not recognisable (which would happen if single atoms were involved.)
How do you know that are not valid?Svein said:How do you know that the wave equations are valid inside the atoms?
Remember what happened when they thought Newtonian mechanics were valid inside the atom?George K said:How do you know that are not valid?
I think it was rather that either you didn't get my point or you weren't clear about what point you were making. (Was I actually disagreeing with you?)George K said:That's why I think that your specific citation (i.e. the light's wavelength compared to the atom's distance, as mentioned in your previous post) was rather pointless.