No. Water waves are waves in water- water does not itself consist of waves. When we say that light can be considered either a wave or a particle we are not talking about wave of light particles.Landru said:Doesn't a prism prove the same thing?
It often seems that the particle/wave question is presented to us layman as an either/or situation, is it a wave, is it a particle, but water and electricty consist of particles that can "wave" so I don't really understand what is the issue at hand.
HallsofIvy said:When we say that light can be considered either a wave or a particle we are not talking about wave of light particles.
Defennder said:How does a prism prove particle-wave duality?
jenzao said:Im just not getting the general concept of what this experiment proves.
Please explain in very simple language --thanks!
Landru said:The photo electric effect proves the particle aspect. I'm not sure that the double slit experiment does that or not. The rainbow of color that comes out of a prism is due to the changing the wave lengths of the light entering it.
Before the Michelson Morley experiment, many people would tell you that EM waves are waves propagating along the ether medium. But no one ever envisaged of the ether as being composed of particles, except perhaps those particle-minded physicists like Newton who subscribed to the Cartesian notion of a mechnanical world.Landru said:If you supposed that light was a particle-free wave then what would that wave be made of? If it was a wave expressed in the particles of something else like a sound wave then how would light traverse a vacuum?
Tw15t3r said:The photoelectric effect, i.e. the jumping of electrons from inner shells to outer shells when shone on by light, proves the particle theory of light because the energy needed to cause an electron to jump say from shell 1 to shell 2, is very specific (quantised), and if light was a wave, light would deliver a continuous energy, not in "bursts" which is needed to jump the electron to an outer shell.
This is incorrect. The photoelectric effect is about electrons being liberated (emitted from metal surface) and not merely promoted from an inner shell to an outer shell. The theory which makes use of the electrons being promoted from the inner shells to the outer is the one on gas absorption of light.Tw15t3r said:The photoelectric effect, i.e. the jumping of electrons from inner shells to outer shells when shone on by light, proves the particle theory of light because the energy needed to cause an electron to jump say from shell 1 to shell 2, is very specific (quantised), and if light was a wave, light would deliver a continuous energy, not in "bursts" which is needed to jump the electron to an outer shell.
The experiment showed that the intensity of incident light on the metal made no difference if there was no photoelectric current observed for a particular frequency of the light source. In classical EM theory, we would expect the energy of the incident light wave to be dependent only on the amplitude of the waveform. The photoelectric effect disproved that notion, because it showed that frequency of light mattered as well. But of course this made no sense in the wave theory, whereby frequency has no influence on the energy of the wave.Cryptonic26 said:I guess my main issue here is that I don't really understand what it is about the mechanics of the photoelectric effect that can't be described with pure wave theory in some form. For example; Could not a summation of two differing wavelengths provide the precise 'burst' of energy required to jump electrons to an outer shell at a consistent frequency? I'm considering that each waveform would provide continuos energy onto the particle independent of each other, but the summation would create a third sine wave oscillation effect onto the particle that would apply the correct energy at one of the infinite crossections of the wave summation to have the observed effect of shell jumping. Which doesn't even account for the great number of wave combinations that a particle experiences from any particular light source.
Defennder said:Before the Michelson Morley experiment, many people would tell you that EM waves are waves propagating along the ether medium. But no one ever envisaged of the ether as being composed of particles, except perhaps those particle-minded physicists like Newton who subscribed to the Cartesian notion of a mechnanical world.
Cryptonic26 said:<snip>
I expect that there is a mathematical reason why this exact explanation couldn't be used to describe the photo-electric effect, however I'm just trying to seek out the reasoning behind why pure-wave form photons were ruled out entirely beyond the shadow of a doubt as a possibility.
I didn't claim so, I was addressing another poster's post on the propagation of EM waves through a medium.Andy Resnick said:The Michaelson-Morley experiment had nothing to do with the wave/particle nature of light (or matter). The corpuscular theory of light was held in prominence well into the 1800's: Poisson, in particular, was one of the last hold-outs.
The aether was assumed to be a continuous field of something that pervaded all the universe.
Andy Resnick said:There's a difference between intensity and energy: a low-frequency (low energy) light source of 10 bizillion watts will not eject a single photo-electron, while a high frequency (high energy) light source of incredibly low intensity will eject photo-electrons at a well-defined rate, proportional to the intensity.
Based on this data, the interpretation is that light comes in discrete packets of energy, in correspondence with a particle-type model.
Defennder said:Well the key point about the photoelectric effect is that the wave theory assumes that energy of a wave is influenced only by amplitude and not by its frequency. Which effectively means that even if the frequency affects the energy, and if an incident light source is really low, you can compensate for it by increasing the intensity of the incident light by a lot. And this was contradicted by photoelectric effect.
Landru said:That's really hard to make sense of.
Think of it this way. Wave theory says that the energy of a wave, such as light, is given by its amplitude, not the frequency. Which means to say that given a particular wave, increasing its amplitude would also increase the amount of energy a light wave could provide to metal surface electrons. The photoelectric effect is the observation that metal surface electrons could be liberated from the surface by shining a light source of sufficient energy on it.Landru said:That's really hard to make sense of.
The keyword here is quantisation of energy. The wave theory doesn't say that energy is quantised, whereas the particle picture strongly suggests so. There are plenty of other observations which complements the wave-particle duality picture as well, such as the double slit experiment.Cryptonic26 said:In other words.. If one could offer an alternative explination as to how a pure wave model could create the same effect, would that throw the fundamentals of particle wave duality theory into question? Or are there other reasons that suggest particle wave duality as well as this observation?
Cryptonic26 said:I think what it is that bothers me the most about these statements is that assumptions are required to make these statements true. At least, that's what I'm gathering so far.
Would it be logically incorrect to say that there is a possibility that photowave mechanics were not fully understood when it was decided that only particle photons could create this photo-electric effect?
In other words.. If one could offer an alternative explination as to how a pure wave model could create the same effect, would that throw the fundamentals of particle wave duality theory into question? Or are there other reasons that suggest particle wave duality as well as this observation?
Lol, yeah. The double slit experiment also proved the wave-particle duality of things at the very small scale, like electrons for example. You would imagine electrons to be "billiard balls", but hey, no, they are probability waves.Defennder said:Anyway it looks like this thread is turning into a discussion of the photoelectric effect instead of the significance of the double slit experiment.
Defennder said:<snip>
The wave theory says that since the amount of energy is determined by the wave amplitude, which is proportionate to the intensity of the light source (Poynting vector), <snip>
Defennder said:This is incorrect. The photoelectric effect is about electrons being liberated (emitted from metal surface) and not merely promoted from an inner shell to an outer shell. The theory which makes use of the electrons being promoted from the inner shells to the outer is the one on gas absorption of light.
Defennder said:Think of it this way. Wave theory says that the energy of a wave, such as light, is given by its amplitude, not the frequency. Which means to say that given a particular wave, increasing its amplitude would also increase the amount of energy a light wave could provide to metal surface electrons. The photoelectric effect is the observation that metal surface electrons could be liberated from the surface by shining a light source of sufficient energy on it.
The wave theory says that since the amount of energy is determined by the wave amplitude, which is proportionate to the intensity of the light source (Poynting vector), if shining a low-frequency light on metal does not liberate any electrons (which are measurable as photo-current), all one needs to do is brighten that light by increasing the intensity and hence amplitude of the incident light wave. Even if we assume that the the frequency of the light affects the amount of energy associated with the wave, it was thought that one could always compensate for the low-frequency light wave by increasing the intensity of the light source.
The photoelectric effect shows that this isn't true. In other words, if one uses a low frequency light source, no matter how great the intensity (or brightness) of the light source is, there would not be any electrons emitted from the surface of the metal. In contrast, if you instead use a low-intensity but high frequency light source there will be electrons ejected from the metal.
Jeff Reid said:How would physicists know if double slit experiment results aren't due to interaction with electrons in the molecules at the edges of the slits? There could be some special interaction between these electrons and either light or a free stream electrons. One property of the molecules at a sharp edge is that they are more "exposed", where both refraction and reflection could take place.
My point was about refraction (the apparent bending of light), not reflection.Landru said:Assuming the edges of the slits could and did reflect light like a perfect mirror the interferance pattern would be extremely faint in comparison to the line of site, but as it is the interference is relatively bright. The number of photons that follow the wave outline greatly outnumbers what would be accounted for due to reflection alone.
Jeff Reid said:My point was about refraction (the apparent bending of light), not reflection.
Landru said:What I don't understand and what I can't find is why a single high energy photon can pop off an electron but lots of low evergy photons won't even if the low evergy photons are collectively more energetic than the single high energy photon.
Cthugha said:Well, in principle they can via two photon absorption, but this process is usually not very likely as it is a nonlinear effect. Those electronic excitation processes happen very fast, somewhere between the attosecond and femtosecond range, so two photons need to "hit" the electron in a very short time window somewhere between 10^{-15} and 10^{-18} seconds.
In an hypothetical example you now have a monochromatic light source of 10 mW power (this is already more than one of the HeNe lasers in my lab has), which emits photons of 1 eV energy (somewhere in the infrared), but you need exactly 2 eV to liberate the electrons. A quick calculation shows, that there are about 6,2 * 10^{16} photons emitted per second, so the number of photons arriving within a femtosecond is about 60 photons. Now you also have to consider, that the time window is smaller than a femtosecond, the beam will be very large compared to the cross section of the electron, conservation rules apply (spin, for example) and by far not every photon will interact with the electron. So even with a VERY optimistic estimate, you will have a few thousand two photon absorptions per second and a few liberated electrons.
So the reason, why the high energy photons work better, is that you have to compare the number of photons arriving within a certain time window (high energy case) to the number of photon pairs arriving within a certain time window (low energy case). So unless you go to very, very high intensities, liberating electrons with low energy photons won't work.
