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having problems trying to understand and derive the equation for intensity in the double-slit interference pattern. any sort of help would be welcomed. thanks
No, a photon is a unit that cannot be divided or spread out. Wherever it is absorbed, it is absorbed as a whole with its entire energy. However, an inverse square rule (like you described) can be used to understand how the probability of that photon striking a given point spreads out with distance.Originally posted by FrankMak
This really doesn't answer the basic question that was posed, and whether the energy of the photon is subject to the square of the distance rule.
Originally posted by FrankMak
Think of the electromagnetic wave as a sphere that keeps expanding and the energy distribution in the wavefront has to spread itself over a greater and greater spherical surface area.
Originally posted by Doc Al
No, a photon is a unit that cannot be divided or spread out. Wherever it is absorbed, it is absorbed as a whole with its entire energy. However, an inverse square rule (like you described) can be used to understand how the probability of that photon striking a given point spreads out with distance.
Yes.Originally posted by David
Wouldn’t this idea match what I just said? Light from a light bulb dims at a distance because we are shining fewer photons per sq cm on a flat white wall at a distance, because the individual photons are spreading out the further away from the source they get?
Does the energy of a single photon get spread out according to an inverse square law? No. (I thought I had answered this, but perhaps I was unclear.)Originally posted by FrankMak
The basic question I was asked, is the energy of a single photon subject to the inverse square of the distance rule?
Originally posted by FrankMak
Once a single photon is emitted from a point source in a particular direction, the photon has no connection to the source, and since it does not in itself expand, why would it lose energy? Once on its way it is self sufficient, correct? /B]
Originally posted by Doc Al
Yes.
You are clearly a perceptive and intelligent person...Originally posted by David
Since you are a smart guy ...
As russ_watters points out, the demo is not to scale. That's not a stupid question at all: the laser must illuminate both slits or it won't work!In this demonstration, why wouldn’t the dot of laser light hit the card the slits are on, right in the middle between the slits, and thus not go through either slit?
Now you are asking the really tough questions that get to the heart of quantum mechanics! The same thinking applies to photons or electrons. In order to "understand" what's going on, you must consider the "wave" or quantum mechanical nature of the photon or electron. The wave function representing the particle must be such that it can be considered to go through both slits.Originally posted by David
What do they mean when they say a single electron is fired through both slits at the same time, or am I misunderstanding what they are saying about the electrons and the slits? Does the electron go through one slit or the other, or is it also split because the slits are so small and so close, and so it goes through both slits at the same time?
Originally posted by Doc Al
Now you are asking the really tough questions that get to the heart of quantum mechanics! The same thinking applies to photons or electrons. In order to "understand" what's going on, you must consider the "wave" or quantum mechanical nature of the photon or electron. The wave function representing the particle must be such that it can be considered to go through both slits.
I put the word "understand" in quotes, because if you mean how is it possible for a particle like an electron to go through both slits at once: as long as you think of a particle as a classical object (like a tiny marble) you cannot understand it!
Originally posted by Doc Al
The great Feynman said: "The basic element of quantum theory is the double-slit experiment. It is a phenomenon which is impossible, absolutely impossible to explain in any classical way and which has in it the heart of quantum mechanics. In reality it contains the only mystery ... the basic peculiarities of all quantum mechanics."
There are several key differences between the watermelon and the electron. When the watermelon splits apart it is essentially destroyed: all we detect are pieces of watermelon. But with electrons we never detect anything but whole electrons. Also, the watermelon does not display those pesky interference patterns.Originally posted by David
Let’s say we throw a watermelon through a picket fence. The watermelon is going to split apart and go through two or more of the slits between the pickets at the same time, yet a watermelon is not a “wave”.
Could a particle-electron be splitting in the same manner, with half of it going through one slit and the other half going through the other slit?
Yes and yes.Is the “mystery” of the electron going through both slits at the same time based on the idea that it is “impossible” for a “particle” to go through both slits at the same time, or is it based on the fact that an “interference pattern” shows up on the screen behind the two slits?
For the reasons I gave above, I do not think this is a viable explanation. Not if by splitting you mean the physical splitting of an electron into two electron "pieces".Yet another question: Could the “interference pattern” actually be a pattern caused by the splitting apart of the particle-electron, with, in some cases, a bigger part going through the right slit, while the smaller part goes through the left slit, and in other cases, because the larger part goes through he left slit while the smaller part goes through the right slit, with the two parts being deflected slightly, in slightly different directions, each time a new electron is split while going through the two slits?
No. The pieces would form a simple pattern (lots of debris where the slits are; little debris where the slits aren't) but it wouldn't be an interference pattern. In an interference pattern you'll get maxima right behind between the slits---where you would expect few particles to reach.I suppose this question leads to this question: Would a series of several watermelons thrown through a picket fence result in a debris splash pattern on the wall behind the fence that resembles an “interference-like pattern” that develops when a series of separate electrons are fired through two narrow slits?
I would expect you to see the same sort of pattern as with the watermelon.This may also sound like a dumb question, but I’ve seen such an “interference-like pattern” on walls and large cards, whenever I’ve spray painted things that have slits, that are located in front of the walls and large cards.
Excellent thinking on your part. You were most likely witnessing the infamous single slit diffraction (interference) pattern. Look it up!Also, before I forget, I know a guy who had a peculiar problem regarding a narrow slit through which light passed in a special type of camera he made. He got an “interference-type pattern” with just one slit. We puzzled over this problem for months, trying to figure out all kinds of reasons for the cause of the pattern. Finally, I remembered the double-slit experiment, and I suggested that he might be getting some type of pattern because the slit was too narrow. So, he widened the slit and the pattern disappeared.
Originally posted by StarThrower
Stage 1: Are photons point particles?
Yes or no?
Originally posted by Doc Al
I would expect you to see the same sort of pattern as with the watermelon.
Originally posted by FrankMak
Once a photon is emitted from a source, does it then proceed without loss of energy?
The difference of energy is due to the relative motion of the reference frames in which it is measured. In the distant galaxy's (moving) frame, the emitted photon has energy hf; but in our frame we measure the photon as being red-shifted. There is no missing energy; it's just that different observers measure different energies.Originally posted by Adrian Baker
... it is worth considering photons from distant galaxies, redshifted, due to the expanding Universe. These set off with a certain amount of energy, E=hf, and arrive at our detectors with less energy (f decreases) than they set out with. So where did the energy go??
Originally posted by Doc Al
The difference of energy is due to the relative motion of the reference frames in which it is measured. In the distant galaxy's (moving) frame, the emitted photon has energy hf; but in our frame we measure the photon as being red-shifted. There is no missing energy; it's just that different observers measure different energies.
Originally posted by Adrian Baker
I can't answer your full question in detail, but as to the bit above, it is worth considering photons from distant galaxies, redshifted, due to the expanding Universe. These set off with a certain amount of energy, E=hf, and arrive at our detectors with less energy (f decreases) than they set out with. So where did the energy go??
This was the question that I wanted answering when I first searched the 'net and discovered this forum.
The answers I got were helpful, but from what I can gather, this energy loss is not easy to account for. But it does answer your question above about 'does a photon proceed without loss of energy?'
The answer must be no, but it will take a better person than me to explain to you why this is.
Originally posted by Adrian Baker
I'm not sure that this is correct Doc Al. If c is the same for all observers, then the energy of a photon must be fixed for all observers surely? I really don't understand this energy loss with redshift, perhaps you could expand your reply to show me how this could be so?
McQueen's reply is very illuminating (no pun intended) but again doesn't give an explanation to where the Energy lost (no matter how small) actually goes.
Originally posted by Doc Al
There are several key differences between the watermelon and the electron. When the watermelon splits apart it is essentially destroyed: all we detect are pieces of watermelon. But with electrons we never detect anything but whole electrons. Also, the watermelon does not display those pesky interference patterns.
Yes and yes.
Originally posted by David
It’s not “lost”, it’s just delayed more. We will receive the same frequencies emitted, but more gradually, over more time. We’ve basically got a light “beam” 12 billion light-years long. It wouldn’t be that long if the galaxies weren’t moving relative to us. 12 billion years of photons would be compressed into a shorter beam if they weren’t moving.