Understanding Intensity in Double-Slit Interference: Help Needed

In summary: Originally posted by Doc Al Yes, it is referring to a group of photons. They lose energy because they are travelling away from the point source. In summary, the photon has no connection to the source, and since it does not in itself expand, why would it lose energy?
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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
 
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  • #2
What kind of problems? Show us what you've tried so far and what you don't understand.
 
  • #3
Point Source

I was asked by an old mechanical engineer how a photon that is emitted from a point source can then give up all its energy at a discrete point many thousands of diameters away.

I gave an answer using the electromagnetic concept of radiation from a point source, thusly:

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. If you place an object somewhere in this expanding wavefront it will interrupt a portion of the expanding sphere (either absorbing or reflecting depending upon the characteristics of the object), the remaining wavefront keeps on going and going and expanding.

The strength of an rf source dissipates with the square of the distance. Radiated energy can be starting
at a point source and "spreading" three dimensionally. Take a volume of a sphere and double the radius, the volume is squared, thus the energy level at any point on the spheres surface is distributed over the
spherical wavefront, and gets smaller and smaller at discrete points.
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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.
 
  • #4


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.
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.
 
  • #5


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.


I’d like to express my opinion about this and then receive critiques from others. I might not be understanding this properly.

It’s my understanding that a single photon or light “wave” goes out only in one direction and not as an “expanding sphere” of light. I think it could be described something like a long single (perhaps wiggling) bullet.

However, when light bulbs radiate light and when radio transmitters radiate radio waves, a lot of these “bullets” are emitted out in all directions at the same time, and when all their different directions are plotted, they all go out in the form of a “sphere of radiation”. So, the sphere of radiation does not result from just one photon going out or a steady stream of them going out such as from a laser.

If I am wrong about this, someone can please let me know.
 
  • #6


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.


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?
 
  • #7
Inverse square law

In the following URL, equation (11) expresses the inverse square law for "intensity" and equation (16) expresses the inverse square law for "density".

http://prancer.physics.louisville.edu/optics/manual/node7.html [Broken]

"Either a detector counting photons or one measuring energy will show an inverse square law with distance from the source."

The following URL contains a "simple" explanation of the Inverse Square Law,
http://www.astro.sunysb.edu/fwalter/CEN511/week2.html [Broken]

"All waves have the property that, once emitted, they propagate outwards from the source. As they do so the energy density in the wave decreases. This is because the total energy in the wave is fixed, while the volume of space the wave is expanding into is continuously increasing."

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? Equation (11) of the first reference implies photons lose energy.

The value of P in equation (11) implies it refers to a group of photons, even though the source power emitted could be that of a single photon.
 
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  • #8


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?
Yes.
 
  • #9
photon point source

Equation (11) mentioned earlier is,

[tex]F = \frac {P} {4 \pi r ^ 2}[/tex]

where F is the detected energy and P is the emitted energy which is divided by what is the equation for the surface of a sphere.

If I am concerned with only one photon from a point source and it is measured at a distance with a detector that is aimed at the incoming photon, there is no spreading out. Since it is not going to be measured with a detector that is spread out, the divisor is eliminated and the photon arrives at a distance with the same energy as emitted. P can be equated to the energy of one photon.

Thus a single photon, whether it travels a thousand diameters or a thousand light years would arrive with the same energy that it started with.

The basic question I was asked, is the energy of a single photon subject to the inverse square of the distance rule?
 
  • #10


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?
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.)
 
  • #11


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]



This was on one of your links:

“Electro-magnetic waves expand spherically out into space.”

I think this might be a common myth which has been confounding radio and electrodynamics students and old guys like me for many years.

Surely a single photon doesn’t “expanding spherically out into space”. And even if a photon is also a kind of “wave” or a small long “wave packet”, a single one wouldn’t necessarily have to “expand spherically out into space”, not any more than a single writhing snake would have to “expand spherically out into space” if he were thrown in the air.

If we throw out about 5,000 snakes in all directions: up, down, North, South, East, West, back, forward, left, and right, the whole group of snakes might “expand spherically out into space”, but all the individual snakes would not do that. Each one would go in only one direction, with that direction being altered only by gravity and air resistance.
 
  • #12


Originally posted by Doc Al
Yes.

Great!

So, I wonder if we could say that sound waves become weaker at a distance because we are hearing or recording or receiving a smaller section of their spherical waves. Whereas light becomes weaker at a distance because we are receiving fewer side-by-side photons, since the photons are spreading apart as they move.
 
  • #13


Doc Al,

Since you are a smart guy, let me ask you something. This might be a very stupid question, but I’ve never understood mystery of the double slit phenomenon. 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?

http://www.colorado.edu/physics/2000/schroedinger/two-slit2.html [Broken]
 
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  • #14
The slits are much smaller and closer together than in the demo you linked - microscopic even. Often, DIFFRACTION GRATINGS are used.

The size/placement of the slits is on the order of the wavelength of the light.
 
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Originally posted by David
Since you are a smart guy ...
You are clearly a perceptive and intelligent person...
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?
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!
 
  • #16


Doc Al and Russ,


Ok, thanks. I have dozens of science books of all kinds that never mention that the slits are so close together a small laser dot of light can go through both slits at the same time, and every time I’ve had a chance to talk to a physicist or a physics professor, during the past 40 years, I’ve forgotten to ask them that question.

Ok, now, the next question. 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?
 
  • #17


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?
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!

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."
 
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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!

Ok, and let me use a simple thought experiment to ask my next dumb question. My thought experiments are usually extremely simple.

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?

Suppose I think of an electron as a 19th Century watermelon that easily splits apart when thrown through a 19th Century picket fence? Apparently 21st Century watermelons might be considered to be “waves” rather than “solid” objects.

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."

Also, another question. 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?

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?

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?

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.

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.
 
  • #19
David,

You asked me about the double slit experiment in another post, I think it is better to discuss the issue over here. At any rate, the double slit explanation given by quantum mechanics has not satisfied me, for many reasons. It appears to me, that quantum mechanics is really a statistical theory. Experiments performed using trillions of atoms in an atomic gas, lead to statements about the structure of a single atom. I don't think the reasoning there is correct.

At any rate, I wouldn't mind really thinking hard about the double slit experiment if you want to.

I will start out the discussion.

Either no photons are point particles, or at least one photon is a point particle. Suppose that at least one photon is a point particle. We now have to deal with the possibility that all photons are point particles. So really then, we are looking for a model of a single photon. If we had a model of photons, as well as a model for the atomic structure of the diffraction grating, we might be able to explain the results of a double slit experiment.

Now, let us set aside photon models, and think about the terms 'wavelength' and 'frequency'. The terminology is from the mathematical description of waves. And there are many different kinds of waves. Spherical, traveling, etc.

Now, suppose that all photons are point particles. Then it follows that the wavelength of a photon and the frequency of a photon cannot be properties of photons. It would thus follow that they are parameters of the motion of a photon through space, and they would be related to orbital parameters of the photon before emission from the atom.

Now trying to construct a model of photons, which models them as point particles is going against most of modern physics, but we can still analyze things logically.

But here is my current point. Suppose that photons are point particles. It will then follow that if you fire a single photon at a diffraction grating, and the slit size is d, and the photon strikes an atom in the grating, that photon should be reflected. Now, if it is reflected with the same frequency, then its energy hasnt changed, and the collision was elastic. On the other hand, suppose that the photon entered the wall, for a ways, and then some random sequence of collisions occurred, resulting in many atoms emitting many photons, all of lower energy than the original photon. Perhaps the diffraction pattern might be explained by this "one causing many" idea. But in order to explain the double slit experiment this way, we would have to really understand what happens to the original photon at the atomic level.

To properly explain the double slit experiment is going to take a whole lot of effort.
I say we analyze things in stages.

Stage 1: Are photons point particles?

Yes or no?
 
  • #20


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?
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.
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?
Yes and yes. :smile:
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?
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".
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?
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.
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.
I would expect you to see the same sort of pattern as with the watermelon.
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.
Excellent thinking on your part. You were most likely witnessing the infamous single slit diffraction (interference) pattern. Look it up!
 
  • #21
Originally posted by StarThrower
Stage 1: Are photons point particles?

Yes or no?


Uhh, my guess is “no”. My guess is they could be “packets” of EM “wavelets”. That is, small, somewhat long, groups of smaller wavelets that travel together as one “package” that we call a “photon”. This type of “wave packet” could produce a “particle-like effect” when it hits something, and it could appear to have “mass” under certain circumstances.

For example, take a big sound hi-fi speaker, move it up close to some of your window curtains, and give it a short loud blast of “sound” (ie vibrating electrical energy inside the coil of the speaker). I am inclined to believe that that loud short sound blast would quickly shake your curtains, just as if you touched your curtains with your hands. So, the short burst of sound waves would be acting like “a particle” or as if it “had mass”. Not like a small “point particle”, but like a large flat surface of a plane, like a big board. You could produce short bursts of sound waves that could appear to give the experimental result as being or resembling a “point particle” if you focused the sound wave packet and concentrated it down to a “point”, such as with a parabolic dish. Then you could hit, let’s say, some soft pudding and put a dent in the center of the pudding, as if it had been hit in the center by a small point particle.
 
  • #22


Originally posted by Doc Al
I would expect you to see the same sort of pattern as with the watermelon.


So you think I wasn’t seeing an “interference pattern” with the paint, huh, just some strips of paint dots?
 
  • #23
StarThrower, Doc Al,

I’ve read that explosion compression waves travel as expanding sphere’s of “wavelet packets” or groups of “sound” waves. I’ve seen seismograph type print-outs of these compression waves, actually ground waves (earth waves), and the print-outs resemble my verbal description of a “photon” packet.

The actual waves in the air and ground are of course longitudinal waves and that is why they expand in a spherical manner in the air, but the graphic print-out gives more of a visual appearance of the wave packets looking like transverse waves on the graph, like the way a sound wave can be turned into an image that looks like a transverse wave on an oscilloscope screen.

I suppose that if we could focus an explosion air sound-wave “packet”, we could make it act something like a “point particle”.

This reminds me of something. Since the molecules of air don’t actually touch each other in sound waves, a sound wave, seems to me, is actually the result of compression and vacuum “field” forces between the molecules, rather than the molecules actually “hitting” each other. The molecules would produce the field, and these fields would compress and expand as the wave moves through the air. The molecules would move a little too, but it’s my understanding that they don’t actually touch each other. Would you happen to know what this “field” effect is called and what is the name of that kind of “field”?
 
  • #24
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.
Have you taken the trouble to calculate the chance of a give photon from a small source (the antenna of voyager 1 or 2 spacecraft ) being detected one and a half billion kilometres from the source. If you have done the calculation (and the geometry ) I would be very anxious to see what you have found.
 
  • #25
photon forever?

Is a single photon subject to the inverse square law (ISL)? Once a photon is emitted from a source, does it then proceed without loss of energy?

I have searched my references and many, many on the internet and I have yet to find an "authoritative" reference that states a single photon is not subject to the ISL. The typical ISL formula is written based upon energy "spreading out" on the surface of a sphere. The divisor is the formula for the area of the surface of a sphere.

https://www.physicsforums.com/latex_images/119455-0.png [Broken]

I noted the internal URL for the .png rather than have the software regenerate the "LaTex" formula. The classic explanation taught to millions is that electromagnetic energy is reduced by the square of the distance. Distance is a key parameter, but stating that without mentioning that is just one parameter misrepresents the concepts of the ISL equation.

If a single photon is not subject to the ISL, that should apply to any single packet of electromagnetic energy. I need an authoritative reference that I can pass on to the mechanical engineer that questioned whether a photon loses energy.
 
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  • #26


Originally posted by FrankMak
Once a photon is emitted from a source, does it then proceed without loss of energy?


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.
 
  • #27


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??
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.
 
  • #28
photon redshifted

The red shift theory was postulated based upon 1920s astronomical instruments. Current instruments with higher optical resolutions show a different redshift picture of the same galaxies observed by Hubble.

It is assumed that space has the same index of refraction as a vacuum, "1", and one wonders with all the "stuff" that has been found in space whether the true index of refraction might be just a hair over 1.

I wonder why I have such trouble finding an authoritative reference that simply states that a single photon does not lose energy no matter how far it travels from its source.
 
  • #29
energy of photon

Is a single photon subject to the inverse square law (ISL)? Once a photon is emitted from a source, does it then proceed without loss of energy?

In order to understand this question , it is necessary to understand the actual manner in which electromagnetic waves are thought to propagate and which knowingly or unknowingly is the issue around which this whole discussion has been revolving. If we take a point source such as a radio antenna and assume that photons are radiating from it and continue to radiate outwards in the direction of propagation , it would not make much difference at small distances , but at large distances the gaps between these radiating photon rays would be huge , approx. 1020 diametres at a distance of 106 Kms. These are distances too dramatically large to be ignored because what we observe is totally different from what is expected , the observations show that photons can be detected at every point even on a sphere having a million kilometre radius. To account for this , quantum physicists have come up with the theory that every point of the Universe is filled with “virtual” photons . These “virtual” photons are not still but are themselves constantly undergoing transformations into “virtual” electrons and positrons and back into “virtual photons , the Universe is a hotbed of “virtual” activity , however because of the extremely small duration and limited energies of this activity it cannot be termed real. When a real photon comes into contact with a “virtual” photon , the “virtual photon , splits up into a positron/electron “virtual” pair which since they represent matter and anti-matter annihilate each other giving rise in this process to a real photon of exactly the same value as the original photon with which they had interacted. Electromagnetic waves are propagated in this way both laterally and in the direction of propagation. The second point which has been raised is , does a photon gain or lose energy as it travels , observation shows that a photon tends to retain its energy or , to put it more accurately , that the loss of energy is extremely limited. One of the photons characteristics is that regardless of how far it travels , it either gives up all its energy intact at its destination or is absorbed along the way. There are numerous instances to prove this . Light from a star moving away from us undergoes a so called red-shift (i.e there is a slight tendency of the light waves to lengthen slightly ) but it never changes so much as to lose its identity . Deep space broadcasts in the 20cm wave length at a hundred million kilomteres out are received . by Earth stations in the 20cm band width , meaning that they have not noticeably lost energy on the journey , and so on.
 
  • #30


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.

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.

So does a Cosmologically red-shifted photon (say from the microwave background radiation, or from a quasar) lose energy or not? If so, where does it go, and if not, why not?
I'd love an answer to this that I could understand.. I've done a fair bit of searching on the 'net and got all kinds of conflicting answers. I look forward to your thoughts.

Perhaps it should be on another thread though?
 
  • #31


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.


The photons are “stretched out” in space. That’s a standard Doppler shift. Doppler predicted this with starlight in 1842. I don’t know if each individual photon is “stretched out” or if the space between them is. The “energy loss” is an apparent thing, due to the Doppler shift. Physicists tend to think of a light redshift as both a frequency shift and an “energy loss”, while a sound redshift is just thought of as a “frequency shift”.
 
  • #32


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.

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.
 
  • #33


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. :smile:


What is "interference pattern"?
 
  • #34


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.


Thanks for the ideas here, and I understand the stretching idea of a photon as the universe expands and this gives a good mental picture of what is happening. But I still have a problem with this. For an observer, the energy of a photon is given by E=hf Where is the time factor in this then? Mathematically, Energy is lost. Is the Maths wrong, or is there a correction factor to E=hf that I am unaware of?
I can't see how Relativty helps either, although that maybe because my Maths isn't good enough.

Normally we get the message in QM; don't visualise it, it just is that way - do the Maths. The answers I'm getting here with Red Shift are forget the Maths - visualise it! This is why I'm still puzzled by this.
I'm sorry if I keep on about this, but as I said, I've done a lot of research on it and always get incomplete explanations. It is something that bothers me every time I teach Cosmology.

Can anyone point me to a mathematical answer to the problem? Or perhaps show mathematically that energy for the emitted photon = the energy of the received one with appropriate correction factors? Is there such a solution?
I just can't find any way to do this.
Thanks for your patience. :smile:
 
  • #35
Sorry. Here comes another unintelligent question: What is the "inverse square law"?
 

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