Photon Energy when Doppler Shifted

• smyri
In summary: This is because the momentum of the photons (and the energy they contain) is the same in both frames. However, when you try to create an electron and photon pair, the energy of the photon is going to be different in the two frames. This is because the momentum of the photon is a function of its energy, and the energy of the electron is a function of its momentum.
smyri
I don't know how to understand doppler shifts and yet conserve energy.

Consider a red photon is emitted by a torch. This photon is let free to travel through space. Any observer at rest relative to the torch (which by now may be millions of miles away) will see it as a red photon.

Another observer aproaching the torch from a distance at a high speed will see this photon as blue shifted; let's say as a UV ray. This observer will conclude that this photon has enough energy to cause a photo-electric emission of a Sodium metal electron.

Still another observer receeding form the torch at high speed will conclude that this photon, severely shifted into the red, say an IR photon, cannot bring about the photo-emission of an electron from the surface of a piece of Sodium.

Questions:
1) It seems that whether or not a photo-electron is emitted depends on which frame of reference you are in...i.e. whether the freed electron exists outside from the metal or not depends on how the incident photon looks...UV..ok!. IR no effect. Well, the electron can and can't exist freely simply by virtue of what frame of reference we look at a photon from.

Worse still; what about using that damned photon for electron pair creation? If I view it as a gamma ray, I'll get the pair creation...not so if I view it as IR! We must all agree, no matter what our inertial frames of references are to each other, that the electron-positron pair either does or doesn't exist!

2) The photon energy will differ from one observer to another because of the relative motion between observers. This is due to the Doppler effect for light. Where does the photon get it's extra energy from if it's seen as UV; or where does the photon loose it's energy to if it absorbed as IR?

..or more simply stated..

If the photon's energy depends on it's frequency; then we can turn a red photon into a UV, gamma, IR or radio wave photon merely by changing our frame of reference to a long departed photon source which can be considered, to all intents and purposes, out of the picture.

This means we can make the energy content of a photon anything we like just by viewing it in different reference frames. So much for the conservation of energy??

I'm obviously missing some essential element in my understanding; but for the life of me I can't see what it is.

Energy is not conserved when changing reference frames; neither is momentum.
The trick is to do your calculations in only one frame; you will find that the change of energy adds up to the same amount in every system - enough to set an electron free.

You need to adjust all the parameters to get the right result. A redshifted photon is both distance and time dilated. When you view a redshifted light source, you are viewing it in slow motion. The individual photons have less energy, but you receive more of them.

No, the explanation is simpler (photon number AFAIK is Lorentz invariant, plus as a photn of a given frequency has a fixed energy you couldn't use this scheme to make energy Lorentz invaraint), as Ich indicates energy is a frame-dependent property, so the total energy of an isolated system is not going to be the same in two frames. The conservation of energy only demands that the total enrgy of an isolated system be constant in all inertial frames (i.e. the energy of an isolated system some inertial frame is not a function of time), not that it should be the same value in all inertial frames).

The frame dependcence of energy is not peculair to relativity either, infact if you could of asked the same question about Doppler shift in a non-relativstic context (well actually looking at your post you didn't mention relativity specifcally, but I assume it was relativity you were thinking of as this is the relativity forum) and you would of gotten exactly the same answer.

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Let's forget about photons for the moment. Stretching space is the same thing as stretching time. One minute of light energy emitted by a Z=1 redshifted object will take two minutes to pass an observer at Z=0. The emitting object will only appear to be half as bright, but the image will persist twice as long.

Ignore power, as that needlessly complicates things, the total enrgy received by the two obsrevers from the source is different.

The question was about a single photon.
When you use m²=E²-p², and calculate the change in m (preferably a small dm), it will always be equal to hf, with f as seen from the absorbing body.

m is zero for photons anyway, so E = p in natural units, but I don't see your point as f is not Lorentz invariant. Energy is not a Lorentz scalar (infact it's a component of the momentum 4-vector for which the associated Lorentz scalar is mass)

m, E, and p refer to the absorbing body. It´s (rest)mass increases by dm=h*f0, with f0 being the photon´s frequency in the frame of the absorber. dm is invariant.

Why do you think that is correct? Remember momentum as well as energy is conserved.

Red shift and blue shift arise from time dilation and conservation of momentum considerations.

juju

I guess I still need to know that if this photon in one frame seems to have enough energy to create a proton-antiproton pair; where did this energy come from if the original photon was emitted by a mere red torch?

smyri said:
I guess I still need to know that if this photon in one frame seems to have enough energy to create a proton-antiproton pair; where did this energy come from if the original photon was emitted by a mere red torch?

This is precisely why single free photons cannot create a proton-antiproton pair.

jcsd said:
Why do you think that is correct? Remember momentum as well as energy is conserved.
I think it´s correct because
1 I did the calculation and I think I did not mix up all those signs
2 it is the only answer consistent with the principle of relativity: it is allowed to view things from the absorber´s frame, and any calculation in a different frame must yield the same result.

smyri said:
I guess I still need to know that if this photon in one frame seems to have enough energy to create a proton-antiproton pair; where did this energy come from if the original photon was emitted by a mere red torch?
As jcsd said, that´s impossible. You need a second particle to absorb the excess momentum. And then the required energy is defined in the frame of this particle (at least if it´s big enough).

Ich said:
I think it´s correct because
1 I did the calculation and I think I did not mix up all those signs
2 it is the only answer consistent with the principle of relativity: it is allowed to view things from the absorber´s frame, and any calculation in a different frame must yield the same result.

1. It can't be true, the actual formula is: dm = sqrt(2Mh*f), (where M is the intial mass of the object, f the frequency of the photon in the frame which the absorber is initally at rest in and h is Planck's constant, all in units such that c = 1).

1) The reason your formula is not correct is that your fromual cosnerves energy, but it doesn't conserve momentum (remember the photon has a momentum of h*f).

2) As enegry and momentum are the tiem and spatial compenents repsectively of the same four-vector you find that not only is moamntum not conserved, but enrgy is not conserved in all frames either if your formula was correct.

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smyri said:
Consider a red photon is emitted by a torch. This photon is let free to travel through space. Any observer at rest relative to the torch (which by now may be millions of miles away) will see it as a red photon.
Then the total energy is conserved. That means that the sum of the energy of the torch plus the energy of the photon remains unchanged.
Another observer aproaching the torch from a distance at a high speed will see this photon as blue shifted; let's say as a UV ray. This observer will conclude that this photon has enough energy to cause a photo-electric emission of a Sodium metal electron.

Still another observer receeding form the torch at high speed will conclude that this photon, severely shifted into the red, say an IR photon, cannot bring about the photo-emission of an electron from the surface of a piece of Sodium.
It appears to mean that you're assuming that the potential energy of the sodium atoms remains constant. Why?
Questions:
1) It seems that whether or not a photo-electron is emitted depends on which frame of reference you are in...i.e. whether the freed electron exists outside from the metal or not depends on how the incident photon looks...UV..ok!. IR no effect. Well, the electron can and can't exist freely simply by virtue of what frame of reference we look at a photon from.
Do not dismiss that which is emitting both photons. There must be something which emitted them. In this case you chose sodium (torch).
Worse still; what about using that damned photon for electron pair creation? If I view it as a gamma ray, I'll get the pair creation...not so if I view it as IR! We must all agree, no matter what our inertial frames of references are to each other, that the electron-positron pair either does or doesn't exist!
That's called "pair production." To create two photons one has an atom around to take up the difference in the momentum and energy that is required to balance the energy/momentum books. The atoms do this and thus act as a sort of catalyst for pair production. So the atom starts moving and two photons are created.

Pete

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jcsd...are you saying one photon would not conserve together the momentum-energy? What about a pair of photons? Can they do it?

Do these two photons then need to be viewed from a frame in which the newly created pair would have a zero COM? That would overcome my problems as together they would need the required total energy; even if the two photons viewed in a frame seeing them with different frequencies would have one red shifted and the other blue shifted.

jcsd said:
1. It can't be true, the actual formula is: dm = sqrt(2Mh*f), (where M is the intial mass of the object, f the frequency of the photon in the frame which the absorber is initally at rest in and h is Planck's constant, all in units such that c = 1).

1) The reason your formula is not correct is that your fromual cosnerves energy, but it doesn't conserve momentum (remember the photon has a momentum of h*f).

2) As enegry and momentum are the tiem and spatial compenents repsectively of the same four-vector you find that not only is moamntum not conserved, but enrgy is not conserved in all frames either if your formula was correct.

M+dm = sqrt(M²+2Mhf (+(hf)²-(hf)²) ); <-- only in M´s rest frame!
using M>>hf (and sqrt(1+x)~=1+x/2) yields
dm = hf.
1) Momentum is conserved, but in M´s frame does not contribute to the increase in rest mass.
2) In different frames neither E nor p stay the same, but mass (which is the length of the four vector) does.

Ich said:
M+dm = sqrt(M²+2Mhf (+(hf)²-(hf)²) ); <-- only in M´s rest frame!
using M>>hf (and sqrt(1+x)~=1+x/2) yields
dm = hf.
1) Momentum is conserved, but in M´s frame does not contribute to the increase in rest mass.
2) In different frames neither E nor p stay the same, but mass (which is the length of the four vector) does.

Why is it imcomplete?

Sorry I made a silly error (i.e. I put M^2 on the wrong side of the brackets), it should be dm = sqrt(M^2 + 2Mhf) - M and the equation is trjue in all frmae sby viture of the fact that f is specifically the frequency of the phton ithe frma ein which the absorber is intially at rest.

1) it isn't conserved by your formula dm = hf and I don't see your problem if you recognizxe that dm = hf is just an approximation anyway that can be a very poor approximation.

2) Yes I know that E and p are not the same, where did I ever imply otherwise and yes I know that m^2 is the sqaure norm of the 4-vector and tehrfore a Lonrentz invaraint, but I fail to see your point.

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And then there's also the question I have with a single photon either having enough energy to cause the photo electric effect or not. If the "absorber" is approaching the red photon fast enough, suddenly the photon will have enough energy to eject an electron but before it didn't? I don't understand!

jcsd said:
Why is it imcomplete?

Sorry I made a silly error (i.e. I put M^2 on the wrong side of the brackets), it should be dm = sqrt(M^2 + 2Mhf) - M and the equation is trjue in all frmae sby viture of the fact that f is specifically the frequency of the phton ithe frma ein which the absorber is intially at rest.
it should be M+dm = sqrt(M^2 + 2Mhf). And for M>>hf, as I explicitly stated, this becomes dm = hf.
jcsd said:
1) it isn't conserved by your formula dm = hf and I don't see your problem if you recognizxe that dm = hf is just an approximation anyway that can be a very poor approximation.
Even if M was a single electron, the approximation would be extremely good. And since we´re talking about the photo effect, where M means a macroscopic piece of metal, there is absolutely no point in arguing that it could be a poor approximation.
jcsd said:
2) Yes I know that E and p are not the same, where did I ever imply otherwise and yes I know that m^2 is the sqaure norm of the 4-vector and tehrfore a Lonrentz invaraint, but I fail to see your point.
It was not me who came up with this four-vector stuff, and I did´nt see the point in the first place.

And to conclude:
For M>>hf, f as seen in M´s frame, dm = hf. full stop.
dm is the increase in internal energy of the absorber. It is invariant and either enough to free an electron or not. In every frame you choose.

smyri said:
And then there's also the question I have with a single photon either having enough energy to cause the photo electric effect or not. If the "absorber" is approaching the red photon fast enough, suddenly the photon will have enough energy to eject an electron but before it didn't? I don't understand!
To make it plausible without math:
Drop a pebble on the windscreen of your car. Nothing happens.
Drop a pebble on the windscreen of your car when it drives by at 100 Mph. Suddenly the pebble has enough energy to smash it.
It´s nothing different with the photon.

Ich said:
it should be M+dm = sqrt(M^2 + 2Mhf). And for M>>hf, as I explicitly stated, this becomes dm = hf.

Even if M was a single electron, the approximation would be extremely good. And since we´re talking about the photo effect, where M means a macroscopic piece of metal, there is absolutely no point in arguing that it could be a poor approximation.

It was not me who came up with this four-vector stuff, and I did´nt see the point in the first place.

And to conclude:
For M>>hf, f as seen in M´s frame, dm = hf. full stop.
dm is the increase in internal energy of the absorber. It is invariant and either enough to free an electron or not. In every frame you choose.

I'm not arguing with you over those points though, infact I don't know where the arguemnt came from: dm --> hf as hf/M --> 0.

Tell me if I misundretsoofd you, but you semed to imply taht as dm is approximately equal to hf that it menas there is something special about the frmae where , now as infact dm = hf can be an exceddingly poor situation when hf>M (an unlikely situation granted), I cannot see you argument.

Ok, last time:
smyri asked about the photo effect, where a photon (E~3 eV) hits a piece of metal (E~10^33 ev). He wanted to know how one would see the same thing (an electron set free) in every frame, when in some frames the photon would have far too little energy for the photo effect to happen.
I pointed out that you will see what happens when you calculate the photon´s energy in the absober´s frame. That´s obvious anyway.
Then I said that the photon´s energy after absorption will appear as an increase in the absorber´s rest mass (dm!) and is therefore the same in every frame, as mass is invariant. That´s why a redshifted photon hitting a moved absorber will supply exacly the same amount of energy (dm) to the absorber as the unshifted photon hitting the resting absorber.
So yes, I said there is something special about the absorber´s frame, because only in this frame hf equals dm (the actual increase of internal energy of the absorber). If you want to decide whether f is large enough to free an electron, you have to look a it in the absorber´s frame.
And forget about hf>M or even hf~M. That´s Compton scattering - nobody asked about it. Of course then my argument wouldn´t hold - and because I´m cautious and aware of nitpickers (no offense intended) I repeated in EVERY post that I presume M>>hf, because I answered to smyri´s question and did not want to cover any eventuality like gamma ray - electron scattering or whatever one could think of because that would unnecessarily complicate things and principally have nothing to do with this thread where some innocent guy asked about the photo effect and not about life, the universe and everything. So I think you should be able to see my point: dm is invariant and so the photo effect is.
You agree?

Thanks for all your thoughts; esp. Ich...the only thing stil puzzling me is the question; what is the photon energy with respect to the absorber? I guess it's determined by the relative intertial speeds between the emitter and absorber and the frequency of the photon in the emitter's frame of reference...I don't think photons are emitted by any particle traveling at speed c...i.e. massless particles don't interact with photons.

I guess then, that there does appear to be an "absolute" frequency for a photon; or an absolute frame of reference for a created photon...namely the frame of reference of the emitter! or that frequency it has upon emission relative to the emitter frame of reference.

Of course, since all photons travel at the speed "c"; and that the frquency observed is "observer" dependent..it does, somehow, seem to suggest to me that there is an absolute preferred frame of reference for photons; namely that of their emitters. This seems to be in contrast to my understanding of what relativity is really about.

Sorry, that was a very confusing description...I mean...

Suppose an emitter sends out a UV photon and it hits an absorber at rest relative to the emitter. In all frames of reference we see an ejected photo-electron.

Suppose now only an IR photon is emitted under otherwise identical situations as described above. Then all frames will see no ejected photon.

Now if I understand Ich correctly; then all frames would agree with the existence or non-existence of that electron even if there was relative motion between the emitter and observer.

However, from the standpoint of the photon alone; traveling at speed c; it can be seen by another observer to have any frequency the observer wishes; depending on it's relative motion with respect to the emitter...but what if the emitter is long gone!

All photons have speed c! It can be arranged so that a photon can appear to have any energy we like it to simply by arranging the detector speed relative to other detector speeds if you like. What then really differentiates one photon from another? How can we talk of a soft photon from a hard photon when it is merely a matter for detector speeds relative to each other? After all, for any photon in the universe there will exist inertial frames making it look UV or IR. And all frames are supposed to be on an equal footing; and all frames agree the photon has a speed c. So what is it then that makes one photon, when travelling, different from another? Thus, it would still seem to be the case that there is somehow a "preferred" frame of reference for the photon; namely that of it's emitter. And there is one preferred "frequency" for a photon, namely that frequency as determined by it's emitter.

Ok, we have to clear something: Photons all travel at c; still, tey are different from frame to frame. They can have different "kinetic" energy even if the all go at the same speed. You´ll have to swallow that.
The photon´s energy is defined by
1) the energy that the emitter lost in it´s own frame - say 13.6 eV for a slow electron binding to a proton.
2) the relative speed of observer and emitter. You could make those 13.6 eV to 100 keV or .001 eV - it just depends on relative speed (see post #24). And exactly this energy is what counts to determine if there is a photo-electron or not.
p.s. it doesn´t matter if the emitter is long gone. The kinetic energy of the photon still is there - even if they all travel at the same speed.

I guess I still need to know that if this photon in one frame seems to have enough energy to create a proton-antiproton pair; where did this energy come from if the original photon was emitted by a mere red torch?

These guys are smarter than me, .. but I think the answer is being glossed over.

The frame of reference that it is a low energy photon is from emission. From your frame of reference it is a high energy photon. If you are to see it make a "high energy" splash, then it will have to splash into something comoving along with you.

It comes from the energy bound up in the momentum of both the photon and what it hits.

Thank you Ich...I can see now that I need to see the photon frequency (or energy) in the rest frame of the absorber to determine the effect. So, if a UV photon was emitted from the emitter and this photon was absorbed by the absorber as an IR photon; I now realize there'l be no photo effect.

I still wonder where that extra photon energy went...even though I accept the absorber, in it's frame of rest sees only an IR photon (because relative to the emitter, it is receeding away at a very high speed). I understand that dm for the absorber will be IR photon energy/c*c and that this is a lorentz invariant. Still, the value of dm seems to depend on how the photon looks to the absorber, not on how much energy was liberated by the emitter at photon creation. So if the frequency of the photon is different for absorber than emitter...I still can't see where the energy difference goes.

Ich.. I see your analogy with the windscreen and the stone. I guess if the windscreen is moving relative to the stone, it supplies some of the energy that gets dissapated in the smashing of the windscreen...i.e. the energy of the collision is not that of the stone alone.

Is it possible this might also be the case with an emitted IR photon that is seen to look like a UV photon by the absorber? I guess then that dm will be greater for the absorber. Maybe the absorber has supplied some of it's kinetic energy for the photon collision to make dm relative to the absorber greater..but then it seems to say to me that the photon sees the absorber with a range of different kinetic energies based on the absorber's "speed" relative to the frame that the "true" frequency the photon had when emitted.

I'm sorry if I appear to be so thick; it's just that until I thought this way, I used to feel very comfortable with the special theory of relativity. I still feel happier with it than the recent New Scientist claim I read the other week saying people may be looking for an "absolute" motion component of the Earth through an aether.

I truly appreaciate you help very much Ich...you've taken my thoughts on this to a deeper level but I still haven't resolved it completely in my own mind. Please don't think I'm stubborn...but you can think I'm dumb!

The reason I started this line of questioning is I was trying to understand where all the radiation energy from the big bang had gone simply because the radiation field has cooled from an unbeleivably high temperature to only 2.7 degrees Kelvin...but as there are many photons involved in this case, berhaps it's a question of entropy...which I don't fully understand anyway. So I tried to think of it in the simpler case of only one photon at a time and came decidedly unstuck! I'm not a physicist...I guess it shows!

The extra energy goes into kinetic energy. A small dm moving at some speed contains the same energy as a larger dm at rest. This works out for every frame, whether you regard it in the emitter´s, the absober´s or some other frame.

<h2>1. What is photon energy when Doppler shifted?</h2><p>Photon energy when Doppler shifted refers to the change in energy of a photon due to the Doppler effect. The Doppler effect is the change in frequency and wavelength of a wave when the source of the wave is moving relative to the observer. This change in frequency results in a change in energy of the photon, which is directly proportional to the frequency of the wave.</p><h2>2. How does the Doppler shift affect photon energy?</h2><p>The Doppler shift affects photon energy by changing the frequency of the wave. If the source of the wave is moving towards the observer, the frequency of the wave will increase, resulting in an increase in photon energy. Conversely, if the source is moving away from the observer, the frequency will decrease, leading to a decrease in photon energy.</p><h2>3. Can Doppler shifting change the color of light?</h2><p>Yes, Doppler shifting can change the color of light. This is because the color of light is determined by its frequency, and the frequency can be changed by the Doppler effect. For example, if a light source is moving towards an observer, the frequency will increase, causing the light to shift towards the blue end of the spectrum (blue shift). If the source is moving away, the frequency will decrease, resulting in a shift towards the red end of the spectrum (red shift).</p><h2>4. What is the equation for calculating photon energy when Doppler shifted?</h2><p>The equation for calculating photon energy when Doppler shifted is E = hf, where E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 joule seconds), and f is the frequency of the wave. This equation shows that photon energy is directly proportional to frequency, meaning that any change in frequency due to Doppler shifting will result in a corresponding change in photon energy.</p><h2>5. How is Doppler shifting used in astronomy?</h2><p>Doppler shifting is used in astronomy to study the movement and properties of astronomical objects. By measuring the Doppler shift of light emitted from stars and galaxies, scientists can determine their relative motion and the presence of objects such as planets and black holes. Doppler shifting is also used to study the expansion of the universe and to detect the presence of exoplanets.</p>

1. What is photon energy when Doppler shifted?

Photon energy when Doppler shifted refers to the change in energy of a photon due to the Doppler effect. The Doppler effect is the change in frequency and wavelength of a wave when the source of the wave is moving relative to the observer. This change in frequency results in a change in energy of the photon, which is directly proportional to the frequency of the wave.

2. How does the Doppler shift affect photon energy?

The Doppler shift affects photon energy by changing the frequency of the wave. If the source of the wave is moving towards the observer, the frequency of the wave will increase, resulting in an increase in photon energy. Conversely, if the source is moving away from the observer, the frequency will decrease, leading to a decrease in photon energy.

3. Can Doppler shifting change the color of light?

Yes, Doppler shifting can change the color of light. This is because the color of light is determined by its frequency, and the frequency can be changed by the Doppler effect. For example, if a light source is moving towards an observer, the frequency will increase, causing the light to shift towards the blue end of the spectrum (blue shift). If the source is moving away, the frequency will decrease, resulting in a shift towards the red end of the spectrum (red shift).

4. What is the equation for calculating photon energy when Doppler shifted?

The equation for calculating photon energy when Doppler shifted is E = hf, where E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 joule seconds), and f is the frequency of the wave. This equation shows that photon energy is directly proportional to frequency, meaning that any change in frequency due to Doppler shifting will result in a corresponding change in photon energy.

5. How is Doppler shifting used in astronomy?

Doppler shifting is used in astronomy to study the movement and properties of astronomical objects. By measuring the Doppler shift of light emitted from stars and galaxies, scientists can determine their relative motion and the presence of objects such as planets and black holes. Doppler shifting is also used to study the expansion of the universe and to detect the presence of exoplanets.

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