do photons have mass?

[Garvi]

photon is a particle. then how come it does not have mass?

[1MileCrash]

The word "particle" doesn't imply mass. I couldn't think of a more vague word.

Furthermore, this is one of the reasons light is sometimes thought of a wave and sometimes thought of a particle. I'd rather it be a wave when talking about its lack of mass, and I'd rather it be a particle when talking about its momentum.

[Naty1]

Why should it have mass??

Anyway, nobody really knows...anymore than we know why an electron has the mass we observe...or the charge.

Or asking "Why is the earth here?"

Some good related discussion here:

http://www.physicsforums.com/showthread.php?t=507973&highlight=photon+mass

[sophiecentaur]

Why should it have mass?
Look at what wiki has to say about the fundamental particles, how they interact and how the forces between particles can be accounted for. It's very involved.

But, personally, I think that referring to a photon as a particle is not helpful. There are many paradoxes involved if you want to look at is as a little bullet - like how big could it be? Would a photon of 200kHz radiation really be 1000 times 'bigger' than a photon of 200MHz radiation (bearing in mind that there would also have to be a thousand times as many 200kHz photons for the same power flow)?
I prefer to stick with an idea of a photon as being just a quantum of energy which only shows up when an em wave interacts with some system of charges, such as an atom, molecule or nucleus. That view doesn't actually run counter to the 'proofs' of it being a particle, such as the photoelectric effect and can save many sleepless nights, worrying about it.

[jishitha]

Photons possess mass, when it is moving., and rest mass of photon is 0....

[DrDu]

In superconductors, photons are massive.

[sophiecentaur]

Photons possess mass, when it is moving., and rest mass of photon is 0....

When is a photon not moving?

[DrDu]

When is a photon not moving?

In the case of massive particles, the mass is the energy of the particle at momentum 0 (divided by c^2). If one extends this definition to the case of photons in vacuo one has to set the mass of the photon to 0 as E=c p and therefore vanishes at p=0.

[sophiecentaur]

So when is a photon not moving? You haven't said. You have merely introduced a property of particles that do have mass. (A circular argument).
Measured mass of a photon, from wiki <1×10−18 eV/c2. Not 'measured' as zero because of the accuracy of the measurement that was possible - hence the upper limit is quoted.

The "non-zero mass" is referred to as "effective rest mass". That is hardly the same thing as mass. It's a bit like discussing semiconductor Holes. They are only a way of describing an observed conduction mechanism.

[DrDu]

So when is a photon not moving? You haven't said. You have merely introduced a property of particles that do have mass. (A circular argument).
Measured mass of a photon, from wiki <1×10−18 eV/c2. Not 'measured' as zero because of the accuracy of the measurement that was possible - hence the upper limit is quoted.

The "non-zero mass" is referred to as "effective rest mass". That is hardly the same thing as mass. It's a bit like discussing semiconductor Holes. They are only a way of describing an observed conduction mechanism.

I don't think this is circular. I just wanted to point out how the concept of (rest-) mass can be extended to objects which can't be at rest.

As to "effective rest mass" vs "true mass". Obviously it is something different to discuss propagation of light in vacuo vs in matter. But I don't think that there is a difference of principle. Electrons in vacuo aren't more fundamental than holes in solids. They both only describe an observed conduction mechanism, nothing more.

[sophiecentaur]

That's a fair enough comment. I rather shot myself in the foot there!
I still think that there is a (quantitative) difference between the behaviour of a photon under most conditions ( i.e., it exhibits no mass pretty much all the time ) and the 'effective' behaviour of a photon under special fringe-quantum conditions in which a mass-like nature reveals itself.
But, as with the Higgs Boson and other frontier-bashing concepts, it may turn out to be very relevant to our better understanding and development of a TOE.
My problem is that such inconsistencies can be taken out of context by the 'less well informed' and applied to everyday situations where they are not relevant and can introduce even more confusion. That little word "effective" can so easily be confused with "it's really there" and could be used in arm waving explanations of such things as Light Pressure, where Momentum is the relevant quantity.

[joyever]

Photons possess mass, when it is moving., and rest mass of photon is 0....

so can we say light has mass?

[sophiecentaur]

Can someone tell me of a time when a photon is not moving? If not, then what has "rest mass" got to do with this discussion?

The one quoted situation when a photon may 'exhibit the property of mass', still doesn't imply that the photon is ever stationary. So where does that take us?

Photons have momentum - they show it all the time. That doesn't imply mass.

[Khashishi]

jishitha: I don't think your physics are wrong. Your disagreement is stemming from a non-conventional definition of the term "mass". "mass" when used by itself is ambiguous, but is conventionally taken to mean "rest mass", which is the energy left over when you've subtracted all the kinetic energy and removing any potential well. If you prefer to mean relativistic mass, you'd better spell it out, but the term isn't used much because it's redundant with the term energy.

[DrGreg]

There are (at least) two competing definitions of "mass" within relativity. One definition is now used by almost all professional physicists; the other definition was used historically by physicists and is still used in some schools, and books aimed at the general public.

The old definition included an object's kinetic energy* as part of its mass (via E = mc2). Thus an object's "mass" (a.k.a. "relativistic mass") varied with velocity*; the object's mass when it was stationary* was called its "rest mass".

The new definition excludes the kinetic energy of the object and is therefore constant. In the old terminology it was "rest mass" and is now called just "mass".

It's unfortunate that both definitions are still in use, so if someone refers to "mass", you need to check whether they mean the old definition (="relativistic mass") or the new definition (="rest mass" or "invariant mass").

The equation relating mass (=rest mass) m to energy* E and momentum* p is
E^2 = m^2 c^4 + |\mathbf{p}|^2 c^2
This applies to particles with non-zero (rest) mass, but also applies to photons if you set m=0. This justifies saying that the mass of a photon is zero (in the new terminology), or that the rest mass of a photon is zero (in the old terminology) even though a photon can never be at rest. This is one case where it was less confusing to use the term "invariant mass" instead of "rest mass".

So, the answer to the original question "do photons have mass?" is "no" under the modern definition, but "yes" under the old definition which some people still use.

There is some more about this in the relativity forum's FAQ Do photons have mass? (http://www.physicsforums.com/showthread.php?t=511175).

____
*relative to some frame of reference

[juanrga]

When is a photon not moving?

The concept of rest mass does not imply that a body is not moving. Electrons have rest mass but does not imply that are always at rest, neither that the concept of rest mass only applies when they are at rest.

[Hootenanny]

The concept of rest mass does not imply that a body is not moving. Electrons have rest mass but does not imply that are always at rest, neither that the concept of rest mass only applies when they are at rest.
Indeed this is only true, but that wasn't sophie's point. In order to define a non-zero rest mass for a particle, there must exist a fame in which the particle is stationary. There exists such a frame for an electron, but not the photon.

[sophiecentaur]

Wouldn't a finite rest mass imply infinite mass at c? Is that not a clincher?

[juanrga]

Indeed this is only true, but that wasn't sophie's point. In order to define a non-zero rest mass for a particle, there must exist a fame in which the particle is stationary. There exists such a frame for an electron, but not the photon.

Rest mass is not the mass that a particle has only when is at rest.

The rest mass of a particle is also well-defined for frames in which the particle is not at rest. Rest mass is one of the properties that defines the particle (together with spin, charge..) and those properties are frame-independent.

From a particle point of view, the rest mass can be obtained from the mass operator in the energy-momentum space.

Maybe the term "rest" is at the root of the confusion. Just substitute the term "rest mass" by invariant mass or simply mass for avoiding it.

[sophiecentaur]

Whatever sort of mass you are talking of, how would it not go infinite for a speed of c?

[dchris]

Hey, i have a question. Photons posses mass only when they are moving, so according to our actual laws of physics, shouldnt they reach an infinite mass by moving with the speed of light? Photons obtain mass while moving, and they are always in motion,so due to our laws of physics anything that obtains mass is not able to move with speed of light in vacumm because it would have an infinite mass, and to move an infinte mass you need infinite energy, and there is nothing in our universe that is infinite (well, maybe excpect the universe itself). So logically light shouldnt be able to move with c. Am i correct? Or is there something wrong with modern physics?

[Phrak]

"particle" is a word used to distingush one field from another having different quantum numbers. Why are we talking about particles as little bee-bees?

Or sometimes I see talk about "waves" as contrart to particles, as if there is some sort of particle-wave duality.

Throwing out these two conceptual crutches may improve ones thought processes.

[cmb]

If a photon's momentum is h/λ and it is travelling at c, then one might naively suppose its mass is h/cλ, no?

Perhaps the issue with the question is understanding what 'mass' means for relativistic particles?

[Hootenanny]

Rest mass is not the mass that a particle has only when is at rest.

The rest mass of a particle is also well-defined for frames in which the particle is not at rest. Rest mass is one of the properties that defines the particle (together with spin, charge..) and those properties are frame-independent.

From a particle point of view, the rest mass can be obtained from the mass operator in the energy-momentum space.

Maybe the term "rest" is at the root of the confusion. Just substitute the term "rest mass" by invariant mass or simply mass for avoiding it.
If you carefully re-read what I wrote, you will find that I never said that the rest mass is only defined when a particle is at rest, nor did I say that the rest mass does no exist in a frame moving relative to the particle.

What I said is that there must exist a frame in which the particle is at rest. In other words, you must be able to associate a frame of reference with the particle. These statements do not mean that the rest mass is undefined in a moving frame.

Do you follow?

[Drakkith]

Hey, i have a question. Photons posses mass only when they are moving, so according to our actual laws of physics, shouldnt they reach an infinite mass by moving with the speed of light? Photons obtain mass while moving, and they are always in motion,so due to our laws of physics anything that obtains mass is not able to move with speed of light in vacumm because it would have an infinite mass, and to move an infinte mass you need infinite energy, and there is nothing in our universe that is infinite (well, maybe excpect the universe itself). So logically light shouldnt be able to move with c. Am i correct? Or is there something wrong with modern physics?

In the equation E=MC^2 the M is equal to an objects REST mass. For a photon this is 0 and the equation is invalid because it is an abbreviated form of the whole equation and is applicable to objects with no velocity or momentum. The rest of the equation adds in the energy due to an objects momentum. In relativity ALL energy contributes to gravity, and so a photon has 0 mass yet it does have gravity and it can, and does, travel at c to the best of our knowledge. Check out wikipedia for more.

If a photon's momentum is h/λ and it is travelling at c, then one might naively suppose its mass is h/cλ, no?

Perhaps the issue with the question is understanding what 'mass' means for relativistic particles?

Unfortunately the term "mass" is taken to refer to both invariant and relativistic mass. To avoid confusion it is usually advised to only use the term mass when referring to the invariant mass. The post I quoted above is a perfect example of this confusion.

[juanrga]

If you carefully re-read what I wrote, you will find that I never said that the rest mass is only defined when a particle is at rest, nor did I say that the rest mass does no exist in a frame moving relative to the particle.

What I said is that there must exist a frame in which the particle is at rest. In other words, you must be able to associate a frame of reference with the particle. These statements do not mean that the rest mass is undefined in a moving frame.

Do you follow?

I think that you missed my point again. As said in the message that you are replying, rest mass of a particle (i.e. its mass) is frame-independent, as is also any other property that defines a particle (I cited spin and charge, above). This independence means that you do not need "to associate a frame of reference with the particle".

[Hootenanny]

I think that you missed my point again. As said in the message that you are replying, rest mass of a particle (i.e. its mass) is frame-independent, as is also any other property that defines a particle (I cited spin and charge, above). This independence means that you do not need "to associate a frame of reference with the particle".
I think that we have crossed wires here. My point was merely this: A necessary and sufficient condition for a particle to have a non-zero rest mass is that there exists a valid frame in which the particle is at rest.

Do you disagree with that statement?

[dchris]

[QUOTE=Drakkith;3547936]In the equation E=MC^2 the M is equal to an objects REST mass. For a photon this is 0 and the equation is invalid because it is an abbreviated form of the whole equation and is applicable to objects with no velocity or momentum. The rest of the equation adds in the energy due to an objects momentum. In relativity ALL energy contributes to gravity, and so a photon has 0 mass yet it does have gravity and it can, and does, travel at c to the best of our knowledge. Check out wikipedia for more.



Thanks Drakkith, now i understand. So a photon always has 0 mass, so its gravity comes from its energy, not mass. But now i have another question. So why does a photon restrict itself to c and not to higher speeds? IS the law that an object at c gains infinite mass somehow similar to energy? I mean can energy be so great at c that in becomes infinite? (meaning that E=M)

[Drakkith]

Thanks Drakkith, now i understand. So a photon always has 0 mass, so its gravity comes from its energy, not mass. But now i have another question. So why does a photon restrict itself to c and not to higher speeds? IS the law that an object at c gains infinite mass somehow similar to energy? I mean can energy be so great at c that in becomes infinite? (meaning that E=M)

Why? Not sure. All we can say is that it simply does. Maybe all the math says more but I don't know. I just know that c is the absolute limit that anything can travel at.
As for energy becoming infinite at c, that only applies for objects with mass.

[Tim_BandTech]

The simplest way to assign the mass of a photon is to marry
e = h v
with
e = m c c
which gives
m = ( h v ) / ( c c )
but this is typically called the 'rest mass' and photons do not seem to rest. Neither do electron's seem to rest, but that does not prevent a claim of measure of the electron's rest mass.

Like the consideration of mass, the consideration of wavelength as a real geometrical factor in the quality of a photon is not confronted within modern physics. The photon model must be oversimplified, and any who are happy with the current state ought perhaps to reconsider this position. The subject is alive and open for new interpretations, though these new interpretations will obviously conflict with the old principles. These old principles are inherently conflicted, such as particle wave duality, and the fact that we treat such terms as fixed and truthful principles exposes a human weakness: we are in danger of making physics into a religion. The subject must remain open to new interpretations. Keep going with your questioning and keep it first order as much as possible.

[sophiecentaur]

C and 0Kelvin needn't be regarded as ' barriers ' in the same way as Escape Velocity and the Sound Barrier. They are 'soft' limits that just require more and more effort as you want to get closer.
They are more like 'definitions' than boundaries to be conquered. Wherever you are and however fast you think you're going, something colder can be overtaking you.

[dchris]

Why? Not sure. All we can say is that it simply does. Maybe all the math says more but I don't know. I just know that c is the absolute limit that anything can travel at.
As for energy becoming infinite at c, that only applies for objects with mass.

You seem very certain about the speed limit, and i understand it. But what do you think about the discovery about those neutrinos that traveled 60 nanoseconds faster than light? Do you think its some kind of mistake or error? Or maybe a revolution?

[Vanadium 50]

I think that we have crossed wires here. My point was merely this: A necessary and sufficient condition for a particle to have a non-zero rest mass is that there exists a valid frame in which the particle is at rest.

Do you disagree with that statement?

It's true, but I don't think it's helpful to think of it as necessary. Rest mass is the norm of the energy-momentum four vector. This extends the concept seamlessly to massless particles, because even if you can't boost to a frame where the particle is at rest, every four vector has a norm.

[phinds]

You seem very certain about the speed limit, and i understand it. But what do you think about the discovery about those neutrinos that traveled 60 nanoseconds faster than light? Do you think its some kind of mistake or error? Or maybe a revolution?

First, I would caution you to develop a habit of not being sloppy in your phrasing of any statement purported to discuss science. "60 nanoseconds faster than light" is a meaningless phrase. What was observed was that neutrinos SEEM to have traveled over a certain distance in an amount of time that was 60 nanoseconds less than what it would have been had they been traveling at the universal speed limit (which is a speed limit that light obeys).

There has not as yet been any discovery that neutrinos travel faster than light. What there HAS been is a large number of experimental trials that give a result that the people who made the observations seem pretty convinced is an observational error, but they have tried six ways from Sunday to figure out where the error is and have not been able to do so, so they published their results with the specific request that other physicists (that is other than the 160 of them that had participated in the trials) see if they can figure out where the error is.

The popular press immediately pounced on this with such inanities as showing a picture of Einstein upside down and claiming a revolution in physics.

It certainly remains an open question until either the error is found or it is concluded that there is no error, in which case, THEN you can talk about faster than light, but it seems likely that an error will be found.

[jetwaterluffy]

A photon does not have mass, but it does have energy, and gravity works on anything that has energy (I think).

[Low-Q]

so can we say light has mass?No.
Say that you have a flashlight that radiates X Watt of light. Say you put frictionless wheels on the flashlight. The flashlight weights Y Newton. Now after a given time, the flashlight should now roll a given distance due to the energy from the light. Since the flashlight actually don't move at all, it cannot be mass in the light that push the flashlight away.

Vidar

[Cantab Morgan]

No.
Say that you have a flashlight that radiates X Watt of light. Say you put frictionless wheels on the flashlight. The flashlight weights Y Newton. Now after a given time, the flashlight should now roll a given distance due to the energy from the light. Since the flashlight actually don't move at all, it cannot be mass in the light that push the flashlight away.

Vidar

Why do you contend that the flashlight will not roll? Of course, a real flashlight will not, but in your thought experiment, the flashlight with frictionless wheels will enjoy an impulse from the light.

[ZapperZ]

You seem very certain about the speed limit, and i understand it. But what do you think about the discovery about those neutrinos that traveled 60 nanoseconds faster than light? Do you think its some kind of mistake or error? Or maybe a revolution?

There should be NO discussion on the OPERA results outside of the existing thread in the Relativity forum. Please confine the discussion only to on-topic subject.

Zz.

[phinds]

In superconductors, photons are massive.

Huh? Could you expand on that?

[jetwaterluffy]

No.
Say that you have a flashlight that radiates X Watt of light. Say you put frictionless wheels on the flashlight. The flashlight weights Y Newton. Now after a given time, the flashlight should now roll a given distance due to the energy from the light. Since the flashlight actually don't move at all, it cannot be mass in the light that push the flashlight away.

Vidar

It would move. That's how solar sails work. Light has momentum, but no mass.

[Hootenanny]

It's true, but I don't think it's helpful to think of it as necessary. Rest mass is the norm of the energy-momentum four vector. This extends the concept seamlessly to massless particles, because even if you can't boost to a frame where the particle is at rest, every four vector has a norm.
You are of course right. Defining the mass as the norm of the energy-momentum vector is more convenient. Perhaps necessary is was too stronger word, I was merely trying to expand on sophie's point for the OP's benefit, who I fairly sure would benefit more from a physical explanation, than a more formal one.

[Phrak]

Do photons have mass?-

On shell photon fields have mass. Off shell fields have nonzero mass.

[sophiecentaur]

On shell? What shell?

[dchris]

First, I would caution you to develop a habit of not being sloppy in your phrasing of any statement purported to discuss science. "60 nanoseconds faster than light" is a meaningless phrase. What was observed was that neutrinos SEEM to have traveled over a certain distance in an amount of time that was 60 nanoseconds less than what it would have been had they been traveling at the universal speed limit (which is a speed limit that light obeys).

There has not as yet been any discovery that neutrinos travel faster than light. What there HAS been is a large number of experimental trials that give a result that the people who made the observations seem pretty convinced is an observational error, but they have tried six ways from Sunday to figure out where the error is and have not been able to do so, so they published their results with the specific request that other physicists (that is other than the 160 of them that had participated in the trials) see if they can figure out where the error is.

The popular press immediately pounced on this with such inanities as showing a picture of Einstein upside down and claiming a revolution in physics.

It certainly remains an open question until either the error is found or it is concluded that there is no error, in which case, THEN you can talk about faster than light, but it seems likely that an error will be found.

Thanks for pointing out my mistake. But also keep in mind im just 15 years old.

[sophiecentaur]

But also keep in mind im just 15 years old.

Perhaps you could also bear that in mind when you feel tempted to jump in with both feet. You can expect to get a good kicking, sometimes, if you appear to be making unfounded assertions. OK as long as you have a thick skin - if you're 'ard enough. :devil:

[phinds]

Thanks for pointing out my mistake. But also keep in mind im just 15 years old.

I was not aware of that and I'd say you're doing GREAT for a 15 year old, but as sophiecentaur pointed out, this is not a forum for the thinskinned and while I'm very serious in saying that you're doing well for a 15 year old, I would further caution you that if you want to play in the deep end of the pool you need to keep in mind that it was your own idea. Your statement "Thanks for pointing out my mistake. But also keep in mind im just 15 years old." would really have been better as just "Thanks for pointing out my mistake". We ALL here make mistakes and part of playing in the deep end of the pool is to just fess up to them when we make them. You'll find my posts littered with them.

[sophiecentaur]

Yes. You may notice that people who start to start to 'wax lyrical' and 'alternative' very often get jumped on, here. There are always ways of saying things that avoid a bad response, at least initially. Asking questions rather than making statements and loads of IMHO's and things will help to oil the wheels. Remember, you may well be conversing with someone who knows a fair bit more about the particular topic than your teacher at School. And you would, of course, be verrrrry respectful about your teacher's knowledge. :wink:

There is a fair smattering of BS, too. You get to distinguish, soon enough, though.

[Low-Q]

It would move. That's how solar sails work. Light has momentum, but no mass.OK. I did not take that into account, but you're right. However the force is very very weak. Vidar

[sophiecentaur]

Maxwell had all this sorted out before anyone discovered photons, you might be interested to know. My old classical electromag theory book derives the amount of 'light pressure' on a surface by just using the fields and how the surface modifies them. Surprise, surprise, old JCM got the same answer as the QM crowd. I bet that made the new boys pretty happy and relieved.

[Phrak]

On shell? What shell?

http://en.wikipedia.org/wiki/On_shell_and_off_shell

[sophiecentaur]

http://en.wikipedia.org/wiki/On_shell_and_off_shell

I don't get it. How does this imply that a photon has mass? That wiki article seems to be dealing with items that 'have mass'. Experiments seem to imply that a photon has none (or, at least they give it a very very low, higher limit - which is the best an experiment could ever do).

Does that equation, relating Energy, momentum and mass, really imply anything other than something about items that actually have mass? The article doesn't seem to mention photons so are you sure it applies here? Are you sure you aren't just using a sort of circular argument?

[Phrak]

I don't get it. How does this imply that a photon has mass? That wiki article seems to be dealing with items that 'have mass'. Experiments seem to imply that a photon has none (or, at least they give it a very very low, higher limit - which is the best an experiment could ever do).

Does that equation, relating Energy, momentum and mass, really imply anything other than something about items that actually have mass? The article doesn't seem to mention photons so are you sure it applies here? Are you sure you aren't just using a sort of circular argument?

There is no restriction in the Wikipedia article that m cannot equal 0 such as the case with photons.

In the pedagogy of quatum field theory, virtual particles, photons included, can be massive--including negative mass. These particles can give rise to Coulombic forces such as the repulsion of two like-charged pith balls, where the virtual photons have spacelike trajectories.

For any given particle the mass shell in energy-momentum space has a one to one correspondence with the 4-velocity vector of that particle in space-time space.

[sophiecentaur]

There is no restriction in the Wikipedia article that m cannot equal 0 such as the case with photons.

In the pedagogy of quatum field theory, virtual particles, photons included, can be massive--including negative mass. These particles can give rise to Coulombic forces such as the repulsion of two like-charged pith balls, where the virtual photons have spacelike trajectories.

For any given particle the mass shell in energy-momentum space has a one to one correspondence with the 4-velocity vector of that particle in space-time space.

OK then. The equation just loses the term with m in it if photons don't have mass. That agrees with experiment. Where did the on-shell / off-shell thing take us with regard to photons?

[jnorman]

inre: "there is no rest frame for a photon"

how about this - when a photon is absorbed by an atom (now at rest), the mass of the atom is increased. thus the photon has added mass to the system. i understand that this is because the photon has added energy to the system. and that e=mc2, implying that energy and mass are essentially the same, and interchangeable.

to me, the reason to believe that photons have no mass is that a photon cannot be accelerated, and that they ALWAYS travel at C (they do not accelerate or decelerate when emitted or absorbed).

i have no f'ing idea what the results of OPERA could mean, other than it is simply a mistake. given the nearly complete inability to detect neutrinos, i would guess they are making some inappropriate assumptions about the measurement setup - but i am an idiot and they are uber-physicists, so iam baffled.

[phinds]

to me, the reason to believe that photons have no mass is that a photon cannot be accelerated, and that they ALWAYS travel at C (they do not accelerate or decelerate when emitted or absorbed).

Yep, that's my understanding as well.


i have no f'ing idea what the results of OPERA could mean, other than it is simply a mistake. given the nearly complete inability to detect neutrinos, i would guess they are making some inappropriate assumptions about the measurement setup.

Well don't feel bad ... all 160 of them are JUST as puzzled, which is why they published their results in the first place ... they are asking the world of physicists to please help them find the error because no one, including them, really believes at this point that neutrinos travel faster than c.

[dchris]

[QUOTE=jnorman;3552723]inre: "there is no rest frame for a photon"

how about this - when a photon is absorbed by an atom (now at rest), the mass of the atom is increased. thus the photon has added mass to the system. i understand that this is because the photon has added energy to the system. and that e=mc2, implying that energy and mass are essentially the same, and interchangeable.

The photon doesn't add mass but increases the atoms gravity by adding energy. Or did i mistaken something?

[Drakkith]

[QUOTE=jnorman;3552723]inre: "there is no rest frame for a photon"

how about this - when a photon is absorbed by an atom (now at rest), the mass of the atom is increased. thus the photon has added mass to the system. i understand that this is because the photon has added energy to the system. and that e=mc2, implying that energy and mass are essentially the same, and interchangeable.

The photon doesn't add mass but increases the atoms gravity by adding energy. Or did i mistaken something?

No, the energy does add to mass. And as such the photon does add to the mass of the system by adding that energy.
I think a key here is that when you talk about a system of particles you can talk about mass increasing. A single particle cannot have energy or mass added without being in a larger system.

[sophiecentaur]

e=mc2, implying that energy and mass are essentially the same, and interchangeable.



"the same" is tantamount to saying that a photon must have mass, though. The 'mass' quality doesn't express itself in the photon so mass and energy are not so much "the same" as equivalent or interchangeable.

[Phrak]

OK then. The equation just loses the term with m in it if photons don't have mass. That agrees with experiment. Where did the on-shell / off-shell thing take us with regard to photons?

Some background may be helpful, maybe... You might recall Feynman's "sum over histories" paradigm. A photon takes all paths to get from point A to B. A good example is a photon bouncing off a reflecting surface. In Feynman's quantum electrodynamics, it bounces off the entire surface. There is only one tiny spot on the mirror where the angle if incidence is equal the angle of reflection.

For all this to work some of the paths, or various parts of some paths will be spacelike and some time like. For a real photon, the interference from the paths with trajectories not on the light cone, will cancel. But not all interactions of particles involve cancelation of the paths off the light cone. So there can be massive photons exchanged between two particles. They are not observed directly, or we would measure a spectum for the mass of photons, but are a necessary part of the theory and called virtual particles.

I'm not sure this answers you; I'm not sure what you were asking.

[Low-Q]

I don't get it. How does this imply that a photon has mass? That wiki article seems to be dealing with items that 'have mass'. Experiments seem to imply that a photon has none (or, at least they give it a very very low, higher limit - which is the best an experiment could ever do).

Does that equation, relating Energy, momentum and mass, really imply anything other than something about items that actually have mass? The article doesn't seem to mention photons so are you sure it applies here? Are you sure you aren't just using a sort of circular argument?This reply made me start thinking on how sound waves are transfered through the air. The energy within the wave itself have no mass, but the air that is "energized" by the wave have mass. Therfor the soundwave can do practical work on an object near by. Could light-waves be just waves which travels through "something" that appears to have mass, and therefor it will appear that light have mass?

If light actually can propell a solar sail, that "mass" can be calculated. If the speed of that light is 299 792 458m/s, and we have 1kW of light pointed directly on a solar sail, this sail will accelerate at a given rate. The mass of the sail is known, so then it would be easy to find out the "mass" of that "somthing" which light travels trough and use to transfer energy into work(?).

Vidar

[Drakkith]

This reply made me start thinking on how sound waves are transfered through the air. The energy within the wave itself have no mass, but the air that is "energized" by the wave have mass. Therfor the soundwave can do practical work on an object near by. Could light-waves be just waves which travels through "something" that appears to have mass, and therefor it will appear that light have mass?

If light actually can propell a solar sail, that "mass" can be calculated. If the speed of that light is 299 792 458m/s, and we have 1kW of light pointed directly on a solar sail, this sail will accelerate at a given rate. The mass of the sail is known, so then it would be easy to find out the "mass" of that "somthing" which light travels trough and use to transfer energy into work(?).

Vidar

The energy in a sound wave is carried as kinetic energy in the particles that make up the air. A collection of moving particles, such as found in a sound wave, do in fact have more mass than they would if they were stationary. As for finding the "mass" of light, it is irrelevant. We have defined mass to mean a specific thing and light does not meet that criteria, therefor it does not have mass. It has ENERGY and as such it does contribute and is effected by gravity, but it does not have mass.

[sophiecentaur]

This reply made me start thinking on how sound waves are transfered through the air. The energy within the wave itself have no mass, but the air that is "energized" by the wave have mass. Therfor the soundwave can do practical work on an object near by. Could light-waves be just waves which travels through "something" that appears to have mass, and therefor it will appear that light have mass?

If light actually can propell a solar sail, that "mass" can be calculated. If the speed of that light is 299 792 458m/s, and we have 1kW of light pointed directly on a solar sail, this sail will accelerate at a given rate. The mass of the sail is known, so then it would be easy to find out the "mass" of that "somthing" which light travels trough and use to transfer energy into work(?).

Vidar
It is the Momentum of the light and not any 'implied mass' that causes the force. It would be better to avoid making things up as you go along and read (from beginning to end and not just the odd sentence here and there) what Wiki has to say about this. Wiki is not wrong on this topic.

[Low-Q]

It is the Momentum of the light and not any 'implied mass' that causes the force. It would be better to avoid making things up as you go along and read (from beginning to end and not just the odd sentence here and there) what Wiki has to say about this. Wiki is not wrong on this topic. I did put a question mark at the end. I had some thoughts. That's all. Wiki might be representing the todays facts untill these facts is changed by new discoveries some time in the future. I just keep my mind open, and do not always trust well established facts - because they change all the time as we learn and discover new things.

Vidar

[sophiecentaur]

If you make sure of understanding, fully, the facts as accepted, at present, then you stand a chance of understanding any new facts, as they emerge. How will you be able to judge any new stuff if you have no basics?

[Low-Q]

If you make sure of understanding, fully, the facts as accepted, at present, then you stand a chance of understanding any new facts, as they emerge. How will you be able to judge any new stuff if you have no basics?Good point! I do consider the basics, at least what we know about the behaviour / appearence of light under given conditions. I just play with some thoughts about WHY it behave/appear like it does. So I ask: Is it momentum in light itself, or is it the medium it travels trough, or the matter which absorbs or reflect light that has transformed light into momentum? Is light itself affected by gravity, or is it the path/eather the light travels through that is affected by gravity? I'll keep my mind open for any explanation, but basics will be important in any case to find answers.

Vidar

[sophiecentaur]

In our present understanding and in the model that works best, we don't consider Space as a 'medium'. If you want to go down a different road then you would have to start from a lot further back than here and build an entirely different model from scratch. Are you capable of that?. Idle speculation may be fun but, unless you accept quite a lot of the present state of knowledge, I can't see you getting very far.
I have a feeling that all this may be a lot harder than you imagine. No one, in history has built a whole model for themselves and that is what you seem to be proposing.

[jetwaterluffy]

Good point! I do consider the basics, at least what we know about the behaviour / appearence of light under given conditions. I just play with some thoughts about WHY it behave/appear like it does. So I ask: Is it momentum in light itself, or is it the medium it travels trough, or the matter which absorbs or reflect light that has transformed light into momentum? Is light itself affected by gravity, or is it the path/eather the light travels through that is affected by gravity? I'll keep my mind open for any explanation, but basics will be important in any case to find answers.

Vidar

I think light itself has momentum, but that is different to it having mass. The medium of light is electromagnetic fields, and to me, it seems even more unlikely that they have mass than light has mass. Light is affected by gravity, that is the basis of the general theory of relativity and the definition of a black hole. The light follows the curves in space.
EDIT: just looked on wikianswers for some reason. They seem to have got it into their head that light does have mass.:confused:

[sophiecentaur]

I think light itself has momentum, but that is different to it having mass. The medium of light is electromagnetic fields, and to me, it seems even more unlikely that they have mass than light has mass. Light is affected by gravity, that is the basis of the general theory of reletivity and the definition of a black hole. The light follows the curves in space.
Light just does what it always does (follows a straight line in space, in very crude terms). It's just space that is messed about by the presence of masses.

[DrDu]

Huh? Could you expand on that?

In a superconductor, local gauge symmetry is broken. The original gauge degree of freedom becomes the longitudinal component of the magnetic vector potential. I.e. the kinetic energy
operator is \frac{1}{2m_e}(p-eA)^2. In a homogeneous superconductor, the ground state wavefunction of the condensate can be written as \psi=\rho^{1/2} \exp(i\phi) where rho is approximately constant. So the expectation value of the kinetic energy is
\rho\frac{e^2}{2m_e}(\frac{\hbar}{e}\frac{\partial \phi}{\partial x_i}+A)^2. Now we re-define \tilde{A} =\frac{\hbar}{e}\frac{\partial \phi}{\partial x_i}+A and \rho\frac{e^2}{2m_e}\tilde{A}^2 looks now like a mass term for the electromagnetic field. Note that the gradient of phi makes a longitudinal contribution to \tilde{A}.
That's the famous Anderson-Higgs mechanism which was found by Anderson in 1958 in superconductors before it was transferred to elementary particle physics by Higgs in the 1960's.
The non-zero effective mass of the electromagnetic field is responsible for the rapid exponential decay of the magnetic field on the surface of a superconductor, i.e. the Meissner effect.
You can find an elementary introduction e.g. at the end of vol. 3 of the Feynman lectures.

That's not the only way to generate effectively massive photons. E.g. about half a year there was a report in Science (I think) where a group obtained a superfluid condensate of (massive) photons who gained there mass by being restricted to a wave-guide.

[phinds]

In a superconductor, local gauge symmetry is broken. The original gauge degree of freedom becomes the longitudinal component of the magnetic vector potential. I.e. the kinetic energy
operator is \frac{1}{2m_e}(p-eA)^2. In a homogeneous superconductor, the ground state wavefunction of the condensate can be written as \psi=\rho^{1/2} \exp(i\phi) where rho is approximately constant. So the expectation value of the kinetic energy is
\rho\frac{e^2}{2m_e}(\frac{\hbar}{e}\frac{\partial \phi}{\partial x_i}+A)^2. Now we re-define \tilde{A} =\frac{\hbar}{e}\frac{\partial \phi}{\partial x_i}+A and \rho\frac{e^2}{2m_e}\tilde{A}^2 looks now like a mass term for the electromagnetic field. Note that the gradient of phi makes a longitudinal contribution to \tilde{A}.
That's the famous Anderson-Higgs mechanism which was found by Anderson in 1958 in superconductors before it was transferred to elementary particle physics by Higgs in the 1960's.
The non-zero effective mass of the electromagnetic field is responsible for the rapid exponential decay of the magnetic field on the surface of a superconductor, i.e. the Meissner effect.
You can find an elementary introduction e.g. at the end of vol. 3 of the Feynman lectures.

That's not the only way to generate effectively massive photons. E.g. about half a year there was a report in Science (I think) where a group obtained a superfluid condensate of (massive) photons who gained there mass by being restricted to a wave-guide.

OK, that's over my head, but I'll check it out further. Thanks for the reference to the Feynman lectures.

[Phrak]

In our present understanding and in the model that works best, we don't consider Space as a 'medium'. If you want to go down a different road then you would have to start from a lot further back than here and build an entirely different model from scratch. Are you capable of that?. Idle speculation may be fun but, unless you accept quite a lot of the present state of knowledge, I can't see you getting very far.
I have a feeling that all this may be a lot harder than you imagine. No one, in history has built a whole model for themselves and that is what you seem to be proposing.

It better be a medium. Relativity theory stips spacetime of Newtonian mechanistic properties but not the ability to support energy/momentum density, charge and current density, and so forth. Fields don't move in space. Charge doesn't move. Energy doesn't move from place to place. Mass doesn't move. The amplitudes of these quantities change from location to location. Like the water waves Low_Q was referring to where the water doesn't move with the wave, fields don't move; their amplitudes change over time.

Low_Q was much closer than you give him credit.

[dchris]

[QUOTE=dchris;3552887]

No, the energy does add to mass. And as such the photon does add to the mass of the system by adding that energy.
I think a key here is that when you talk about a system of particles you can talk about mass increasing. A single particle cannot have energy or mass added without being in a larger system.

But previously you said that photons dont have mass, just energy that contributes to gravity. So how can something that has no mass add mass to a system?

[harrylin]

I think light itself has momentum, but that is different to it having mass. The medium of light is electromagnetic fields, and to me, it seems even more unlikely that they have mass than light has mass. Light is affected by gravity, that is the basis of the general theory of relativity and the definition of a black hole. The light follows the curves in space.
EDIT: just looked on wikianswers for some reason. They seem to have got it into their head that light does have mass.:confused:

It just depends on the definition of mass that you prefer. As far as we know, light does not have rest mass.

[harrylin]

Good point! I do consider the basics, at least what we know about the behaviour / appearence of light under given conditions. I just play with some thoughts about WHY it behave/appear like it does. So I ask: Is it momentum in light itself, or is it the medium it travels trough, or the matter which absorbs or reflect light that has transformed light into momentum? Is light itself affected by gravity, or is it the path/eather the light travels through that is affected by gravity? I'll keep my mind open for any explanation, but basics will be important in any case to find answers.

Vidar

What do you mean with "light itself"? Compare "sound itself". Momentum is a property of the phenomenon that we call "light". Your question has been partly answered by Einstein: he modeled light as a wave that is bent due to the properties of space, using the Huygens construction - this is also called "gravitational lensing".

Einstein considered Space as a kind of medium, but apparently sophiecentaur uses a different, more modern model. It could be interesting to compare the two models.

[sophiecentaur]

It better be a medium. Relativity theory stips spacetime of Newtonian mechanistic properties but not the ability to support energy/momentum density, charge and current density, and so forth. Fields don't move in space. Charge doesn't move. Energy doesn't move from place to place. Mass doesn't move. The amplitudes of these quantities change from location to location. Like the water waves Low_Q was referring to where the water doesn't move with the wave, fields don't move; their amplitudes change over time.

Low_Q was much closer than you give him credit.

Bit of a problem with that one. This is contrary to experience, isn't it? Some of the statements are clearly true and other 'Zen' statements, I can go along with because you probably just need to look at things differently. But how do you square the idea of Energy not going anywhere? Are we re-defining what we mean by 'go anywhere'?

[Phrak]

Bit of a problem with that one. This is contrary to experience, isn't it? Some of the statements are clearly true and other 'Zen' statements, I can go along with because you probably just need to look at things differently. But how do you square the idea of Energy not going anywhere? Are we re-defining what we mean by 'go anywhere'?

We define electric and magnetic fields as fixed vectors, rather than free or sliding vectors. I'm not sure what sort of entities one would have with free vector fields. Taking Maxwell's equations at face value without metaphysical interpretation, charge and current are aspects--particular derivatives--of these fields, so comprise a fixed tensor in spacetime. Current density is a fixed tensor in space. Charge density is a fixed scalar (it's really a 3-form, but that's another matter).

Mass of a system can be defined as the norm of the energy momentum vector, or more properly the norm of theenergy momentum strength vector. This is a fixed tensor field over spacetime. I imagine one could find some consistent system using free vectors, or "free tensors", but these shouldn't have the same equations relating definitions of mass, momentum and energy, I would think. I'm not sure how it would all blend together without a great deal of mess, where vectors are free in space but where 4-vectors need to be fixed in spacetime.

[juanrga]

I think light itself has momentum, but that is different to it having mass. The medium of light is electromagnetic fields, and to me, it seems even more unlikely that they have mass than light has mass.

Light is composed of photons and photons have momentum p and zero mass m, so you are right here.

Photons are the quanta of free electromagnetic fields, therefore the fields are not «The medium».

[sophiecentaur]

We define electric and magnetic fields as fixed vectors, rather than free or sliding vectors. I'm not sure what sort of entities one would have with free vector fields. Taking Maxwell's equations at face value without metaphysical interpretation, charge and current are aspects--particular derivatives--of these fields, so comprise a fixed tensor in spacetime. Current density is a fixed tensor in space. Charge density is a fixed scalar (it's really a 3-form, but that's another matter).

Mass of a system can be defined as the norm of the energy momentum vector, or more properly the norm of theenergy momentum strength vector. This is a fixed tensor field over spacetime. I imagine one could find some consistent system using free vectors, or "free tensors", but these shouldn't have the same equations relating definitions of mass, momentum and energy, I would think. I'm not sure how it would all blend together without a great deal of mess, where vectors are free in space but where 4-vectors need to be fixed in spacetime.

Am I dumb and is that an answer to my question? :smile:

[Phrak]

Am I dumb and is that an answer to my question? :smile:

Yes, it's an answer. Fields don't move in field theory and energy is a component of a field.

[sophiecentaur]

I can disagree with that ok. Whilst a field can be regarded as storing energy (energy is required to produce a field), the essence of a wave is fields continually changing - not moving. So we can accept Energy moving from place to place as the fields change but don't need to move.
How's that.?

[Phrak]

It's rare in physics that the word "move" is not associated with velocity.

[sophiecentaur]

Incontravertable.
But apropos of what?

[Low-Q]

What do you mean with "light itself"? Compare "sound itself". Momentum is a property of the phenomenon that we call "light". Your question has been partly answered by Einstein: he modeled light as a wave that is bent due to the properties of space, using the Huygens construction - this is also called "gravitational lensing".

Einstein considered Space as a kind of medium, but apparently sophiecentaur uses a different, more modern model. It could be interesting to compare the two models.
What I ment is that sound energy appears to move from one place to another, but the air itself does not go anywhere. So by saying "light itself" I meant the waves that appears to travel throug space, without the actual space going anywhere. I see space as a "carrier" of the energy of light. This carrier might as well have the momentum, and not the energy of light itself. I am therefor questioning the theory that light have momentum, when it MIGHT not be the case at all. The recent experiments with neutrinos, which preliminary concludes that they travel faster than light, they could likely use another "carrier" that has different properties than the "carrier" of the energy of light.

Just assumtions from my side, because I do not fully understand space, time, and the transmission of energy/waves over distance. I just try to make sense of all this.

Vidar

[sophiecentaur]

Just assumtions from my side, because I do not fully understand space, time, and the transmission of energy/waves over distance. I just try to make sense of all this.

Vidar

Despite what you may read on these pages, you are not alone!
It's a brave person who claims to understand 'all' or even 'any' of this stuff. If you can feel 'comfortable' or, maybe, 'familiar' with some of it then you are doing well. Remember - many people in this world don't even think about these matters at all.

Many people rant away, using their own personal model, and come to amazing conclusions (me. too, at times). Hopefully they take on board the sharp replies that people can dish out in response, without being too offended.:smile:

It often comes down to semantics. You can choose to call light the thing that moves or the thing that carries energy. I think it's reasonable to say that the waves move but the fields don't. - they just change.

[harrylin]

[..]
It often comes down to semantics. You can choose to call light the thing that moves or the thing that carries energy. I think it's reasonable to say that the waves move but the fields don't. - they just change.

I agree that light is like a wave that moves ("propagates") through space. Even using the photon concept, photons are sometimes called "wave packets" that carry momentum.

[Low-Q]

I agree that light is like a wave that moves ("propagates") through space. Even using the photon concept, photons are sometimes called "wave packets" that carry momentum.
Interesting. I must investigate further :smile:

[DrDu]

That's not the only way to generate effectively massive photons. E.g. about half a year there was a report in Science (I think) where a group obtained a superfluid condensate of (massive) photons who gained there mass by being restricted to a wave-guide.

Here is the link to the paper I mentioned (not Science but Nature):
http://www.nature.com/nature/journal/v468/n7323/full/nature09567.html

[sophiecentaur]

I agree that light is like a wave that moves ("propagates") through space. Even using the photon concept, photons are sometimes called "wave packets" that carry momentum.

The issue that I was addressing was whether Fields Move. Waves / Energy move - in anybody's book, I think but fields don't need to move in order for the wave to move.

[juanrga]

Here is the link to the paper I mentioned (not Science but Nature):
http://www.nature.com/nature/journal/v468/n7323/full/nature09567.html

From the article (bold style from mine):

The cavity mirrors provide both a confining potential and a non-vanishing effective photon mass, making the system formally equivalent to a two-dimensional gas of trapped, massive bosons.

Effective mass is the mass that a particle would carry in a semiclassical model of transport. The article does not say/suggest that photons have any mass.

[DrDu]

Effective mass is the mass that a particle would carry in a semiclassical model of transport. The article does not say/suggest that photons have any mass.

Nevertheless below the article there is some interesting discussion as to whether these photons are to be considered free photons or merely some polariton. The authors seem to be of the first opinion. So it depends somehow on your definition of a photon. This definition does not necessarily coincide in high energy physics and in condensed matter physics.

[sophiecentaur]

What worries me about this sort of discussion is the fact that people tend to ignore the caveats. There is 'implied mass', 'effective mass' and other descriptions, which apply in certain 'bound states' of photons. The innocent (young) reader will ignore the qualifying words and run away shouting "Photons have mass - yah boo" and suchlike, thinking their teachers (and all other elderly geezers) are totally wrong when they tell them that photons are massless.

The articles that deal with occasions where photons display the quality of having mass are all describing situations in which photons (even if they still can be called photons at the time) when interacting with massive entities are seen to produce mass-like effects. This must be taken into consideration and the whole thing put in proportion. Whether or not they 'really' have mass is quite irrelevant to whether photons 'really' have mass when they are buzzing around under normal conditions.

Fools rush in where angels fear to tread.
(I'm not, of course, referring to anyone who could possibly be reading this. haha)

[ZealScience]

I also have the same question. People rarely use this photon mass, instead they use photon momentum. But I think that it has effect of mass like gravitation.

And since energy in other particles give mass, then photons should also have similar effect to my perspective. I am not sure either.

[sophiecentaur]

I also have the same question. People rarely use this photon mass, instead they use photon momentum. But I think that it has effect of mass like gravitation.

And since energy in other particles give mass, then photons should also have similar effect to my perspective. I am not sure either.

They don't use it because it's not there. Momentum doesn't imply mass. You're doing exactly what I was warning about. Momentum is only mv where there is mass involved. For photons its h/λ.'

[ZealScience]

They don't use it because it's not there. Momentum doesn't imply mass. You're doing exactly what I was warning about. Momentum is only mv where there is mass involved. For photons its h/λ.'

I didn't say that momentum implies mass. I know that E/c=p where E=pc could be derived for massless particles.

But what is the exact definition of mass? Photons have gravitation. You can say that photon momentum is involved in the gravitation according to Einstein's equations. But isn't energy contributing to gravitation? And what is the difference between this energy and energy of a particle? It is a fact that energy in a fermion contributes to its mass (relativistic), right?

[DrDu]

I didn't say that momentum implies mass. I know that E/c=p where E=pc could be derived for massless particles.

But what is the exact definition of mass? Photons have gravitation. You can say that photon momentum is involved in the gravitation according to Einstein's equations. But isn't energy contributing to gravitation? And what is the difference between this energy and energy of a particle? It is a fact that energy in a fermion contributes to its mass (relativistic), right?

To describe correctly the coupling of particles which move at relativistic speeds (like free photons) to gravitational fields you have to use the equations of general relativity. There, it is the energy-momentum tensor which couples to the field, and no longer mass.

[ZealScience]

To describe correctly the coupling of particles which move at relativistic speeds (like free photons) to gravitational fields you have to use the equations of general relativity. There, it is the energy-momentum tensor which couples to the field, and no longer mass.

Yes, you need to apply Einstein's Equations. I think excluding gravitational waves, all other particles are included. Thus, fermions should be the same as bosons. Fermions have increase in mass when having more energy, so I think that bosons should also have more mass when having more energy.

[romsofia]

Even using the photon concept, photons are sometimes called "wave packets" that carry momentum.

What?? As far as I'm concerned, photons have a definite energy and wave packets are a range of possible energies and frequencies.

If I'm incorrect, please correct me!

[sophiecentaur]

Definite energy?
If you could tell me the exact frequency of one photon then I could tell you the energy, definitely.

"Wave packet" is just a rather tawdry attempt to give a photon some sort of a size.

[theman2000]

When is a photon not moving?

in its own rest frame...

[sophiecentaur]

is it valid to talk of a 'rest frame' for something travelling at c?

[Phrak]

Nevertheless below the article there is some interesting discussion as to whether these photons are to be considered free photons or merely some polariton. The authors seem to be of the first opinion. So it depends somehow on your definition of a photon. This definition does not necessarily coincide in high energy physics and in condensed matter physics.

A wave guide can be modeled as the interference generated by an array of stub antennae between parallel conducting plates or a defraction gratting at optical frequencies. In any case, we can say that all that is required to produce 'massive' photons is a little interference. This is the situation common in nature, outside the laboratory, where incoherent states are the rule. They're not really massive.

Mass can be inferred from the increased wave velocity (or reduced group velocity). But these velocities are directionally depended. The inferred mass, greater than 0, is a function of a direction vector.

There is nothing really interesting going on with this. Simply take the energy momentum 4-vector of a photon with a norm of zero--meaning zero mass--and project it onto any prefered spacetime ray--like down the axis of a wave guide. The projection will, in general, be nonzero.

[Phrak]

There have been some misconceptions floating about, on this thread. Energy and mass should not be confused unless talking about the center of mass frame, where we can use E=mc^2.

In other cases, the nontensoral equation for regions of spacetime applies (for a sufficiently flat spacetime), m^2 = E^2/c^4 - p^2/c^2.

m is the mass within a region, E is the energy, and p is the momentum. For some region containing a particle, the equation applies to the particle.

E = h \nu and p = k \lambda, where \nu is the angular frequency, and k is the reduced wave number. To find the mass in terms of \nu and k, (or lambda and omega) do the substitutions.

[dchris]

[QUOTE=Drakkith;3553101]

But previously you said that photons dont have mass, just energy that contributes to gravity. So how can something that has no mass add mass to a system?

Hey, will some of you smart guys answer my question?

[jaketodd]

When is a photon not moving?

In the reference frame of another photon traveling along side it. However, I've heard that photons are considered not to have a reference frame, so ya.

Oh, and how about at the center of a black hole?

[sophiecentaur]

[QUOTE=dchris;3557915]

Hey, will some of you smart guys answer my question?

I don't think there is an answer which involves just familiar concepts. It's outside the set of things that we are used to.

[dchris]

[QUOTE=dchris;3573294]

I don't think there is an answer which involves just familiar concepts. It's outside the set of things that we are used to.

Does my previous question outgrow you all? Cause its funny how you go on talking about stuff and cant answer such an easy question.

[Phrak]

Does my previous question outgrow you all? Cause its funny how you go on talking about stuff and cant answer such an easy question.

Which question?

[sophiecentaur]

[QUOTE=sophiecentaur;3573553]

Does my previous question outgrow you all? Cause its funny how you go on talking about stuff and cant answer such an easy question.
If it's easy as you say then why can't you answer it yourself? Or perhaps you don't understand all its ramifications.
Does a mocking tone give a post gravitas?

[dchris]

[QUOTE=dchris;3574190]
If it's easy as you say then why can't you answer it yourself? Or perhaps you don't understand all its ramifications.
Does a mocking tone give a post gravitas?

My opinion is that photons dont add mass to any system, as they dont posses any mass. They just add their energy to the system, which increases the systems gravity (curvature of spacetime).

Drakkith wrote:

"No, the energy does add to mass. And as such the photon does add to the mass of the system by adding that energy.
I think a key here is that when you talk about a system of particles you can talk about mass increasing. A single particle cannot have energy or mass added without being in a larger system."

and then i replied:
"But previously you said that photons dont have mass, just energy that contributes to gravity. So how can something that has no mass add mass to a system?"

[dchris]

So we are emerging with two different opinions, Drakkith's and mine, and im simply asking which is correct.

[Astronuc]

See - Do photons have mass? (http://www.physicsforums.com/showthread.php?t=511175)

[phinds]

So we are emerging with two different opinions, Drakkith's and mine, and im simply asking which is correct.

Yours is based on a personal theory and is incorrect, his is based on currently accepted theory and is correct.

[sophiecentaur]

"Correct" doesn't necessarily come into it. Why do you expect it to be all cut and dried, like the Victorians did?

There are many ideas about the interaction between em radiation and matter. Sometimes it is convenient to think in terms of particle-like photons, sometimes it is convenient to think in terms of waves. I think there is a lot to be said for totally ignoring the photon particle model and just think of the photon as a quantum of energy which only exists during the interaction process. I think it has to be true that the photon completely loses its identity once the energy has been absorbed into an Aton / charge system.
So what happens inside some resonating / waveguide system when a photon 'enters it' could be looked at in the same terms as what happens in other absorption situations. I.e. the photon no longer has an identity and its energy is just part of the system. So you don't need a frame of reference for a photon, ever, because it is onle in existence at the point and time of energy transfer into or out of a system.

[dchris]

Yours is based on a personal theory and is incorrect, his is based on currently accepted theory and is correct.

care to explain why?

[dchris]

His opinion based on current theories says that photons add mass to a system (thats what he wrote) but in his previous posts he wrote that photons dont have mass. So how can something without mass add mass to any system?

[Redbelly98]

Anything with energy, like a photon, can add mass to a system, in accordance with Einstein's equation E=mc2.

When people say that a photon itself has no mass, they mean it has zero rest mass, as discussed earlier in this thread.

[DrGreg]

His opinion based on current theories says that photons add mass to a system (thats what he wrote) but in his previous posts he wrote that photons dont have mass. So how can something without mass add mass to any system?Because mass isn't conserved in relativity, only energy and momentum.

In relativity, the mass of an object is given by
m = \sqrt{\frac{E^2}{c^4} - \frac{|\textbf{p}|^2}{c^2}}
where E is the total energy of the object and p is the total momentum of the object.

Now consider a photon with momentum p, and therefore energy |p|c (and zero mass), that is absorbed by an object with momentum 0 and energy m1c2. By conservation of momentum the object's momentum becomes p and by conservation of energy its energy becomes |p|c + m1c2. Then its mass becomes
m_2 = \sqrt{\frac{\left(|\textbf{p}|c + m_1c^2 \right)^2}{c^4} - \frac{|\textbf{p}|^2}{c^2}} = \sqrt{\frac{2|\textbf{p}|m_1}{c} + m_1^2} > m_1

[phinds]

I think there is a lot to be said for totally ignoring the photon particle model and just think of the photon as a quantum of energy which only exists during the interaction process.

I don't know enough about this stuff to have a "professional" opinion on your point of view but I sure do like it and I think it would avoid lots of apparently pointless discussions (it's the "pointless" that I'm not 100% sure about)

[Astronuc]

His opinion based on current theories says that photons add mass to a system (thats what he wrote) but in his previous posts he wrote that photons dont have mass. So how can something without mass add mass to any system? It's more than a theory. A deuteron can absorb a photon of a certain minimum energy and dissociate into its constitutive nucleons, a proton and a neutron.

The sum of the proton mass and neutron mass exceeds that of the deuteron.

m(p) = 1.007276466812(90) u ref: http://physics.nist.gov/cgi-bin/cuu/Value?mpu
m(n) = 1.00866491600(43) u ref: http://physics.nist.gov/cgi-bin/cuu/Value?mnu

m(d) = 2.013553212724(78) u ref: http://physics.nist.gov/cgi-bin/cuu/Value?mdu

The difference in energy is the equal to the energy given off as a gamma ray when a neutron combines with a proton to form a deuteron, or the minimum energy of a gamma ray required to dissociate a deuteron into the two nucleons. Actually the energy is slightly higher than the mass difference because one must account for the kinetic energies of the masses involved.

[sophiecentaur]

This is a typical example of people trying to 'explain' or come to terms with an advanced concept whilst still holding on to 'Victorian' ideas of Science. Why do you think the changes in the last hundred years or so have been call 'Earth Shattering'? It's because you just can't make do with the old notions if you want to get nearer to Scientific 'truth'.
To get a grasp of modern Science one has, constantly, to hurt the brain considerably and go with the flow. There's one paradox after another to deal with on this road.
If you think you've understood it then you haven't.

[juanrga]

Because mass isn't conserved in relativity, only energy and momentum.

In relativity, the mass of an object is given by
m = \sqrt{\frac{E^2}{c^4} - \frac{|\textbf{p}|^2}{c^2}}
where E is the total energy of the object and p is the total momentum of the object.

Now consider a photon with momentum p, and therefore energy |p|c (and zero mass), that is absorbed by an object with momentum 0 and energy m1c2. By conservation of momentum the object's momentum becomes p and by conservation of energy its energy becomes |p|c + m1c2. Then its mass becomes
m_2 = \sqrt{\frac{\left(|\textbf{p}|c + m_1c^2 \right)^2}{c^4} - \frac{|\textbf{p}|^2}{c^2}} = \sqrt{\frac{2|\textbf{p}|m_1}{c} + m_1^2} > m_1


It seems that you are taking the absorption of one photon by one electron as an elastic collision.

For the object at rest

m = \sqrt{\frac{E_0^2}{c^4}}

For the object in motion with momentum \textbf{p}

m = \sqrt{\frac{E_1^2}{c^4} - \frac{\textbf{p}^2}{c^2}}

With evidently E_1 > E_0 and m the same because is an invariant. The mass of an electron m=m_e is always the same when it is at rest or when it is moving at 0.99 c.

[DrDu]

Consider a highly reflecting cavity where some photons are captured and reflected on and off the walls. The mass of the whole system cavity+photons will depend on the number of photons present, as the photons do increase the rest mass of the whole system.

[sophiecentaur]

Consider a highly reflecting cavity where some photons are captured and reflected on and off the walls. The mass of the whole system cavity+photons will depend on the number of photons present, as the photons do increase the rest mass of the whole system.

But does that scenario imply that the photons 'have' mass any more than when you get a measurable Mass Defect after a nuclear reaction? Can one afford to be so literal in these circumstances?

[Phrak]

I don't think it's a good idea to use E=mc^2, in general, for the definition of mass as some seem to be doing. This is not really done by most, except within the center of mass frame.

Using m^2 = E^2 - p^2, good in special relativity in cartesian coordinates, then promoting to general relativity, obtains the oriented tensor density consisting of energy and momentum density, invariant in both curved spacetime and curvilinear coordinates, having a norm equal to the mass density.

The four-momentum tensor density, \mathbb{P}_a dx^a is the most fundemental object in this discussion of photon mass. Fortunately, in special relativity, P_\mu dx^\mu \Leftarrow \mathbb{P}_a dx^a, so we can talk about the 4-momentum intensity of a system having a norm equal to the invariant mass as long as we stick to right handed coordinate systems.

[DrDu]

But does that scenario imply that the photons 'have' mass any more than when you get a measurable Mass Defect after a nuclear reaction? Can one afford to be so literal in these circumstances?

No, but it shows clearly how photons can contribute to the mass of an object. Furthermore, it should be clear that a standing wave in a cavity is not made up of virtual but rather real photons in contrast e.g. to the photons which are responsible for the Coulomb attraction and add to the mass of an atom.

[sophiecentaur]

No, but it shows clearly how photons can contribute to the mass of an object.

I don't think there was ever any question about that. But 'contribution' is not necessarily the same as 'having' or 'being'. A drop of ink is just a drop of ink but it may contribute to a written word.


Furthermore, it should be clear that a standing wave in a cavity is not made up of virtual but rather real photons in contrast e.g. to the photons which are responsible for the Coulomb attraction and add to the mass of an atom.

Except that the photons are bound to a structure. They are interacting with the walls of the cavity - unlike when they are in free space.

[Phrak]

No, but it shows clearly how photons can contribute to the mass of an object. Furthermore, it should be clear that a standing wave in a cavity is not made up of virtual but rather real photons in contrast e.g. to the photons which are responsible for the Coulomb attraction and add to the mass of an atom.

Why do you say photons rather than electromagnetic field?

Pervect had something interesting to say about the mass of a confined electomagnetic field. Apparently, the calculation of the field energy will yield a value exactly twice the energy put into the system. The field pressure put on the cavity walls puts the material elements in tension, reducing their mass by the correct amount.

[dchris]

Because mass isn't conserved in relativity, only energy and momentum.

In relativity, the mass of an object is given by
m = \sqrt{\frac{E^2}{c^4} - \frac{|\textbf{p}|^2}{c^2}}
where E is the total energy of the object and p is the total momentum of the object.

Now consider a photon with momentum p, and therefore energy |p|c (and zero mass), that is absorbed by an object with momentum 0 and energy m1c2. By conservation of momentum the object's momentum becomes p and by conservation of energy its energy becomes |p|c + m1c2. Then its mass becomes
m_2 = \sqrt{\frac{\left(|\textbf{p}|c + m_1c^2 \right)^2}{c^4} - \frac{|\textbf{p}|^2}{c^2}} = \sqrt{\frac{2|\textbf{p}|m_1}{c} + m_1^2} > m_1


thanks Greg, now i get it. Can you just tell me why do we take |p|c in the place of E?

[DrGreg]

Now consider a photon with momentum p, and therefore energy |p|c (and zero mass), that is absorbed by an object with momentum 0 and energy m1c2....It seems that you are taking the absorption of one photon by one electron as an elastic collision.You have misread my mind. I never said the second object was an electron. It could be an atom, or, for that matter, an apple.

[DrGreg]

In relativity, the mass of an object is given by
m = \sqrt{\frac{E^2}{c^4} - \frac{|\textbf{p}|^2}{c^2}}
where E is the total energy of the object and p is the total momentum of the object.

Now consider a photon with momentum p, and therefore energy |p|c (and zero mass)...thanks Greg, now i get it. Can you just tell me why do we take |p|c in the place of E?For a photon, E = pc. If you didn't already know that, you can get it from the equation above with m = 0. See photon (http://en.wikipedia.org/wiki/Photon).

[juanrga]

You have misread my mind. I never said the second object was an electron. It could be an atom, or, for that matter, an apple.

I read what your wrote. You wrote general claims for an «object», in general, and I showed that your claims are not valid for a kind of objects, including electrons.

If your object is an atom or apple then the total energy E is not the kinetic energy {*} but includes U and V (if any) and your equation is not valid.

{*} Including the rest term.

[DrGreg]

I read what your wrote. You wrote general claims for an «object», in general, and I showed that your claims are not valid for a kind of objects, including electrons.What I said was:

Now consider a photon with momentum p, and therefore energy |p|c (and zero mass), that is absorbed by an object with momentum 0 and energy m1c2.It is a matter of simple verbal logic that whatever follows applies only to those objects that have the capability to absorb a photon (and therefore excludes electrons). (If I'd said "consider a red object", then obviously I'd be excluding blue objects.)If your object is an atom or apple then the total energy E is not the kinetic energy {*} but includes U and V (if any) and your equation is not valid.

{*} Including the rest term.I stand by what I said. You haven't defined what you mean by "U" and "V", but I'm guessing you are referring to the internal potential energies and kinetic energies of the object's constituent particles, but I don't care about that. All I am interested in is the total momentum 0 and total energy m1c2 of the entire composite object, in which case I can apply the equation of post #117 and conclude that the total "system mass" of the object is (by definition) m1.

In case there is still confusion, the total system mass is not the sum of the individual particle masses, it's defined as the total energy (as measured in the zero-momentum frame) divided by c2.

[sophiecentaur]

What I said was:

It is a matter of simple verbal logic that whatever follows applies only to those objects that have the capability to absorb a photon (and therefore excludes electrons). (If I'd said "consider a red object", then obviously I'd be excluding blue objects.)I stand by what I said. You haven't defined what you mean by "U" and "V", but I'm guessing you are referring to the internal potential energies and kinetic energies of the object's constituent particles, but I don't care about that. All I am interested in is the total momentum 0 and total energy m1c2 of the entire composite object, in which case I can apply the equation of post #117 and conclude that the total "system mass" of the object is (by definition) m1.

In case there is still confusion, the total system mass is not the sum of the individual particle masses, it's defined as the total energy (as measured in the zero-momentum frame) divided by c2.

People seem to think that electrons absorb photons because they are seem unaware that it is the electron as part of a charge system that is doing the 'absorbing'. This may account for the confusion.

[juanrga]

What I said was:

It is a matter of simple verbal logic that whatever follows applies only to those objects that have the capability to absorb a photon (and therefore excludes electrons). (If I'd said "consider a red object", then obviously I'd be excluding blue objects.)

Electrons can absorb photons, otherwise they could not be accelerated.

Moreover, what you wrote in previous posts do not apply to other objects as atoms, nuclei, apples...

I stand by what I said. You haven't defined what you mean by "U" and "V", but I'm guessing you are referring to the internal potential energies and kinetic energies of the object's constituent particles, but I don't care about that. All I am interested in is the total momentum 0 and total energy m1c2 of the entire composite object, in which case I can apply the equation of post #117 and conclude that the total "system mass" of the object is (by definition) m1.

In case there is still confusion, the total system mass is not the sum of the individual particle masses, it's defined as the total energy (as measured in the zero-momentum frame) divided by c2.

The case of an elementary particle was considered before and I will not repeat.

For a composite object being a collection of free particles the expression for its rest mass is

m = \sqrt{\frac{E_0^2}{c^4}} = \sqrt{\frac{E^2}{c^4} - \frac{\textbf{p}^2}{c^2}}

Where E and p are total energy and momentum, respectively, of the object.

Of course, the object mass m is not just «the sum of the individual particle masses» as you correctly state, but m does not vary with the total momentum p of the object, as you affirm. The mass of the object at rest, p=0, is the same than the mass of the object with a nonzero p.

For a free microscopic object being a collection of bound particles, the expression for its rest mass is the same

m = \sqrt{\frac{E_0^2}{c^4}} = \sqrt{\frac{E^2}{c^4} - \frac{\textbf{p}^2}{c^2}}

where again m is not just the sum of the individual particle masses and, again, m is not a function of the total momentum p, as you affirm. The mass of a nucleus is independent of its p. The mass defect, i.e. the difference between nuclei m and the sum of the mass of the nucleons, is a function of the binding energy, neither of E nor p of the nuclei.

For objects as apples, the above expression does not apply, because the total energy of an apple is not just kinetic, it is lacking U and V contributions.

For an apple falling with non-relativistic speeds within Earth gravitational potential, the mass is

m = \frac{p^2}{2(E-E_0-V)}

Once again this m is not just the sum of the individual particle masses and, again, m is not a function of the total momentum p, as you affirm. The mass of the apple is independent of p. In group theory it is often said that m is a central charge, i.e. one of the basic invariants.

It seems that you are confounding the rest mass m as defined above with the mass M

M = \frac{E}{c^2}

In some (old?) texts you can find this definition of mass, which of course varies with the total momentum p of the object.

M = M(p)

[Phrak]

Because mass isn't conserved in relativity, only energy and momentum.

I don't know what you mean by this. The mass of a system is conserved in special relativity. We can form a mass continuity equation in special relativity, and promote it to general relativity with the use of densities.

[Phrak]

Electrons can absorb photons, otherwise they could not be accelerated.

DrGreg was speaking of an electron in isolation. In isolation, an electron is in uniform motion. It will not radiate. The converse is equally true; it will not absorb a quanta of light and not accelerate [speaking in the lingo where elementary particles are presumed little bee bee balls of stuff].

The rest of your post is incoherent.

[sophiecentaur]

Not light, of course, because the energy change would have to be too high. There would be a quantum of very low energy electromagnetic energy, though, in the case of interaction with a near-static electric field - which is something I had forgotten in my last post. It is a good idea not to restrict one's ideas about photons to those of light.
You can often shed more light on the nature of photons by looking beyond light.

[juanrga]

DrGreg was speaking of an electron in isolation. In isolation, an electron is in uniform motion. It will not radiate. The converse is equally true; it will not absorb a quanta of light and not accelerate [speaking in the lingo where elementary particles are presumed little bee bee balls of stuff].

The rest of your post is incoherent.

If you read the posts and check the context of the quote that you extract, you will see that was that of an electron initially at rest, absorbing a photon so that finally it is moving with momentum p.

[DaleSpam]

Electrons can absorb photons, otherwise they could not be accelerated.An electron can be accelerated by scattering. Absorption is not the only process that can result in acceleration.

[sophiecentaur]

An electron can be accelerated by scattering. Absorption is not the only process that can result in acceleration.

Does that mean that the scattered photon has a different frequency? (If it's transferred some KE to the electron) It could be a very small difference, I suppose.

[DaleSpam]

Does that mean that the scattered photon has a different frequency? (If it's transferred some KE to the electron) It could be a very small difference, I suppose.In the COM frame the scattered photons frequency is the same. In all other frames it changes.

[sophiecentaur]

COM of the electron?

[DaleSpam]

No, COM of the electron+photon.

[sophiecentaur]

But, if the photon has no mass . . . .?
(Reasonable question?)

[jtbell]

COM = center of momentum; the frame in which the total momentum is zero.

[DaleSpam]

But, if the photon has no mass . . . .?
(Reasonable question?)Oops, sorry, when you asked your follow-up question I should have realized that the acronym was unclear. As jtbell mentioned COM means center of momentum, not center of mass.

[sophiecentaur]

All clear now. Thanks.

[juanrga]

An electron can be accelerated by scattering. Absorption is not the only process that can result in acceleration.

Compton scattering, Møller scattering, Bhabha scattering... are all based in the existence of a photon absorption step.

But my point was really other. I was said that electrons cannot absorb photons, which is not true.

[Phrak]

I find it cleaner and more obvious to avoid square roots where possible and use the vector equations that are good in Minkowski coordinates for adding masses.

define\mu for a particle.

\mu = (E/c^2, \textbf{p}/c)

The bases vectors are dropped for convenience.

For a two particle system.
\mu_1 = (E_1/c^2, \textbf{p}_1/c)
\mu_2 = (E_2/c^2, \textbf{p}_2/c)
Vector addition.
\mu_1+\mu_2 = ([E_1 + E_2 ]/c^2, [\textbf{p}_1 + \textbf{p}_2 ]/c)
The particle masses.
m_1 = |\mu_1|
m_2 = |\mu_2|
The combined mass.
m_\Sigma = |\mu_1 + \mu_2 |

[DaleSpam]

Compton scattering, Møller scattering, Bhabha scattering... are all based in the existence of a photon absorption step.In a Feynman diagram scattering is indeed represented as an absorption and an emission, but there is a key difference between absorption and scattering. The absorption step, by itself, cannot conserve momentum, so by itself it is a "virtual" event which can never be observed directly, much like virtual photons. On the other hand scattering can be directly observed. In fact, an isolated electron has never been observed to change momentum due only to the absorption of a real photon.


But my point was really other. I was said that electrons cannot absorb photons, which is not true.I wasn't responding to your other point. Merely correcting the assertion that the only way for an electron to change momentum is by absorption of a photon.

[juanrga]

I find it cleaner and more obvious to avoid square roots where possible and use the vector equations that are good in Minkowski coordinates for adding masses.

define\mu for a particle.

\mu = (E/c^2, \textbf{p}/c)

The bases vectors are dropped for convenience.

For a two particle system.
\mu_1 = (E_1/c^2, \textbf{p}_1/c)
\mu_2 = (E_2/c^2, \textbf{p}_2/c)
Vector addition.
\mu_1+\mu_2 = ([E_1 + E_2 ]/c^2, [\textbf{p}_1 + \textbf{p}_2 ]/c)
The particle masses.
m_1 = |\mu_1|
m_2 = |\mu_2|
The combined mass.
m_\Sigma = |\mu_1 + \mu_2 |

Beautiful, only to remark that if the particles are not free, neither m_1 nor m_2 represent the masses of the particles.

[juanrga]

In a Feynman diagram scattering is indeed represented as an absorption and an emission, but there is a key difference between absorption and scattering. The absorption step, by itself, cannot conserve momentum, so by itself it is a "virtual" event which can never be observed directly, much like virtual photons. On the other hand scattering can be directly observed. In fact, an isolated electron has never been observed to change momentum due only to the absorption of a real photon.

Therefore, scattering is only possible if the electron absorb a photon.

I wasn't responding to your other point. Merely correcting the assertion that the only way for an electron to change momentum is by absorption of a photon.

See above.

[tom.stoer]

In a Feynman diagram scattering is indeed represented as an absorption and an emission, but there is a key difference between absorption and scattering. The absorption step, by itself, cannot conserve momentum, so by itself it is a "virtual" event ...
This is wrong. At each vertex in a Feynman diagram 4-momentum is exactly conserved.

[DaleSpam]

... In fact, an isolated electron has never been observed to change momentum due only to the absorption of a real photon.Therefore, scattering is only possible if the electron absorb a photon.That is an impressive non-sequitir.

[DaleSpam]

This is wrong. At each vertex in a Feynman diagram 4-momentum is exactly conserved.Oops, sorry, you are right. I meant to say that the absorption of a real photon by an isolated electron cannot conserve momentum.

[tom.stoer]

Oops, sorry, you are right. I meant to say that the absorption of a real photon by an isolated electron cannot conserve momentum.
Here I agree :-)

[logics]

... if the photon has no mass . . . .? (Reasonable question?...)The rationale of this thread is right there: question is not reasonable, it is pointless. If it is so, all answers become pointless too, even though discussion may be interesting; and "light has no mass" may seem more appropriate.
Milk has cheese?, water has ice? can we answer yes or no? The logical problem that makes question meaningless looks like "http://wikipedia.org/wiki/Present_King_of_France has hair?", but it has a different cause :
there the subject "king" is not, here the subject "EMR" is also the predicate.
Misunderstanding arises because of the inadequacy of term annihilation of a photon or of mass +e -e : EMR transforms itself into mass and vice versa.
They say mass is "trapped", "transformed", "a form of" energy ": just like ice of water. water is mobile, ice is not.
does water have ice?

[Phrak]

The rationale of this thread is right there: question is not reasonable, it is pointless. If it is so, all answers become pointless too, even though discussion may be interesting; and "light has no mass" may seem more appropriate.
Milk has cheese?, water has ice? can we answer yes or no? The logical problem that makes question meaningless looks like "http://wikipedia.org/wiki/Present_King_of_France has hair?", but it has a different cause :
there the subject "king" is not, here the subject "EMR" is also the predicate.
Misunderstanding arises because of the inadequacy of term annihilation of a photon or of mass +e -e : EMR transforms itself into mass and vice versa.
They say mass is "trapped", "transformed", "a form of" energy ": just like ice of water. water is mobile, ice is not.
does water have ice?

Maybe. But physics does not advance in this way.

[Phrak]

Beautiful, only to remark that if the particles are not free, neither m_1 nor m_2 represent the masses of the particles.

There are as many as two approches to this that I can decern. One is that \mu_n represent the integrals of energy and momentum density over a sufficiently large region of space where a single quanta is to be found and where the values are propperly signed so that they combine as a vector rather than a co-vector. That is,
E = \int \rho dx dy dz
p_i = \int P_{jkt} dx dy dt
where \rho is energy density, and P_{jkt} is the momentum density; the infinitessimal flux of mass per time dt through an area dx^i dx^j.

The other, is that we presume isolated billiard ball-like particles that have an associated energy-momentum intensity vector. Both representations are elements of a vector space (not the same one) and transform as displacement vectors in Minkowski spacetime.

Unlike the unintegrated density, neither is viable in curved spacetime.

[juanrga]

... In fact, an isolated electron has never been observed to change momentum due only to the absorption of a real photon.
Therefore, scattering is only possible if the electron absorb a photon.
That is an impressive non-sequitir.

I was really replying to the part of you message that you have not quoted:
In a Feynman diagram scattering is indeed represented as an absorption and an emission...
Therefore, scattering is only possible if the electron absorb a photon.

In my response I ignored the last part that you quoted, because, as already emphasized in #138, I am not considering an isolated electron but an interacting one. And I did not consider that I may be repeating the same again and again in each new post.

I only want to add that if the electron was isolated, it could not participate in a scattering. And if the electron is being scattered then it is not isolated.

If you want believe that an electron can be scattered (e.g. Compton scattering) without absorbing a photon whereas, at the same time, accepting that in QED that scattering involves a step where a photon is absorbed by an electron, you are welcome.

[juanrga]

Beautiful, only to remark that if the particles are not free, neither m_1 nor m_2 represent the masses of the particles.
There are as many as two approches to this that I can decern. One is that \mu_n represent the integrals of energy and momentum density over a sufficiently large region of space where a single quanta is to be found and where the values are propperly signed so that they combine as a vector rather than a co-vector. That is,
E = \int \rho dx dy dz
p_i = \int P_{jkt} dx dy dt
where \rho is energy density, and P_{jkt} is the momentum density; the infinitessimal flux of mass per time dt through an area dx^i dx^j.

The other, is that we presume isolated billiard ball-like particles that have an associated energy-momentum intensity vector. Both representations are elements of a vector space (not the same one) and transform as displacement vectors in Minkowski spacetime.

Unlike the unintegrated density, neither is viable in curved spacetime.

Valid only if the particles are free.

In previous posts I have considered the mass of generic objects including non-isolated apples.

[DaleSpam]

If you want believe that an electron can be scattered (e.g. Compton scattering) without absorbing a photon whereas, at the same time, accepting that in QED that scattering involves a step where a photon is absorbed by an electron, you are welcome.Yes, this is exactly the point. QED models the scattering that way, also involving a virtual particle which by definition can never be detected. So it is an unfalsifiable part of QED. However QED is not the only model which accurately describes scattering, and other models do not involve an abbsorption. It is an unfalsifiable, model dependent claim. Therefore, scattering does not imply absorption.

Btw I am not objecting to your overall claim, just the logic that the observed acceleration of free electrons implies absorption. It is simply not true of real photons and free electrons, to make it true requires one specific model in which it is unfalsifiably true.

[Phrak]

Valid only if the particles are free.

I don't know what you mean, unless you missed my use of the conditional "isolated".

[juanrga]

Valid only if the particles are free.
I don't know what you mean, unless you missed my use of the conditional "isolated".

It was only a redundant emphasis of what I wrote just after of the paragraph that you quoted above:

In previous posts I have considered the mass of generic objects including non-isolated apples.

[juanrga]

However QED is not the only model which accurately describes scattering, and other models do not involve an abbsorption.

I am curious, could you say what models please?

[DaleSpam]

I am curious, could you say what models please?Rutherford scattering, Mott scattering, Compton scattering, for example.

[juanrga]

However QED is not the only model which accurately describes scattering, and other models do not involve an abbsorption.

I am curious, could you say what models please?

Rutherford scattering, Mott scattering, Compton scattering, for example.


In #148 I cited «Compton scattering», «Møller scattering», «Bhabha scattering» and others. And remarked as they «are all based in the existence of a photon absorption step».

When you said me that scattering could be explained with the same accuracy by other models different than QED, I asked you waiting the name of that alternative to QED that also explains scattering... but your response has been «scattering». Fine :yuck:.

[Phrak]

Try other formulations of quantum field theory.

[logics]

See - Do photons have mass? (http://www.physicsforums.com/showthread.php?t=511175)
The quick answer: NO
question is ...pointless.
does water have ice?
Maybe. But physics does not advance in this way.
Physics, Science in general, advances looking for, finding, and accepting the "truth".
what is your quick "yes/no...because" answer?,
what is your argument?

[juanrga]

Try other formulations of quantum field theory.

Well I was said above:
However QED is not the only model which accurately describes scattering, and other models

It seems that he was talking about alternatives to QED model, not about another formulation of the same QED, but in any case after reading your reply:

what other formulation of QED(QFT) you claim that does not use photons to describe electron electron scattering for instance?

[logics]

I am curious, could you say what models please?... Compton scattering, for example.
and what is your argument?
Compton scattering is an elastic collision between light and a "free" electron", with a transfer of Energy from γ to -e: can we say this is "absorption" in any model?
If we can accept that E [=]→ M, a fortiori we must accept that E [=]→ KE, that "EMR" can transform itself not only into mass but also into kynetic energy.
The "absorption" of extra energy causes an increase of KE and then this causes an increase of velocity and then this causes an increase of momentum.
[EMR]E → Ek → v → p is not an identity but a chain of causal relationships
Is there any problem? do photons have mass? do photons have momentum?