0 Members and 2 Guests are viewing this topic.

Why do you need a photon for this? A light pulse wouldn't work?

Ok, anyway, if you want to use the term "photon" because it's emitted or absorbed by an atom, you can't say it can be localized in flight. Either is a photon, with its quantum properties, or is a classical pulse of light; you can't mix the two in a sort of "chimeric" beast.

Quote from: lightarrow on 19/01/2013 14:13:05Ok, anyway, if you want to use the term "photon" because it's emitted or absorbed by an atom, you can't say it can be localized in flight. Either is a photon, with its quantum properties, or is a classical pulse of light; you can't mix the two in a sort of "chimeric" beast.I'm afraid that you’ve made the same mistake here that you’ve made in your previous posts. You say it can’t be done but give no proof. In this case you claim “you can't say it can be localized in flight” but don’t explain what that means or why you can’t say it and what it means not to be able to say something when in practice (i.e. in practical examples, math and all) it works just fine.

Lightarrow, is there any reason why you can't take a classical limit of the quantum theory to come up with classical photons like Pmb claims? That approach is certainly valid for electrons (and explains why we have "classical" electrons" when we know they also behave like waves in the 2 slit experiment).

All matter and energy has mass. Even the kinetic energy of an object has mass.

In fact, when you compress a spring, its mass increases... not detectably, but the potential energy added to the spring has its own contribution to the total mass.

Quote from: JP on 21/01/2013 16:43:56Lightarrow, is there any reason why you can't take a classical limit of the quantum theory to come up with classical photons like Pmb claims? That approach is certainly valid for electrons (and explains why we have "classical" electrons" when we know they also behave like waves in the 2 slit experiment).When you take the classical limit for a photon (h --> 0) it gets zero energy, so it disappears.

When you take the classical limit for a photon (h --> 0) it gets zero energy, so it disappears.

If you mean relativistic mass, ok.

Recall that in classical electrodynamics one can have a very small packet/burst of radiation (which can be described by a Fourier integral) which has enegy and momentum p. The relationship between them is a non-quantum mechanical relationship, i.e. E = pc. The shape of the light pulse can be selected such that the spatial extention is small enough for all practical purposes.

Yep, and a photon is usually defined similar to a Fourier component (monochromatic plane wave) of a pulse.

Quote from: lightarrowWhen you take the classical limit for a photon (h --> 0) it gets zero energy, so it disappears.(sigh!) I'm clearly sorry that I asked. That's wrong. If you were right then no classical particles exist in the classical limit. Don't forget what h physically means. I means that for every quantum mechanical particle that has inertial energy (defined as E = K + E_{0} = Kinetic Energy + Rest Energy) has an associated frequency given by the relationship E = hf. What does it mean to take h -> 0 for an electron? It means that there is no associated wavelength.

Quote from: AndroidNeox on 21/01/2013 17:44:02All matter and energy has mass. Even the kinetic energy of an object has mass. If you mean relativistic mass, ok. If you mean "mass", with this term it's usually intended "invariant" mass and then it's false

Quote from: lightarrow on 21/01/2013 19:12:37Quote from: AndroidNeox on 21/01/2013 17:44:02All matter and energy has mass. Even the kinetic energy of an object has mass. If you mean relativistic mass, ok. If you mean "mass", with this term it's usually intended "invariant" mass and then it's falseNaturally I'm referring to relativistic mass, since that's what the question is about. The rest mass doesn't change because it's never in motion and has no kinetic energy. I was specifically referring to kinetic energy having mass.

Certainly. However we should be more precise when we discuss this subject because it's very easy to make confusion. For example, saying "All matter and energy has mass. Even the kinetic energy of an object has mass" is very confusing: in the first sentence, matter has invariant mass, "energy has mass" is incorrect, since energy is "a property" of a body, and a property cannot have mass (as if I would say that a number has a colour); we should say instead that "a body which has energy has mass", but in this case is not always invariant mass...As you see, things are not so simple.

Quote from: lightarrow on 23/01/2013 21:19:32Certainly. However we should be more precise when we discuss this subject because it's very easy to make confusion. For example, saying "All matter and energy has mass. Even the kinetic energy of an object has mass" is very confusing: in the first sentence, matter has invariant mass, "energy has mass" is incorrect, since energy is "a property" of a body, and a property cannot have mass (as if I would say that a number has a colour); we should say instead that "a body which has energy has mass", but in this case is not always invariant mass...As you see, things are not so simple.Energy does have gravitational mass.

Put a kilogram of matter and one of antimatter into an impregnable box, like a Schrödinger cat box, and the mass of the box (any category of mass you care to choose) will not change when the contents annihilate each other. Even if the box only contains light, the mass(es) will not change.

That's wrong. I have already replied you (in a previous post of this or another similar thread, don't remember) that a photon is different because has zero mass.

For an electron, you can have non-zero momentum p even at very low speeds, because of its non-zero mass. Then De-Broglie relationship: = h/p tells that, in the limit h --> 0, = 0, that is, frequency should be infinite.

I can see that you’re not one who is much for agreeing to disagree, huh? Okay.

Quote from: lightarrow linkFor an electron, you can have non-zero momentum p even at very low speeds, because of its non-zero mass. Then De-Broglie relationship: = h/p tells that, in the limit h --> 0, = 0, that is, frequency should be infinite.And since the product of wavelength and momentum has to remain the same

In any case you keep forgetting that we’re treating it as a classical particle for which the relationship: = h/p ignored.

Thus it means nothing in the classical sense to take a classical limit of a photon.

JP and myself explained how it works in a classical sense.

You certainly can’t change the textbooks

or change the usage of the center of mass of a two photon system in the textbooks. As I have asked too many times now with no response – why do people insist in using the scenario I’ve described and never came to a wrong answer.Please explain, in detail if you please, what it means when you say “You can’ do that!” when so many people do it,

and very successfully too I might ad. You imply that you can’t draw a worldline of a photon in a spacetime diagram and yet physicists do it all the time.

E.g. referring to finding the center of mass of two photons you claimed “You can't localize a photon, so you can't do that.” When in fact derivations based on the localizability of property of a photon that plays a crucial role in its use in the derivation, you know, the one you claimed can’t be done. While you’re at it please state in detail what you mean whey you say “You can’t localize a photon”. Again you claimed that just because I was speaking about classical physics you claimed that I was speaking about quantum mechanics, again with no proof provided. I suggest that you take a look at the derivation again ...sorry, you cannot view external links. To see them, please REGISTER or LOGIN just pas Eq. (. Nothing about a quantum nature about anything is mentioned in there. Easy as cake. Over and over again you keep making the unfounded assertion that you can’t localize a photon but then never state what that means in practice. Then you later claim that you “so you can’t speak of photons”. Then you go on an claim I don’t know a lot about QM just because you didn’t understand what I wrote and you didn’t appear to go back to JPs post to see the context in which I made the statement which makes you think I don’t understand QM. As JP says and I believe him Maybe there's a reason why you can't, but based on my (admittedly limited) understanding of the Standard Model, I don't see why this would be.

I've already explained in simple terms why you can't localize a photon, but you don't accept it because, you say, it works only for photons described in quantistic sense; I tried to show you that this is the only description for the term "photon" and so we are in a loop...I sincerely don't know what else I could say.

Quote from: lightarrowI've already explained in simple terms why you can't localize a photon, but you don't accept it because, you say, it works only for photons described in quantistic sense; I tried to show you that this is the only description for the term "photon" and so we are in a loop...I sincerely don't know what else I could say.If that is your response then you weren't paying attention to what I was saying. You consistently keep forgetting the approximation and what it would mean to put the photon's position vector inside the area of uncertainty according to how the wave function would average the position. I gave you an example of a pixel of 0.001 mm in width a length and when it detects the photon then its localized in that area and the location of the photon is the location of the pizel) e.g. geometric center.You youy insist on ignoring every single thing I've said regarding approximation then there is no use to continue this conversation. Why should I post something I know you're going to igore?While you're at it it wouln't hut you to finally state what it is you mean by saying something can or can't be localized. E.g. find a QM texbook and quote the definition of "localized" or "localize" so you won't be vauge anymore.

Is there a good reason why you keep refusing to provide a definition of the term "localized" as you insist on using it? BTW photons are always localized when their position is measured

But a photon that's been measured is no longer a photon.

The entirety of this argument comes down to the fact that Lightarrow uses photon to mean a single photon Fock state, which cannot be localized by definition, and you're using photon to mean a classical pulse..

I guess this is where this second quantinization stuff I hear about so often is addressed. I've never studied QM at this level before. Thanks for mentioning it. It'll give me a goal to reach after I refreshen my quantum mechanics in the next few months.

Quote from: Pmb on 26/01/2013 17:05:54I guess this is where this second quantinization stuff I hear about so often is addressed. I've never studied QM at this level before. Thanks for mentioning it. It'll give me a goal to reach after I refreshen my quantum mechanics in the next few months. It's interesting stuff. I've studied the basics, but haven't really applied it to anything. A single Fock state can't be localized, but physics can and do write approximate photon wave functions when considering photons emitted by a source and absorbed by a detector. I briefly read a section on this in Optical Coherence and Quantum Optics in Mandel and Wolf when this question came up previously, but I didn't have the time to really dig into the details. Again, I'm not going to wade into the argument of whether we should call localized wave functions that aren't single-photon Fock states "photons," but some physicists definitely do so in practice.

Millikan demonstrated very directly what he called the "unitary nature of electricity," the fact that electric charge is quantized and transferred in multiple integrals of e.

Recall the definition that I used. A classical photon is a particle whose inertial energy is related to its momentum by E = pc and interacts with charges via the electromagnetic interaction.

QuotePut a kilogram of matter and one of antimatter into an impregnable box, like a Schrödinger cat box, and the mass of the box (any category of mass you care to choose) will not change when the contents annihilate each other. Even if the box only contains light, the mass(es) will not change.Correct, but it doesn't confirm your statement.By the way, there is no need of matter and antimatter and not even of light in a box, two photons are enough, because such a system have a non-zero mass (I mean invariant mass, not relativistic mass), I have already showed it in a recent thread and in several others, during the years.

Quote from: lightarrow on 24/01/2013 09:04:56QuotePut a kilogram of matter and one of antimatter into an impregnable box, like a Schrödinger cat box, and the mass of the box (any category of mass you care to choose) will not change when the contents annihilate each other. Even if the box only contains light, the mass(es) will not change.Correct, but it doesn't confirm your statement.By the way, there is no need of matter and antimatter and not even of light in a box, two photons are enough, because such a system have a non-zero mass (I mean invariant mass, not relativistic mass), I have already showed it in a recent thread and in several others, during the years.It's both correct and does prove my statement.

proper mass (i.e. what lightarrow is always referring to when he sees the word "mass") has little or nothing to do with the defining characteristics of mass.

Quote from: Pmbproper mass (…) has little or nothing to do with the defining characteristics of mass.That's because you're defining mass to preclude the use of invariant mass.

proper mass (…) has little or nothing to do with the defining characteristics of mass.

Plenty of physicists (mostly in high energy physics) use "mass" to mean invariant mass, which also has desirable properties. (1) Invariant mass properly satisfies m=p/v when you use 4-momentum and 4-velocity, while inertial mass doesn't(2) Invariant mass is invariant when you change inertial reference frames, which is a very elegant property.

I can't comment on your points (3) and (4), since I'm not up to speed on my general relativity. They probably don't matter much to most particle physicists, since they don't deal with gravity.

It's also important to note that we all agree on the properties of mass in the non-relativistic limit, but all these definitions of mass agree in that limit, so we can't use that as a basis for picking the "best" one.

There is legitimate controversy in the teaching of physics over which definition to use.

Quote from: JPQuote from: Pmbproper mass (…) has little or nothing to do with the defining characteristics of mass.That's because you're defining mass to preclude the use of invariant mass.Not at all. I’ve chosen to state the definitions (not to define them since they were defined waaaay before I was even a gleam i m'daddy's eye! lol!) of mass that describe dynamics. I’m not choosing one particular definition of mass over another to state because I wish to preclude the use of invariant mass.

I'm just taking issue your one line above that I quoted.

Quote from: JPI'm just taking issue your one line above that I quoted. I smell a debate about proper mass vs rest mass in the air. That's when I must leave the room. Methinks it be bad juju!

Quote from: Pmb on 28/01/2013 06:13:29Quote from: JPI'm just taking issue your one line above that I quoted. I smell a debate about proper mass vs rest mass in the air. That's when I must leave the room. Methinks it be bad juju!I wasn't the one telling posters that proper mass has little or nothing to do with the definition of mass! I'm content to call them "invariant/proper mass" and "inertial/relativistic mass" and skip the arguing phase over which meets the definition of mass.

Perhaps the solution should be to elect me the President of Physics

and I'll rewrite all the textbooks to clear this up?

Quote from: lightarrowI've already explained in simple terms why you can't localize a photon, but you don't accept it because, you say, it works only for photons described in quantistic sense; I tried to show you that this is the only description for the term "photon" and so we are in a loop...I sincerely don't know what else I could say.If that is your response then you weren't paying attention to what I was saying. You consistently keep forgetting the approximation and what it would mean to put the photon's position vector inside the area of uncertainty according to how the wave function would average the position. I gave you an example of a pixel of 0.001 mm in width a length and when it detects the photon then its localized in that area and the location of the photon is the location of the pizel) e.g. geometric center.

You youy insist on ignoring every single thing I've said regarding approximation then there is no use to continue this conversation. Why should I post something I know you're going to igore?While you're at it it wouln't hut you to finally state what it is you mean by saying something can or can't be localized. E.g. find a QM texbook and quote the definition of "localized" or "localize" so you won't be vauge anymore.

Sorry, but I don’t know what a Fock state is.In any case, that’s not what I meant by a “classical photon.” Recall the definition that I used.

A classical photon is a particle whose inertial energy

is related to its momentum by E = pc and interacts with charges via the electromagnetic interaction. There is no associated wavelength since that’s a quantum property just as a classical electron has no wavelength. By this definition it moves on a classical trajectory, has a position vector, etc.

This is what they use in the derivations for the mass-energy equivalence relationship where they use the conservation of the center of momentum. It’s also what relativists use when they draw a worldline of a photon.

Forget what he's been saying. He has a way of confusing the poperties of mass with those of proper mass.

There are three aspects of mass given three names and each are merely just called "mass" because they all have the same value(1) inertial mass - m = p/v. The higher the inerial mass the harder it is to change its momentum.(2) passive gravitational mass - The property of matter to respond to a gravitational force.(3) active graivtational mass - The property of matter to generate a gravitational field.proper mass (i.e. what lightarrow is always referring to when he sees the word "mass") has little or nothing to do with the defining characteristics of mass. A photon has zero proper mass but has inertial mass, passive gravitational mass and active gravitational mass.

This one does a nice job of explaining the history of mass, and how the idea of passive and active mass came to be. ...sorry, you cannot view external links. To see them, please REGISTER or LOGIN