Post by: cheryl j on 27/11/2012 18:32:17
Post by: simplified on 28/11/2012 09:36:38
I'm sorry if this is a dumb question but physics is not really my area. I've been listening to the CBC Massey lectures by physicist Neil Turok, which I quite like. Anyway, when he talks about mass increasing at higher speeds, how does that happen? Is there actually an increase in the amount of matter or atoms or particles? Or does it just take more force to accelerate it? I had always thought that mass and matter were the same thing.If you stop photon then mass increases.
Post by: Soul Surfer on 28/11/2012 11:00:51
The reason for this is that Einstein realised that the laws of physics cannot change when you are moving, because you can only find out if you are moving (or something else is moving ) by looking at something else. Also if you only have one other thing to look at to tell you that you are moving you cannot tell which of you (or both of you ) are moving.
This results in the fact that light must travel at the same speed to you whatever speed you are doing because a famous experiment done 125 years ago "The Michelson Morley experiment" was designed to measure the true velocity of the earth through space by measuring the velocity of light in different directions throughout the year failed to measure the change brought about by the velocity of the earth in its orbit round the sun. This should have easily been detectable and came as a total shock to physicists at the time.
The non variability of the velocity of light with motion had has been proved experimentally to be true to one part in 10**-17 in recent years
Post by: yor_on on 28/11/2012 17:43:54
Locally, as long as you're uniformly moving, it shouldn't change anything in your local experiments, although you might find starlight outside your solar system to blue shift. Accelerations are another thing in that they are locally (intrinsically?) measurable whereas a uniform motion always will need a referent to measure a speed.
Post by: Pmb on 02/12/2012 15:27:42
I'm sorry if this is a dumb question ...Don't worry. It's not a dumb question, so your safe. :)
Note: relativistic mass is often also referred to by the name inertial mass.
Anyway, when he talks about mass increasing at higher speeds, how does that happen? Is there actually an increase in the amount of matter or atoms or particles? Or does it just take more force to accelerate it? I had always thought that mass and matter were the same thing.
[/quote]
Yes. It does require an increased force by an amount which is greater than the
Newtonian expression.
For a list of jouirnal articles on this subject please see
http://home.comcast.net/~peter.m.brown/ref/mass_articles/mass_articles.htm
Meanwhile I can give you the gist of it. The increase in mass is a result of the properties of spacetime. The combination of the relationship between space and time between two different inertial frames, one the rest frame of the observer. m and the other the rest frame of the particle M, shows that M = m/sqrt(1 - v^2/c^2).
The derivation can be found at my website which is at
http://home.comcast.net/~peter.m.brown/sr/inertial_mass.htm
If you're really randy about this subject I studied this aspect of relativity in great deal and summarized it in this (unpublished) paper. Why would you consider reading an unpublished paper? For the same reason you'd consider reading a post I'd post in a thread. With that in mind the article is at http://arxiv.org/abs/0709.0687
Post by: Phractality on 02/12/2012 20:08:39
Post by: a_dark_knight on 05/12/2012 06:14:51
Post by: Pmb on 05/12/2012 06:22:23
I vote that it's actually inertia which increases, not mass.These are not independant things. Mass is the measure of something's inertia.
Post by: lightarrow on 07/12/2012 18:36:29
How does mass increase at higher speeds?It doesn't.
Post by: Bill S on 08/12/2012 17:05:45
I would be fascinated to know your thoughts.
Post by: JP on 08/12/2012 17:25:19
I vote that it's actually inertia which increases, not mass.These are not independant things. Mass is the measure of something's inertia.
How does mass increase at higher speeds?It doesn't.
Both are right. Like many things in modern physics, we've realized that when we go to extremes that are way outside of our daily experience, seemingly simple concepts like mass can have multiple definitions.
In paticular, at high speeds two types of mass can be defined: invariant mass and relativistic (or inertial) mass. At slow speeds, both are equal and reduce to our everyday idea of mass.
When someone says mass increases at high speeds, they're talking about relativistic mass. Both Soul Surfer and pmb gave good explanations of why relativistic mass increases: it gets more energy to boost something's speed by the same amount if it's going fast than if it's going slow.
When someone says mass doesn't change at high speeds, they're talking about invariant mass, which by definition stays the same at all speeds. Both have their uses in physics, and so long as you're clear about which one you're using, you won't have problems. You can get to one from the other. However, most physics courses and physicists mean invariant mass when they say mass, since that's become the standard concept taught in class.
Post by: lightarrow on 08/12/2012 20:29:37
Post by: JP on 08/12/2012 21:46:15
I actually tend to agree with you, lightarrow, that invariant mass is the better definition to use for many things, including when teaching students relativity. It's certainly the only definition I found useful in a bit of work I did in graduate school. I don't oppose experts in the field using relativistic mass if they tell me it's useful for them. I don't know enough cosmology to intelligently criticize or support its use there, for example.
Post by: a_dark_knight on 11/12/2012 10:09:21
These are not independant things. Mass is the measure of something's inertia.
But mass creates a gravitational field. Whereas inertia doesn't, in my opinion. That's the distinction I'm referring to. Mass also implies the amount of "stuff" (or matter) whereas inertia is just resistance to a force. So does that mean that things moving near the speed of light have a larger gravitational field than they would otherwise? Maybe it would normally be negligible but the whole point of science is to be accurate.
Post by: yor_on on 11/12/2012 15:51:22
Relativistic mass and 'gravity' are not the same.
But if we use 'energy' as our measure then?
=
Although, you have uniform constant accelerations to consider too?
But if we're talking 'speeds' as something uniformly moving.
Post by: JP on 11/12/2012 16:17:06
That's a really good point Dark :)
Relativistic mass and 'gravity' are not the same.
But if we use 'energy' as our measure then?
You have to use the stress-energy tensor: http://en.wikipedia.org/wiki/Stress%E2%80%93energy_tensor
This is a reason why arguments that only involve mass break down, e.g. why don't moving masses become black holes since they get heavier?
A simpler argument is that it's really the invariant mass that matters if it's possible to catch a ride on the mass. Then you'd see the mass at rest with respect to you and its field would be entirely due to its invariant mass. You could then figure out its gravitational field in any reference frame you chose by using GR to change reference frames.
Post by: yor_on on 11/12/2012 19:41:11
Consider yourself heating up a gram of some, very, temperature resistant metal. You've weighted it before but after it gets heated you weight it again, finding it to weight more. One way to describe it might be to consider the particles making the material accelerating inside the metal as they gain 'energy' from heat, moving agitatedly. Can we then discuss those particles as gaining a relativistic mass, or not?
If they have we also find that this relativistic mass indeed have changed the 'proper' mass, that is if we would confine a proper mass to be whatever constitutes of that piece metal macroscopically..
And is it 'accelerations' that do it, or 'uniform motion' :)
Post by: JP on 11/12/2012 23:12:49
But photons individually have no mass? How did it gain mass?
Post by: cheryl j on 12/12/2012 02:47:36
Hi Cheryl J; having worked through this thread I find myself wondering if your original question was answered. I think it may have been, but that could be because I had my own pre-conceived idea as to what it should be.
I would be fascinated to know your thoughts.
I think I'm more confused than ever.
Post by: cheryl j on 12/12/2012 03:02:15
Post by: JP on 12/12/2012 03:14:16
Hi Cheryl J; having worked through this thread I find myself wondering if your original question was answered. I think it may have been, but that could be because I had my own pre-conceived idea as to what it should be.
I would be fascinated to know your thoughts.
I think I'm more confused than ever.
At the risk of confusing matters more...
If you understand mass as a measure of the "resistance" of something to being pushed faster, then it does increase as the speed increases. In brief, since the speed of light is a limit, it takes more and more work to push something and speed it up as you get closer to the speed of light. The number of particles stays the same, so the amount of matter doesn't increase in that sense.
Any time this gets brought up, it'll start an argument over the correct definition of mass. The problem is there are other definitions which don't agree with the above. For slowly moving things all definitions DO agree, but at high speeds they differ. The standard, textbook definition that's taught these days does not increase as speed increases, but I'll leave that to others to debate.
Post by: JP on 12/12/2012 03:15:14
okay, what about that weird meter stick thought experiment where different observers pass it going different relative speeds. Does the meter stick really become shorter as the observers approach the speed of light. Is there "less" of the meter stick? Because in this experiment it doesn't sound like the meter stick's mass, matter, or inertia has changed at all, the observers are different.
No more matter is created or destroyed in the meter stick. The number of particles in it stays the same. The shape of the particles will change as they'll now be measured to be short.
Post by: lightarrow on 12/12/2012 15:15:54
Let's take a example that gives us a totally new, and well earned, headache :)Not to me [:)]
Quote
Correct.
Consider yourself heating up a gram of some, very, temperature resistant metal. You've weighted it before but after it gets heated you weight it again, finding it to weight more.
Quote
One way to describe it might be to consider the particles making the material accelerating inside the metal as they gain 'energy' from heat, moving agitatedly.Correct.
Quote
Can we then discuss those particles as gaining a relativistic mass, or not?You can discuss it, but there is no need of relativistic mass to describe that fact.
In physics is Very important to give attention to which is the physical system considered. The system of particles inside the gram of metal *it's not* the simple "sum" of the particles. Do you remember when I showed that a system of 2 photons with opposite velocities has non zero invariant mass?
There is a way of saying that "the whole is more than the sum of its parts". Something similar in physics (or, at least, in this case).
Post by: yor_on on 12/12/2012 22:47:03
I can see you and Pete gearing up to a argument Lightarrow :)
But I think you can use the idea to argue a 'relativistic mass', if you want?
You can translate the particles agitations into accelerations, and a added '(mass)energy' confined inside the metal.
It depends on how you define 'gravity' though. Using the original equivalence it must be a constant uniform acceleration, but to me all accelerations should present you with a inertia/gravity. The squid those guys used to create a dynamical Casimir effect should have weighted a 'little' more if I'm thinking right there.
Post by: Bill S on 12/12/2012 23:33:52
I think it all depends on the frame of reference. In trying to get my head round this I resorted to a thought experiment that can easily be turned into a real one. All you need is a blank wall, a directional light and a metre (OK, the spell checker doesn't like UK spelling, but I'm too old to change :P ) stick. Hold the stick parallel to the wall and stick and shadow are the same length. Rotate the stick and the shadow becomes shorter.
In its own F of R, the stick is the same length, but in the F of R of the wall (or the observer) it is measurably shorter.
The scientists on the forum will probably throw up their hands in horror, but it helped me.
Post by: Bill S on 12/12/2012 23:39:21
Assuming that the Higgs particle has really been found, this would seem to herald a new era in physics. How will it influence our understanding of mass?
Post by: yor_on on 12/12/2012 23:44:50
Post by: bizerl on 12/12/2012 23:52:34
Is the increase in mass phenomenon the same as the length contraction phenomenon, and dependent on how it is measured (or indeed, defined)?
What type of mass does the "m" represent in the good old E=mc2?
Post by: Phractality on 13/12/2012 00:07:24
Do you remember when I showed that a system of 2 photons with opposite velocities has non zero invariant mass?
If the two photons are somehow bound together, the pair would act like a particle. Since we're in the mainstream forum I'll refrain from speculating on what force or field could conceivable bind a pair of photons together. A change in relative velocity, dv, of the center of the pair (whether they're bound or not) is equivalent to looking at the pair from a different reference frame, having velocity dv relative to the center of the pair. SR gives the ratio of the pair's energy and momentum in the two reference frames. At non-relativistic speeds, the momentum ratio for a given velocity difference is the mass of the pair. That's what inertial mass is ... M = dp/dv.
I don't accept the claim that a photon has no mass. A bound pair of photons (if there is such a thing) would have a rest mass. The radiant energy of the photons would become the rest mass of the pair (or particle).
In the reference frame centered on the pair, the mass would be M = E/c2. In a different reference fame it would be greater by the SR factor, gamma. Mathematically, you would get the same result if the photons are not bound together, but considered as the sum of the two free photons.
If you consider only one photon, it also has different momenta in different reference frames. A change of reference frame gives a change of momentum. If dv is in the direction of v, then dp/dv is the inertial mass of the photon. So a photon has mass. If dv is not parallel to the path of the photon, you have to apply the relativistic form of M = dp/dv. At relativistic speeds, M varies; dp = Mdv +vdM, which, I believe, turns M into a hyperbolic function of v. (Above my pay grade.) Perhaps the mass of an individual photon is different in different directions; I lack the math skill to settle that question.
Post by: yor_on on 13/12/2012 05:52:24
m= can be relativistic mass, or rest mass depending, as far as I've seen.
Einstein is said to not have used that formula. It's a abbreviation made later, by others.
Take a look here for one side of the discussion :) Rest Mass Versus Relativistic Mass. (http://physicsandphysicists.blogspot.se/2009/04/rest-mass-versus-relativistic-mass.html)
I find the idea of a rest mass simpler, as I don't have to consider accelerations relative different uniform motions, because even though you could, if considering a Higgs field, assume that a acceleration then results in more 'energy' locally expressed, what about the changed uniform motion after the acceleration? Gaining a different (although still 'relative') speed, but not a different energy, locally? All as I interpret relativity naturally.
Post by: Spacetectonics on 13/12/2012 08:11:35
Particle gets mass when interacting with higgs field(H.F),higgs field is through out the universe and it is an energy field.
particles considered as wave"behavior" in QM >
wave/Pare. interacting with H.F creating mass >
Nature always wants to be in its lowest state and that is why H.F born>
In QM Wave Interacting with the lowest known state of energy in universe creating mass>
mass is a quantitative measure of an object's resistance to acceleration>
If acceleration reaches C mass will be Infinite.
!!Based on this apparently, argument will be "Resistance"!!
Quantitative measurements are those which involve the collection of numbers.I put this in a post
And yet I don't know the answer ,as everyone comes up with different solutions!
Cheers
Post by: JP on 13/12/2012 15:05:26
[quote = JP] If you understand mass as a measure of the "resistance" of something to being pushed faster, then it does increase as the speed increases.
Assuming that the Higgs particle has really been found, this would seem to herald a new era in physics. How will it influence our understanding of mass?
[/quote]
I'm not a Higgs expert, but from what I understand it won't be too revolutionary to have found the standard Higgs. This is because it's been part of the Standard Model for a while (since the 1960s, I think) and people have spent a lot of time thinking about its implications. If we find the standard Higgs particle, it validates the model, but doesn't introduce new physics.
Now, if we find a Higgs that isn't exactly as the model predicts, find more than one Higgs-like particle or don't find it at all, we'd need to retool our theories. It would, however, save people time in looking for alternatives to the Higgs mechanism.
There are other reasons why the Higgs won't completely revolutionize our idea of mass. As Yor_on mentioned earlier it explains the inertial mass, or resistance to pushing, of simple particles. But most matter is due to particles bound together with forces, and the energy of these bonds also creates mass which the Higgs doesn't explain (as far as I understand it, at least).
Another reason is that the Higgs provides a mechanism for inertial mass--the Higgs field makes particles resist changes in velocity. However the magnitude of a particle's mass is determined by how strongly it interacts with the Higgs field. There's no theoretical reason why an electron couples to the field with one strength and a muon with another. We determine the strengths experimentally. Presumably, if the Higgs is discovered (and even now that there's very strong evidence for it), physicists will start working on theories to explain the interaction strength.
Post by: lightarrow on 13/12/2012 15:27:49
There is no need of binding and, anyway, two photons don't bind together, but I see what you want to say.
Do you remember when I showed that a system of 2 photons with opposite velocities has non zero invariant mass?
If the two photons are somehow bound together, the pair would act like a particle.
Quote
Since we're in the mainstream forum I'll refrain from speculating on what force or field could conceivable bind a pair of photons together.Good choice [:)]
Quote
A change in relative velocity, dv, of the center of the pairHow do you define the centre in a system of two photons?
Quote
(whether they're bound or not) is equivalent to looking at the pair from a different reference frame, having velocity dv relative to the center of the pair. SR gives the ratio of the pair's energy and momentum in the two reference frames. At non-relativistic speeds, the momentum ratio for a given velocity difference is the mass of the pair. That's what inertial mass is ... M = dp/dv.It's not a claim: if it had (invariant) mass, it would have infinite energy.
I don't accept the claim that a photon has no mass.
Quote
A bound pair of photons (if there is such a thing) would have a rest mass.But even unbound, the system has invariant mass, what do you want more? [:)] If two photons escape one from the other, or if staied close each other as in a sort of atomic system, it wouldn't make difference: the system has/would have invariant mass for the same reason.
Post by: lightarrow on 13/12/2012 15:30:01
What type of mass does the "m" represent in the good old E=mc2?Invariant mass (the one sometimes also called "rest" mass or "proper" mass).
Post by: Phractality on 14/12/2012 00:35:27
How do you define the centre in a system of two photons?
Tough question! [::)] I'll have to ponder this for a while. Let's not even think about the effect of the expansion of space; just confine the discussion to short distances and times where space does not expand appreciably.
For openers, let's consider only inertial reference frames whose relative motion is restricted to the x direction, which is the direction of relative motion of the two photons, which are moving in opposite directions parallel to the x-axis. (Later we may try to generalize to other reference frames.) Let's define the origins of all these reference frames as the point in space-time where the two photons pass closest to one another, at x = 0, t = 0. In all such reference frames, the center of the two-photon system is the origin, but the origins of different reference frames only coincide at the instant when the photons pass one another.
I guess you have to start with a reference frame in which both photons have equal energy and equal and opposite momenta. In that reference frame, the energy of each photon is E, and the momentum of each photon is p = E/c. Since the momenta are equal and opposite, the momentum of the system is zero.
Next, consider a reference frame moving in the +x direction at .866 c relative to the first reference frame. Gamma = 2; so in this reference frame, the photon moving in the +x direction has energy E'1= E/2, and the other photon has energy E'2 = 2E. Do I have that correct? My brain is about to trip a circuit breaker, here. Someone, please let me know if I got the photons' energy right before I proceed to dig myself into a deeper hole.
My postulate is that, at low velocities, the two photon system has inertial mass M = dp/dv = E/c2. At higher velocities, I suspect that formula turns into a hyperbolic function. Rats! I hate hyperbolic functions! [xx(]
Thinking ahead: In any one reference frame, the center of the two-photon system is fixed, so a two-photon system can't have a non-zero velocity in an inertial reference frame. However, when you change to another reference frame, you move that center. (This gets into a gray area between SR and GR. I'm not sure I'll be able to handle the math.) A gradual change from one reference frame to another by small increments, dv, will gradually move the center of the system. So the rate of change of reference frame's velocity relative to the first reference frame (dv/dt) imparts motion to the system's center. I don't know yet whether the systems center has acceleration or uniform velocity while the reference frame's velocity is changing at a constant rate. If the momentum of the system changes at the rate dp/dv, that is the inertial mass of the system.
Post by: lightarrow on 14/12/2012 10:25:57
No. In the first case:How do you define the centre in a system of two photons?
Tough question! [::)]
Next, consider a reference frame moving in the +x direction at .866 c relative to the first reference frame. Gamma = 2; so in this reference frame, the photon moving in the +x direction has energy E'1= E/2, and the other photon has energy E'2 = 2E. Do I have that correct?
E' = E*sqrt[(1-beta)/(1+beta)] = E*sqrt[(1-sqrt(3)/2)/(1+sqrt(3)/2)] = [2 - sqrt(3)]E ~ 0.268E;
in the second case:
E' = E*sqrt[(1+beta)/(1-beta)] = E*sqrt[(1+sqrt(3)/2)/(1-sqrt(3)/2)] = [2 + sqrt(3)]E ~ 3.73E.
Post by: Phractality on 15/12/2012 04:46:09
Thanks for the correction. I had a feeling I got it wrong. I know how to do the math, but the math corner of my brain was in full revolt.No. In the first case:How do you define the centre in a system of two photons?
Tough question! ::)
Next, consider a reference frame moving in the +x direction at .866 c relative to the first reference frame. Gamma = 2; so in this reference frame, the photon moving in the +x direction has energy E'1= E/2, and the other photon has energy E'2 = 2E. Do I have that correct?
E' = sqrt[(1-beta)/(1+beta)] = sqrt[(1-sqrt(3)/2)/(1+sqrt(3)/2)] = 2 - sqrt(3) ~ 0.268E;
in the second case:
E' = sqrt[(1+beta)/(1-beta)] = sqrt[(1+sqrt(3)/2)/(1-sqrt(3)/2)] = 2 + sqrt(3) ~ 3.73E.
I'm wondering, now, if I should consider the center of the two-photons to be the center of energy; equivalent to center of mass. Applying inverse square law to the energy of each photon to get a ratio of each photon's distance from the center. My brain hurts; maybe I'll just play solitaire, instead. :-\
Post by: lightarrow on 15/12/2012 15:12:26
Thanks for the correction. I had a feeling I got it wrong. I know how to do the math, but the math corner of my brain was in full revolt.You can't localize a photon, so you can't do that.
I'm wondering, now, if I should consider the center of the two-photons to be the center of energy; equivalent to center of mass. Applying inverse square law to the energy of each photon to get a ratio of each photon's distance from the center. My brain hurts; maybe I'll just play solitaire, instead. :-\
Post by: Phractality on 15/12/2012 22:49:01
You can't localize a photon, so you can't do that.Suppose you know that a pair of equal photons were emitted in opposite directions from a certain point in space-time (as determined by detected particles left behind). Define the origin of an inertial frame at that point, with the photons on the x-axis. Then you can assume they exist at points x = ct and x = -ct.
It gets more interesting and challenging when you try to describe those same photons in a different inertial frame at relativistic speed relative to the first frame. If you reflect the two photons off of perfect mirrors to make their paths parallel but not collinear, it gets a bit more challenging. I don't even want to think about accelerating reference frames, but I'm afraid they are necessary to describe the inertia of the two-photon system, which is given by dp/dv (for small values of dv). Accelerating an observer relative to the two photon system is equivalent to accelerating the center (of perhaps center of mass/energy) of the system in the observer's frame. But acceleration takes us out of the realm of SR, and the math gets way to hairy for my puny brain to handle. [:-\]
Post by: Bill S on 19/12/2012 19:13:57
Quote from: JP
I'm not a Higgs expert, but from what I understand it won't be too revolutionary to have found the standard Higgs. This is because it's been part of the Standard Model for a while (since the 1960s, I think) and people have spent a lot of time thinking about its implications. If we find the standard Higgs particle, it validates the model, but doesn't introduce new physics.
Point taken, but if we have found the standard Higgs particle, then, presumably, we have found the Higgs field.
If we have found the Higgs field, must we not have co-ordinates for an "absolute space".
I could be wide of the mark, but the thinking goes something like this: If we are saying that a particle's mass arises from its motion through the Higgs field, the Higgs field must be stationary, in an absolute sense, or the masses of particles would vary, depending on their direction of movement through the Higgs field.
Post by: JP on 19/12/2012 20:21:55
Post by: jopie64 on 19/12/2012 23:56:43
Yor_on, here's the same question, but in a slightly more extreme form: If you have a box made of perfect mirrors and you inject some light into it, the box's energy has now increased. If it's sitting still next to you, its mass increases (by E=mc2, which holds for stationary objects). So clearly its mass, measured at rest, went up. Since invariant mass is supposed to not change with reference frame, and the rest frame is a reference frame, its invariant mass also went up. Additionally, if you try to push it, you'll find its inertial mass went up.
But photons individually have no mass? How did it gain mass?
To what I've learned from physics, photons dont have mass, but they do have impulse. Photons can 'push' things when something absorbs or reflects it. The mirrors reflect the photons. So when you push the box and accelerate it a bit, photons push harder against the side you push. So you feel resistance. Even so photons are following the spacetime curve of gravity, so they tend to move more down then up in the box. So more photons reflect to the bottom then to the top of the box. This way the box feels heavier.
My question is, does this increased mass you feel, because of the 'impulse pressure' (or whatever you call it), also generate gravity?
Post by: bizerl on 20/12/2012 00:42:17
To what I've learned from physics, photons dont have mass, but they do have impulse. Photons can 'push' things when something absorbs or reflects it. The mirrors reflect the photons. So when you push the box and accelerate it a bit, photons push harder against the side you push. So you feel resistance. Even so photons are following the spacetime curve of gravity, so they tend to move more down then up in the box. So more photons reflect to the bottom then to the top of the box. This way the box feels heavier.I like this idea, however it would only increase the weight of the box if it was already in a gravitational field, but not necessarily the mass.
My question is, does this increased mass you feel, because of the 'impulse pressure' (or whatever you call it), also generate gravity?
It seems like one of those ideas that depend on how "mass" is defined and measured.
Post by: JP on 20/12/2012 02:00:37
Post by: Pmb on 20/12/2012 05:00:11
Quote from: a_dark_knight
But mass creates a gravitational field. Whereas inertia doesn't, in my opinion.By definition, the quantity which generates a gravitational field is an objects active gravitational mass. Since the active gravitational field of a body increases with the speed of a body it follows that an objects active gravitational mass also increases. Since the active gravitational mass of a body equals the body’s inertial mass it follows that the body’s inertial mass also increases with speed.
Quote from: a_dark_knight
That's the distinction I'm referring to.You weren’t the person that post was responding to.
Quote from: a_dark_knight
Mass also implies the amount of "stuff" …The term “matter” does is not well defined and is only be used in a vague sense. Inertia really refers to the idea that Newton referred to when he spoke of “quantity of motion” which refers to the quantity m in the relation p = mv where p is defined as in F = force = dp/dt.
Quote from: a_dark_knight
So does that mean that things moving near the speed of light have a larger gravitational field than they would otherwise?The faster a body moves the stronger its gravitational field
Post by: yor_on on 20/12/2012 19:46:45
Or do you mean that it is strictly observer dependent, and so needs two bodies in relative motion versus each other? But that way 'gravity' would 'fluctuate' with what observer we have in relative motion, relative what body's gravitational field he measures. My thought has been, and still is, that uniform motion no matter its speed, as measured relative something else, has no effect on the gravitational field surrounding it?
Assuming that we have a buildup of gravity depending on uniform motion wreaks havoc to relativity as I think, because to me it implies a 'global speed definition' in where you locally do have a 'absolute definition' of what a speed is, not relative.
Post by: Pmb on 22/12/2012 00:53:32
Quote from: yor_on
But that doesn't count for uniformly moving bodies Pete, right?Sure it does. Why wouldn't it?
Quote from: yor_on
Or do you mean that it is strictly observer dependent, and so needs two bodies in relative motion versus each other?That question is not clear. Please rephrase. Why d you need two bodies? What function do these bodies serve?
Do you do you think that, for some reason, something that is observer dependant requies two bodies in relative motion?
Quote from: yor_on
But that way 'gravity' would 'fluctuate' with what observer we have in relative motion, relative what body's gravitational field he measures.I don't understand what you're talking about wqen you speak of "gravity would fluctuate". Do mean that the gravitational field changes with the speed of the source? Its an odd thing but the fact that the strength of a gravitational source depends on the velocity of the source is a fundamental fact in GR but its such a little known fact on these internet forums. I suppose its because so many people who use the concept of mass being independant of its motion in SR also assume that they should also imploy that same definition in the mass of a gravitational source in GR. Just look at how much confuses it causes??? :(
Quote
My thought has been, and still is, that uniform motion no matter its speed, as measured relative something else, has no effect on the gravitational field surrounding it?Why? What led you to that conclusion? It certainly isn't one that arrives at through calculation, that's for sure.
Consider quantities which define the strength of the gravitational field such as the Christofell symbols or the components of the metric tensor. Since these quantities become velocity dependant when one invokes a coordinate transformation from one inertial frame to another then the new field strengths will become velocity dependant.
Here is a few examples of gravitational fields for which the gravitational field in the "rest frame" is given as well as in a frame moving relative ti the object
object whose gravitational field is velocity dependant.
http://home.comcast.net/~peter.m.brown/gr/grav_moving_rod.htm
http://home.comcast.net/~peter.m.brown/gr/grav_moving_sheet.htm
http://home.comcast.net/~peter.m.brown/gr/grav_moving_sheet.htm
Post by: Pmb on 22/12/2012 00:59:41
Physicists like to find the smallest common nominator for things, and when we (they) talk about mass then that should be 'mass-energy'.This phrase makes no sense to me at all! In what sense are you saying that “Physicists like to find the smallest common nominator for things” and what does that have to do with mass-energy? By the way, when one is speaking of mass-energy one is speaking about the kinetic energy of a body that is moving. E.g. If T is the stress-energy-momentum tensor of, say, a gas then T^00 (the energy density) is also referred to as the mass-energy density. T^00 includes kinetic energy and not just rest energy.
Post by: Pmb on 22/12/2012 01:18:39
Quote from: jopie64
But photons individually have no mass? How did it gain mass?If you keep thinking like this you’re going to keep confusing yourself. I fail to understand why you folks like to make life hard for yourself with this bizarre definition of mass?
Two ways of looking at this; 1) photons have mass m according to E = mc^2. This is the relativistic mass of the photon. The mass of a system of photons is then the sum of the relativistic masses of all the photons.
(2) mass = invariant mass of photons – I.e. for a box of photons the box itself allows there to be a frame of reference in which all the photons are confined. The total energy in the box is the sum of all the energies of all the photons in the box. In the rest frame of the box the total momentum of all the photons is zero. That means that the energy in the box is the rest energy. The mass of the photons in the box is therefore defined through E = mc^2
Post by: Pmb on 22/12/2012 01:20:47
What type of mass does the "m" represent in the good old E=mc2?Invariant mass (the one sometimes also called "rest" mass or "proper" mass).
The term “invariant mass” typically refers to a system of particles and not to a single particle. In anycase the concept of invariant mass can only be used in flat spacetime
Post by: Pmb on 22/12/2012 01:23:46
Quote from: Phractality
I don't accept the claim that a photon has no mass.Smart man! :)
If that is your opinion then you might enjoy reading my article on the subject. It’s at
http://arxiv.org/abs/0709.0687
Post by: Pmb on 22/12/2012 01:27:23
Quote from: lightarrow
You can't localize a photon, so you can't do that.Everything I’ve seen in this thread speaks mostly about classical physics, e.g. relativity. In relativity one uses classical photons, which is basically a point particle having a classical trajectory but zero proper mass. Such a thing can be localized.
Post by: yor_on on 22/12/2012 06:17:40
Are you telling me that gravity is observer dependent?
Post by: yor_on on 22/12/2012 06:30:05
Post by: yor_on on 22/12/2012 07:08:30
Post by: yor_on on 22/12/2012 22:46:41
If now that was what you meant?
Post by: Pmb on 23/12/2012 00:09:01
Maybe I am unclear?Depending on what youi mean by "gravity" yes.
Are you telling me that gravity is observer dependent?
Post by: lightarrow on 23/12/2012 15:59:12
Classical photons? Which movie is it? [:D]Quote from: lightarrowYou can't localize a photon, so you can't do that.Everything I’ve seen in this thread speaks mostly about classical physics, e.g. relativity. In relativity one uses classical photons, which is basically a point particle having a classical trajectory but zero proper mass. Such a thing can be localized.
Post by: Pmb on 27/12/2012 06:25:21
Quote from: lightarrow link
Classical photons? Which movie is it? [:D]I don't understand what you mean, "Which movie is it?"
lightarrow - Have you ever heard of the terms "classical photon" and "classical electron"? Perhaps some use the term “scalar photon” or “scalar electron” instead. I read an article where similar such term(s) were used, rather than forcing someone to explain that what it means. While you may be using it as some sort as slang, I have no idea what it means which means that other people don’t either. It you mean photon then please say photon and he same with electrons. We then won’t have to waste space by trying to explain terms or explain what was a joke and then razz the person who didn’t get the joke.
Post by: lightarrow on 27/12/2012 18:32:10
"classical electron": yesQuote from: lightarrow linkClassical photons? Which movie is it? [:D]I don't understand what you mean, "Which movie is it?"
lightarrow - Have you ever heard of the terms "classical photon" and "classical electron"?
"classical photon": no. The reason is because of qm history: a classical electron was a starting point for Bohr and Sommerfeld when they described the atom. But a classical photon couldn't have any meaning, because m = 0 in this case.
But that was history. Now we know that a precise trajectory of particles is impossible, they don't have at all. Wavefunctions are wat replaced them.
Phractality wrote about defining the centre of a system of two photons: you don't even know where is a photon, and you want to find such a thing?
Post by: JP on 28/12/2012 01:18:49
Post by: lightarrow on 28/12/2012 21:13:46
Post by: JP on 29/12/2012 00:37:48
Post by: Pmb on 29/12/2012 00:58:03
A classical photon is defined as a point particle whose energy i related to its momentum by E = pc, whose speed is c, and whose momentum is p = Mv = mc where M = E/c^2. M is defined as the photon's inertial mass."classical electron": yesQuote from: lightarrow linkClassical photons? Which movie is it? [:D]I don't understand what you mean, "Which movie is it?"
lightarrow - Have you ever heard of the terms "classical photon" and "classical electron"?
"classical photon": no. The reason is because of qm history: a classical electron was a starting point for Bohr and Sommerfeld when they described the atom. But a classical photon couldn't have any meaning, because m = 0 in this case.
Worrying about m = 0 is confusing inertial mass (a pre-relativistic notion) with proper mass (a relativistic notion)
Post by: Pmb on 29/12/2012 01:00:36
I believe a "classical photon" would be a ray. You get ray optics from light waves usingA ray would be approximated as a stream of classical photons. Picture a laser beam as an approximation
the same procedure that you can use to get particle-like electrons from a more thorough
quantum wave theory. But it sounds like Pmb's classical photons are like little bullets, not rays.
I'm not sure how to get to those from the wave theory.
of a ray.
Post by: JP on 29/12/2012 14:47:50
Post by: lightarrow on 29/12/2012 18:29:47
A ray would be approximated as a stream of classical photons. Picture a laser beam as an approximation of a ray.It's the same mistake that one makes stating that an electron's track in a bubble chamber means that elementary particles have a precise trajectory. QM teaches us they actually don't have.
Post by: Pmb on 30/12/2012 01:11:14
But a beam isn't a point particle.Since nobody suggested or inferred otherwise this comment confuses me. Can you elaborate please?
I understand that you can express a beam roughly as a density of classical point particle photons moving along raytrajectories at the speed of light, but the case of two classical photons ...Where did you get the idea of using only two photons from?
Post by: Pmb on 30/12/2012 01:20:25
Hence the term "approximation". In classical mechanics we're most often concerned with physics only down to, perhaps, the micron level. At that level and over small distances such as a mile we don't much need to worry about the non-exact nature of the trajectory of the photons.A ray would be approximated as a stream of classical photons. Picture a laser beam as an approximation of a ray.It's the same mistake that one makes stating that an electron's track in a bubble chamber means that elementary particles have a precise trajectory. QM teaches us they actually don't have.
I've been thinkig of this stream of photons as a laser beam and when I studied ray optics as an undergrad I always had a laser beam in mind.
I never worried about the cross sectional area of the laser beam either.
This is getting off course of the reason I mentioned classical photons.
As I said, in relativity one uses classical photons, which is basically a point particle having a classical trajectory but zero proper mass. Such a thing can be localized.
I had in mind this notion when I was reading Exploring Black Holes and working with GR and SR and when they used photons. Never in that work does anybody ever worry about te photon being localized or it not having a classical trajectory.
Remember this is an analogy which, by definition, means that its alike in some ways and not alike in other ways.
Let is not forget why it was brought up rather than dwell on what each of us knows all to well about how real photons behave,shall we? Otherwise it gets over pedantic and that's when I go back to watch my brand new 60" plasma TV :)
Post by: Pmb on 30/12/2012 01:32:10
Heh :)Ain't gonna happen. I had several months of too much quite to come out and worry about such trivia.
I can see you and Pete gearing up to a argument Lightarrow :)
There are many terms in physics for which you must know the context and perhaps the authors views in order to determine the precise definition of a term. E.g when some people use the term "weight" they refer to the quantity mg. Others refer to weight only when the object of mass m is being supported at rest in the field while only defining weight as mg could mean that a body in free-fall has weight.
Then there is the definition of momentum. In Newtonian physics it means p = mv. In analytical mechanics and quantum mechanics it means canonical momentum. And you have to know that when reading QM material - the authors won't just tell you.
So nope. Nore more debating things anymore for me. I have no desire to let people know the complete picture of things. If they think they're gonna get a complete picture on the internet then they deserve whatever it is they get.
Post by: Pmb on 30/12/2012 01:42:53
[/quote]
If you're speaking of the geometric center of the two photons then no. The center of enegy is not the same as the geometric center of the two photons.
Note - See http://home.comcast.net/~peter.m.brown/sr/center_of_mass.htm
I see that this is where all this photon stuff started. This was a discussion of a classical photon and not a quantum mechanical one. Don't sweat the small stuff. Look how far off track it took this thread even though everyone knew what was meant by a classical and quantum photons.
In relativity we speak of point particles such as electrons, protons etc. even when we're speaking about classical relativity (since there is no quantum theory of relativity yet). When talking classical relativity and one is speaking of photons then what is being discussed is trivially simple: A classical particle having a well-defined trajectory, having its energy related to its mometum by E = pc and whose mass is given by p = mv = mc or E = mc^2.
This is how we speak of all particles in relativity. If you see a relativity text or article speaking of electrons, protons etc then they are treating them as classical particles. There is no difference with photons.
Post by: lightarrow on 30/12/2012 12:52:35
No. When interactions among elementary particles are involved, you have to use the full wavefunctions treatize of QM, so: diffraction, interference, non existence of an exact trajectory. Ask particle physicists. Classical limits are used in macroscopic system *of non zero mass*, so singular photons are excluded by definitions. About lasers, you are using the geometrical optics approximation, so you are not talking of singular photons by definition.It's the same mistake that one makes stating that an electron's track in a bubble chamber means that elementary particles have a precise trajectory. QM teaches us they actually don't have.Hence the term "approximation". In classical mechanics we're most often concerned with physics only down to, perhaps, the micron level.
--
lightarrow
Post by: Pmb on 30/12/2012 15:50:20
Quote from: lightarrow
No. When interactions among elementary particles are involved, ...Then you have to use QM. You did understand, didn't you, that I was speaking about classical relativity? I wasn't speaking of quantum relativity since it doesn't exist yet. If you have a problem which requires QM then you have to use QM. The context of my post tells you that I was referring to classical mechanics problems, not QM problems.
In relativity one speaks of light cones. In a 2-D spacetime diagram a photon moves on a straight worldline, i.e. a straigh in spacetime. We don't speak about quantum trajectories in classical relativity. When QM is required then we've gone outside the realm of classical relativity. In classical relativity one assumes that whatever we are speaking above can be approximated to move on worldlines, and that includes elementary particles too.
Clearly, when one is speaking of a particle moving on a null geodesics one is thinking about classical luxons. And that's what I've been explaining here.
Post by: lightarrow on 31/12/2012 10:30:58
So you can't speak of photons...Quote from: lightarrowNo. When interactions among elementary particles are involved, ...Then you have to use QM. You did understand, didn't you, that I was speaking about classical relativity?
Quote
I wasn't speaking of quantum relativity since it doesn't exist yet.You are joking! It was created in the 30' of the previous century, one of the first was Dirac, have you haver heard of "Dirac equation"? Have you ever heard of QED quantum electrodynamics?
Maybe you intended "quantum gravity", that is, a theory who would unify quantum mechanics and *general* relativity. But we are talking of laser beams, so we don't need gravity here.
Quote
Clearly, when one is speaking of a particle moving on a null geodesics one is thinking about classical luxons. And that's what I've been explaining here.Don't know what they are, but certainly they are not photons.
Post by: Pmb on 31/12/2012 13:27:53
Quote from: lightarrow
So you can't speak of photons...While I disagree with you last post I've decided to end my contribution of quantum/photon thing since I've basically lost interest (of course I'm always available inPM for anything).
Plus I don't like endless debates on subjects, especially when most of it contains rephrasing of statements already made.
I'll end this with one reference: Inertia of energy and the liberated photon by Adel F. Antipa, Am. J. Phys., Vol (44) No.(9) Sep. 1976
Quote
AbstractSee section 7. The Radiating Atom
We follow through the different variants of Einstein's intuitive photon-in-a-box derivation of the inertia of the inertia of energy, then end with a very simple "radiating atom" derivation
This article is basically a modern derivation of Einstein's famous "Photon in a Box" experiment. See my web site at
http://home.comcast.net/~peter.m.brown/sr/einsteins_box.htm
In the first section I outline Einstein's derivation. In the later section I outline Atipa's derivation.
Einstein uses a pulse of radiation to prove the mass-energy relation while Antipa uses an atom and a photon.
In both cases the center-of-mass is calculated and used.
The more general definition of center of mass of radiation is given in my web page at
http://home.comcast.net/~peter.m.brown/sr/conservation_laws.htm
The center of mass is in Eq. (15).
Post by: Pmb on 01/01/2013 17:35:36
(1) If the speed of the particle is always v < c then the particle is classified as a Tardyon.
(2) If the speed of the particle is always v > c then the particle is classified as a Tachyon.
(3) If the speed of the particle is always v = c then the particle is classified as a Luxon.
There shouldn't be any problems now since these are well defined terms and used in the classical relativity literature.
Post by: JP on 01/01/2013 22:22:58
Post by: Pmb on 01/01/2013 22:41:33
Would you folks be interested if I started and/or split off a thread with a title along the lines of "Is there such a thing as a "classical" photon?" or are you OK with this thread being used for discussion?Makes no difference to me, JP. I think we’ve all said what we wanted to say and learned what we wanted to learn at this point. But if you want to start a new thread on it then please feel free to do so.
Post by: lightarrow on 02/01/2013 17:19:25
Post by: JP on 02/01/2013 22:23:43
Post by: Bill S on 03/01/2013 17:09:43
Post by: Pmb on 04/01/2013 11:12:31
JP, as a "hitch-hiker" where science is concerned, I got a bit bogged down in this thread. In general, I find your explanations reasonably easy to follow, as long as you don't wax too technical, so if you are going to say a bit more about the idea of "classical photons" I'll follow with interest.Regarding the term I coined, i.e. “classical photon.” There is nothing special or new about this concept. In the physics literature you’ll find two kinds of “particles” spoken of. One type of particle has its position and momentum simultaneously determined and moves on a classical trajectory. Such particles are called “classical particles.” The other kind of particle has its position and momentum constrained by the Heisenberg uncertainty relation. Such particles are called "quantum particles."
A classical particle whose energy E and momentum p is related by E = pc and whose nature is electromagnetic in origin (i.e. can scatter off a classical charged particle such as a classical electron) is what I had in mind when I referred to a “classical photon”.
Pete
Post by: JP on 04/01/2013 19:31:36
I think this debate has been over two things: whether you can accurately model light with classical particles traveling at the speed of light and whether, if you can, you should call them "classical photons."
To answer the first question, you can in some cases. Whether you use Maxwell's equations (which are accurate, but don't include quantum effects) or quantum optics (in which light is treated more rigorously as individual quantum mechanical photons), you're dealing with waves. Either the light itself is a continuous wave (Maxwell) or the photons are waves (quantum optics). In certain cases, you can approximate waves over space and time by classical trajectories through space and time. This is an approximation whose validity depends on how rapidly the waves are oscillating over space and time (faster oscillation makes this approximation more accurate).
In the case of Maxwell's equations, these trajectories are rays which carry energy at some rate. You could divide this energy up into packets, each moving at the speed of light, and you'd have something like a "classical photon." Alternatively, if you use the Feynman path integral formulation of quantum mechanics, you can compute the effect of the emission and absorption of a photon by accounting for all possible paths the photon can take between emission and absorption. The analogous trick to finding rays should be to find the most probable path for the photon to take. I believe you could call this the "classical photon" trajectory. Its important to remember that these are approximations to more rigorous models, but they should be accurate enough in many cases.
As for whether "classical photon" is a good term for these approximations--It probably isn't, as is evident from the confusion here. However, it is clear that what Pmb means is a model for light as particles moving along a definite trajectory at the speed of light, and as I noted above, that model is an approximation, but can be accurate in some cases.
Post by: Pmb on 04/01/2013 21:01:30
Post by: JP on 05/01/2013 15:51:21
I'll spare you the details, but the Wigner distribution function is one way of thinking of a wave as particles moving along classical trajectories. If you read through that, you'll find that the particles in non-classical states (those where wave effects are important) have some odd properties such as appearing to carry negative energy or probability, which ends up being required for particles to model waves: http://en.wikipedia.org/wiki/Wigner_quasiprobability_distribution
Post by: Pmb on 12/01/2013 11:52:39
Quote from: fertilizerspike
It does not happen.In this forum we've beat that horse dead many times over. In the end whether "mass" = m, depends on speed ultimately depends on how one defines the term. And there are two well-known, but different definitions used in relativity today. One were m is a scalar. It's then given the full name with the qualifier "proper" to end up with "proper mass". When one starts with the laws of physics where in the 3+1 viewpoint F = dp/dt and m is defined as the m in p = mv where p = 3-momentum (linear mechanical) and t = coordinate time. It's then given the full name with the qualifier "inertial" to end up with "inertial mass"
So mass can and does mean either (1) inertial mass or (2) proper mass. While you seem to have used it as a conceptual crutch I know that I haven't. That's not why I prefer it when appropriate.
Post by: lightarrow on 12/01/2013 12:10:40
JP - When the wave nature of light is important I can't see how photons can be used. In fact that's what led to the wave-particle duality.I hope not to offend you, but this sentence means you haven't understood a lot of QM, or about photons, at least.
1. A photon, as well as any other elementary particle, it's not a corpuscle; when in QM we talk of a "particle" we intend an entity which is described with a wave equation and which is detected as a corpuscle, so it has *both* a corpuscolar and a wave behaviour *by definition* (I'm referring to the QM postulates).
2. You want to see light waves and light corpuscles in a single experiment? Send individual photons trough a doble slit and detect them in a screen. You will see individual flashes on the screen (light corpuscles) distributed according to a diffraction pattern (light waves).
3. In particle/particle collision experiments in a collider, you have individual particles colliding and the result of that (particles, energies, ecc.) is something that you can compute *only* using the wave nature of them (= wavefunctions = quantum mechanics).
Conclusion: if you use the word "photon" you are talking of quantum mechanical description by definition.
Post by: Pmb on 14/01/2013 02:55:46
Quote from: lightarrow
Conclusion: if you use the word "photon" you are talking of quantum mechanical description by definition.I disagree.
First off JP knows his QM quite well so one would be quite off the mark and misleading themselves to think that he doesn’t understand QM a great deal.
The term photon is often used in areas of physics where it’s treated like a classical particle. The same holds for any other particle that we know since what you can say about a photon regarding its properties as a particle you can say about any elementary particles in nature. That’s why we quit often use the term photon in relativity,
What you spoke about above also holds true for electrons and when you use the term “electron” it doesn’t mean that you’re talking about a quantum mechanical description. We use the term “electron” in classical electrodynamics which by definition is a classical, not a quantum, branch of physics. .
Post by: lightarrow on 14/01/2013 13:17:11
What you spoke about above also holds true for electrons and when you use the term “electron” it doesn’t mean that you’re talking about a quantum mechanical description. We use the term “electron” in classical electrodynamics which by definition is a classical, not a quantum, branch of physics.But in those cases you neglet the wave aspect of the particle. Instead you intended to discuss exactly the case of the wave aspect of particles, infact you wrote:
"When the wave nature of light is important I can't see how photons can be used".
Post by: Pmb on 14/01/2013 18:01:30
Quote from: lightarrow
But in those cases you neglet the wave aspect of the particle.Of course. That’s entirely the point. Maybe that’s what has you confused. You must have thought that somewhere someone was making the claim that under all conceivable scenarios the photon can always be treated as a classical particle. Not true. If you thought that someone made such an assertion then you were mistaken.
When one is considering only those experiments in which the limits of the experiments/observations are such that one can describe the motion of a photon using a classical trajectory and thus a worldline in spacetime (null worldline to be exact).
Not all experiments and observations pertain to the wave aspects of photons. Those experiments and/or observations which treat the photon as a particle and for which the limits of accuracy of the experiment are such that one can ignore the limitations imposed by the uncertainty principle can one treat the photon as a classical particle.
When one is, say, analyzing photons in Young's double slit experiment then one is going outside the scope of classical relativity since a classical trajectory cannot be used to describe the motion of the photon.
However in other circumstances, say, when one is tracing a photon as a particle bouncing off mirrors which are moving in a vacuum or are being emitted and detected by photon emitters and detectors then one can use classical relativity and classical trajectories. If you want a good example of a photon being treated as a classical particle in special relativity then crack open “Gravitation” by Misner, Thorne and Wheeler. I know of a particular example of a photon being treated as a classical particle in that text. If you have the text and want to read it then I’ll try to find if for you.
Another treatment of a photon as a classical particle (in fact in a derivation where a photon is used in a center of mass calculation) see the article Inertia of energy and the liberated photon, Adel F. Antippa, Am. J. Phys. 44(9), September 1976. I have this article and if you’d like to read it I can make it available on my website for you to download and read.
The new version “Exploring Black Holes – Second Ed.” has examples where photons are treated as classical particles. This version of the book should be out later this year.
Post by: lightarrow on 14/01/2013 19:34:05
I have never thought it. The point is that the photon can never be treated as a classicle particle [:)]Quote from: lightarrowBut in those cases you neglet the wave aspect of the particle.Of course. That’s entirely the point. Maybe that’s what has you confused. You must have thought that somewhere someone was making the claim that under all conceivable scenarios the photon can always be treated as a classical particle.
When at a certan point you talked of "luxons" I believed to have understood that you intended this, with "classical photon", and I accepted to end the discussion. But I see you insist [:)]
Quote
Not all experiments and observations pertain to the wave aspects of photons.If you want to talk about photons in that sense, you can do it only if you talk of the interaction energy and not when you talk of "something in fly between source and detector" which was what I was discussing with (I don't remember who) because he was talking of "the centre of a two photons system".
You can talk of the centre of two heavy atoms, if they are not cooled down to low temperatures, because they are still almost classical particles, maybe in some cases you could even talk of classical electrons, but absolutely not about photons; if there is an example you can't talk of classical particles is just that...
Quote
When one is, say, analyzing photons in Young's double slit experiment then one is going outside the scope of classical relativity since a classical trajectory cannot be used to describe the motion of the photon.Yes. But then you talk of *light rays and geometrical optics or classical electromagnetism*, *not* photons, until they are detected.
However in other circumstances, say, when one is tracing a photon as a particle bouncing off mirrors which are moving in a vacuum or are being emitted and detected by photon emitters and detectors then one can use classical relativity and classical trajectories.
Post by: Pmb on 14/01/2013 19:58:05
Quote from: ]
[
[
I have never thought it. The point is that the photon can never be treated as a classicle particle [:)]Your point is wrong. And its often treated as such.
Please post a proof demonstrating that a photon can never be treated as a classical particle.
While I'm waiting for that I'll show you examples of how physicists actually do treat photons as classical particles.
Examples are under my website at
See - http://home.comcast.net/~peter.m.brown/sr/light_clock.htm
and - http://home.comcast.net/~peter.m.brown/sr/einsteins_box.htm
and - http://home.comcast.net/~peter.m.brown/sr/lorentz_contraction.htm (substitute "photon" where you see "light"
and - http://home.comcast.net/~peter.m.brown/sr/lorentz_contraction_2.htm
and - http://home.comcast.net/~peter.m.brown/sr/spacetime.htm
and - http://home.comcast.net/~peter.m.brown/sr/relativistic_optics.htm (substitute "photon" where you see "beam")
Here's an example of how one treats an electron as a classical particle
http://home.comcast.net/~peter.m.brown/sr/cyclotron.htm
In real life, strictly speaking that is, electrons are quantum particles just like in real life, strictly speaking that is, photons are quantum particles.
Those are examples of how this is actually done by physicists (including myself), contrary to your claim that it can't be done.
You have yet to post a valid reason why one can't treat photons as classical particles in certain cases. You've made assertions to that effect but statements of facts are not considered to be a proof of that fact.
When I write the expression for a null worldline, i.e. x(t) = cos(theta) ct, y(t) = sin(theta) ct. These are the parametric equations of a photon moving in the xy-plane at an angle of theta from the x-axis. That's a null worldline and this is a classical statement.
Anytime a relativist is speaking of the null geodesic that a photon is moving on then he's speaking of a classical photon.
And that, my dear lightarrow, is how its done. :)
Post by: Pmb on 15/01/2013 07:11:42
Quote from: Phractality
I'm wondering, now, if I should consider the center of the two-photons to be the center of energy; equivalent to center of mass.Yes. In fact please see http://en.wikipedia.org/wiki/Two-photon_physics
Post by: lightarrow on 16/01/2013 17:15:53
It's very simple, and I let you do it: define "photon".Quote from: ]
[I have never thought it. The point is that the photon can never be treated as a classicle particle [:)]Your point is wrong. And its often treated as such.
Please post a proof demonstrating that a photon can never be treated as a classical particle.
Post by: Pmb on 16/01/2013 19:09:39
Quote from: lightarrow
It's very simple, and I let you do it: define "photon".I gotta tell ya, lightarrow, I'm quite baffled as to why you’ve been ignoring the examples I've posted for you. I'm not posting them for my health ya know. :) They are meant as counter arguments to the assertions that you've been sticking to. If you ignore them then I see no reason to discuss the physics anymore.
You also seem to be ignoring the fact when I explained that when photons are used in classical derivations its always an approximation to real life, i.e. quantum effects or precision is either ignored or not important.. Such approximation are not unique to photons or the examples I've posted but apply to ever classical treatment in physics that there exists since all real life particles have real life uncertainties in their locations. Why you've singled out photons among other particles such as electrons and protons and why you've been ignoring the usage of classical trajectories such as those in particle physics like tacks in bubble chambers of elementary particles is something I don’t understand.
Anyway … asking for the definition of the term "photon' is not a proof that they can’t be used in the manner which you say they can’t. When photons were first discovered by Einstein and Planck there was no uncertainty principle in existence and there was no need for such even though the particle nature of photons has been used long before the term "photon" was invented. Einstein and Planck came up with the concept of "bundle of energy" long after the first thoughts of "particles of light" were first being thought of and before they knew of the exact nature of them.
In any case the definition of Photon is - a discrete bundle (aka "quantum") of electromagnetic radiation where its kinetic energy E is related to its momentum p by E = pc. Einstein called these discrete bundles "light quanta". When the term “photon” was actually first defined by Gilbert Lewis they didn't have the same properties that we now assign to them.
Here the term "bundle" does not mean that they have a finite extension in space that one might visualize when one hears the term "bundle." The term “bundle” means "finite amount," nothing more and nothing less.
Also I’ve used the term “classical photon” or referred to the use of photons in a classical setting as approximations to the more exact expressions given in quantum mechanics. In your responses its like you’re not reading the parts of my posts which clarifies this point. When the wave aspect of the photon is under consideration or more precise measurements are being considered then the classical approximation of the photon is no longer being used.
I've already given you many examples of how and where photons are used in derivations using the conservation of center-of-mass. In particular, they are used in the article
Inertia of energy and the liberated photon by Adel F. Antippa, Am. J. Phys. 44(9), September 1976
A similar derivation is given in Special Relativity by A.P. French on pages 17 and 27.
In each version of the derivation the correct results are obtained.
I'd like to remind you that it was you who first claimed that the center of energy of a system of photons couldn’t be defined because you claimed that photons couldn’t be localized. In that response you had your mind fixed on quantum mechanics and were ignoring the sense in which that person was using the concept of photons. To be 100% accurate, nothing in the universe can be localized to a mathematical point. What you failed to mention was the fact that any particle can be localized to a finite region of space. If that weren’t true then nobody would have ever referred to anything in physics, quantum or classical, as a “particle.” In fact even in quantum mechanics photons can be localized to an arbitrarily small region of space. One just has to give up some uncertainty in the knowledge of the photons momentum. If we were to take your response literally then you would have said that the center of mass of electrons, protons, neutrons etc. couldn’t be defined because no elementary particle can be localized in space. I’m rather baffled at where you got the notion that photons can’t be localized. No treatment of quantum mechanics would ever make such an assertion. Perhaps you made the mistake of thinking that “localized” meant “located at a mathematical point.” If so then no elementary particle can be localized. But that’s not what is meant by “localized” in quantum mechanics.
I think this point needs to be clarified more (Let’s use the double slit experiment and for purposes of illustration we’ll only look at the x-component). When one is considering experiments like Young’s double slit experiment then what is observed under low enough intensity of the light source are photons hitting the screen at finite locations. That means that if the screen were made of a huge amount of very small pixels (let’s say 0.001 mm) then only one pixel at a time is illuminated at one time. The location of that pixel is then recorded. After a large number of photons hit the array of pixel what you’ll have a bunch of numbers that denote locations where the photons were localized to a region 0.001 mm long. If you were to plot the number of photons at a particular x-location verses the x-coordinate what you’ll have is a graph which has the general shape of a bell curve. Let the width of that curve be dx. This is the dx that the uncertainty principle uses, not the size of a pixel. That is precisely what it means to localize a photon and precisely why photons are said to behave like particles in this experiment.
In case you ignore my previous posts and explanations (since you failed to explain why you’re claiming that all these physicists who use these explanations got them all wrong) I’ll have to repeat myself. I showed you how physicists are and have used photons in a classical context. You didn’t acknowledge the derivations that I’ve shown you. They are at - http://home.comcast.net/~peter.m.brown/sr/einsteins_box.htm
Go down to below Eq. (8) and you'll see how they are used in the classical sense, i.e. by using them in the center of mass theorem. This is how Antippa and French used them.
Other uses are when physicists draw spacetime diagrams and show the worldliness of photons. Such worldliness are classical approximations to what photons actually do on a small scale. When such diagrams are drawn the worldliness used as example are often of the order of a meters. The uncertainty of the photon’s location along such paths is negligible when drawing such diagrams. Thus the width of the worldline is approximated to be zero rather than the real life finite width one would get in experimental data. When one draws a classical trajectory like this for a particle then, in all cases, one is using an approximation. Just like when the path of an electron is drawn as a circle in a cyclotron. That path is also an approximation to real life.
From now on I’ll have to wait until you give a proof and a valid reason why the above approximations cannot be used. Until then I won’t respond to any other post of yours in this thread since all you’ve been doing is making a claim that they can’t be used with no acknowledgement that (1) I’ve been referring to approximations and (2) that they are indeed used by physicist and appropriated so and without error in their derivations.
One last comment before I end this post: To drive this point home let me remind you that what I’ve been, i.e. what I’ve labeled a “classical photon” which you seem to keep mistaking for a “quantum photon” which has a wavelength L related to its momentum p by p = h/L and whose position and momentum satisfy the uncertainty relationship. Now recall what a “classical electron” is. It’s a particle with charge e = 1.60x10^(-19) C, a proper mass of 9.11x10^(-31)kg which moves on a well-defined trajectory. This is different than a “quantum electron” which has a wavelength L related to its momentum p by p = h/L and whose position and momentum satisfy the uncertainty relationship.
With this in mind recall the definition of "classical photon" a quantum of radiation whose kinetic energy is related to its momentum by E = pc which moves on a well-defined trajectory. From this definition it follows that the proper mass of a photon is zero.
Do you understand what I meant by "classical photon" now?
Post by: Pmb on 17/01/2013 00:06:40
... "Is there such a thing as a "classical" photon?" …The answer is “No.” Does that mean that the concept is not useful or that it’s not being used in the physics/relativity literature? The answer to that is “No” as well. Then again there is no such thing as a “classical electron” since no particle exists that has the same properties of a “quantum electron” and also does not move on a well-defined trajectory. Also no “classical electron” satisfies the uncertainty principle. That also doesn’t mean that the concept is not useful or that it’s not being used in the physics/relativity literature.
A classical photon is an approximation to a real photon just as a classical electron is an approximation to a real electron.
Compare this with the Earth’s gravitational field. Does a particle of mass m moving in the Earth’s (Earth mass = M) gravitational field a distance r from its center have a force on it given by F = GMm/r^2. The answer is no. This is true because the Earth is not a perfect sphere and F = GMm/r^2 only holds for a particle moving in the gravitational field of a perfectly spherical body. To say that this is not a useful approximation would be quite wrong.
Post by: lightarrow on 18/01/2013 11:31:42
But there is a difference because you can treat electrons as classical particles in some cases, for example in Maxwell's electromagnetic description you can consider electrons as point charges. Furthermore, considering the electron as a classical particle, you can describe the most important features of the hydrogen atom; in which case you can do something similar with the photon?... "Is there such a thing as a "classical" photon?" …The answer is “No.” Does that mean that the concept is not useful or that it’s not being used in the physics/relativity literature? The answer to that is “No” as well. Then again there is no such thing as a “classical electron” since no particle exists that has the same properties of a “quantum electron” and also does not move on a well-defined trajectory. Also no “classical electron” satisfies the uncertainty principle. That also doesn’t mean that the concept is not useful or that it’s not being used in the physics/relativity literature.
http://en.wikipedia.org/wiki/Photon
<<A photon is an elementary particle, the quantum of light and all other forms of electromagnetic radiation,>>
just the term "quantum" removes every "classical" description *by definition*. A photon is the quantum of the electromagnetic field, that is a quantum object that comes from the QED "quantum electrodynamics", it doesn't have a classical origin as the electron, it borns quantum....
Post by: Pmb on 18/01/2013 14:37:13
Quote from: lightarrow
But there is a difference because you can treat electrons as classical particles in some cases, ...Just as you can treat a photon as a classical particle in some cases.
Quote from: lightarrow
Furthermore, considering the electron as a classical particle, you can describe the most important features of the hydrogen atom; in which case you can do something similar with the photon?Already done in the examples which you continue to ignore.
Quote from: lightarrow
You’re mistaken. All the term “quantum” means is discreteness. Period. That’s all. That photons first arose in the photoelectric effect in no can be interpreted that they can only be used in a quantum context. Electrons arose in a classical context but it was later shown that they can’t be used in all classical situations. And it’s quite clear that an electron can be referred to as a quantum of charge.
http://en.wikipedia.org/wiki/Photon
<<A photon is an elementary particle, the quantum of light and all other forms of electromagnetic radiation,>>
just the term "quantum" removes every "classical" description *by definition*. A photon is the quantum of the electromagnetic field, that is a quantum object that comes from the QED "quantum electrodynamics", it doesn't have a classical origin as the electron, it borns quantum....
So I see that you once more have nothing to add other than making unfounded assertions and you keep ignoring the counter examples given to you which prove you wrong. You certainly can’t back up your position merely by repeating yourself and making statements that you can’t back up.
Post by: JP on 18/01/2013 16:57:37
Post by: Pmb on 18/01/2013 22:09:30
If you look at the Standard Model of particle physics, photons are a quantum of the electromagnetic field, and electrons are a quantum of the electron field: this symmetry between matter and fields is a major part of the Standard Model. We know that we can take a classical limit of the quantum description for electrons and arrive at classical electrons. I've never seen it done explicitly, but I suspect you could do the same for photons and arrive at a classical limit for photons--i.e. the classical photons Pmb brings up. 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.The reason I know why we can is because I see it done all the time with success. Especially when it comes to calculating the deflection of light past the sun and deriving E = mc^2 using the conservation of the center of mass of a atom-photon system. In the case of the former one generates a null worldline a particle of zero proper mass that is passing by the sun and calculates its trajectory and the correct value of deflection is obtained. In the later case one assigns a position vector to both the atom and the photon and the results yield E = mc^2. I’ve never done the former on my web site (perhaps I did it on paper) but the later I did in two ways on my website which is at http://home.comcast.net/~peter.m.brown/sr/einsteins_box.htm
Scroll to just past Eq. (8) and follow the derivation. You’ll notice that the properties of the photon are used and the result is valid
Post by: lightarrow on 19/01/2013 02:03:54
Post by: Pmb on 19/01/2013 03:28:53
Why do you need a photon for this? A light pulse wouldn't work?Who said a photon was needed? Einstein used a "burst of radiation". In any case it's sufficient, to use a photon, not neccesary. If you read the derivation I pointed to by Antippa below Eq. (8) they you'd see that it uses an atom emitting a photon. You can't have an atom emitting light, pe se
Post by: lightarrow on 19/01/2013 14:13:05
Post by: Pmb on 20/01/2013 14:05:14
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.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.
I feel like I’ll just be repeating myself here so I’m ending my contribution in this thread and am as such agreeing to disagree. That means I won’t respond to any further assertions you make even if I know them to be wrong.
I rest easy in the knowledge that has been done successfully on numerous occasions by many physicists in many textbooks and at least some journal articles. Most notably its now being used in the new version of Exploring Black Holes. I’d like to note that one of the authors of that text, Edwin Taylor, is also co-author of the MIT Introductory Physics Series Quantum Mechanics text.
Post by: lightarrow on 21/01/2013 12:43:57
When I answered to the question of what is the "centre" of a system of two photons (question intended in the sense that the photons were emitted but not yet absorbed) I said that it's not possible because you can't say where a photon is exactly in flight, that is after emission and before detection.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.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.
The proof is very simple, it only needs the Young experiment with two slits and photons emitted one at a time: if you can say which slit the photon passed through, the interference pattern disappears...
Post by: JP on 21/01/2013 16:43:56
Post by: AndroidNeox on 21/01/2013 17:44:02
Post by: lightarrow on 21/01/2013 19:09:31
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).When you take the classical limit for a photon (h --> 0) it gets zero energy, so it disappears.
Post by: lightarrow on 21/01/2013 19:12:37
All 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, unless the object rotates around a fixed point.
Quote
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.Correct.
Post by: JP on 21/01/2013 20:32:40
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).When you take the classical limit for a photon (h --> 0) it gets zero energy, so it disappears.
Good point. After reading that and a bit more thinking, I believe that what's going on with these "classical photons" is two limits. If we have a field made of photons, the classical limit does not correspond to taking h->0, but rather to taking many photons. In this limit, you recover Maxwell's equations, but you've lost information about the behavior of individual photons.
Then you take a second limit corresponding to wavelength->0 which gets you ray optics. So essentially, your "classical photon" is an arbitrary packet of energy assigned to propagate along a ray. It's related to real photons only insofar as the sum over many photons gets you the classical field, which you then use to define rays.
Post by: Pmb on 21/01/2013 22:57:47
Quote from: lightarrow
When 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 + E0 = 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.
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.
Post by: Pmb on 21/01/2013 23:02:06
Quote from: lightarrow
If you mean relativistic mass, ok.You know that its relativistic mass from the context in which he used it. Its not wise to assume that everyone who uses the description that he does needs to be reminded of these facts. To assume so is condescending to the poster because we're assuming that he doesn't know what he's talking about and we need to correct them.
I recommend that if someone says "mass depends on speed/energy etc." then we simply assume that by "mass" they mean relativistic mass. Having to remind people all the time is a waste of everybody's time. People can get irritated when someone has to comment on it every time they use the concept, don't you think
Post by: JP on 22/01/2013 03:57:26
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. It has a well-defined frequency, momentum and polarization, but no simple position representation (much as a plane wave has a well-defined frequency, direction and polarization, but exists over all space). Just like a pulse which has a confined position can be expressed as a superposition of many plane waves, a classical beam which has a limited area in space can be expressed as a state consisting of many photons. Once you have that state, you can invoke geometrical optics or similar approximations to make it appear like a classical particle, but you've taken two approximations: many photons->classical pulse->geometrical/partical approximation to Maxwell's equations.
I've no doubt that you can make these two approximations in that order because I know the math fairly well. So if "classical photon" means (essentially) classical approximation to a solution of Maxwell's equations, which are themselves a many-photon approximation to a quantized field, then yes--"classical photons" are a thing.
What I'm less sure of is this: can you go directly from the mathematical description of a photon and, by expanding in powers of h, (presumably by expanding the action?) get to a classical particle description? Or is there something about photons that prevents you from doing this directly? I know, for example, that photon wave-functions are controversial and unlike those of massive particles (http://arxiv.org/pdf/quant-ph/0508202v1.pdf).
Post by: Pmb on 22/01/2013 13:41:39
Quote from: JP
Yep, and a photon is usually defined similar to a Fourier component (monochromatic plane wave) of a pulse.Huh? What is a "Fourier component (monochromatic plane wave) of a pulse."?
Post by: JP on 22/01/2013 15:06:20
where k is wave vector (2*pi/lambda*direction), x is position, omega is (angular) frequency and t is time. Each of these solutions satisfies a wave equation, so any pulse built from these satisfies the wave equation. Since they're exponential solutions, the integral is a Fourier integral that takes a weighting function in k and omega to x and t.
Post by: lightarrow on 23/01/2013 17:24:46
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: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 + E0 = 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.
= h/p tells that, in the limit h --> 0, = 0, that is, frequency should be infinite.For a photon you can't do it, because its momentum p too would vanish, in the limit h --> 0.
Post by: AndroidNeox on 23/01/2013 20:38:27
[/quote]All 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
Naturally 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.
Post by: lightarrow on 23/01/2013 21:19:32
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...Naturally 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.All 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
As you see, things are not so simple.
Post by: AndroidNeox on 23/01/2013 21:32:15
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.
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.
Post by: lightarrow on 24/01/2013 09:04:56
1. Energy *cannot* have mass, or charge, or lenght, colour, as a number cannot have mass, charge, lenght, or colour. I've already written it but you probably still haven't understood it. Energy is *a property* of a body, and *not a body itself*. What would you answer to someone who stated that his "age" has "weight"?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...Energy does have gravitational mass.
As you see, things are not so simple.
Please, answer this question, before stating another time that energy has mass.
2. Even when your statement is correctly written, that is: "a body which has energy
also have gravitational mass" is not exact because a photon, by itself, doesn't have gravitational mass (before you contest this, think to a physics book where this is written). Instead "a region of space which has electromagnetic energy" does have.
Quote
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.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.
Post by: Pmb on 24/01/2013 13:50:43
Quote from: lightarrow link
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.I can see that you’re not one who is much for agreeing to disagree, huh? Okay.
Quote from: lightarrow link
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:And since the product of wavelength and momentum has to remain the same in that relationship it means that it’s an improper limit since as ones going down, the other is going up.
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 insist on ignoring its defining properties. It’s a classical description of a Fourier integral description of a sharp pulse of radiation. Since JP and I both have our convictions I’m wondering what it is that you hope to gain into continuing this? 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 http://home.comcast.net/~peter.m.brown/sr/einsteins_box.htm just pas Eq. (8). 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.
Post by: lightarrow on 24/01/2013 17:11:34
I can see that you’re not one who is much for agreeing to disagree, huh? Okay.It depends. Because of my nickname, I like to talk about light and photons [:)], so I've studied this subject and discussed it in the forums, for several years.
Quote
? We are making the limit h-->0, so it doesn't remain the same!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:And since the product of wavelength and momentum has to remain the same
Quote
In any case you keep forgetting that we’re treating it as a classical particle for which the relationship:But the point is that *you have to prove* treat it can be treated as a classical particle.
Quote
Thus it means nothing in the classical sense to take a classical limit of a photon.Exactly! Read again what you have written here [:)]
Quote
JP and myself explained how it works in a classical sense.JP wrote this:
<<Good point. After reading that and a bit more thinking, I believe that what's going on with these "classical photons" is two limits. If we have a field made of photons, the classical limit does not correspond to taking h->0, but rather to taking many photons. In this limit, you recover Maxwell's equations, but you've lost information about the behavior of individual photons.
Then you take a second limit corresponding to wavelength->0 which gets you ray optics. So essentially, your "classical photon" is an arbitrary packet of energy assigned to propagate along a ray. It's related to real photons only insofar as the sum over many photons gets you the classical field, which you then use to define rays.>>
So JP, essentially, says: OR you recover classical electromagnetism, and you have lost information about the behaviour of individual photons, OR you come up with the geometrical optics approximations and you can talk about packets of energy assigned to propagate along a ray.
But this is exactly what I've already written to you in previous posts [:)]
Quote
You certainly can’t change the textbooksSincerely I have never read "classical photon" in a book of physics.
Quote
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."So many people" are able to localize a photon in flight? Forget it...
Please explain, in detail if you please, what it means when you say “You can’ do that!” when so many people do it,
Quote
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.How do you draw a photon's worldline between (0,0,0,0) and (1,1,0,0)? Just for curiosity.
Quote
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 http://home.comcast.net/~peter.m.brown/sr/einsteins_box.htm just pas Eq. (8). 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.
Post by: JP on 24/01/2013 18:49:36
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.
This sums it up nicely: you two disagree on what you call a photon. Pmb's photons are not at all the quantized photons of quantum optics. If I were writing the textbooks, I'd side with Lightarrow on this and reserve photon to mean a very specific thing in quantum mechanics that has no classical analogue: the state resulting from an exitation of hbar*omega of energy in the EM field. However, as I'm not the President of Physics, I can't do this, and I've noticed that many physicists use the term in a sense similar to what Pmb is doing, and this works out pretty well. The problem arises when compare the properties of these different uses of "photon" and realize they're talking about two completely different things with completely different properties, and that there is no classical limit of a single-photon Fock state.
Perhaps the solution should be to elect me the President of Physics and I'll rewrite all the textbooks to clear this up? :P
Post by: Pmb on 25/01/2013 02:32:57
Quote from: lightarrow
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...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.
I sincerely don't know what else I could say.
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.
Post by: lightarrow on 25/01/2013 11:39:01
It's you that don't pay attention to what I wrote. I've already written that "saying to have exactly localized a photon is the same mistake of saying to have perfectly localized a particle' path in a cloud chamber".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...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.
I sincerely don't know what else I could say.
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.
And, again, you don't need to talk of photons, if you intend to make such coarse approximations, you can talk of pulses of light.
Post by: Pmb on 26/01/2013 15:45:19
Is there a good reason why you keep refusing to provide a definition of the term "localized" as you insist on using it? I case your response is to claim that you already have please post the time and day of your response.
Post by: JP on 26/01/2013 15:49:41
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. A measured photon is no longer in a Fock state, so it is no longer a photon by that definition.
Post by: Pmb on 26/01/2013 16:51:32
Quote from: JP
But a photon that's been measured is no longer a photon.I’m aware of that, of course. I wasn’t sure he was aware that when a photon’s position is measured that it was localized.
Quote from: JP
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..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.
It doesn’t matter at this point. I don’t want to discuss it anymore. I merely wanted to see what lightarrow thinks “localized” means. I've already had enough of his broken record responses.
Post by: Pmb on 26/01/2013 17:05:54
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. :)
Post by: JP on 26/01/2013 19:20:57
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. :)
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.
Post by: Pmb on 26/01/2013 21:55:15
What we were talking about was the position vector of a single photon in a system of only a few photons. Can that be done by using some sort of centroid?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. :)
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.
Post by: JP on 26/01/2013 23:43:47
http://books.google.com/books?id=FeBix14iM70C&pg=PA480&lpg=PA480#v=onepage&q&f=false
They come back to the plane wave description I was giving earlier: if you have a single plane wave it is not localized in space and time, but if you have many added together, it can be localized.
In the same way, if you have a single Fock state, it cannot be localized, but if you add many single-photon Fock states together, it can be. But since this is quantum, the addition of many Fock states can be a single-photon state, since each Fock state represents a configuration of a photon, and the whole state describes the probability of finding that photon in each configuration.
But as they say, you have to be careful when interpreting "localized" photons, since they are defined from second quantization of the field, and do not have all the properties of a particle such as an electron, which can be described in terms of first quantization of a particle.
Post by: Pmb on 27/01/2013 00:12:23
Quote
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.So lightarrows attempt to use the term "quantized" as part of the definition of photon has the defining "quantum mechanical" property is flawed since we know that referred only to it being a discrete amount of something and not as "quantum mechanical" also finds its use as in the defining property of the electron as well. In fact they use it to first describe the electron and only much later to describe the photon. As I said before, you have to be careful of how you interpret the use of the term "quantized" or "quantum." For the most part it refers to discrete lumps of matter more that it does to the name of a theory. :)
Post by: Pmb on 27/01/2013 02:24:00
Quote from: Pmb
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.This has been bothering me recently. It appears to me now that such a classical photon cannot interact with other particles in such a way that it would change its energy E. There’s no way for such a particle to change its energy so when a charged particle interacts with it it can’t change its energy. So a classical photon can’t change its kinetic energy, since E = kinetic energy. A quantum photon can change its kinetic energy by changing its wavelength. So a classical photon cannot change its speed.
Since the uses I've seen for such a particle never invoke a change of energy this isn't a serious limitation.
Post by: AndroidNeox on 27/01/2013 17:04:34
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.
It's both correct and does prove my statement.
Post by: Pmb on 27/01/2013 17:50:48
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 valueQuotePut 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.
(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.
Post by: JP on 27/01/2013 20:10:03
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.
That's because you're defining mass to preclude the use of invariant 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. It's my impression that invariant mass is generally winning insofar as it's being adopted in introductory textbooks. http://en.wikipedia.org/wiki/Mass_in_special_relativity#Controversy
Post by: Pmb on 27/01/2013 23:20:36
Quote from: JP
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. Invariant mass is quite useful. I’d hate to try to work out your typical particle physics problems without it. Proper mass is also a defining property of a particle in that it is the unique limit of inertial mass for low v, i.e. m = M(v) as v->0.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.
Quote from: JP
Plenty of physicists (mostly in high energy physics) use "mass" to mean invariant mass, which also has desirable properties.Absolutely (Noting that #1 fails for all luxons). Please don’t misinterpret my comments to mean that proper mass is not useful. That would be an outrageously inaccurate assumption.
(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.
Quote from: JP
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.Its my opinion that there are many more particle physicists than other kinds of users of relativity so that the majority of use is mass = proper mass. However that is merely a game of numbers and what’s currently a popular area of research.
Quote from: JP
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.Except when it comes to photons where there is no such limit, of course.
Quote from: JP
There is legitimate controversy in the teaching of physics over which definition to use.
Oh yes. That horse has been thoroughly beat upon in virtully every physics forum on the internet ever time the subject of mass comes up. Its become a virus which derails most threads merely because when people say “mass depends on speed” the non rel-mass people have to change it to debate against rel-mass rather than jus think to themselves “Okay. By ‘mass’ that person means rel-mass.” and then moves on with the discussion. Never happens in practice thought. The problem is that it causes tons of confusion because you then have to explain why light can generate a gravitational field. When it comes to defining mass density for a uniform magnetic field it becomes very dicey.
By the way, I’ve changed my mind on the only properties of a classical photon being its energy and momentum. It’s mass is the real physical property. While in QM the wavelength changes giving a corresponding change in energy in classical mechanics it’s the mass that changes giving a corresponding change in energy, same as in QM but phrased in terms of classical properties.
Post by: JP on 28/01/2013 05:02:24
Quote from: JPNot 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.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.
I'm just taking issue your one line above that I quoted. Invariant mass has plenty to do with the definition of mass, since it agrees completely with non-relativistic mass in the non-relativistic limit, just as inertial mass does. It's all a matter of the way you choose to extend that definition into a relativistic framework, and both invariant and inertial mass are useful extensions.
Post by: Pmb on 28/01/2013 06:13:29
Quote from: JP
I'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!
Post by: JP on 28/01/2013 14:31:45
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!
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.
Post by: Pmb on 28/01/2013 16:48:34
Don't get me wrong.I wasn't blaming anyone for anything about that. More later. Gotta go to pain clinic.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!
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.
Post by: lightarrow on 28/01/2013 21:57:50
Perhaps the solution should be to elect me the President of Physics[:)]
Quote
and I'll rewrite all the textbooks to clear this up? :PI like your kind of humour.
Post by: lightarrow on 28/01/2013 22:22:34
Ah, yes, localized after detection, of course. The problem is, and this is not the first time I write it, I was talking of localizing it in flight, between source and detector.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...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.
I sincerely don't know what else I could say.
Quote
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?Have a look also here:
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.
http://stochastix.files.wordpress.com/2008/05/what-is-a-photon.pdf
<<What is a photon?
Rodney Loudon
University of Essex, Colchester, UK>>
...
<<A one-photon excitation in such a mode again carries an energy
quantum ¯hω distributed over the entire interferometer,
including both internal paths. Despite the absence of any localization
of the photon, the theory provides expressions for
the distributions of light in the two output arms, equivalent to
a determination of the interference fringes.>>
Post by: lightarrow on 28/01/2013 22:45:31
Sorry, but I don’t know what a Fock state is.Do you mean it's a definition invented by you? It's just a question.
In any case, that’s not what I meant by a “classical photon.” Recall the definition that I used.
Quote
A classical photon is a particle whose inertial energy"inertial energy"? Sorry but I've never read this term; it's not another of "your definitions", isnt'it?
Quote
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.Ok. What you have described here is simply a classical pulse of light: an electromagnetic wavepacket. Why do you call it "classical photon"? Well, if I will find it in books of physics, I will conform to it, no problem; don't see any problem in using that term, as soon as it will be defined.
Quote
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.If they use the term "photon" it's a misuse and they could simply talk of an electromagnetic wavepacket. I suspect some relativists don't actually now what "a photon" exactly is; I don't mean I know it well, but it's a lot of time I discuss this specific subject with physicists, at university, first and in the forums.
Post by: lightarrow on 28/01/2013 23:06:11
If you say so...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.
It's both correct and does prove my statement.
Post by: lightarrow on 28/01/2013 23:13:21
Forget what he's been saying. He has a way of confusing the poperties of mass with those of proper mass.[;D] Sorry, I wasn't.
Quote
There are three aspects of mass given three names and each are merely just called "mass" because they all have the same valueJust "Four" kinds? My God, where has gone your creativity? From you I expected at least a hundreds kinds [:)]
(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.
You still have a lot of work to do, if you want to write since-fiction books [:)]
Post by: yor_on on 29/01/2013 04:06:03
Post by: lightarrow on 29/01/2013 08:40:06
This one does a nice job of explaining the history of mass, and how the idea of passive and active mass came to be. The Equivalence Principle: A Question of Mass (http://astronomy.swin.edu.au/sao/downloads/HET625-M04A01.pdf)History is interesting, but once physics has established the equivalence of those masses, there is no need to talk about them any longer, there is just one.
Post by: yor_on on 29/01/2013 17:34:02
Post by: yor_on on 29/01/2013 17:43:57
==
Locally I would say that we all have a equivalent arrow, as proved when being in a same frame of reference with what you measure. From a point of locality we're all 'equal' :) regarding the arrow. The same goes for distances. From that point of view the universe consist of one unchanging base, same for us all, observer dependencies created through 'c' (combined with energy/mass/'motion'). But the fact is that your reality is defined through your experiment, so finding someone else's watch to go slower than yours relativistically and experimentally must be true to/for you.
So locality is where we're all equal as I see it. And where all arrows are equivalent.
And one more thing, accelerations.
The weirdest example, and proof, of a 'change' that I know. Everything that exist seems to me to somehow (be able too?) accelerate? What is a virtual particle? If it is not there, but then is, can you see that as a 'acceleration' from a probability into a 'outcome' ('real' particle) ?
Post by: Pmb on 29/01/2013 23:53:20
Quote from: JP
I wasn't the one telling posters that proper mass has little or nothing to do with the definition of mass!Inertial Mass - Defines momentum. Quantifies resistance to changes in momentum.
Passive Gravitational Mass - The property on which gravity acts
Active Gravitational Mass - That which generates a gravitational field.
Those are what characterizes mass, by definition. How does invariant mass simply fit in there? We all know that it does, but in what direct manner?
Post by: JP on 30/01/2013 05:01:14
The last two of your points have to do with general relativity, so I'd add that relativistic quantum mechanics uses invariant mass, in the place of where mass would enter in the non-relativistic equations, mostly because of the simplicity with which it enters the equations in place of non-relativistic mass when promoting 3-vectors to 4-vectors. This isn't as elegant or long-standing as "it generates gravity," but that's because we happened to be born into a world where we experience gravity daily, and not because it's any less fundamental. If we were born as quantum objects, we'd find it very fundamental indeed!
Post by: waytogo on 30/01/2013 09:38:10
I'm sorry if this is a dumb question but physics is not really my area. I've been listening to the CBC Massey lectures by physicist Neil Turok, which I quite like. Anyway, when he talks about mass increasing at higher speeds, how does that happen? Is there actually an increase in the amount of matter or atoms or particles? Or does it just take more force to accelerate it? I had always thought that mass and matter were the same thing.
Hi there, Its NOT a dumb question, I still have no answer of that. Anyway, you have to consider that its just a ancient theory and new ones should be replaced in next decades.
Post by: Pmb on 30/01/2013 18:51:51
Mass is the length of the energy-momentum four ..That is merely an equality relating inertial energy, momentum and proper mass. It's not a definition of mass, at least it shouldn't be. The onlyway you can arrive at that relationship is by relating inertial mass to velocity of an isolated system. It doesn't work in general by the way. It only works for competely isolated systems. It wouf fail for a drop of water in an electric field. The definitions are as I gave them above. They are what characterize mass.
Post by: Pmb on 30/01/2013 18:53:01
Its hardly ancient since its found in even the most recently published textbooks.I'm sorry if this is a dumb question but physics is not really my area. I've been listening to the CBC Massey lectures by physicist Neil Turok, which I quite like. Anyway, when he talks about mass increasing at higher speeds, how does that happen? Is there actually an increase in the amount of matter or atoms or particles? Or does it just take more force to accelerate it? I had always thought that mass and matter were the same thing.
Hi there, Its NOT a dumb question, I still have no answer of that. Anyway, you have to consider that its just a ancient theory and new ones should be replaced in next decades.
Post by: JP on 30/01/2013 19:17:09
The definitions are as I gave them above. They are what characterize mass.
Ah, the old "I chose to characterize mass by these definitions, therefore they characterize mass" argument. That works for the definition I gave, too. :)
I thought up another interesting question about invariant vs. inertial mass, though. I don't understand the Higgs mechanism enough to give a definite answer, but from the descriptions I've heard of it by scientists, it explains inertial mass in terms of a field, and the strength of a particle's inertial mass by the coupling to that field. That could mean a coupling constant would probably be the best definition of inertial mass, which seems like invariant mass. The Higgs mechanism could provide a means to explain inertial mass in terms of a single constant plus a field of nature. If that's true, invariant mass is presumably more fundamental than inertial mass, which is another good reason to consider it a definition of mass. This is speculation, so take it with a large grain of salt unless a Higgs specialist on the forum wants to confirm or refute it. :p
This horse, however, has been beaten to a fine paste at this point. Arguing our opinions on the internet doesn't change the fact that all these definitions are valid extensions of the concept of classical mass, and all find widespread use, albeit in different fields of study. I guess we'll keep on arguing until someone figures out quantum gravity and ties all the concepts of mass into one fundamental definition.
Post by: Pmb on 31/01/2013 22:22:42
Quote from: JP
Ah, the old "I chose to characterize mass by these definitions, therefore they characterize mass" argument. That works for the definition I gave, too. :)Not really. It was never I who chose them. They've been around long before I was born, and still are.
Post by: Pmb on 02/02/2013 00:13:07
Quote from: JP
I wasn't the one telling posters that proper mass has little or nothing to do with the definition of mass!I just realized that you misquoted me here. I never said that. What I said was that proper mass has little or nothing to do with the defining characteristics of mass.
I would never say that it had nothing to do with mass, never! The reason being, that it’s simply not true. Did you miss that part or were you confused? [we all have our days :) ] I think you mistook me for saying that proper mass has nothing to do with the definition of mass. That’s now what I said simply because I don't believe it to be. The defining characteristics being the whole point I was making.
If you'll notice, I do my best to avoid getting into discussions about the definition of mass. There's just way too much to the concept than can be gotten to in a discussion in a discussion forum. In fact it was for just that reason that wrote that article on the subject. Everything I believe regarding the concept of mass is in the paper at http://arxiv.org/abs/0709.0687
I recommend that you read it. I’d enjoy your feedback and any comments or constructive criticism that you might have on it.
I can pretty much guarantee that every aspect about mass that you could come up with are tell me about mass is contained in that paper, including the most obvious notion that the “mass” of a particle is the magnitude of the particle’s 4-momentum. If you read the paper you’ll find what I consider to be a much better definition of proper mass, i.e. as the quantity m such that the quantity P = mU (where P = 4-momentum and U = 4-velocity) for a system of particles which interact only by contact forces, is conserved.
Note: What's with this stupid editor? Everytime I try editing in this window it keeps popping up to the top so I have to keep scrolling down to see what I'm editing!
Post by: JP on 02/02/2013 16:43:55
My point still stands in roughly the same form. If you list a bunch of defining characteristics of mass that preclude another definition of mass, then of course it has little or nothing to do with those characteristics. Though I would argue that the use of invariant mass does meet the kinematical characteristics of mass, since classically, inertial mass in Newton's second law, F=ma, can be replaced in the four-vector version with invariant mass, and it's particularly elegant to view the transition to special relativity geometrically in terms of 4-vectors.
The horse is still very dead, though. We all agree (I hope!) on the physics involved, and that describing it requires 4-vectors and/or tensors. In the end, we're arguing about a single scalar value pulled out of these equations, so of course there's multiple ways to do that! Similar debates come up a lot in physics, where one has vectors/tensors and tries to use a single scalar to describe the physics. It's always insufficient and very often just as contentious (look up degree of entanglement in quantum mechanics for a similar debate), and obscures the fact that everyone involved agrees on the physics. I imagine this argument over mass will go on unless, some day, a unified theory comes along that does involve a single scalar value in the place of mass (for example, a coupling constant to some field describing inertial and gravitational masses) from which all other definitions follow.
I'll check out your paper when I have a chance. I've got a backlog of reading to do at the moment. :(
Post by: Pmb on 02/02/2013 18:36:17
Quote from: JP
Ah true, I did misquote you. Sorry for that.Yup. And all of that is stated in my paper. :)
My point still stands in roughly the same form. If you list a bunch of defining characteristics of mass that preclude another definition of mass, then of course it has little or nothing to do with those characteristics. Though I would argue that the use of invariant mass does meet the kinematical characteristics of mass, since classically, inertial mass in Newton's second law, F=ma, can be replaced in the four-vector version with invariant mass, and it's particularly elegant to view the transition to special relativity geometrically in terms of 4-vectors.
This is related to a new idea that I came up with. I call it the principle of controlled igorance. It states that physicists write with given assumptions that are agreed among their peers. E.g. what one person uses as the definition of the term momentum might be different that what someone else uses. Someone who works in classical mechanics will define momentum, when unqualified by anything, as p = mv. But another person working in quantum mechanics will define it as being identical to canonical momentum.
Quote from: JP
I'll check out your paper when I have a chance. I've got a backlog of reading to do at the moment. :(Very cool. Thanks!
Post by: Pmb on 02/02/2013 18:49:56
Clifford M. Will wrote a wonderful article called The Confrontation between General Relativity and Experiment
It’s online at http://arxiv.org/abs/gr-qc/0510072