Photon Energy

[martu]

I have never quite understood how or why a photon’s energy is related to the wavelength and thought about it this way:

Picture the photon as a particle with a very small mass that travels in a wave like manner where all the waves have an equal peak-to-peak amplitude and all photons have the same wave speed (c in a vacuum). Though wave speed is fixed (in a medium) the speed perpendicular to travel can vary so that blue photons are oscillating faster than red photons, 1.6 times faster in fact. The faster the photon oscillates the more energy it has and, as amplitude is fixed, the wavelength varies with oscillation speed so that blue photons have a shorter wavelength than red photons.

This amplitude is so small that to us photons travel in a straight line at c.

What observation rules this picture out?

(Ozziemate’s recent threads have made me reread a lot of Feynman’s books recently and I mentioned something in the ‘proof of a photon’ thread that no one commented on so would like some input from the physicists here showing what me what is wrong the picture, thanks in advance)

[blutoski]

My limited exploration of quantum mechanics leads me to believe that attempts to model quantum events in classical terms is more likely to lead to frustration than understanding.

Just as an exercise to see if the explanation generalizes, does your analogy work for the wavelength of a thrown baseball?

Oh, and one nitpick (sorry) - I'm not sure why you're assuming that amplitude is fixed. Amplitude variability is important to assigning probability to locality.

Or, are you saying that the amplitude of a single photon is fixed?

[Ziggurat]

Picture the photon as a particle with a very small mass

Why? There's no reason or need to assign any mass to photons.

that travels in a wave like manner where all the waves have an equal peak-to-peak amplitude and all photons have the same wave speed (c in a vacuum). Though wave speed is fixed (in a medium) the speed perpendicular to travel can vary so that blue photons are oscillating faster than red photons, 1.6 times faster in fact. The faster the photon oscillates the more energy it has and, as amplitude is fixed, the wavelength varies with oscillation speed so that blue photons have a shorter wavelength than red photons.

This amplitude is so small that to us photons travel in a straight line at c.

Photons do NOT wiggle from side to side in space. The magnetic and electric fields point from side to side, but the field itself is still located along a straight path. Thinking of this in terms of a spatial displacement is wrong. So it's not that photons "appear" to move in a straight line, they do move in a straight line. Waves don't need to involve sideways displacements. In fact, even many physical waves involving displacement of atoms do not involve transverse displacement (for example, sound waves in air).

[martu]

My limited exploration of quantum mechanics leads me to believe that attempts to model quantum events in classical terms is more likely to lead to frustration than understanding.

Just as an exercise to see if the explanation generalizes, does your analogy work for the wavelength of a thrown baseball?

Oh, and one nitpick (sorry) - I'm not sure why you're assuming that amplitude is fixed. Amplitude variability is important to assigning probability to locality.

Or, are you saying that the amplitude of a single photon is fixed?

All photons have the same maximum amplitude. Probability of location is defined by the perpendicular oscillation. Too small for us to see.

[martu]

Why? There's no reason or need to assign any mass to photons.



Photons do NOT wiggle from side to side in space. The magnetic and electric fields point from side to side, but the field itself is still located along a straight path. Thinking of this in terms of a spatial displacement is wrong. So it's not that photons "appear" to move in a straight line, they do move in a straight line. Waves don't need to involve sideways displacements. In fact, even many physical waves involving displacement of atoms do not involve transverse displacement (for example, sound waves in air).

At any point in time a photon's direction is a straight line defined by velocity in direction of travel and velocity of the oscillation.

[Reality Check]

I have never quite understood how or why a photon’s energy is related to the wavelength and thought about it this way:

Picture the photon as a particle with a very small mass that travels in a wave like manner where all the waves have an equal peak-to-peak amplitude and all photons have the same wave speed (c in a vacuum). Though wave speed is fixed (in a medium) the speed perpendicular to travel can vary so that blue photons are oscillating faster than red photons, 1.6 times faster in fact. The faster the photon oscillates the more energy it has and, as amplitude is fixed, the wavelength varies with oscillation speed so that blue photons have a shorter wavelength than red photons.

This amplitude is so small that to us photons travel in a straight line at c.

What observation rules this picture out?

(Ozziemate’s recent threads have made me reread a lot of Feynman’s books recently and I mentioned something in the ‘proof of a photon’ thread that no one commented on so would like some input from the physicists here showing what me what is wrong the picture, thanks in advance)
The observation that photons have no mass?

But perhaps you can give us a list of predictions from your picture?

[Ziggurat]

At any point in time a photon's direction is a straight line defined by velocity in direction of travel and velocity of the oscillation.

"Velocity of oscillation" is meaningless for a photon. Velocity is distance/time. But what's oscillating is the field. It's got a frequency to that oscillation, but it's not a displacement, so you CANNOT assign it a velocity. It has a field strength/time, which one might be tempted to label as a velocity, but 1) it has the wrong units, 2) it adds when multiple photons are present, and 3) even for a single photon it depends on normalization requirements (ie, how spread out the photon is spatially - and no, that's NOT the same thing as the photon wavelength), not just frequency.

Edit to add: there are only two velocities which ever make sense for light: a group velocity and a phase velocity. In a vacuum, they are identical, and they are in the direction of propagation.

[Vorpal]

Photons are pointlike. A photon's energy and momentum are related through E = pc as required by relativity, and the wavelength is connected to the uncertainty of the photon's position. The frequency of a photon is not really in terms of anything it itself is "doing" (although it does measure energy), but rather the frequency of its probability amplitude. If you want a (relatively) intuitive picture, look up Feynman's "arrows" in both his book "QED" and some video lectures available online.

[Acleron]

Photons are pointlike. A photon's energy and momentum are related through E = pc as required by relativity, and the wavelength is connected to the uncertainty of the photon's position. The frequency of a photon is not really in terms of anything it itself is "doing" (although it does measure energy), but rather the frequency of its probability amplitude. If you want a (relatively) intuitive picture, look up Feynman's "arrows" in both his book "QED" and some video lectures available online.

Good. A very clear description. But may I ask a question which has perplexed me for some time? A light sail (http://en.wikipedia.org/wiki/Solar_sail) works by the photon imparting momentum to the sail. My question is this: What measurable physical property of the photon changes after losing momentum to the sail?

[Scazon]

Its frequency. E=hf where f is the frequency and h is the size of the Planck. On hitting the sail, it is reflected, and imparts some of its energy to the sail, and goes off at a different angle with a redder face.

[Acleron]

Its frequency. E=hf where f is the frequency and h is the size of the Planck. On hitting the sail, it is reflected, and imparts some of its energy to the sail, and goes off at a different angle with a redder face.

Ah, I think I see what I was getting wrong. I was thinking that light reflected from a light sail would be the same as from a bathroom mirror. Thanks for that.

[martu]

The observation that photons have no mass?

But perhaps you can give us a list of predictions from your picture?

What observation that photons have no mass? We have observed a limit (http://www.aip.org/pnu/2003/split/625-2.html) but that is all.

Here are a couple:

We wouldn't be able to predict exactly where a photon is due to the oscillation until we observed it.
The equations for energy of a photon would include a constant related to the amplitude.

As both of these phenomena are related to the amplitude it should be the same constant that defines both the Uncertainty and the Energy. Planck’s constant.

[martu]

"Velocity of oscillation" is meaningless for a photon. Velocity is distance/time. But what's oscillating is the field. It's got a frequency to that oscillation, but it's not a displacement, so you CANNOT assign it a velocity. It has a field strength/time, which one might be tempted to label as a velocity, but 1) it has the wrong units, 2) it adds when multiple photons are present, and 3) even for a single photon it depends on normalization requirements (ie, how spread out the photon is spatially - and no, that's NOT the same thing as the photon wavelength), not just frequency.

Edit to add: there are only two velocities which ever make sense for light: a group velocity and a phase velocity. In a vacuum, they are identical, and they are in the direction of propagation.

How can something have a frequency that isn’t due to displacement? If it isn’t displacement what is it? The only other thing I can think of is a change in size\mass.

[martu]

Its frequency. E=hf where f is the frequency and h is the size of the Planck. On hitting the sail, it is reflected, and imparts some of its energy to the sail, and goes off at a different angle with a redder face.

Indeed which can be explained by the photon having mass to give it a momentum and variable perpendicular velocity which is affected by the sail. If it slows down the wavelength gets longer and hence a ‘redder face’. h defines the amplitude and mass which is equal for all photons.

Another question related to the OP is how a photon can have momentum if it doesn’t have mass?

[Vorpal]

How can something have a frequency that isn’t due to displacement? If it isn’t displacement what is it? The only other thing I can think of is a change in size\mass.
The frequency is related to the oscillation of the position wavefunction rather than the photon itself. If you've read Feynman, as you claimed to have done in the OP, you should have a clearer picture of the probability amplitude and its frequency.

Indeed which can be explained by the photon having mass to give it a momentum and variable perpendicular velocity which is affected by the sail.
Uh... sure, but why would it need mass to give it momentum?

Another question related to the OP is how a photon can have momentum if it doesn’t have mass?
Because the stage is that of spacetime, where the conserved quantity is the momentum four-vector. Its time component gives the energy of the particle and the spatial components the ordinary three-momentum. In ordinary Euclidean space, the length of vectors is given by the distance formula: sqrt[x²+y²+z²]. But in spacetime, four-vectors may be nonzero but still have zero length due to a curious signature of the Minkowski distance formula: sqrt[t²-(x²+y²+z²)]. The four-momentum is actually proportional to the particle four-velocity, and the length is the mass of the particle, or in ordinary units mc² = sqrt[E² - (pc)²].

[Dancing David]

Oh, mighty sages, thank you for this discussion, I am glad to see that there are answers to these questions.

[Ziggurat]

How can something have a frequency that isn’t due to displacement? If it isn’t displacement what is it? The only other thing I can think of is a change in size\mass.

Easily. I already told you: the field is changing with time in a periodic manner. This is an oscillating field, so it has a frequency of how fast the field is changing, but it is not a displacement. You keep trying to picture the photon in terms of a massive particle rather than a field, but that's not what it is.

[Ziggurat]

h defines the amplitude and mass which is equal for all photons.

Another question related to the OP is how a photon can have momentum if it doesn’t have mass?

Some people talk about photons having "relativistic mass", which is nonzero but is NOT constant for all photons. But relativistic mass is redundant with energy, and it's a pointless concept which isn't used anymore. Rest mass (or invariant mass) is the only mass one ever needs to use, and it's zero for photons. As previously indicated, relativity allows for non-zero momentum even for massless particles. The equation given above is usually written as E2=m2c4+p2c2. Putting in a zero for m still gives you p=E/c.

[The Man]

Ah, I think I see what I was getting wrong. I was thinking that light reflected from a light sail would be the same as from a bathroom mirror. Thanks for that.


Refection of the photons results in transferring twice the momentum to the sail as absorption see radiation pressure (http://en.wikipedia.org/wiki/Radiation_pressure).

[martu]

The frequency is related to the oscillation of the position wavefunction rather than the photon itself. If you've read Feynman, as you claimed to have done in the OP, you should have a clearer picture of the probability amplitude and its frequency.

I didn’t say I fully understood it.... :)

Feynam describes phenomena then the maths that can predict behaviour to certain probabilities. It doesn’t rule out a particle travelling like a wave.

Uh... sure, but why would it need mass to give it momentum?

Because p = mv. Or to put it another way how can something have momentum if it doesn’t have mass?

Because the stage is that of spacetime, where the conserved quantity is the momentum four-vector. Its time component gives the energy of the particle and the spatial components the ordinary three-momentum. In ordinary Euclidean space, the length of vectors is given by the distance formula: sqrt[x²+y²+z²]. But in spacetime, four-vectors may be nonzero but still have zero length due to a curious signature of the Minkowski distance formula: sqrt[t²-(x²+y²+z²)]. The four-momentum is actually proportional to the particle four-velocity, and the length is the mass of the particle, or in ordinary units mc² = sqrt[E² - (pc)²].

Yes I just about follow that Six Not So Easy Pieces was a joy to read again. How exactly does this rule out a particle travelling along a wave?

[martu]

Easily. I already told you: the field is changing with time in a periodic manner. This is an oscillating field, so it has a frequency of how fast the field is changing, but it is not a displacement. You keep trying to picture the photon in terms of a massive particle rather than a field, but that's not what it is.

What property of the field is changing?

[martu]

Some people talk about photons having "relativistic mass", which is nonzero but is NOT constant for all photons. But relativistic mass is redundant with energy, and it's a pointless concept which isn't used anymore. Rest mass (or invariant mass) is the only mass one ever needs to use, and it's zero for photons. As previously indicated, relativity allows for non-zero momentum even for massless particles. The equation given above is usually written as E2=m2c4+p2c2. Putting in a zero for m still gives you p=E/c.

Thanks.

This is the bit I don’t get how can something without mass have momentum? Is it merely stipulated?

I have an idea why rest mass is zero for a photon but I don’t want to derail this at the moment.

[martu]

Oh, mighty sages, thank you for this discussion, I am glad to see that there are answers to these questions.

Seconded, it is fantastic being able to discuss this with experts to be honest.

[martu]

Refection of the photons results in transferring twice the momentum to the sail as absorption see radiation pressure (http://en.wikipedia.org/wiki/Radiation_pressure).

Which could be explained by massive particles hitting an object.

[Ziggurat]

What property of the field is changing?

Its amplitude and/or its direction. But remember: we're talking about the field at a fixed point. There's no displacement involved. It's hard to visualize without drawing arrows to represent the field at each point, as in this picture (http://www.geo.mtu.edu/rs/back/spectrum/):
http://www.geo.mtu.edu/rs/back/spectrum/e_mag.gif
and it can be tempting to think of those arrows as being a displacement, but they are not. They represent the field at a point. The field at the point changes, but the point doesn't move.

[Ziggurat]

Which could be explained by massive particles hitting an object.

Could be, but isn't. People have done experiments to search for a mass for photons, and they always come up empty. You don't need mass to have momentum - it's not intuitive, but the math works out beautifully, and experiments all seem to confirm it.

[The Man]

Which could be explained by massive particles hitting an object.

It could also be explained by invisible elves blowing into the sail, but neither of those explanations is relevant. The only relevant explanation is the conservation of momentum.

[sol invictus]

This is the bit I don’t get how can something without mass have momentum? Is it merely stipulated?

No, it's not merely "stipulated". It follows immediately from fundamental principles (Lorentz invariance, for example) and it's also measured experimentally with extreme accuracy.

Think of it like this. Take a massive particle, and accelerate it by imparting a certain total amount of energy. The more energy you give it, the faster it will end up going, but as the energy gets very large the speed will simply approach c, the speed of light.

At speeds close to the speed of light the momentum is no longer given by mass times velocity. That would approach a maximum m*c, which is not what experiments observe. Instead, the momentum continues to grow as you add more and more energy, and for speeds close to the speed of light the momentum is given by a formula that gets close and closer to p=E/c. The same formula works perfectly for photons.

If you don't like that you can try to think of the momentum of a photon as zero times infinity, where zero is the rest mass and the infinity is related to the amount of energy you'd need to accelerate something of finite mass to the speed of light. That's a distasteful (and rather confusing) way to say it, but it might make clear why the momentum doesn't have to be zero.

[martu]

Its amplitude and/or its direction. But remember: we're talking about the field at a fixed point. There's no displacement involved. It's hard to visualize without drawing arrows to represent the field at each point, as in this picture (http://www.geo.mtu.edu/rs/back/spectrum/):
http://www.geo.mtu.edu/rs/back/spectrum/e_mag.gif
and it can be tempting to think of those arrows as being a displacement, but they are not. They represent the field at a point. The field at the point changes, but the point doesn't move.

It is damn tempting I can tell you, it is also damn tempting to consider that to be a visual representation of two streams of photon particles travelling as waves and interfering with each other as they collide. But maybe that is just me.

So the field has a property which is amplitude and a property which is direction but this does not relate to a spatial displacement is that correct? Why do we use the terms amplitude and direction if no displacement is involved?

[martu]

Could be, but isn't. People have done experiments to search for a mass for photons, and they always come up empty.

Yet you agree the mass could be greater than zero? The link I gave above defines a limit which is very small but it doesn’t prove the mass is 0.

You don't need mass to have momentum - it's not intuitive, but the math works out beautifully, and experiments all seem to confirm it.

What experiments confirm it? Is there anything else with a zero mass that has momentum?

The maths does work I agree but it also works for a non-zero ‘relativistic mass’ but a zero rest mass.

[martu]

It could also be explained by invisible elves blowing into the sail, but neither of those explanations is relevant.

Yes absolutely I am aware this idea is practically not falsifiable hence I was asking for observations that contradict it. I think you can model the photoelectric effect, the double slit experiment, light interference and light diffraction if you assume this to be true. Consider it a thought experiment.

The only relevant explanation is the conservation of momentum.

Again agreed and on that note I have a couple of questions:

1) Consider De Broglie’s hypothesis that every moving particle has a wave why can’t this be interpreted as an electron, to use his original example, actually moves in a wave? Do we ‘know’ how electrons move?
2) Neutrinos change flavour as they travel how is momentum conserved?

[Ziggurat]

Yet you agree the mass could be greater than zero?

It is zero in our theories Our theories could, in principle, be wrong. I have confidence in our theories, but there is no absolute disproof. Nor can there be. But the upper limits on any possible mass (which experiments can provide) are amazingly small.

What experiments confirm it?

As mentioned above, experiments can only set an upper bound for the mass. Here is one such experiment (http://prola.aps.org/abstract/PRL/v73/i4/p514_1).

Is there anything else with a zero mass that has momentum?

gluons (excitations in the strong force field) are massless, and if they exist, gravitons should be as well. Neutrinos were once suspected to be massless, but experiments have shown that they do have mass.

The maths does work I agree but it also works for a non-zero ‘relativistic mass’ but a zero rest mass.

Rest mass is the only mass that matters. Relativistic mass is an outdated and thoroughly useless concept, and is mathematically redundant with energy. It was introduced to maintain a superficial similarity between some relativistic mechanics equations and their Newtonian counterparts, but that doesn't work for all of them anyways, and there's really no reason or need to try to maintain that superficial similarity. Modern relativity texts frequently don't even introduce it, because it leads to confusion more often than enlightenment.

[martu]

No, it's not merely "stipulated". It follows immediately from fundamental principles (Lorentz invariance, for example) and it's also measured experimentally with extreme accuracy.

Think of it like this. Take a massive particle, and accelerate it by imparting a certain total amount of energy. The more energy you give it, the faster it will end up going, but as the energy gets very large the speed will simply approach c, the speed of light.

At speeds close to the speed of light the momentum is no longer given by mass times velocity. That would approach a maximum m*c, which is not what experiments observe. Instead, the momentum continues to grow as you add more and more energy, and for speeds close to the speed of light the momentum is given by a formula that gets close and closer to p=E/c. The same formula works perfectly for photons.

If you don't like that you can try to think of the momentum of a photon as zero times infinity, where zero is the rest mass and the infinity is related to the amount of energy you'd need to accelerate something of finite mass to the speed of light. That's a distasteful (and rather confusing) way to say it, but it might make clear why the momentum doesn't have to be zero.

I am not multiplying anything by infinity, isn’t this bizarre enough? :)

Ok so we know photons have momentum, all the experiments work and we know QM is the most rigorously tested theory we have. So now we have the problem of how an object can have momentum but no rest mass yet all the equations work. Or to put it another way why doesn’t

p = mv

work for a photon?

Again thought experiment time. Consider rest mass to be the amount of energy released as photons when a mass is ‘converted’ into energy:

E = mc2

(The energy is proportional to c2 as the photons travel in a sphere at c from the mass, an area effect)

For a photon this would be 0 as they couldn’t be ‘converted’ into energy they are energy. So

E2 = (mc2) 2 + (pc) 2

Can be read as

ETotal = Efrom converting into photons + Efrom momentum

For a photon:

ETotal = 0 + Efrom momentum

I don't think this idea invalidates anything you have written above but I am sure someone will point out my error soon enough.

[Ziggurat]

1) Consider De Broglie’s hypothesis that every moving particle has a wave why can’t this be interpreted as an electron, to use his original example, actually moves in a wave? Do we ‘know’ how electrons move?

Electrons do move as waves. But what oscillates is NOT the position of the electron, but its phase. Unlike for a photon, there is no classical quantity to describe the phase of an electron: it is a purely quantum property. Even for a massive particle like an electron, the picture of sideways oscillations is wrong. In other words, assigning the photon a nonzero rest mass wouldn't make your original picture any more accurate.

[sol invictus]

Ok so we know photons have momentum, all the experiments work and we know QM is the most rigorously tested theory we have. So now we have the problem of how an object can have momentum but no rest mass yet all the equations work. Or to put it another way why doesn’t

p = mv

work for a photon?

But p=mv doesn't work for anything, as I just explained in detail. It's approximately correct only when v<<c, which is not true for a photon.

Again thought experiment time.

I don't understand what you're trying to say there - sorry.

[Reality Check]

It is damn tempting I can tell you, it is also damn tempting to consider that to be a visual representation of two streams of photon particles travelling as waves and interfering with each other as they collide. But maybe that is just me.

So the field has a property which is amplitude and a property which is direction but this does not relate to a spatial displacement is that correct? Why do we use the terms amplitude and direction if no displacement is involved?
Beacause fields have an amplitude and direction.
Displacement is not the only thing that can have an amplitude .
Displacement is not the only thing that can have a direction.

[Reality Check]

Yet you agree the mass could be greater than zero? The link I gave above defines a limit which is very small but it doesn’t prove the mass is 0.



What experiments confirm it? Is there anything else with a zero mass that has momentum?

The maths does work I agree but it also works for a non-zero ‘relativistic mass’ but a zero rest mass.
There are no experiments I know of that have measured a zero mass for light. But the upper limit is extremely small, e.g. see What is the mass of a photon? (http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html)
The gluon also has no mass (according to quantum field theory).

[sol invictus]

There are no experiments I know of that have measured a zero mass for light. But the upper limit is extremely small, e.g. see What is the mass of a photon? (http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html)
The gluon also has no mass (according to quantum field theory).

The only particle which one knows for sure is massless is the graviton. You can add a tiny mass for the photon without violating either experimental or theoretical consistency, and same goes for the gluon (although in both cases the mass must be extraordinarily and unnaturally small). The reason is that the limit where mass goes to zero is smooth, i.e. the results for the theory with very small mass are very close to the results for the theory with zero mass.

This, however, does not hold for gravity. The angle through which light is lensed by a massive object is not a smooth function of the graviton's mass, and that makes it possible to rule out anything other than zero mass (at least for the simplest and most obvious way to add one).

[The Man]

Yes absolutely I am aware this idea is practically not falsifiable hence I was asking for observations that contradict it. I think you can model the photoelectric effect, the double slit experiment, light interference and light diffraction if you assume this to be true. Consider it a thought experiment.

Ok, can you define such a “thought experiment”, as Zig has noted actual experiments have failed to show any rest mass for the photon (within the limits of those experimets)



Again agreed and on that note I have a couple of questions:

1) Consider De Broglie’s hypothesis that every moving particle has a wave why can’t this be interpreted as an electron, to use his original example, actually moves in a wave? Do we ‘know’ how electrons move?

The wave is a quantum wave function (http://en.wikipedia.org/wiki/Wave_function) relating to the complex probability amplitude of finding the electron in some particular state.

The best description we have for how an electron travels is the path integral ( http://en.wikipedia.org/wiki/Path_integral_formulation).


2) Neutrinos change flavour as they travel how is momentum conserved?

Well, as the Neutrino oscillations (http://en.wikipedia.org/wiki/Neutrino_oscillation) in flavor are currently modeled as oscillations of varing mass and flavor states, the momentum would be maintained by those combined states, if I’m getting it correctly.

[Reality Check]

Yes absolutely I am aware this idea is practically not falsifiable hence I was asking for observations that contradict it. I think you can model the photoelectric effect, the double slit experiment, light interference and light diffraction if you assume this to be true. Consider it a thought experiment.
It is not a thought experiment just the idea of photons being a "very small mass that travels in a wave like manner where all the waves have an equal peak-to-peak amplitude".

Double-slit experiment (in fact any slit experiment): How do photon "waves" fit through slits smaller then their amplitude?

What about radio waves and their large wavelength (and so presumably amplitude of mass oscillation). How do they get detected by aerials that are much smaller?
Especially consider the extremely low frequency radio waves with wavelengths of 100,000 km – 10,000 km.

[martu]

But p=mv doesn't work for anything, as I just explained in detail. It's approximately correct only when v<<c, which is not true for a photon.

If it is approximately correct it must be linked somehow surely or is it just a coincidence?

I don't understand what you're trying to say there - sorry.

OK I may try again.

[martu]

Beacause fields have an amplitude and direction.
Displacement is not the only thing that can have an amplitude .
Displacement is not the only thing that can have a direction.

This is what confuses me, how\why is direction used if it doesn't relate to 'going somewhere'?

[martu]

Ok, can you define such a “thought experiment”, as Zig has noted actual experiments have failed to show any rest mass for the photon (within the limits of those experimets)

Just picture a wave of light as many individual photons oscillating perpendicular to the direction of travel. All photon interactions can now be considered collisions between objects.

The wave is a quantum wave function (http://en.wikipedia.org/wiki/Wave_function) relating to the complex probability amplitude of finding the electron in some particular state.

The best description we have for how an electron travels is the path integral ( http://en.wikipedia.org/wiki/Path_integral_formulation).

Why can’t the state be related to a displacement? Again another pixie question I guess.

Well, as the Neutrino oscillations (http://en.wikipedia.org/wiki/Neutrino_oscillation) in flavor are currently modeled as oscillations of varing mass and flavor states, the momentum would be maintained by those combined states, if I’m getting it correctly.

So what changes when a Neutrino changes state?

[martu]

It is not a thought experiment just the idea of photons being a "very small mass that travels in a wave like manner where all the waves have an equal peak-to-peak amplitude".

Double-slit experiment (in fact any slit experiment): How do photon "waves" fit through slits smaller then their amplitude?

Individual photons can travel through any slit as long as it is bigger than they are of course. Picture it like water only replace water molecules with photons, when a wave of photons travels through a slit the consequent wave will be changed.

What about radio waves and their large wavelength (and so presumably amplitude of mass oscillation). How do they get detected by aerials that are much smaller?
Especially consider the extremely low frequency radio waves with wavelengths of 100,000 km – 10,000 km.

You have misunderstood my fault I expect. Radio waves have the same peak-to-peak amplitude as light waves. The difference between light waves and radio waves is that the individual photons for light are oscillating faster hence more energy and shorter wavelengths.

How about my predictions?

[sol invictus]

If it is approximately correct it must be linked somehow surely or is it just a coincidence?


It's not a coincidence at all - it all follows from one principle. Lorentz invariance implies that $E^2 = m^2c^4 + p^2 c^2$, and that $p = mv/\sqrt{1-v^2/c^2}$. For a photon, m=0 and you get E = p*c. For a massive particle moving slowly you get E = m*c^2 + p^2/(2m), and p=m*v. For a fast moving massive particle you can see that E is again close to p*c.

[Ziggurat]

This is what confuses me, how\why is direction used if it doesn't relate to 'going somewhere'?

Then this is the heart of your confusion. I'm not sure if my explanation will fix that for you, but it's what you'll need to figure out a way to understand. The direction of an electric field tells you which direction a charge would be pushed if it was at that point. But the field exists at that point even if there's no charge there (which there isn't in a vacuum), and even if there is a charge there, if can be moving in a different direction than it's being pushed (for example, a car can be moving forward even while it's being pushed backwards, which is what happens when you're braking).

[martu]

It's not a coincidence at all - it all follows from one principle. Lorentz invariance implies that $E^2 = m^2c^4 + p^2 c^2$, and that $p = mv/\sqrt{1-v^2/c^2}$. For a photon, m=0 and you get E = p*c. For a massive particle moving slowly you get E = m*c^2 + p^2/(2m), and p=m*v. For a fast moving massive particle you can see that E is again close to p*c.

Why does only one equation relate to a photon? What I mean is why is it only E = p * c and not:

$p = mv/\sqrt{1-v^2/c^2}$

putting the numbers in for a photon we get:

p = 0 / 0

If you've answered this already and I have missed it my apologies.

[martu]

Then this is the heart of your confusion. I'm not sure if my explanation will fix that for you, but it's what you'll need to figure out a way to understand. The direction of an electric field tells you which direction a charge would be pushed if it was at that point. But the field exists at that point even if there's no charge there (which there isn't in a vacuum), and even if there is a charge there, if can be moving in a different direction than it's being pushed (for example, a car can be moving forward even while it's being pushed backwards, which is what happens when you're braking).

OK so it is a property of a field at a certain point that defines what would happen if a charge was put there is that it? We also have to take into account the charge's properties obviously.

How do we know a field exists in a vacuum without 'seeing' what it does to a charge?

[Reality Check]

Why does only one equation relate to a photon? What I mean is why is it only E = p * c and not:

http://www.randi.org/latexrender/latex.php?$p = mv/\sqrt{1-v^2/c^2}$

putting the numbers in for a photon we get:

p = 0 / 0

If you've answered this already and I have missed it my apologies.
The m=0 really refers to the first equation not the second equation (which is the momentum for a massive particle, not a massless particle).

[sol invictus]

Why does only one equation relate to a photon? What I mean is why is it only E = p * c and not:

$p = mv/\sqrt{1-v^2/c^2}$


That equation applies to the photon in a sense. Consider a particle with mass m and speed v. Now imagine you could reduce its mass while increasing its velocity in such a way that p remains constant. As you do that, you'll find that E gets closer and closer to p*c. You can think of massless particles like the photon as the limit that m->0 of that process.

[Reality Check]

Individual photons can travel through any slit as long as it is bigger than they are of course. Picture it like water only replace water molecules with photons, when a wave of photons travels through a slit the consequent wave will be changed.



You have misunderstood my fault I expect. Radio waves have the same peak-to-peak amplitude as light waves. The difference between light waves and radio waves is that the individual photons for light are oscillating faster hence more energy and shorter wavelengths.

How about my predictions?
What observation that photons have no mass? We have observed a limit (http://www.aip.org/pnu/2003/split/625-2.html) but that is all.

Here are a couple:

We wouldn't be able to predict exactly where a photon is due to the oscillation until we observed it.
The equations for energy of a photon would include a constant related to the amplitude.

As both of these phenomena are related to the amplitude it should be the same constant that defines both the Uncertainty and the Energy. Planck’s constant.
Can you give us the mathematics behind these predictions so that we can check them?

The first is exactly what we observe and predict using existsing theory.
The second is not observed (the energy of a photon matches the existing theory) but you need to define how amplitude is going to be measured.

The OT has "Picture the photon as a particle with a very small mass that travels in a wave like manner where all the waves have an equal peak-to-peak amplitude".
What is waving and in what? This sounds like aether theory (http://en.wikipedia.org/wiki/Aether_theory).

The nice thing about electromagnetic waves is that they do not need a medium to wave in.

[Ziggurat]

OK so it is a property of a field at a certain point that defines what would happen if a charge was put there is that it?

Yes.

We also have to take into account the charge's properties obviously.

Why? For a given field, we would need to know the charge in order to calculate the force acting on it, but if we're only interested in the field itself, then the details of our hypothetical test charge are irrelevant. Even if we're doing an actual measurement, we should be able to measure the same field (and get the same result) using different test charges.

How do we know a field exists in a vacuum without 'seeing' what it does to a charge?

This is essentially the same question of whether or not the fields are real at all. Suppose we've got a source charge, and we probe its field by seeing what it does to some small test charge. By moving the test charge around, we map out a field. Can you accept that the field from the source charge will remain even if we remove the test charge? If you can, then that's basically your answer: by doing exactly that, we can deduce the equations we need to predict magnetic and electric fields, we can test them with test charges, and if all our tests work out (and they do), then we assume that they are correct even when we're not doing measurements. If you cannot accept that the field remains when the test charge is gone, well, we've hit an impasse.

[Reality Check]

martu: A problem with your picture - it assumes that photons just wander around in space. Photons are also created, e.g. by the transition of electrons between energy levels in an atom. But Special Relativity states that it takes an infinite amount of energy to get a massive particle to the speed of light. So your pictured photons cannot be travelling at the speed of light. Thus your picture is wrong.

[martu]

The m=0 really refers to the first equation not the second equation (which is the momentum for a massive particle, not a massless particle).

Why? Why can we use the first equation for the photon but not the second?

[martu]

Can you give us the mathematics behind these predictions so that we can check them?

The first is exactly what we observe and predict using existsing theory.
The second is not observed (the energy of a photon matches the existing theory) but you need to define how amplitude is going to be measured.

It is an interpretation of what we see currently, all the existing maths works with it. What if the photon amplitude was less than the Planck Length? I know, what ifs...

The OT has "Picture the photon as a particle with a very small mass that travels in a wave like manner where all the waves have an equal peak-to-peak amplitude".
What is waving and in what? This sounds like aether theory (http://en.wikipedia.org/wiki/Aether_theory).

The nice thing about electromagnetic waves is that they do not need a medium to wave in.

Yes I know no medium is required for this. I’ll try an analogy though that might make matters worse…

Imagine a flow of water from one place to another. Now zoom in and picture an individual water molecule in that flow. Now restrict each molecule so that it travels in a straight line in the direction of flow with a small oscillation perpendicular to the direction of travel.
Now consider that water molecule to be a photon and the flow to be a beam of light. The more intense the beam of light the more photons in the flow. The more energy an individual photon has the faster it is oscillating.

[Reality Check]

It is an interpretation of what we see currently, all the existing maths works with it. What if the photon amplitude was less than the Planck Length? I know, what ifs...



Yes I know no medium is required for this. I’ll try an analogy though that might make matters worse…

Imagine a flow of water from one place to another. Now zoom in and picture an individual water molecule in that flow. Now restrict each molecule so that it travels in a straight line in the direction of flow with a small oscillation perpendicular to the direction of travel.
Now consider that water molecule to be a photon and the flow to be a beam of light. The more intense the beam of light the more photons in the flow. The more energy an individual photon has the faster it is oscillating.
Your anology has the water as the medium.

There is no medium needed for an electromagnetic wave (which does not have mass).
There is a meduim neded for your picture of a photon "that travels in a wave like manner" (and infinite energy needed to get it up to the speed of light)

[Reality Check]

Why? Why can we use the first equation for the photon but not the second?
Beacuse m=0 ... and the second equation is for a particle with m not 0.

[The Man]

Just picture a wave of light as many individual photons oscillating perpendicular to the direction of travel. All photon interactions can now be considered collisions between objects.

It would need something to contain that oscillation. In water surface waves it is gravity that brings the water molecules into the trough of the wave and higher pressure in the water that drives the crest up. No such mechanism for positional oscillation exists for photons. Even sound waves which are longitudinal result from an increase and decrease in pressure (also contained by gravity) along the direction of travel, were you to graph the changes in pressure you would get a sine wave representing the pressure variations yet the air does not move up and down like waves on the surface of water but instead move back and forth. Both sound and water waves represent pressure oscillation in a media (water or air), electromagnetic waves were once thought to be similar oscillations in the a ‘luminiferous aether’ but experimentation has demonstrated that this is not the case


Why can’t the state be related to a displacement? Again another pixie question I guess.

It is related to position (or the probability of detecting an electron at some position) but not a displacement like waves on water or even longitudinal wave like sound.



So what changes when a Neutrino changes state?

It is not so much that the Neutrino changes state, but that the traveling neutrino is a superposition of mass and flavor states, it is the probability of which flavor you will most likely detect at any give time and location that changes or oscillates.

[martu]

Your anology has the water as the medium.

There is no medium needed for an electromagnetic wave (which does not have mass).
There is a meduim neded for your picture of a photon "that travels in a wave like manner" (and infinite energy needed to get it up to the speed of light)

OK then photons arte the medium, I thought you meant a medium is required for photons to travel through. A beam of light in a vacuum is a stream of photons travelling as described, what characteristic of light can’t be explained this way?

Which equation are you using for the infinite energy consequence?

[martu]

Beacuse m=0 ... and the second equation is for a particle with m not 0.

Why does it only work if m > 0, that is my question? Is it merely because?

[martu]

Why? For a given field, we would need to know the charge in order to calculate the force acting on it, but if we're only interested in the field itself, then the details of our hypothetical test charge are irrelevant. Even if we're doing an actual measurement, we should be able to measure the same field (and get the same result) using different test charges. .

I only meant we’d have to factor in the value of the charge any speed and what not.

This is essentially the same question of whether or not the fields are real at all. Suppose we've got a source charge, and we probe its field by seeing what it does to some small test charge. By moving the test charge around, we map out a field. Can you accept that the field from the source charge will remain even if we remove the test charge? If you can, then that's basically your answer: by doing exactly that, we can deduce the equations we need to predict magnetic and electric fields, we can test them with test charges, and if all our tests work out (and they do), then we assume that they are correct even when we're not doing measurements. If you cannot accept that the field remains when the test charge is gone, well, we've hit an impasse.

Yes absolutely.

Any reason why these fields can’t be photons?

[Ziggurat]

Any reason why these fields can’t be photons?

The fields are the photons, which is the point. But as I've said before, it's the field transverse to the propagation direction which oscillates. That oscillation of the field direction and amplitude is not a displacement of the field. Which means the photon is not moving side to side.

[Reality Check]

Why does it only work if m > 0, that is my question? Is it merely because?
Because the second equation is the momentum of a particle with mass travelling at less than the speed of light. Put in zero mass and there is no momentum. Put in c and the momentum is undefined.

As sol said:
For a photon, m=0 and you get E = p*c. For a massive particle moving slowly you get E = m*c^2 + p^2/(2m), and p=m*v. For a fast moving massive particle you can see that E is again close to p*c.
Note that the first sentence does not refer to the second equation. The second equation only comes in for massive particles.

[martu]

The fields are the photons, which is the point. But as I've said before, it's the field transverse to the propagation direction which oscillates. That oscillation of the field direction and amplitude is not a displacement of the field. Which means the photon is not moving side to side.

No I am not sure that follows, if the photons were oscillating in a range smaller than the Planck length we couldn’t know they weren’t moving in a side to side direction could we? It could also explain why a charge’s direction is modified by the field, it hits a photon (or many) which influences it.

[martu]

Because the second equation is the momentum of a particle with mass travelling at less than the speed of light. Put in zero mass and there is no momentum. Put in c and the momentum is undefined.

The bold bit is the problem for me, consider a photon travelling through a medium other than a vacuum and it will have a speed smaller than c am I correct? So how can it have momentum, put in zero mass and you get p = 0?

Or have I missed the point completely?

[martu]

It would need something to contain that oscillation. In water surface waves it is gravity that brings the water molecules into the trough of the wave and higher pressure in the water that drives the crest up. No such mechanism for positional oscillation exists for photons. Even sound waves which are longitudinal result from an increase and decrease in pressure (also contained by gravity) along the direction of travel, were you to graph the changes in pressure you would get a sine wave representing the pressure variations yet the air does not move up and down like waves on the surface of water but instead move back and forth. Both sound and water waves represent pressure oscillation in a media (water or air), electromagnetic waves were once thought to be similar oscillations in the a ‘luminiferous aether’ but experimentation has demonstrated that this is not the case

Yes no mechanism for this very true. I still think the picture is consistent assuming an unknown mechanism. I understand this is non science territory.

It is related to position (or the probability of detecting an electron at some position) but not a displacement like waves on water or even longitudinal wave like sound.

How do you know it isn’t a displacement like waves?

As you say it is the probability of finding an electron in a position and, as I understand it, that position has a boundary defined by a constant h. Picture an electron oscillating perpendicular to travel such that the amplitude is constant and is less than the Planck Length. What we have then is an electron where we do not know the position exactly (variable due to oscillation) but it is in a boundary as defined by the amplitude.

It is not so much that the Neutrino changes state, but that the traveling neutrino is a superposition of mass and flavor states, it is the probability of which flavor you will most likely detect at any give time and location that changes or oscillates.

As a neutrino is a massive particle we can use the equations already used (OK mangled by me) in this thread to show momentum is conserved, am I right?

[sol invictus]

Why does it only work if m > 0, that is my question? Is it merely because?

It does work for m=0, it just doesn't tell you anything.

Look - take the equation p = gamma*m*v, where gamma is that inverse square root. Now it is a fact that for any value of m, there is some velocity v which gives you any value of p you want. When m is very small, v must be very close to c. So you can take the limit that m goes to zero, v goes to c, and p is fixed. The equation is valid everywhere in that limit, and the limit is smooth (i.e. there's nothing wrong at the point m=0).

But all that says is that one can have particles with m=0, v=c, and any p.

[sol invictus]

The bold bit is the problem for me, consider a photon travelling through a medium other than a vacuum and it will have a speed smaller than c am I correct? So how can it have momentum, put in zero mass and you get p = 0?

Or have I missed the point completely?

Think of a photon propagating in a medium as being occasionally absorbed and re-emitted by the molecules making up the medium. The photon always travels at c, but each absorption/emission process takes a little bit of time, which slows it down on average.

[Reality Check]

The bold bit is the problem for me, consider a photon travelling through a medium other than a vacuum and it will have a speed smaller than c am I correct? So how can it have momentum, put in zero mass and you get p = 0?

Or have I missed the point completely?
A photon traveling through a medium has a speed that is not c. A photon traveling through vacuum has a speed of c.
The magnitude of the momentum of a photon (http://en.wikipedia.org/wiki/Photon) is
http://upload.wikimedia.org/math/9/f/2/9f2b178529c44cacf86f78115512f9e1.png
This does not depend on any mass.

So you have missed the point entirely :D.

[Ziggurat]

No I am not sure that follows, if the photons were oscillating in a range smaller than the Planck length we couldn’t know they weren’t moving in a side to side direction could we?

There is no model of a photon in which it displaces side to side as it travels through a vacuum. If you want to propose such a model, feel free to do so. But you'll have to explain what force is making it move sideways, and I think you will search in vain for any such force. Hell, you'll have to start out defining what it even means for a photon to move sideways. I cannot categorically rule out such possibilities (in the same sense that I cannot categorically rule out the possibility that gnomes sniff my dirty laundry at night), but given that such a model will solve no outstanding problems and will only add unnecessary complexity, Occam's razor indicates that we should discard it. I'm at a loss as to why you cling to the notion of transverse displacement for photons. Such transverse displacements are unecessary even for massive particles, which also travel as waves.

It could also explain why a charge’s direction is modified by the field, it hits a photon (or many) which influences it.

Uh, no. The charge moves in response to a photon for the exact same reason that it moves in response to a static field: that's what the field does, it produces a force on charges.

[The Man]

Yes no mechanism for this very true. I still think the picture is consistent assuming an unknown mechanism. I understand this is non science territory.

Well I am glad you understand that, but what is the point of introducing “an unknown mechanism” to make something consistent with observations when known mechanisms will suffice?



How do you know it isn’t a displacement like waves?

In a sense we don’t, did you read the link about the path integral ( http://en.wikipedia.org/wiki/Path_integral_formulation)? However the experimental evidence that supports the path integral means that we can not limit the electron to any one path (or history) even one with a wavelike displacement, but must consider all possible paths (or histories). Given the implications of the path integral a wave like displacement path, due to virtual interactions, is certainly a possibility, but it is only the probability of those interactions that need to be considered a not any one path (or history). We can however assert, based on the success of the path integral, that an electron does not have to travel in a path with “displacement like waves”.

If I recall correctly according to Quantum electrodynamics (http://en.wikipedia.org/wiki/Quantum_electrodynamics
) an electron in an atom might be considered to recoil towards the nucleus upon absorbing a virtual photon and away from it upon emitting a virtual photon, a somewhat wavelike motion, but that would be a gross oversimplification. Also it would tend more to a classical interpretation of electrons in an atom (having a well defined position at any given time) and not to the quantum aspects that QED and modern physics is based upon.



As you say it is the probability of finding an electron in a position and, as I understand it, that position has a boundary defined by a constant h. Picture an electron oscillating perpendicular to travel such that the amplitude is constant and is less than the Planck Length. What we have then is an electron where we do not know the position exactly (variable due to oscillation) but it is in a boundary as defined by the amplitude.

Actually the Compton wavelength (http://en.wikipedia.org/wiki/Compton_wavelength) is considered the fundamental limitation on measuring the position of a particle.

Again read the link on the path integral (http://en.wikipedia.org/wiki/Path_integral_formulation) and you might see that the boundary condition you suggest on a traveling electron is not applicable.




As a neutrino is a massive particle we can use the equations already used (OK mangled by me) in this thread to show momentum is conserved, am I right?

Your question was…

2) Neutrinos change flavour as they travel how is momentum conserved?

and momentum is conserved by those combined mass and flavor states. If a theory of Neutrino flavor oscillations was not consistent with the conservation of momentum I doubt it would be taken seriously, without substantial definitive experimental evidence.

[martu]

Thanks everyone who contributed to this thread, I have a lot of reading to do.

To answer one question, (might be able get to go back to the specifics after a bit of reading or more likely I’ll have answered my own questions) I was thinking how unsatisfactory particle wave duality is. It isn’t very elegant is it? Or perhaps I don’t get it yet and at some level, way above me currently, it is.

[Reality Check]

Thanks everyone who contributed to this thread, I have a lot of reading to do.

To answer one question, (might be able get to go back to the specifics after a bit of reading or more likely I’ll have answered my own questions) I was thinking how unsatisfactory particle wave duality is. It isn’t very elegant is it? Or perhaps I don’t get it yet and at some level, way above me currently, it is.
Actually I think of wave/particle duality is one of the most elegant concepts in physics since it is a consequence of the beautiful Schrödinger equation. It also has the advantage of explaining all of the experimental results.

[The Man]

What I have found to be an easier way of considering wave particle duality is somewhat similar to adding a water molecule to a full container. If we consider a container of water full to the brim (such that it can not hold even one more molecule) then drop in an additional molecule it would produce a traveling wave of displacement until at some point along the brim the extra molecule (not necessarily the one that was dropped in) is expelled. We can think of vacuum fluctuations much the same way, balanced particle anti-particle interactions. When we introduce an additional particle or anti-particle that balance is perturbed, that perturbation travels out from the introduction point becoming more distributed. At some point and time distant, the excess that caused the perturbation can be extracted (again not necessarily the exact same particle as was initially introduced). It is a very simplistic view and I do not profess it as an accurate representation (and I may get some static just for brining it up), but I do find it a helpful way of thinking about wave particle duality.

[martu]

Actually I think of wave/particle duality is one of the most elegant concepts in physics since it is a consequence of the beautiful Schrödinger equation. It also has the advantage of explaining all of the experimental results.

I find the maths elegant don't get me wrong, it is the explanation that I find less so. Something is a particle to explain one thing and a wave to explain another?

Hilbert said he was good at maths because he found it so difficult, he was always looking for the simpler proof or explanation. More often than not he found one. Is it likely we could find a simpler explanation that didn't require us to use different pictures?

[martu]

What I have found to be an easier way of considering wave particle duality is somewhat similar to adding a water molecule to a full container. If we consider a container of water full to the brim (such that it can not hold even one more molecule) then drop in an additional molecule it would produce a traveling wave of displacement until at some point along the brim the extra molecule (not necessarily the one that was dropped in) is expelled. We can think of vacuum fluctuations much the same way, balanced particle anti-particle interactions. When we introduce an additional particle or anti-particle that balance is perturbed, that perturbation travels out from the introduction point becoming more distributed. At some point and time distant, the excess that caused the perturbation can be extracted (again not necessarily the exact same particle as was initially introduced). It is a very simplistic view and I do not profess it as an accurate representation (and I may get some static just for brining it up), but I do find it a helpful way of thinking about wave particle duality.

I like it thanks. The big problem of course is that water doesn't interact as individual particles.

[Reality Check]

I find the maths elegant don't get me wrong, it is the explanation that I find less so. Something is a particle to explain one thing and a wave to explain another?

Hilbert said he was good at maths because he found it so difficult, he was always looking for the simpler proof or explanation. More often than not he found one. Is it likely we could find a simpler explanation that didn't require us to use different pictures?
But the mathematics is the explanation!
I think what you mean is that people have not succeeded in explaining wave particle duality to you in an elegant way, perhaps using inelegant analogies or examples.

[blutoski]

But the mathematics is the explanation!
I think what you mean is that people have not succeeded in explaining wave particle duality to you in an elegant way, perhaps using inelegant analogies or examples.

I come across this a lot, unfortunately. Student wants an explanation with existing concepts, when really, relativity and quantum mechanics is quite clearly new concepts.

eg: "Baker to doctor: please explain cardiac bioelectric potentials and EKGs in terms of muffins."

Ultimately, I just have to admit that my inability to fully grasp non-Euclidean geometry a la Riemann tensors may be a personal challenge, but not actually be a deficiency in the general theory of relativity.

[The Man]

I come across this a lot, unfortunately. Student wants an explanation with existing concepts, when really, relativity and quantum mechanics is quite clearly new concepts.

eg: "Baker to doctor: please explain cardiac bioelectric potentials and EKGs in terms of muffins."

Ultimately, I just have to admit that my inability to fully grasp non-Euclidean geometry a la Riemann tensors may be a personal challenge, but not actually be a deficiency in the general theory of relativity.


My personal challenge as well, I would probably find it easier a la Ramen noodles then Riemann tensors.

[sol invictus]

I find the maths elegant don't get me wrong, it is the explanation that I find less so. Something is a particle to explain one thing and a wave to explain another?

Why do you expect the physics of sub-atomic particles to be intuitive to you?

Elegance is far better defined by mathematical simplicity - which quantum mechanics has - than by your sense of what seems right.

[The Man]

Why do you expect the physics of sub-atomic particles to be intuitive to you?

Elegance is far better defined by mathematical simplicity - which quantum mechanics has - than by your sense of what seems right.

Well that is what it comes down to Sol, eating, defecating, urinating and crying (prior to eating or after the other two) are the primary things intuitive to us, that we are born with, all the rest we learn. Once we start depending more on our intuition then improving by learning, we might actually revert back to and limit ourselves to just those base intuitive considerations.