Photon Energy

[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 is

This does not depend on any mass.

So you have missed the point entirely .

 

[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? 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 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 is considered the fundamental limitation on measuring the position of a particle.

Again read the link on the path integral 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.