Incredibly huge inconsistency of modern physics

[wogoga]

Mass-Energy equivalence means that even though a photon is massless it has energy


This could be the most concise formulation of a central contradiction of modern orthodox physics: Equivalence between mass and energy on the one hand, but energy without mass on the other hand. How is it possible that educated persons accept such a huge inconsistency?

See also:

uncertainty principle (http://groups.google.com/group/sci.physics.relativity/msg/08b57c2f826fd3e1)
Mass-energy-equivalence (was: uncertainty principle) (http://groups.google.com/group/sci.physics.relativity/msg/183d17a9597e8d55)
Comments on FAQ - Does light have mass? (post 1 (http://groups.google.com/group/sci.physics.relativity/msg/b364a59c96c03cce), post2 (http://groups.google.com/group/sci.physics.particle/msg/f6cc9251ca380a84))
Plädoyer für die Masse-Energie-Äquivalenz (http://groups.google.com/group/de.sci.physik/browse_frm/thread/e9948d4a21e442b2)
Geschwindigkeitsabhängige Masse tabuisiert: (post 1 (http://groups.google.com/group/de.sci.physik/msg/70713424b42490f0), post 2 (http://groups.google.com/group/de.sci.physik/msg/f1b263419779d751), post 3 (http://groups.google.com/group/de.sci.physik/msg/336ace0301482146))

Cheers, Wolfgang

[Gravy]

You seem to be confusing rest mass with relativistic mass. Thanks for the web posts in German, though. I always enjoy reading that language.

Sincerely,
A semi-educated person

[Henners]

...and also confusing "equivalence" and "equality".

How is it that educatated people...

Oh, never mind.

Life is too short.

[wollery]

No.

[Reality Check]

Incredibly huge inconsistency of your knowledge of modern physics. Photons (http://en.wikipedia.org/wiki/Photon) have an energy that is related to their frequency. They also have zero rest mass and so no rest energy.

Educated people know that there is no inconsistency.

[LordoftheLeftHand]

The idea that a photon has no mass while at rest (even though they apparently aren't ever at rest) but has a mass while moving at c is certainly weird. But that does not mean it isn't true.

LLH

[The Man]

The idea that a photon has no mass while at rest (even though they apparently aren't ever at rest) but has a mass while moving at c is certainly weird. But that does not mean it isn't true.

LLH

E = M * C2, so mass and energy are related by the square of the speed of light. Total energy equals potential energy plus kinetic energy. For an object with rest mass that rest mass is potential energy so EREST = MREST * C2. The kinetic energy of an object is related to its relative velocity but that kinetic energy also represents mass EKINETIC = MKINETIC * C2. This is often referred to as the change in or delta mass. The total mass, often referred to as relativistic mass or MV (mass at velocity), is just the rest mass plus the kinetic (or delta) mass or the total energy (potential plus kinetic) divided by the speed of light squared. As a varying amplitude of electromagnetic fields or a wave packet of electromagnetic radiation a photon has no rest energy, it is purely kinetic and therefore has no rest mass only relativistic mass, so its total (or relativistic) mass equals its kinetic (or delta) mass . What is weird or inconsistent about that?

[Dancing David]

This could be the most concise formulation of a central contradiction of modern semnantic mistakes: Equivalence between fur and dogs on the one hand, but dogs without fur on the other hand. How is it possible that educated persons accept such a huge inconsistency?



Hmmm?

[LordoftheLeftHand]

E = M * C2, so mass and energy are related by the square of the speed of light. Total energy equals potential energy plus kinetic energy. For an object with rest mass that rest mass is potential energy so EREST = MREST * C2. The kinetic energy of an object is related to its relative velocity but that kinetic energy also represents mass EKINETIC = MKINETIC * C2. This is often referred to as the change in or delta mass. The total mass, often referred to as relativistic mass or MV (mass at velocity), is just the rest mass plus the kinetic (or delta) mass or the total energy (potential plus kinetic) divided by the speed of light squared. As a varying amplitude of electromagnetic fields or a wave packet of electromagnetic radiation a photon has no rest energy, it is purely kinetic and therefore has no rest mass only relativistic mass, so its total (or relativistic) mass equals its kinetic (or delta) mass . What is weird or inconsistent about that?

Yeah yeah, i've read all that before. Even though it's true it's still weird. :p

LLH

[Henners]

To be fair, wogoga's "claims" make just as much sense in German.

(I don't read German.)

[elgarak]

E = m c^2
is just a special case that applies only for particles without momentum, or systems without net momentum. Photons are outside of its validity since they're never at rest.

The full equation is:

E^2 - (p c)^2 = (m c^2)^2

If you plug in rest mass m=0, you'll get the correct value of a photon's energy as

E = pc

[Henners]

Doesn't all this equation-writing rather miss the misunderstanding inherent in the claim Equivalence between mass and energy on the one hand, but energy without mass on the other hand....where wogoga clearly indicates that the word "equivalence" is the problem.

English, not Physics.

[fuelair]

Either can be corrected if there is sufficient intelligence and access to data in the correctee.

[Henners]

Somehow I get the impression that wogaga would be happier to stick with the idea that everybody else is stupid.

[fuelair]

Ignorance is bliss......(or is that piss, I always forget that one).

[The Man]

is just a special case that applies only for particles without momentum, or systems without net momentum. Photons are outside of its validity since they're never at rest.

No there is nothing special about E = M*C2 and you demonstrate that.

Sorry but for some reason I could not quote you directly so I had to resort to a cut and paste, loosing the equations and formatting you posted.

Regardless, the final equation of you post is E = pc or the energy of the photon is equal to its momentum times the speed of light. Since the momentum of the photon is p = MV * C or the relativistic mass of the photon times the speed of light. This makes the equation for the energy of a photon as E = MV * C2 just as valid as the form you presented. None of these variations in the form of the equation says anything different or is in any way special.

ETA: You could also write the momentum of the photon as Planck’s constant over the wavelength but that is just another form that is equally valid.

[The Man]

Doesn't all this equation-writing rather miss the misunderstanding inherent in the claim ...where wogoga clearly indicates that the word "equivalence" is the problem.

English, not Physics.


Equivalence between mass and energy on the one hand, but energy without mass on the other hand.

The real misunderstanding I see in the claim is the bit about “energy without mass”, so the problem seems to be with wogoga’s “other hand”.

[JoeEllison]

The irony, as usual, is that someone is arguing against basic principles of physics on the Internet... which they could not do unless the basic principles of physics are exactly what scientists say they are.

[Fredrik]

The real misunderstanding I see in the claim is the bit about “energy without mass”, so the problem seems to be with wogoga’s “other hand”.
Those of us who prefer to let the word "mass" refer to the rest mass of a particle would prefer to say it's the other way round. I'm not saying that you're wrong. It's a matter of taste.

This is how I've been thinking about these things since I read chapter 2 of Weinberg's QFT book: The (four-)momentum operators are defined as the generators of translations in time and space. They commute with each other, so we can find simultaneous eigenstates of all four. Those states are also eigenstates of

\footnotesize P^2=-(P^0)^2+(P^1)^2+(P^2)^2+(P^3)^2=-H^2+\vec P^2

The vectors in a subspace that's invariant under Lorentz transformations represent the possible states of a physical system that consists of exactly one particle. Every momentum eigenstate in such an invariant subspace must have the same eigenvalue of P2, since P2 commutes with all the generators of the Poincaré group. This eigenvalue apparently represents an intrinsic property of the partice. Let's write the eigenvalue as -m2 and call m "mass". This means that

\footnotesize H^2=\vec P^2+m^2

where the last term must be interpreted as m2 times the identity operator. When this operator acts on a momentum eigenstate, the result is

\footnotesize E^2=\vec p^2+m^2

This is in units such that c=1. Restoring factors of c, we get

\footnotesize E^2=\vec p^2 c^2+m^2 c^4

Now what's "relativistic mass"? If we simply write the left-hand side as (something)2c4, then the "something" must have dimensions of mass. Let's call the magnitude of this something "relativistic mass".

When you look at it from this point of view, the concept of relativistic mass seems pretty pointless and irrelevant.

[hgc]

What is "orthodox" physics?

[fuelair]

What is "orthodox" physics?

Higher than conservative physics, more higher even than reform physics.:D

[a_unique_person]

Somehow I get the impression that wogaga would be happier to stick with the idea that everybody else is stupid.

;)

[zosima]

I'll admit my original statement was poorly phrased and I should have said 'rest mass', I apologize if my statement was confusing. Although, I don't think it was particularly problematic if taken in the context of the discussion.

[Beausoleil]

If you replace mass with rest mass, where is the inconsistency? A photon has mass, energy and momentum, but not rest mass. What's the problem?

[Reality Check]

If you replace mass with rest mass, where is the inconsistency? A photon has mass, energy and momentum, but not rest mass. What's the problem?
A photon (http://en.wikipedia.org/wiki/Photon#Physical_properties) has no rest mass or mass. It has energy and momentum. These can be considered to as "relativistic mass".

[a_unique_person]

A photon (http://en.wikipedia.org/wiki/Photon#Physical_properties) has no rest mass or mass. It has energy and momentum. These can be considered to as "relativistic mass".

I love this stuff, even if it is hard to wrap my head around it. :boxedin:

Isn't momentum mass * velocity? (OK, I'm going all the way back to my high school physics). I get the feeling from what has been said in other posts, my high school definition is lacking, but the definition in high school was p=mv.

[Reality Check]

I love this stuff, even if it is hard to wrap my head around it. :boxedin:

Isn't momentum mass * velocity? (OK, I'm going all the way back to my high school physics). I get the feeling from what has been said in other posts, my high school definition is lacking, but the definition in high school was p=mv.
You may want to read the link (http://en.wikipedia.org/wiki/Photon#Physical_properties) which explains all of this. p=mv is only for particles with mass and as the article states:

In empty space, the photon moves at c (the speed of light) and its energy E and momentum p are related by E = cp, where p is the magnitude of the momentum.

[Dancing David]

I think that it bother some people that a photon can have mass because it moves at the speed of light but no rest mass.

They want the universe to agree with their preconceived notions.

[a_unique_person]

I think that it bother some people that a photon can have mass because it moves at the speed of light but no rest mass.

They want the universe to agree with their preconceived notions.

Nothing bothers me, except my inability to understand it all.

[Fredrik]

Isn't momentum mass * velocity? (OK, I'm going all the way back to my high school physics). I get the feeling from what has been said in other posts, my high school definition is lacking, but the definition in high school was p=mv.
That's the pre-relativistic definition of the momentum of a massive particle. Special relativity changes that definition to

\footnotesize $\vec{p}=\gamma m \vec v$

where

\footnotesize \gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}

Note that we recover the pre-relativistic definition in the limit \footnotesize c\rightarrow\infty.

The m above is the particle's mass. Some people prefer to call it the rest mass. They would call the quantity \footnotesize\gamma m the mass (or relativistic mass). They would also write m0 instead of m, and they would write m instead of \footnotesize\gamma m.

[The Man]

Those of us who prefer to let the word "mass" refer to the rest mass of a particle would prefer to say it's the other way round. I'm not saying that you're wrong. It's a matter of taste.



Oh, I agree completely and that was primarily my point, the matter of taste. wogoga taking the mass energy equivalence on one hand and then only considering invariant mass on the other. Kind of like playing classical music and rap at the same time and then wondering why it sounds so bad. Mass is simply a value having the units of (or equivalent) Newton Second2 Meter-1. As far as tastes go my taste varies depending on what I am tying to find out. However, although I agree that relativistic mass can seem irrelevant, as long as the math works I dislike discarding anything as irrelevant since you could miss an important relationship, but that is just my taste. It is in fact that aspect which can also show us a photon has a frame dependent mass. With wavelength (and thus momentum) dependent on relative motion between the receiver and the light source and with the speed of light frame invariant then with the p = m * V relationship (which may be technically incorrect for a photon but the math still works) we know that the mass (or that Newton Second2 Meter-1 something) of the photon must also be frame dependent.

[a_unique_person]

That's the pre-relativistic definition of the momentum of a massive particle. Special relativity changes that definition to
http://www.randi.org/latexrender/latex.php?%5Cfootnotesize%20$%5Cvec%7Bp%7D=%5Cgamm a%20m%20%5Cvec%20v$where
http://www.randi.org/latexrender/latex.php?%5Cfootnotesize%20%5Cgamma=%5Cfrac%7B1%7 D%7B%5Csqrt%7B1-%5Cfrac%7Bv%5E2%7D%7Bc%5E2%7D%7D%7DNote that we recover the pre-relativistic definition in the limit http://www.randi.org/latexrender/latex.php?%5Cfootnotesize%20c%5Crightarrow%5Cinfty .

The m above is the particle's mass. Some people prefer to call it the rest mass. They would call the quantity http://www.randi.org/latexrender/latex.php?%5Cfootnotesize%5Cgamma%20m the mass (or relativistic mass). They would also write m0 instead of m, and they would write m instead of http://www.randi.org/latexrender/latex.php?%5Cfootnotesize%5Cgamma%20m.

So what happens when v=c?

[Reality Check]

So what happens when v=c?
It never does for a particle with mass.

[a_unique_person]

It never does for a particle with mass.

I was thinking of the photon, but I didn't realise that the equation was only for particles with a resting mass.

[Dave Rogers]

So what happens when v=c?

p becomes infinite. Since this means that an infinite impulse would be required to impart sufficient momentum for a particle to travel at c, this shows that a particle of non-zero rest mass cannot travel at the speed of light.

Just expanding a little on Reality Check's answer.

Dave

[wogoga]

I wrote:

Equivalence between mass and energy on the one hand, but energy without mass on the other hand.

Dancing David brought forward this analogy:

Equivalence between fur and dogs on the one hand, but dogs without fur on the other hand.

If we take seriously the equivalence between fur and dogs we conclude:


a fur without corresponding dog or a dog without corresponding fur is impossible
a fur corresponds nevertheless to a dog without a fur
fur and fur are two different concepts



I'll admit my original statement was poorly phrased and I should have said 'rest mass', ...


Zosima's original statement:

Mass-Energy equivalence means that even though a photon is massless it has energy

If we replace mass with restmass we get:

Restmass-Energy equivalence means that even though a photon has no restmass it has energy

This version is obviously wrong, because there is no equivalence between restmass and energy. Equivalence only reigns between (relativistic) mass and energy. The alternative

Equivalence between (relativistic) mass and energy means that even though a photon has no restmass it has energy

doesn't make a lot of sense, because of the confusion between two different mass-concepts (mass and restmass).


Cheers, Wolfgang

Comments on FAQ - Does light have mass? (post 1 (http://groups.google.com/group/sci.physics.relativity/msg/b364a59c96c03cce), post 2 (http://groups.google.com/group/sci.physics.particle/msg/f6cc9251ca380a84), post 3 (http://groups.google.com/group/sci.physics.relativity/msg/a98c0f66353cff05))

[Reality Check]

I wrote:
Equivalence between mass and energy on the one hand, but energy without mass on the other hand.
..snip..
Zosima's original statement:
Mass-Energy equivalence means that even though a photon is massless it has energy
If we replace mass with restmass we get:
Restmass-Energy equivalence means that even though a photon has no restmass it has energy
This version is obviously wrong, because there is no equivalence between restmass and energy. Equivalence only reigns between (relativistic) mass and energy. The alternative
Equivalence between (relativistic) mass and energy means that even though a photon has no restmass it has energy
doesn't make a lot of sense, because of the confusion between two different mass-concepts (mass and restmass).


Cheers, Wolfgang

Comments on FAQ - Does light have mass? (post 1 (http://groups.google.com/group/sci.physics.relativity/msg/b364a59c96c03cce), post 2 (http://groups.google.com/group/sci.physics.particle/msg/f6cc9251ca380a84), post 3 (http://groups.google.com/group/sci.physics.relativity/msg/a98c0f66353cff05))

There is no such thing as "restmass". A particle with mass has a "rest mass". A particle with mass that is moving has a relativistic mass that is its rest mass multiplied by the Lorentz factor.

The original posting by zosima was "Mass-Energy equivalence means that even though a photon is massless it has energy" and that is inexact since mass-energy equivalence is usually taken to be E=mc2 and that is the rest energy of a particle. A photon is never at rest.

There is no confusion between mass and rest mass, just in the expected conventions of their their use.
A photon (http://en.wikipedia.org/wiki/Photon#Physical_properties) has energy. We know that through their interactions with matter, especially the photoelectric effect (http://en.wikipedia.org/wiki/Photoelectric_effect).

[Gravy]

I wrote:Equivalence between mass and energy on the one hand, but energy without mass on the other hand.Dancing David brought forward this analogy:Equivalence between fur and dogs on the one hand, but dogs without fur on the other hand.If we take seriously the equivalence between fur and dogs we conclude:
a fur without corresponding dog or a dog without corresponding fur is impossible
a fur corresponds nevertheless to a dog without a fur
fur and fur are two different concepts



Zosima's original statement:Mass-Energy equivalence means that even though a photon is massless it has energyIf we replace mass with restmass we get:Restmass-Energy equivalence means that even though a photon has no restmass it has energyThis version is obviously wrong, because there is no equivalence between restmass and energy. Equivalence only reigns between (relativistic) mass and energy. The alternativeEquivalence between (relativistic) mass and energy means that even though a photon has no restmass it has energydoesn't make a lot of sense, because of the confusion between two different mass-concepts (mass and restmass).


Cheers, Wolfgang

Comments on FAQ - Does light have mass? (post 1 (http://groups.google.com/group/sci.physics.relativity/msg/b364a59c96c03cce), post 2 (http://groups.google.com/group/sci.physics.particle/msg/f6cc9251ca380a84), post 3 (http://groups.google.com/group/sci.physics.relativity/msg/a98c0f66353cff05))Ah, now you make complete sense! Have you considered writing a book in which you use color to elucidate the principles of physics, like Oliver Byrne did with Euclidean geometry?

[Almo]

Educated people know that there is no inconsistency.

I have to say this isn't quite true. People educated in physics know there's no inconsistency. But I would say that at least many people educated in other fields would trust the physics dudes when they say there is none.

[The Man]

I think some of the problem and weirdness wogoga is experiencing comes from the applied units vs the actual meaning of what those units represent. Let’s take Energy for example measured in units of Newton Meters, or Joules as such it is a force multiplied by the distance that force is applied through. Torque also takes the same form of units Newton Meters but has a different meaning, in this case it is a force multiplied by the distance from the rotational center (or radius) that the force is applied. Same units different meanings, to convert torque to energy you would need the rotational angle during the force application in radians and multiply it by the radius to get the cord subtended by that angle or in other words the actual distance the force was applied then multiply that times the force.

A similar thing is occurring here, Mass is measured in units of Kilograms or Newton Second2 Meter-1, some values can take that form of units yet may not adhere to the specific aspects that generate those units and thus not technically be mass. The property of mass that defines these units is Newton’s second law of motion F = m*a or Force equals mass times acceleration. We can convert this to represent mass as m = F / a or mass equals Force divided by acceleration. With the units of mass as Kilograms, force as Newtons, and the inverse of acceleration as Second2 Meter-1 the result is Kg = Newton Second2 Meter-1. More importantly though, is the meaning of the units or Force per acceleration. As a photon does not have the capability of acceleration (changing velocity) this specific aspect of mass does not apply. The energy of the photon can still be represented in this form of units that we use for mass (since it is just a frame invariant conversion) and all the normal equations will work out mathematically, but since the photon lacks the ability to accelerate the true meaning of the units is not applicable. With the units of acceleration as Meter Second-1 the “acceleration” of a photon becomes the speed of light times the frequency of the photon. So what we might represent as the units of “acceleration” for a photon is simply the Doppler shift of frequency due to relative motion and not the result of any real force applied to the photon.

[ponderingturtle]

There is no such thing as "restmass". A particle with mass has a "rest mass". A particle with mass that is moving has a relativistic mass that is its rest mass multiplied by the Lorentz factor.


This statement is dependant ont he definition of mass being rest mass and not relativistic mass.

Yes that is what is currently popular, but really as long as you are being clear as to what you define mass as being, either can be viewed as equaly true, unless you have some experiment to show that reletivistic mass doesn't exist(seems unlikely given that they are just different formulations of the same equations and as such you are not going to get a difference in predictions)

[Fredrik]

Zosima's original statement:

Mass-Energy equivalence means that even though a photon is massless it has energy

If we replace mass with restmass we get:

Restmass-Energy equivalence means that even though a photon has no restmass it has energy

This version is obviously wrong, because there is no equivalence between restmass and energy. Equivalence only reigns between (relativistic) mass and energy. The alternative

Equivalence between (relativistic) mass and energy means that even though a photon has no restmass it has energy

doesn't make a lot of sense, because of the confusion between two different mass-concepts (mass and restmass).

I agree with most of what you're saying about Zosima's statement, but it's not an inconsistency of modern physics. He just chose his words poorly. He admitted that himself. I don't see what the big deal is.

All particles satisfy \footnotesize $E^2=\vec p^2c^2$+m^2c^4. This means that massive particles have energy even in a coordinate system where their velocity is zero, and that this energy comes from the particle's mass.

When m>0, we can express the right-hand side as \footnotesize (\gamma m)^2c^4. The quantity in parentheses is often called the "relativistic mass". So for massive particles, the equation expresses an equivalence between "relativistic mass" and energy.

I don't have a problem with that. What I don't like is when the equation is interpreted as an equivalence between relativistic mass and energy for all particles. The only way to interpret it that way is to define the relativistic mass m (using your trick with the colors) of a particle by setting the left-hand side equal to m2c4. This makes the equivalence between relativistic mass and energy true by definition, so it doesn't really say anything about the physics.

As you can see, there are several things that can be described as an equivalence between mass and energy: The equivalence between mass and energy in the rest frame. The equivalence between relativistic mass and energy, for massive particles. The equivalence between relativistic mass and energy, for massless particles. The first two describe properties of our universe. The last one doesn't really say anything.

Note also that by setting m=0, we see that the equation doesn't prevent massless particles from having energy or momentum.

[The Man]

Try this equation P2 = EKINETIC*(Mrel + Mrest) an algebraic extension of the one equation that I have mostly been confronted with.

Sorry I do not know how to use the formatting to make it look significant.

I would appreciate any input on the formatting or the physics.

[zosima]

Jesus Christ. My original comment was a throw-away comment about the flaws in a model that was completely awful. Mainly to say that it did not account for the behavior of objects with respect to relativity. It was terribly imprecise, but despite what Wogoga may claim, I am in no way a spokes-person for the theory of general relativity. My incompetence at explaining physics in no way impeaches the theory of general relativity, nor does Wogoga's capacity to misunderstand and/or misconstrue my words.

If I were going to phrase my statement precisely. I would probably say:
"Even though a photon has no rest mass it does contribute to the stress-energy tensor of general relativity and thus can be expected to exert a tiny gravitational force on the objects around it."

This is true. But honestly the mathematics is complex enough that any attempt to explain it is going to involve a certain amount of compromise. If you are truly interested I would recommend you order one of the many books available on the topic from amazon.

So this property of photons is a consequence of the theory of general relativity, a theory which has been far better tested than the soul juice theory of reincarnation.(Which you have no trouble believing, Wogoga)

[The Man]

Recycling is a bitch, what you throw away might be used to build someone’s house. A very crappy house, but some people aren’t that concerned about where they live.

[Fredrik]

Try this equation P2 = EKINETIC*(Mrel + Mrest) an algebraic extension of the one equation that I have mostly been confronted with.

Sorry I do not know how to use the formatting to make it look significant.

I would appreciate any input on the formatting or the physics.
Your equation says that the kinetic energy is

$\frac{\vec p^2}{(\gamma+1)m}$

If my algebra is correct (I did it very fast without double-checking anything) this can be simplified to \footnotesize\gamma m-m. So your equation expresses the same thing as \footnotesize$E^2=\vec p^2+m^2.

(I'm using units such that c=1).

[wogoga]

... soul juice ...


Soul juice

is a contradictio in terminis (http://en.wikipedia.org/wiki/Contradiction_in_terms) in the same way as

the quadrangle-area of a triangle (i.e. the area of a triangle with four angles and four straight sides)
the rest-mass of a photon (i.e. the mass of photon at rest).

Juice is a paragon (an ideal incarnation) of divisibility or continuous changeability in quantity. Soul on the contrary is an embodiment of indivisibility and discreteness (see also: The relationship between quanta and psychons (http://forums.randi.org/showpost.php?p=3458950&postcount=51)).

Interestingly, according to QED, photons in a transparent medium propagate always at c, and are regularly absorbed and reemitted so that a propagation speed of c/n is the result (n is the refractive index (http://en.wikipedia.org/wiki/Refractive_index)). There has been an interesting discussion on this aspect of QED on sci.physics.research: The speed of a photon (http://groups.google.com/group/sci.physics.research/browse_frm/thread/d2a8cdb76df6f383). On my opinion see: Was QED intended as a joke? (http://groups.google.com/group/talk.origins/msg/d70e88ad545df532)

Cheers, Wolfgang

A modern ('skeptic') variant of dogmatic belief: "the basic principles of physics are exactly what scientists say they are"

[Tubbythin]

Yup. The most precisely tested and verified theory in the history of physics is just a big joke.:rolleyes:
So could you explain to me how Huygens explains, I dunno, Bose-Einstein condensate?

[Reality Check]

...snipped woo...
Interestingly, according to QED, photons in a transparent medium propagate always at c, and are regularly absorbed and reemitted so that a propagation speed of c/n is the result (n is the refractive index (http://en.wikipedia.org/wiki/Refractive_index)). There has been an interesting discussion on this aspect of QED on sci.physics.research: The speed of a photon (http://groups.google.com/group/sci.physics.research/browse_frm/thread/d2a8cdb76df6f383). On my opinion see: Was QED intended as a joke? (http://groups.google.com/group/talk.origins/msg/d70e88ad545df532)
Cheers, Wolfgang


You have it wrong. You do not really need QED to understand this.
Special Relativity that states that photons always travel at the speed of light in a vacuum. Matter (your "transparent medium") is mostly vacuum and photons travel at c in that vacuum. But photons are absorbed and emitted by atoms. This takes time and slows the measured speed of the photons.

P.S. "reemitted" suggests that the same photon that was absorbed is emitted. This is not correct. It is a different photon or even multiple photons, e.g. an electron absorbs a photon with a certain energy and ends up in an energy state that can decay via intermediate energy states back to the original energy state. Each decay emits a photon.

Quantum electrodynamics (http://en.wikipedia.org/wiki/Quantum_electrodynamics) is definitely not a joke. It is the among the most precisely experimentally verified scientific theories (http://en.wikipedia.org/wiki/Precision_tests_of_QED) in physics.



Your posted opinion is nonsense. A couple of points shows this:
QED does not say that photons always travel at c. Special Relativity states that photons that are not interacting with anything (in a vacuum) always travel at c. When the absorption and emission by atoms is added then the effective speed of photons is changed . This is where QED comes in.
The path-integral method of QED is not essentially the same as Huygens principle. The path-integral method integrates over all possible paths and is a quantum theory. Huygens principle adds up the waves from each point of an advancing wave front and is a classical theory.

[The Man]

Your equation says that the kinetic energy is

$\frac{\vec p^2}{(\gamma+1)m}$

If my algebra is correct (I did it very fast without double-checking anything) this can be simplified to \footnotesize\gamma m-m. So your equation expresses the same thing as \footnotesize$E^2=\vec p^2+m^2.

(I'm using units such that c=1).


Spot on, Fredrik, on all points, as usual, and your algebra is non the worse for ware, but what this form of the equation can do is give us a way to explore the differences and similarities of relativistic and rest mass and perhaps some of the physics involved. For a photon, rest mass equals zero, so the proportion of momentum squared over kinetic energy equals its relativistic mass. Both momentum and kinetic energy are well understood, momentum a force times the time that force is applied and kinetic energy a force times the distance that forces is applied. Now the only time and distance that we can ascribe to a photon are its cyclic period and its wavelength. The resulting force that you can calculate from this relationship does not represent any real force applied to a traveling photon ( as I mentioned before) but does represent the maximum force that photon can apply to anything it interacts with, within the constraints of its cyclic period and its wavelength.

Although we can certainly conduct experiments that demonstrate a lack of rest mass (as in the case of the photon) we can conduct no experiments that demonstrate a lack of relativistic mass since even in a co-moving reference frame relativistic mass is just equal to rest mass. If we want to consider a reference frame co-moving with a photon we then loose all ability to perform any experiments.

If we consider an electron having rest mass but no discernable spatial extent at rest (as far as we can currently detect) other then its Compton wavelength, the same considerations apply. If we consider a proton or neutron with rest mass and a discernable spatial extent at rest (other then their Compton wavelengths) again the same consideration applies to those Compton spatial and temporal extents. Although we do now have something different in that discernable rest spatial extent (other then Compton wavelength) and its relationship to relativistic mass in a co-moving reference frame. Let’s remember that a co-moving reference frame is still a relativistic frame, just one with 0 relative velocity. That difference, in the non-Compton spatial extent, also provides us with a density related to the stress energy tensor from which we can calculate the frame invariant curvature of spacetime. A critical difference, no doubt, but in this aspect the force that we might calculate form that non Compton (or perhaps Compton as it is also frame invariant) spatial extent is actually (or apparently) applied to the curving of spacetime and/or the bodies within that curved spacetime, much like the photonic calculated force can be applied, in interactions only, and in this case an (considered resting) interaction with spacetime. From this consideration, it seems there are more similarities between rest mass and relativistic mass then there are differences.

[wogoga]

...


You do not really need QED to understand this. ... QED does not say that photons always travel at c. Special Relativity states that photons that are not interacting with anything (in a vacuum) always travel at c. When the absorption and emission by atoms is added then the effective speed of photons is changed. This is where QED comes in.

The attempt to attribute the explanation of the photon speed in transparent materials (http://en.wikipedia.org/wiki/Speed_of_light_in_vacuum#Interaction_with_transpar ent_materials) by absorption and emission to Special Relativity must be considered deliberate disinformation. Special Relativity is in this respect based on Maxwell's theory and Maxwell's theory explains different propagation speeds of electromagnetic waves by different permittivity and permeability values.

By claiming that "the light has an amplitude to go faster or slower than the speed c, but these amplitudes cancel each other out over long distances" (QED, p89-90), Richard Feynman somehow repeated the error of Niels Bohr (BKS theory (http://en.wikipedia.org/wiki/BKS_theory)), who had opposed the assumption of orderliness in emission and absorption processes of electromagnetic wave quanta by denying conservation of momentum and energy in single events. Both Bohr and Feynman were scientists in the theological tradition, preferring impressive counter-intuitive explanations to unprejudiced realistic ones.

P.S. "reemitted" suggests that the same photon that was absorbed is emitted. This is not correct.

Whether we consider the reemitted photon the same or a different photon, is irrelevant. Relevant however is that the reemitted photon must have exactly the same energy and momentum as the absorbed photon.

It is a different photon or even multiple photons, e.g. an electron absorbs a photon with a certain energy and ends up in an energy state that can decay via intermediate energy states back to the original energy state. Each decay emits a photon.

You confuse the extremely regular behaviour of light in optical lenses with scattering of photons e.g. in the atmosphere (http://math.ucr.edu/home/baez/physics/General/BlueSky/blue_sky.html), where momentum and energy of photons actually are changed.

If continuous absorption and reemission were a genuinely physical explanation of the reduced speed in transparent media, then it would be possible to answer such questions as: How often (on average) is a 450-nm-photon absorbed and reemitted when traversing 1 cm of diamond?

By the way, in principle it is possible to label each atom of a given diamond with a unique number. If photons traversing this diamond actually were more than half of the time (n_diamond > 2) at rest (i.e. transformed into energy and momentum of carbon atoms), then it would in principle be possible to determine the path of a photon by enumerating the label-numbers of the atoms where the photon rested.

Yet on the other hand, QED (http://en.wikipedia.org/wiki/Quantum_electrodynamics) makes the absurd claim (http://groups.google.com/group/sci.physics.relativity/msg/767bad1256aea053) that photons somehow use all possible paths.

Cheers, Wolfgang

Science of yesterday degenerates into religion of today and superstition of tomorrow

[Reality Check]

You do not really need QED to understand this. ... QED does not say that photons always travel at c. Special Relativity states that photons that are not interacting with anything (in a vacuum) always travel at c. When the absorption and emission by atoms is added then the effective speed of photons is changed. This is where QED comes in.
The attempt to attribute the explanation of the photon speed in transparent materials (http://en.wikipedia.org/wiki/Speed_of_light_in_vacuum#Interaction_with_transpar ent_materials) by absorption and emission to Special Relativity must be considered deliberate disinformation. Special Relativity is in this respect based on Maxwell's theory and Maxwell's theory explains different propagation speeds of electromagnetic waves by different permittivity and permeability values.


You really do not understand Special Relativity or QED. SR is not based on Maxwell's equations which are for electromagnetic fields. SR applies to all things in motion and not only light.


By claiming that "the light has an amplitude to go faster or slower than the speed c, but these amplitudes cancel each other out over long distances" (QED, p89-90), Richard Feynman somehow repeated the error of Niels Bohr (BKS theory (http://en.wikipedia.org/wiki/BKS_theory)), who had opposed the assumption of orderliness in emission and absorption processes of electromagnetic wave quanta by denying conservation of momentum and energy in single events. Both Bohr and Feynman were scientists in the theological tradition, preferring impressive counter-intuitive explanations to unprejudiced realistic ones.


Both Bohr and Feynman were scientists in the scientific sense - proposing theories on the evidence, making falsifiable predictions and only accepting their theories if the predictions were tested and found true.

I do not have the book so the quote may be correct and in context. But in QED it is the phases that cancel out or reinforce not the amplitudes.


P.S. "reemitted" suggests that the same photon that was absorbed is emitted. This is not correct.
Whether we consider the reemitted photon the same or a different photon, is irrelevant. Relevant however is that the reemitted photon must have exactly the same energy and momentum as the absorbed photon.
It is a different photon or even multiple photons, e.g. an electron absorbs a photon with a certain energy and ends up in an energy state that can decay via intermediate energy states back to the original energy state. Each decay emits a photon.
You confuse the extremely regular behaviour of light in optical lenses with scattering of photons e.g. in the atmosphere (http://math.ucr.edu/home/baez/physics/General/BlueSky/blue_sky.html), where momentum and energy of photons actually are changed.

The momentum and energy of photons is not changed when multiple photons are emitted.

If continuous absorption and reemission were a genuinely physical explanation of the reduced speed in transparent media, then it would be possible to answer such questions as: How often (on average) is a 450-nm-photon absorbed and reemitted when traversing 1 cm of diamond?

By the way, in principle it is possible to label each atom of a given diamond with a unique number. If photons traversing this diamond actually were more than half of the time (n_diamond > 2) at rest (i.e. transformed into energy and momentum of carbon atoms), then it would in principle be possible to determine the path of a photon by enumerating the label-numbers of the atoms where the photon rested.

Yet on the other hand, QED (http://en.wikipedia.org/wiki/Quantum_electrodynamics) makes the absurd claim (http://groups.google.com/group/sci.physics.relativity/msg/767bad1256aea053) that photons somehow use all possible paths.

Cheers, Wolfgang

Science of yesterday degenerates into religion of today and superstition of tomorrow[/quote]

You are correct about the labelling. In classical mechanics it is possible to work out the path.

QED makes the valid claim that photons use all possible paths. For some strange reason this is verified by experiments. Look up quantum entanglement some time. If you really want to blow your mind then look up the Elitzur-Vaidman bomb-tester (http://en.wikipedia.org/wiki/Elitzur-Vaidman_bomb-tester) and its experimental verification: Interaction-free Measurement (http://prola.aps.org/abstract/PRL/v74/i24/p4763_1) (PDF (http://www.quantum.univie.ac.at/publications/pdffiles/1994-08.pdf)). Note that this experiment was done in 1994 so there are probably more recent ones (I will leave finding them as an easy exercise for you).

[The Man]

Both Bohr and Feynman were scientists in the theological tradition, preferring impressive counter-intuitive explanations to unprejudiced realistic ones.


Ah yes the old Science is a theology claim. Do you know why these “counter-intuitive explanations” are so impressive? Because they work, please provide any “unprejudiced realistic ones” that work better, along with the supporting experimental evidence for them.

[wogoga]

"The situation is especially strange as the path-integral method of QED is essentially the same as Huygens principle. So in this respect, QED is not even self-consistent!" (See (http://groups.google.com/group/talk.origins/msg/d70e88ad545df532))

The path-integral method of QED is not essentially the same as Huygens principle. The path-integral method integrates over all possible paths and is a quantum theory. Huygens principle adds up the waves from each point of an advancing wave front and is a classical theory.


A quote from Wikipedia (http://en.wikipedia.org/wiki/Huygens_principle):

Huygens principle follows formally from the fundamental postulate of quantum electrodynamics – that wavefunctions of every object propagate over any and all allowed (unobstructed) paths from the source to the given point. It is then the result of interference (addition) of all path integrals that defines the amplitude and phase of the wavefunction of the object at this given point, and thus defines the probability of finding the object (say, a photon) at this point.

We could also invert the order and say that the path-integral method of QED follows from Huygens' principle by formalization. Even the deficiencies are the same. Two quotes from mathpages.com (http://www.mathpages.com/home/kmath242/kmath242.htm):

From this simple principle Huygens was able to derive the laws of reflection and refraction, but the principle is deficient in that it fails to account for the directionality of the wave propagation in time, i.e., it doesn't explain why the wave front at time t + Dt in the above figure is the upper rather than the lower envelope of the secondary wavelets. Why does an expanding spherical wave continue to expand outward from its source, rather than re-converging inward back toward the source?

In any case, the Huygens-Fresnel Principle has been very useful and influential in the field of optics, although there is a wide range of opinion as to its scientific merit. Many people regard it as a truly inspired insight, and a fore-runner of modern quantum electro-dynamics, whereas others dismiss it as nothing more than a naive guess that sometimes happens to work. For example, Melvin Schwartz wrote that to consider each point on a wavefront as a new source of radiation, and to add the radiation from all the new sources together, "makes no sense at all", since (he argues) "light does not emit light; only accelerating charges emit light". He concludes that Huygens' principle "actually does give the right answer" but "for the wrong reasons". However, Schwartz was expressing the classical (i.e., late 19th century) view of electromagnetism. The propagation of light in quantum field theory actually is consistent with the very interpretation of Huygens' principle that Schwartz regarded as nonsense.

A third quote from the same page:

It could be argued that the "path integral" approach to quantum field theory – according to which every trajectory through every point in space is treated equivalently as part of a possible path of the system – is an expression of Huygens' Principle.

Cheers, Wolfgang

[Ziggurat]

By the way, in principle it is possible to label each atom of a given diamond with a unique number.

No. It is in principle possible to label each carbon nuclei in a sample of diamond. It is most definitely NOT possible, even in principle, to label each electron in a sample of diamond. And electrons, not nuclei, dominate the interaction of light with matter. So you're simply wrong.

[ponderingturtle]

You have it wrong. You do not really need QED to understand this.
Special Relativity that states that photons always travel at the speed of light in a vacuum. Matter (your "transparent medium") is mostly vacuum and photons travel at c in that vacuum. But photons are absorbed and emitted by atoms. This takes time and slows the measured speed of the photons.


In SR why view a photon as a particle and not an electromagnetic wave?

Electrodynamics already take all the effects of SR into acount anyway.

[Reality Check]

In SR why view a photon as a particle and not an electromagnetic wave?

Electrodynamics already take all the effects of SR into acount anyway.Because the subject of the threrad is photons.

[godless dave]

So, wogoga, where's the incredibly huge inconsistency?

[ponderingturtle]

Because the subject of the threrad is photons.

But photons are not an SR object, but a QM object. Their SR counterpart is EM waves.

And EM is already in compliance with SR.

[Reality Check]

But photons are not an SR object, but a QM object. Their SR counterpart is EM waves.

And EM is already in compliance with SR.
Ask the original poster.

But IMHO wave/particle duality means that it does not matter whether you treat photons as particles or waves. For example EM waves in media also do not travel at the speed of light (interference from scattering means that their effective speed changes).

[The Man]

So, wogoga, where's the incredibly huge inconsistency?


It would seem the inconstancy may be that he does not understand wave particle duality. Huygens principle is about wave theory only and therefore lacks the quantum elements so essential to modern physics and our modern technology. QED is a quantum theory and by that nature includes both wave and particle aspects. Regardless of their similarities (in wave aspects) Huygens principle is simply a principle about wave fronts and not a description of electromagnetic interactions (incomplete or otherwise). QED on the other hand is a complete and very precisely experimentally verified description of electromagnetic interactions.

[ponderingturtle]

Ask the original poster.

But IMHO wave/particle duality means that it does not matter whether you treat photons as particles or waves. For example EM waves in media also do not travel at the speed of light (interference from scattering means that their effective speed changes).

I know, the point is that if you are really talking about special relativity then you are looking at a more classical theory

[The Man]

I know, the point is that if you are really talking about special relativity then you are looking at a more classical theory


No, please remember that it was Einstein in one of the papers preceding Special Relativity that confirmed the quantum nature of light (photo-electric effect). Special Relativity is certainly not a classical theory (although classical theories of relative motion had been proposed before that).

[a_unique_person]

It would seem the inconstancy may be that he does not understand wave particle duality.

It is amazing how often that attitude is mistaken for a 'skeptical' stance on an issue. "I can't understand it, therefore you can't prove it".

[Skwinty]

Photons have mass!!
I didn't even know they were Catholic.
"Woody Allen"

[ponderingturtle]

No, please remember that it was Einstein in one of the papers preceding Special Relativity that confirmed the quantum nature of light (photo-electric effect). Special Relativity is certainly not a classical theory (although classical theories of relative motion had been proposed before that).

I know that, it is also what won him the nobel prize. But what everyone is talking about here is from "On the Elctrodynamics of Moving Bodies".

E&M already fits perfectly with relativity and describes what happens to light perfectly.

The problem is that someone introduced the unnessacary concept of relativistic mass that it confuses people who don't really understand that it has fallen out of favor with physicists in general and so talking about massless particles is perfectly acceptable.

[sol invictus]

Special Relativity is certainly not a classical theory (although classical theories of relative motion had been proposed before that).

That's not really correct, given the way these terms are used in modern physics. By a classical theory we mean one in which Planck's constant h is zero. Many classical theories - EM is a good example - are relativistic.

The point really is that SR is not exactly a theory - it's a set of rules which a theory can either be consistent or inconsistent with.

[Viper Daimao]

If you really want to blow your mind then look up the Elitzur-Vaidman bomb-tester (http://en.wikipedia.org/wiki/Elitzur-Vaidman_bomb-tester) and its experimental verification: Interaction-free Measurement (http://prola.aps.org/abstract/PRL/v74/i24/p4763_1) (PDF (http://www.quantum.univie.ac.at/publications/pdffiles/1994-08.pdf)). Note that this experiment was done in 1994 so there are probably more recent ones (I will leave finding them as an easy exercise for you).

What's your address? I need to send you the repair bill for my brain.

[The Man]

That's not really correct, given the way these terms are used in modern physics. By a classical theory we mean one in which Planck's constant h is zero. Many classical theories - EM is a good example - are relativistic.

The point really is that SR is not exactly a theory - it's a set of rules which a theory can either be consistent or inconsistent with.

Sorry, I was referring to "classical" both in the sense of non-quantum (zero h value) and as applying to fixed universal space and time (upon which theories of relative motion prior to SR were based). I did not make that clear and my use of the term “classical” might not have been understood in both of those contexts.

[wogoga]

By the way, in principle it is possible to label each atom of a given diamond with a unique number. If photons traversing this diamond actually were more than half of the time (n_diamond > 2) at rest (i.e. transformed into energy and momentum of carbon atoms), then it would in principle be possible to determine the path of a photon by enumerating the label-numbers of the atoms where the photon rested.


No. It is in principle possible to label each carbon nuclei in a sample of diamond. It is most definitely NOT possible, even in principle, to label each electron in a sample of diamond. And electrons, not nuclei, dominate the interaction of light with matter. So you're simply wrong.


Your defense of the weird QED explanation of sub-luminal propagation speed of photons in transparent media is in the lines of Werner Heisenberg and Niels Bohr, who introduced immunization stratagems (uncertainty principle, complementarity) in order to defend their prejudices against genuinely physical objections (e.g. from Albert Einstein).

"Bohr rejected Einstein's particle of light, proposed in 1905, until well into the 1920s" (see (http://www.nysun.com/arts/overestimation-of-niels-bohr/58224/)). After evidence had shown more and more that Einstein was right, Bohr (and others) succeeded in transforming Einstein's physical photon-concept into a theological concept. Einstein then fought (http://groups.google.com/group/sci.physics.relativity/msg/6cba81ecbc0067e4) the theological properties attributed to quanta. Nevertheless, the supporters of the Bohr-fraction succeeded in convincing the world that Einstein (instead of Bohr) was the obstructionist in recognizing the reality of quanta.

What Ziggurat actually means is this:

I concede that physical reasoning (e.g. conservation laws, possibility of images in space and time) is applicable in the case of carbon atoms. However, such physical reasoning is not applicable in the case of electrons, because electrons are quantum objects, and quantum objects are not subject to physical reasoning.

But why should it not be possible to label the electrons of a diamond? It may be difficult or impossible to distinguish between the electrons of an electron pair. But from the fact that (normal, ideal) diamonds have a very regular crystal structure and are excellent electrical insulators, we conclude (by physical reasoning) that the two electrons of each carbon bond are essentially stationary. So if it is possible to label the carbon atoms, then it is also possible to label the carbon bonds and therefore the electron pairs.

So what Ziggurat advocates is this:

A photon traversing a diamond takes on the one hand all paths, i.e. the photon comes close to every single electron pair of the diamond. On the other hand, the photon is absorbed and reemitted by a sequence of (labeled) photon pairs.

Such a physical absorption and reemission is obviously inconsistent with the photons-use-all-paths claim. But in theology (as opposed to physical reasoning), there is always a solution. In analogy to virtual particles one can introduce virtual absorption and reemission. So photons are virtually absorbed and reemitted by all electron pairs of the diamond. In media with refractive index (http://en.wikipedia.org/wiki/Refractive_index) n > 1 in general, such virtual stops happening everywhere reduce the speed from c between two stops to c/n over many stops.

Cheers, Wolfgang

[Reality Check]

I see Wolfgang is back on his "science is theology" hobby horse. Any yet more arguments from authority!

Where have I seen this before? I know - in every thread that he has posted in!

P.S. Wolfgang: QCD is not weird.

[Reality Check]

Wolfgang: Just in case you are interested here is one of the tests of the fact that photons follow all paths in a system: Elitzur-Vaidman bomb-tester (http://en.wikipedia.org/wiki/Elitzur-Vaidman_bomb-tester)
Note that this test has been experimentally verified. Here is a paper on it (PDF (http://www.quantum.univie.ac.at/publications/pdffiles/1994-08.pdf))

[Ziggurat]

Your defense of the weird QED explanation of sub-luminal propagation speed of photons in transparent media is in the lines of Werner Heisenberg and Niels Bohr, who introduced immunization stratagems (uncertainty principle, complementarity) in order to defend their prejudices against genuinely physical objections (e.g. from Albert Einstein).

Uh, no. The uncertainty principle is very physical, and can be experimentally observed all over the place. One of the simplest examples is a simple Stern-Gerlach experiment. Establish the spin about one axis, and measure it about another, and you'll see rather directly that the uncertainty principle holds, with direct and unambiguous consequences.

What Ziggurat actually means is this:

Do not tell me what I mean. And you're still completely wrong. Nuclei are very much subject to quantum mechanics, but their larger masses and charges serve to localize them very well. The same cannot be said for electrons. That they do not behave the way your intuition suggests they should does not mean that they do not behave according to the laws of physics. That's your problem, not the universe's.

But why should it not be possible to label the electrons of a diamond?

Because they are identical particles whose wave functions experience significant overlap.

It may be difficult or impossible to distinguish between the electrons of an electron pair. But from the fact that (normal, ideal) diamonds have a very regular crystal structure and are excellent electrical insulators, we conclude (by physical reasoning) that the two electrons of each carbon bond are essentially stationary.

You can conclude whatever you like by whatever reasoning you think is applicable, but you're still be wrong. There are electrons in diamond which are delocalized. Diamond is a transparent insulator not because all the electrons are localized, but because the bands are full, and you cannot excite an electron without adding a significant amount of energy. Were electrons simply stationary and inert as you suggest, with every electron just stuck on one atom, carbon wouldn't even solidify, let alone turn into the hardest substance known.

But in theology (as opposed to physical reasoning), there is always a solution.

You keep using the phrase "physical reasoning". I can only assume you mean hand-waving analogies, because what you're saying has absolutely no connection to any mathematical theory of the physical world. And here's a clue for you, since you seem to be in short supply: if you can't express your theory mathematically, nobody who actually knows anything about physics is going to take you seriously.

[hgc]

But in theology (as opposed to physical reasoning), there is always a solution.


You said it, Brother! Loosy-goosy gets the job done every time.

Theology: The study of the unknowable.

[sol invictus]

Sorry, I was referring to "classical" both in the sense of non-quantum (zero h value) and as applying to fixed universal space and time (upon which theories of relative motion prior to SR were based). I did not make that clear and my use of the term “classical” might not have been understood in both of those contexts.

OK. FYI, the more common term for fixed space-time is "Galilean" or (less specifically) "Newtonian". When physicists say "classical", they always (in my experience at least) mean not quantum.

[sol invictus]

Your defense of the weird QED explanation of sub-luminal propagation speed of photons...
<nonsense snipped>

Wolfgang: do you see that computer you're reading this on? Do you know how it was made? Do you realize it's full of transistors, semiconductors, electron beams (if you have an old monitor), and tons of other extremely sophisticated parts? Do you realize that all that would never had been possible if physicists hadn't understood quantum mechanics?

Of course I'm talking to the guy that claimed the declining birthrate in Japan was due to a shortage of souls, so I'm not sure why I'm bothering.

[The Man]

Such a physical absorption and reemission is obviously inconsistent with the photons-use-all-paths claim. But in theology (as opposed to physical reasoning), there is always a solution. In analogy to virtual particles one can introduce virtual absorption and reemission. So photons are virtually absorbed and reemitted by all electron pairs of the diamond. In media with refractive index (http://en.wikipedia.org/wiki/Refractive_index) n > 1 in general, such virtual stops happening everywhere reduce the speed from c between two stops to c/n over many stops.

Cheers, Wolfgang


Why is physical absorption and reemission obviously inconsistent with the photons-use-all-paths claim? Sum over histories and the path integral refer to the paths from emission to absorption (or detection). A photon takes all paths from a physically emitting electron to the next physically absorbing electron within the lattice structure, not all paths through all valence electrons of the entire lattice structure. Real or physical photons require real and physical absorption and emission. What do you mean by virtual absorption? If the photons are not really absorbed or reemitted then they can not really be delayed. Even the exchange of virtual photons between the electrons and the protons of the carbon atoms involve real emission and absorption resulting in a real exchange of momentum. It seems that you only want physical absorption and reemission to be inconsistent with the photons-use-all-paths claim so you can solve that inconsistency by proposing your own principle of virtual absorption.

ETA:
http://en.wikipedia.org/wiki/Virtual_particle

It is sometimes said that all photons are virtual photons. This is because the world-lines of photons always resemble the dotted line in the above Feynman diagram: the photon was emitted somewhere (say, a distant star), and then is absorbed somewhere else (say a photoreceptor cell in the eyeball). Furthermore, in a vacuum, a photon experiences no passage of (proper) time between emission and absorption. This statement illustrates the difficulty of trying to distinguish between "real" and "virtual" particles as mathematically they are the same objects and it is only our definition of "reality" which is weak here. In practice, a clear distinction can be made: real photons are detected as individual particles in particle detectors, whereas virtual photons are not directly detected; only their average or side-effects may be noticed, in the form of forces or (in modern language) interactions between particles.

[The Man]

"Bohr rejected Einstein's particle of light, proposed in 1905, until well into the 1920s" (see (http://www.nysun.com/arts/overestimation-of-niels-bohr/58224/)). After evidence had shown more and more that Einstein was right, Bohr (and others) succeeded in transforming Einstein's physical photon-concept into a theological concept. Einstein then fought (http://groups.google.com/group/sci.physics.relativity/msg/6cba81ecbc0067e4) the theological properties attributed to quanta. Nevertheless, the supporters of the Bohr-fraction succeeded in convincing the world that Einstein (instead of Bohr) was the obstructionist in recognizing the reality of quanta.



For those interested in the Einstein Bohr debates and not just what wogoga writes about them on other forums.

http://en.wikipedia.org/wiki/Bohr-Einstein_debates


Also check out the links in the “see also” section.

[wogoga]

There are electrons in diamond which are delocalized. Diamond is a transparent insulator not because all the electrons are localized, but because the bands are full, and you cannot excite an electron without adding a significant amount of energy. Were electrons simply stationary and inert as you suggest, with every electron just stuck on one atom, carbon wouldn't even solidify, let alone turn into the hardest substance known.


Look at any visualisation (http://newton.ex.ac.uk/research/qsystems/people/sque/diamond/structure/structure.html) of the cristal structure of diamond. The nuclei have a strong positive charge (6 protons versus 2 electrons), whereas each of the four bonds surrounding a carbon atom consists of an electron pair. The negative charge of such a pair is a strong adhesive between two neighbouring nuclei. The hardness of diamonds is due to the short distance between the positive nuclei and the negative electron pairs, resulting in huge electrostatic attraction. Because the electron pairs are (physically) stationary, a diamond is a good insulator and can only be deformed by breaking the bonds between atoms and thus resulting in fragmentation of the crystal.

Cheers, Wolfgang

[Henners]

The hardness of diamonds is due to the short distance between the positive nuclei and the negative electron pairs

...which also explains why hydrogen is far harder than diamond!

[wollery]

Look at any visualisation (http://newton.ex.ac.uk/research/qsystems/people/sque/diamond/structure/structure.html) of the cristal structure of diamond. The nuclei have a strong positive charge (6 protons versus 2 electrons), whereas each of the four bonds surrounding a carbon atom consists of an electron pair. The negative charge of such a pair is a strong adhesive between two neighbouring nuclei. The hardness of diamonds is due to the short distance between the positive nuclei and the negative electron pairs, resulting in huge electrostatic attraction. Because the electron pairs are (physically) stationary, a diamond is a good insulator and can only be deformed by breaking the bonds between atoms and thus resulting in fragmentation of the crystal.

Cheers, WolfgangAh, I see you're using the simplified model they teach in high school to people who aren't sophisticated enough to grasp the elegant weirdness of quantum mechanics.

[Ziggurat]

Ah, I see you're using the simplified model they teach in high school to people who aren't sophisticated enough to grasp the elegant weirdness of quantum mechanics.

Indeed. It's not a bad handwaving model, and it's close enough to the real thing to get some of the essentials across, but as you point out, it's still wrong. In more technical terms, I don't think you can actually form an orthogonal basis set of electron states that way.

[wogoga]

Just in case you are interested here is one of the tests of the fact that photons follow all paths in a system: Elitzur-Vaidman bomb-tester (http://en.wikipedia.org/wiki/Elitzur-Vaidman_bomb-tester)


Wikipedia on Elitzur-Vaidman bomb-tester (http://en.wikipedia.org/wiki/Elitzur-Vaidman_bomb-tester):

Start with a Mach-Zehnder interferometer and a light source which emits single photons. When a photon emitted by the light source reaches a half-silvered plane mirror, it has equal chances of passing through or reflecting. On one path, place a bomb for the photon to encounter. If the bomb is working, then the photon is absorbed and triggers the bomb.

On Mach-Zehnder interferometer (http://en.wikipedia.org/wiki/Mach-Zehnder_interferometer):

The Mach-Zehnder interferometer is a device used to determine the phase shift caused by a small sample which is placed in the path of one of two collimated beams (thus having plane wavefronts) from a coherent light source.

On collimated light (http://en.wikipedia.org/wiki/Collimated):

Collimated light is light whose rays are nearly parallel, and therefore will spread slowly as it propagates. The word is derived from "collinear" and implies light that does not disperse with distance. ... Collimated light is sometimes said to be focused at infinity. Thus as the distance from a point source increases, the spherical wavefronts become flatter and closer to plane waves, which are perfectly collimated.

On coherence length (http://en.wikipedia.org/wiki/Coherence_length)

In physics, coherence length is the propagation distance from a coherent source to a point where an electromagnetic wave maintains a specified degree of coherence. The significance is that interference will be strong within a coherence length of the source, but not beyond it. ... Helium-neon lasers have a typical coherence length of 20 cm, while semiconductor lasers reach some 100 m. Fiber lasers can have coherence lengths exceeding 100 km.

In addition to an arbitrary use of statistics, the central fallacy of such experiments lies in the fact that photons are 'social particles', which tend to come into existence and travel together in the same state. Wikipedia on stimulated emission (http://en.wikipedia.org/wiki/Stimulated_emission):

In optics, stimulated emission is the process by which, when perturbed by a photon, matter may lose energy resulting in the creation of another photon. The perturbing photon is not destroyed in the process (cf. absorption), and the second photon is created with the same phase, frequency, polarization, and direction of travel as the original.

So already the central premise of this QM thought experiment, namely that one single photon functions as two collimated beams of coherent light, is an impossibility. Coherence is a property only of groups of photons and not of single photons. A single photon cannot have different phases, frequencies, polarizations, and directions of travel. (Yes, I know, Heisenberg's authority ...).

On the one hand, it is very astonishing how little research has been done on coherence of light. On the other hand, this is understandable, because the acknowledgement that photons normally appear as coherent groups undermines even on the theoretic side the beloved strangeness of the world of quanta. Interference effects which are now assumed to result from interference of photons with themselves can then easily be explained by interference between photons belonging to a same coherence group, but having taken different paths before reuniting again.

So the meaning of this interesting quote (Wikipedia on Bell test experiments (http://en.wikipedia.org/wiki/Bell_test_experiments))

Nevertheless, despite all these deficiencies of the actual experiments, one striking fact emerges: the results are, to a very good approximation, what quantum mechanics predicts. If imperfect experiments give us such excellent overlap with quantum predictions, most working quantum physicists would agree with John Bell in expecting that, when a perfect Bell test is done, the Bell inequalities will still be violated.

actually is this (see also (http://groups.google.com/group/sci.physics.relativity/msg/bf9267d4323acb32)):

We must admit that strictly speaking, these experiments are imperfect (i.e. not good enough in order to decide the question). However, we as working quantum physicists devoutly believe that also perfect experiments would confirm that Bohr is right and Einstein wrong.

Cheers, Wolfgang

[Reality Check]

Wikipedia on Elitzur-Vaidman bomb-tester (http://en.wikipedia.org/wiki/Elitzur-Vaidman_bomb-tester):
Start with a Mach-Zehnder interferometer and a light source which emits single photons. When a photon emitted by the light source reaches a half-silvered plane mirror, it has equal chances of passing through or reflecting. On one path, place a bomb for the photon to encounter. If the bomb is working, then the photon is absorbed and triggers the bomb.
On Mach-Zehnder interferometer (http://en.wikipedia.org/wiki/Mach-Zehnder_interferometer):
The Mach-Zehnder interferometer is a device used to determine the phase shift caused by a small sample which is placed in the path of one of two collimated beams (thus having plane wavefronts) from a coherent light source.
On collimated light (http://en.wikipedia.org/wiki/Collimated):
Collimated light is light whose rays are nearly parallel, and therefore will spread slowly as it propagates. The word is derived from "collinear" and implies light that does not disperse with distance. ... Collimated light is sometimes said to be focused at infinity. Thus as the distance from a point source increases, the spherical wavefronts become flatter and closer to plane waves, which are perfectly collimated.
On coherence length (http://en.wikipedia.org/wiki/Coherence_length)
In physics, coherence length is the propagation distance from a coherent source to a point where an electromagnetic wave maintains a specified degree of coherence. The significance is that interference will be strong within a coherence length of the source, but not beyond it. ... Helium-neon lasers have a typical coherence length of 20 cm, while semiconductor lasers reach some 100 m. Fiber lasers can have coherence lengths exceeding 100 km.
In addition to an arbitrary use of statistics, the central fallacy of such experiments lies in the fact that photons are 'social particles', which tend to come into existence and travel together in the same state. Wikipedia on stimulated emission (http://en.wikipedia.org/wiki/Stimulated_emission):
In optics, stimulated emission is the process by which, when perturbed by a photon, matter may lose energy resulting in the creation of another photon. The perturbing photon is not destroyed in the process (cf. absorption), and the second photon is created with the same phase, frequency, polarization, and direction of travel as the original.
So already the central premise of this QM thought experiment, namely that one single photon functions as two collimated beams of coherent light, is an impossibility. Coherence is a property only of groups of photons and not of single photons. A single photon cannot have different phases, frequencies, polarizations, and directions of travel. (Yes, I know, Heisenberg's authority ...).

On the one hand, it is very astonishing how little research has been done on coherence of light. On the other hand, this is understandable, because the acknowledgement that photons normally appear as coherent groups undermines even on the theoretic side the beloved strangeness of the world of quanta. Interference effects which are now assumed to result from interference of photons with themselves can then easily be explained by interference between photons belonging to a same coherence group, but having taken different paths before reuniting again.

So the meaning of this interesting quote (Wikipedia on Bell test experiments (http://en.wikipedia.org/wiki/Bell_test_experiments))
Nevertheless, despite all these deficiencies of the actual experiments, one striking fact emerges: the results are, to a very good approximation, what quantum mechanics predicts. If imperfect experiments give us such excellent overlap with quantum predictions, most working quantum physicists would agree with John Bell in expecting that, when a perfect Bell test is done, the Bell inequalities will still be violated.
actually is this (see also (http://groups.google.com/group/sci.physics.relativity/msg/bf9267d4323acb32)):
We must admit that strictly speaking, these experiments are imperfect (i.e. not good enough in order to decide the question). However, we as working quantum physicists devoutly believe that also perfect experiments would confirm that Bohr is right and Einstein wrong.
Cheers, Wolfgang

What a waste of your time - did you look at the actual experiment that verified the bomb tester?

There are plenty of single photon emitting devices around. They are a standard part of physics. They are usually diodes not lasers (devices that use simulated emission).

Bell tests have nothing to do with the Elitzur-Vaidman bomb-tester.

But if you want somthing more to misinterpret completely then have a look at Wheeler's delayed choice experiment (http://en.wikipedia.org/wiki/Wheeler%27s_delayed_choice_experiment) (also experimentally verified).

[Ziggurat]

So already the central premise of this QM thought experiment, namely that one single photon functions as two collimated beams of coherent light, is an impossibility. Coherence is a property only of groups of photons and not of single photons. A single photon cannot have different phases, frequencies, polarizations, and directions of travel.

Not so. A photon can very easily be put in a superposition state, with different components of that superposition state having different phases and directions of travel. That, in fact, is the heart of the two-slit single-photon interference effect.

Interference effects which are now assumed to result from interference of photons with themselves can then easily be explained by interference between photons belonging to a same coherence group, but having taken different paths before reuniting again.

Except you can observe those same effects even when you only use one photon at a time. But perhaps you weren't aware that it's actually fairly easy to do such an experiment, and it's been done lots of times. (http://ophelia.princeton.edu/~page/single_photon.html) How, pray tell, do you propose to explain that?

[The Man]

<sinped preceeding BS>.... experiments would confirm that Bohr is right and Einstein wrong.

Cheers, Wolfgang


So are you inferring that Einstein was correct about something?

From one of the links you provided to your writings on other forums



Simple black hole paradox refuting General Relativity:
http://members.lol.li/twostone/E/paradoxGR.html
Why SR does not explain MMX:
http://members.lol.li/twostone/E/refutationSR.html
Spaceship paradox refuting Special Relativity:
http://members.lol.li/twostone/E/paradox.html


It would seem that the only person you think is always correct is you.

[wogoga]

There are plenty of single photon emitting devices around. ... But if you want something more to misinterpret completely then have a look at Wheeler's delayed choice experiment (http://en.wikipedia.org/wiki/Wheeler%27s_delayed_choice_experiment) (also experimentally verified).


All the 'delayed choice' stuff of such quantum interference experiments only serves as a distraction from the essential weaknesses of such experiments. The latest experiment seems to be Experimental Realization of Wheeler's Delayed-Choice Gedanken Experiment, Science, 2007 (http://www.sciencemag.org/cgi/content/abstract/315/5814/966?etoc) (preprint (http://arxiv.org/abs/quant-ph/0610241v1)). The authors use a 'single-photon source' "based on the pulsed, optically excited photoluminescence of a single N-V color center in a diamond nanocrystal". The following quotes are from a page (http://www.physique.ens-cachan.fr/franges_photon/single_photon_source.htm) dealing with exactly this 'single-photon source':

We consider the emission of a single nitrogen-vacancy (N-V) colour centre in diamond, a system which has an unsurpassed photostability even at room temperature.

The N-V centres consist of a substitionnal nitrogen atom (N) and a vacancy (V) in an adjacent lattice site. They are created by irradiation of a diamond sample with high-energy electrons followed by annealing at 800°C. At small electron exposure doses, the N-V centre density is small enough so that single N-V colour centres can be spatially isolated and detected using standard confocal microscopy. Their fluorescence then appears as bright spots when the sample is scanned with strongly focused green laser radiation.

The fluorescence spectrum of the colour centre consists of a narrow zero phonon line (ZPL) at approximatively 1.945 eV (wavelength 637.7 nm) and a broad phonon wing with a width of about 300 meV (wavelength of about 100 nm FWHM (http://en.wikipedia.org/wiki/FWHM)).

So 'single-photon sources' can be bright spots with a continuous spectrum primarily in the range between 600 nm to 800 nm.

... Furthermore, the small volume of diamond excited by the pumping laser yields very low background light. Such property is also of crucial importance for single photon emission, since residual background light will contribute to a non-vanishing probability of having more than two photons within the emitted light pulse.

Nanostructured samples are prepared by starting with type Ib synthetic diamond powder (de Beers, Netherlands). The diamond nanocrystals are size-selected by centrifugation, yielding a mean diameter of about 90 nm.

Diamond is said to be a Type 1b diamond, if the nitrogen atoms are evenly spread out throughout the carbon lattice (source (http://24carat.co.uk/diamondtypes1a1b2a2bframe.html)). This type contains 500 ppm of nitrogen (source (http://www.mapsofworld.com/referrals/metals/diamond/types-of-diamond.html)). The density of diamond is 3.5 g/cm3 and the weight of 6*1023 diamond atoms is 12 g. A spherical diamond with a diameter of 90 nm has a volume of 0.38*10-15 cm3 and a weight of 1.34*10-15 g.

So typical nano-crystals consist of around 67 million atoms and around 33 thousand potential 'color centers'. In addition to that, it seems that even more than one nano-crystal is used:

A polymer solution containing selected diamond nanocrystals is deposited by spin-coating onto the surface of a dielectric mirror, resulting in a 30-nm-thick polymer layer holding the nanocrystals.

'Single photons' emitted by the N-V colour centre are produced in this way:

Under pulsed excitation with a pulse duration shorter than the excited-state lifetime, a single dipole emits photon one by one. To excite the N-V colour centre in such conditions, we use a home-built pulsed laser at a wavelength of 532 nm. The laser system delivers 800 ps pulses with energy 50 pJ.

The energy of a 532-nm-photon is 3.7*10-19 Joule. So a pulse of 5*10-11 Joule still consists of 1.3 * 108 photons. The belief that such a pulse results in the emission of exactly one suitable photon is rather wishful thinking, especially if we also take into account (quotes from (http://en.wikipedia.org/wiki/Population_inversion#Stimulated_emission)):

If an atom is already in the excited state, it may be perturbed by the passage of a photon which has a frequency v corresponding to the energy gap of the excited state to ground state transition. In this case, the excited atom relaxes to the ground state, and is induced to produce a second photon of frequency v. The original photon is not absorbed by the atom, and so the result is two photons of the same frequency. This process is known as stimulated emission. The rate at which stimulated emission occurs is proportional to the number of atoms in the excited state, and the radiation density of the light. The base probability of a photon causing stimulated emission in a single excited atom was shown by Albert Einstein to be exactly equal to the probability of a photon being absorbed by an atom in the ground state.

The critical detail of stimulated emission is that the induced photon has the same frequency and phase as the inducing photon. In other words, the two photons are coherent. It is this property that allows optical amplification, and the production of a laser system.

If the higher energy state has a greater population than the lower energy state (N1 < N2), then the emission process dominates, and light in the system undergoes a net increase in intensity.

Cheers, Wolfgang

[Reality Check]

Hi Wolfgang ,
You should have read the first paragraph:
We use a single nitrogen-vacancy (N-V) color center in a diamond nanocrystal. The N-V centers are created by irradiation of type Ib diamond sample with high-energy electrons followed by annealing at 800◦C1. Under a well controlled irradiation dose, the N-V center density is small enough to allow independent addressing of a single center using standard confocal microscopy.
and the description of the source:
Confocal microscopy setup. The 532 nm pulsed excitation laser beam is tightly focused on a diamond nanocrystals with a high numerical aperture (NA=0.95) microscope objective. The photoluminescence of the N-V color center is collected by the same objective and then spectrally filtered from residual pumping light. Following standard confocal detection scheme, the collected light is focused onto a 100 μm diameter pinhole. To identify a well isolated photoluminescent emitter, the sample is first raster scanned. For the center used in the experiment, a signal over background ratio of about 10 is achieved.


The source collects the emission from one (a single) nitrogen-vacancy (N-V) color center in a diamond nanocrystal. A laser system is used to put this one color center into an excited state using 800 picosecond pulses. The color center has time to emit 1 (one, single, not more than 2, an integer between 0 and 2, etc.) photon to go into a stable state before the next pulse arrives.
The laser system delivers 800 ps pulses with energy 50 pJ, high enough to ensure efficient pumping of the color center in its excited level.


There is no simulated emission since there is not a population of excited states to be simulated (just 1 color center is excited).

A single photon source emits single photons. That is a fact. If it was an apparatus producing multiple photons at a time then they would not call it a single photon source.
You are essentially calling the authors liars.