# The magical creation of the photon.

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### #18 CraigD

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Posted 19 December 2007 - 11:22 PM

What particle carries the force that holds the electron to the proton?

More photons.

That’s my understanding, also.

What I’ve long wanted to know, and haven’t been able to find or calculate on my own, is what energies are typical of the virtual photons of magnetic force in various atoms and collections of atoms – that is, where they would fit in the spectrum – their “color” - if they were real photons of EM radiation?

### #19 Erasmus00

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Posted 20 December 2007 - 11:39 AM

What I’ve long wanted to know, and haven’t been able to find or calculate on my own, is what energies are typical of the virtual photons of magnetic force in various atoms and collections of atoms – that is, where they would fit in the spectrum – their “color” - if they were real photons of EM radiation?

First, a caveat- virtual photons can be quite different then "real photons" because often the thing that makes them virtual is that they are "off-shell." Off shell is a funky phrase that sort of means "has the wrong mass." Which for photons means some of the virtual guys have non-zero mass.

Because some of the virtual photons have non-zero mass, some of the photons can be longitudinal (a spin z of 0, instead of +1 or -1), and so are very different then the photons we often think about.

That being said, virtual photons are exchanged across the spectrum. Often, virtual photons are inside loops in feynman diagrams which means their momentum and energy are unconstrained, which means integrating over every momentum, and every energy. However, they are weighted so that the further "off shell" (the more massive the photon) the less likely the probability of that virtual photon.
-Will

### #20 snoopy

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Posted 21 December 2007 - 02:21 AM

Because some of the virtual photons have non-zero mass, some of the photons can be longitudinal (a spin z of 0, instead of +1 or -1), and so are very different then the photons we often think about.

-Will

If they are longitudinal does that mean they can exceed C ??

### #21 Erasmus00

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Posted 21 December 2007 - 09:13 AM

If they are longitudinal does that mean they can exceed C ??

Nope, the upper speed limit of c is true regardless of the whether or not they have mass. Quantum field theory IS relativistic.

However, since these virtual photons have mass they may travel at LESS then c.
-Will

### #22 snoopy

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Posted 23 December 2007 - 05:21 AM

Nope, the upper speed limit of c is true regardless of the whether or not they have mass. Quantum field theory IS relativistic.

However, since these virtual photons have mass they may travel at LESS then c.
-Will

Thanks for the reply I was only asking as longitudinal waves in an iron bar travel faster than transverse waves.

### #23 Little Bang

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Posted 23 December 2007 - 09:58 PM

Will, is there experimental proof that photons spend some of their time as a particle? If there is why can't we put a mass to them?

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Posted 28 December 2007 - 01:58 AM

Little Bang,

You started this thread asking of a description of slamming two Hydrogen
atoms together producing an infrared photon.

Depending on the input energy of the two systems of particles
(Hydrogen = H == proton (p) + electron (e)), you may get more than just
Infrared photons.

To simply produce Infrared photons, one only needs a single Hydrogen ion
(proton) and a free electron. If the energy of the electron were such that
the proton captured it though not to the lowest energy level, a photon
would be released on capture. The energy or wavelength of the photon
could be then determined on output.

The Schroedinger equation mentioned by Qfwfq will definitely calculate the
energy level of the output photon. I agree the full Standard Model need
not be considered here. No nuclear forces are involved.

However, Virtual particles are a different matter. In Feynman Diagrams,
any particle which is not directly observed being intermediate process,
no reference to its speed or travel time or distance be directly inferred.
In fact under some circumstances not in violation of the Standard Model,
such a Virtual Particle may borrow time or energy in the exchange. A
better description than mine can be found in a fantastic recent book
"About Time" (I forget the Author). This does not make "superluminal"
speed (> c) possible. [emphasis]

The last point is more about Duality: both particle and wave description
coexist for both mass and massless particles (i.e. Fermions, Bosons).
De Broglie showed this for electrons, Einstein showed this for photons.

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### #25 CraigD

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Posted 28 December 2007 - 12:43 PM

Will, is there experimental proof that photons spend some of their time as a particle?

First, a semantic correction: photons do not spend some time as a particle, then some time as a wave. They always appear to be particles to experiments that measure their particle-like behavior, and always appear to be waves to experiment that measure their wave-like behavior.

Experiments conducted for most of the last half of the 19th century and first decade of the 20th demonstrated the particle nature of photons, leading up to what most people consider the definitive explanation, Einstein’s 1905 paper “A Heuristic Point of View [Model] of the Creation and Transformation of Light".

These experiments are commonly reproduced in Modern Physics classes. A very simple one involves using a detector with single-photon sensitivity and a monochromatic (or nearly) light source, such as hot filament lamp, uncoated gas vapor lamp, or laser, and many “smoked glass” or similar photographic filters. Adding filters while keeping the light source constant, you eventually reach a step where the detected light drops from low to zero. Plotting this, you find a “notchiness” in the graph, where the energy represented by the smallest notch agrees with the famous formula $E = h v$, where $v$ is the photon’s frequency, and $h$ is the Planck constant.

It’s important to understand that this “wave-particle duality” is not limited to photons. Every particle with momentum ($p$), massless particles such as photons or massive ones such as electrons, has an associated wave nature.

Even big, composite bodies like cannon balls and planets have an associated de Broglie wave with wavelength give by $\lambda = \frac{h}{p}$, but it’s so small compared to their size and the associated wavelengths of their constituant particles that it’s physically irrelevant. Thus, the only particles for which experiments show wave-like effects, such as interference, are very small – to the best of my knowledge, nothing larger than the nuclei of metals, such as gold.

If there is why can't we put a mass to them?

In modern particle physics, particles are allowed to have zero or non-zero rest mass. Particles with zero rest mass must travel at the speed of light. Because they have finite, measurable energy and momentum, it’s possible to calculate their relativistic mass. However, because this mass is very small – about $10^{-32} \,\mbox{kg}$ for a photon of visible light, it’s not significant in most calculations involving mass, so is little discussed.

### #26 TruthChaser

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Posted 03 January 2008 - 01:36 AM

I love Heisenberg's Uncertainty Principle... whatever happens to the photons, just the fact that we are trying to observe it, makes the answer unknowable. Quaint isn't it!

"Heisenberg's Uncertainty Principle tells us it is impossible for science to understand the universe completely. No matter how carefully we experiment, no matter how accurately our scientific instruments are, some things in the universe will always be hidden from us. In an effort to find the laws that tell us how our universe works, scientists will continue to experiment. But in a sense, Heisenberg's uncertainty principle may be the last law of the universe. No matter how wonderful human brains may be, we can't know everything." Paul Fleisher (Book: Relativity and Quantum Mechanics: Principles of Modern Physics.

### #27 Farsight

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Posted 04 January 2008 - 07:41 AM

I can give you a partial picture using analogies, Little Bang. I can't say it's the correct picture, because I really need some computer modelling to validate it. But here goes:

A photon is a travelling warp in space. If you think about a rubber sheet, then twang it by flicking your finger, a ripple will run through the rubber sheet. But this is a rather special ripple, in that it doesn't spread out like a ripple in a pond. It travels in some given direction and stays together. It's something like the ripple in a long rubber mat that you've just shaken. The thing to note here is that when the photon is travelling through some given area of space, that area of space is bigger than it was. It's like three-dimensional ghostly rubber, but the rubber and the rubber sheet is only an analogy. It isn't a perfect picture.

An electron is another travelling warp in space. This is different to the photon, in that it's travelling round and round in a twisted circle. It's rather a strong warp, and this warping affects the direction that the warp itself travels in. It ends up travelling in a "clockwise" direction, and this is constant.

The proton is yet another travelling warp in space. It something like the electron, but it's travelling a more tortuous twisted-circle path, and it's a very strong warp. It travels in an "anticlockwise" direction, and again this is constant.

All this warping means an electron will move towards a proton, because each causes a distortion that affects the warp direction. But the electron is bigger than the proton, because the latter is a stronger tighter warp. They can't meet and annihilate one another like an electron annihilates with a positron to produce photons. Instead the electron orbits round the proton like a tyre rolling around and around in a depression.

But there's got to be a neatness to this. It's something like this: the orbit of the electron has to be an integer number of turns. You can't reduce the orbit of the electron by just a little, like you can with a planet. The travelling warping of space that is the rolling electron has to match the travelling warping of space that is the proton. They're something like meshing gears, only there's nothing there other than warped space. And to reduce the electron orbit, some warp has to leave the system. So twang, out goes a photon, and the electron has dropped down a gear.

I really need some computer modelling to show you this properly, but you should get the drift, and until somebody can come up with something better, it's got to be better than nothing.

### #28 Heisenberg

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Posted 11 January 2008 - 12:37 AM

I can only think of a very simple explaination to this besides using dirac eqaution but when two atoms collide, they create energy. This energy is transfered to the respective electrons of each atom. With knowledge of atomic structure, the atom is consisted of electrons, protons, and neutrons. The electrons (according to the wave-mechanical atomic model) do not have orbits but rather encircle the nucleus (sp?) and form an 'electron cloud'. To be quite frank, the 'excited' (energized) electrons have enough power to escape the valence 'shell', there is no energy released (light photon) until the electron returns to its former position during which it emits the energy as a light photon. The energy of the photon is dependant upon the amount of 'shells' it jumped. E=hv; E, energy is equal to, h, the constant times, v, the frequency of the photon. h [planck's constant]=6.626068 × 10^-34 m^2 kg / s[econds]. Good luck .
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### #29 Little Bang

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Posted 12 January 2008 - 06:59 PM

The energy of a particular frequency is given by E=hf and of course E=MC^2. If I substitute and solve for f I have f=MC^2/h. If I plug the mass of an electron into the equation does that frequency have any meaning? How do we know there is no relationship between EMR and mass?

### #30 CraigD

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Posted 13 January 2008 - 04:17 PM

The energy of a particular frequency is given by E=hf and of course E=MC^2. If I substitute and solve for f I have f=MC^2/h. If I plug the mass of an electron into the equation does that frequency have any meaning? How do we know there is no relationship between EMR and mass?

We know – that is, best accepted theory predicts – that photons do have mass. This is relativistic mass, applicable only when a photon is moving, as it always is, at the speed of light ©. Photons have no rest mass, but are never at rest.

The equation Little Bang gives appears a valid one from which to derive and calculate the mass of a photon. For a photon of green-yellow visible light ($5 \times 10^{14} \,\mbox{hz}$), it evaluates to around $3.6 \times 10^{-36} \,\mbox{kg}$, about 4 millionths of the mass of an electron.

We have ample, clear experimental evidence that they have momentum and constant velocity.

To the best of my knowledge, what we lack experimental evidence for is another major theoretical prediction: that photons exert gravitational force, as opposed to only experiencing it, of which we do have ample, clear evidence.

For example, a single photon of visible light ($5 \times 10^{14} \,\mbox{hz}$) passing a single hydrogen atom at a distance of 1 cm should be accelerated by about $10^{-46} \,\mbox{m/s}$ – an undetectably small but physically real amount.

### #31 Little Bang

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Posted 15 January 2008 - 08:25 AM

The equation may have some bigger implications than just calculating the mass of the photon. BTW Craig good job on the mass of the photon. We could calculate a frequency for the electron or even a proton. We could never have the energy to produce the frequency of an electron let alone that of a proton. It also shows a relationship between the speed of light and what we as observers see in the universe. For instance, if the expansion of the universe is accelerating as it appears, and there is a frequency related to the proton then the matter outside the visible universe would have red shifted almost out of existence with respect to us but it’s apparent mass would be approaching infinity. This would create a huge gravity well that could be causing the accelerated expansion of the visible universe. As my other thread “ My Turn To Guess” implies, I think that mass is some form of magnetic or electromagnetic energy that only appears to the observer as mass.

### #32 Moontanman

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Posted 15 January 2008 - 10:51 AM

Just to remind everyone, while proper spelling and grammar might help, no one here--well no one with any *manners* (and those that lack them don't last very long in most cases!)--cares if you can spell. Most folks around here have English as their second or third language, so its really unimportant...

Now I know absolutely nothing about this topic, so I'll yield the floor to someone (with good manners!) who does...

At the Derek Zoolander Center For Children Who Can't Read Good And Wanna Learn To Do Other Stuff Good Too, we teach you that there's more to life than being really, really good looking,
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thank goodness, with out spelling check I can hardly spell my own name!

### #33 Little Bang

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Posted 21 January 2008 - 06:13 PM

If I calculate the mass of a gamma ray it turns out to be 2.21 X 10^-30 kg. It's relatively close to the mass of the electron at 9.11 X 10^-31kg. Someone check my numbers on the mass of the gamma ray please.

### #34 freeztar

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Posted 21 January 2008 - 07:41 PM

If I calculate the mass of a gamma ray it turns out to be 2.21 X 10^-30 kg.

How did you determine this?

Since gamma rays are EMR and travel at c, would they not be massless (or a rest mass of zero [M0])?

Here's a post by Erasmus that explains it nicely.
http://hypography.co...otons have mass