# What exactly is a photon

1. Dec 18, 2005

### debeng

what exactly is photon? chemical , physical properties? how could they survivr insppeds of light?

2. Dec 18, 2005

### Dr.Brain

Photons are not particles as in physical world. They represent the particle nature of light in the sense , they are 'bundles' of energy , since each photon is associated with some particular amount of energy , therefore each one of them has an individuality associated with themselves , thus counting them as one 'particle' , not in the sense that they are point-masses or something , they are simply 'bundles of energy' which transforms to wave-energy when we talk about wave-nature of light.

BJ

3. Dec 18, 2005

### cosmik debris

A photon is a quantum object and is thus not describable in classical terms. The best description is that a photon is a term in a mathematical model or a squiggly line in a Feynman diagram.

4. Dec 18, 2005

### Mk

Photons are light, that's how they survive. Which is the easy explanation.

Photons, are like little pieces of light. Quanta of light, quanta coming from quantization, meaning to express something with numbers. Turning something qualitative, such as light, into qualitative, by cutting it up, into different pieces, a number of pieces. Light/photons have no chemical properties.

5. Dec 18, 2005

### debeng

thanks a lot

6. Dec 19, 2005

### Bob3141592

How's this explanation?

Photons are electromagnetic energy with momentum. Since you cannot have a stable electric field without a charge to anchor it, the electric field of a photon decays, creating a magnetic field as described by Maxwell's equations. However, you cannot sustain a magnetic field without anything to anchor it either, so this also decays in it's turn into an electric field. As long as nothing absorbs these fields, the photon continues to oscillate between one and the other. Since the photon has momentum, these fields are asymmetric so the fields regenrate each other in a specific direction, giving a particulate appearance to the energy packet in motion.

Is that more or less correct?

7. Dec 21, 2005

### vanesch

Staff Emeritus
In their barest definition, photons are a bookkeeping device in the numbering of the different quantum states the EM field can be in. If you apply quantum theory to the electromagnetic field (meaning, if you want to find out the quantum description of the electromagnetic field), then it turns out that the different possible stationary quantum states the EM field can be in takes on the following structure:
In each classical mode of the EM field, with frequency f_i, there can be quantities of energy which equal E = (n+1/2) h f_i
So a stationary state of the EM field has its quantum description completely specified if you give all the numbers n for each classical mode.
So a set {n_i} of natural numbers, i running over all the possible modes, determines one stationary state.
Now, what people call a photon, is an INCREASE BY 1 of ONE of these n_i. In other words, we can say that there are n_i photons in each mode.

If all the n_i are zero, except one n_i, which equals 1, we have that the EM field is in a so-called one-photon state.
If there are two different n_i's equal to 1, and all the rest 0, or if ONE n_i equals 2, we have a 2-photon state. In the latter case, we have two photons in the same mode, in the former case, we have 2 photons in 2 different modes.

Now, the EM field interacts with other systems (say, an atom), by changing its {n_i} state ; usually by changing ONE SINGLE value n_i with an increase ("emission of a photon") or a decrease ("absorption of a photon") of 1, but multi-photon processes can also occur.

cheers,
Patrick.

8. Dec 21, 2005

### vaishakh

This is something that has kept me puzzles for long. How can photons that have 0 mass have mommentum-Bob

9. Dec 21, 2005

### debeng

is there any difference between their oscillaiton in the vacum and in the air ? when they pass through a prism, why would they disperse? and why not disperse in the air?

10. Dec 22, 2005

### Harmony

To determine the momentum of photon, we do not use the equation momentum=mass*velocity
Quote from Wikipedia
"Although they lack mass, photons have both energy and momentum proportional to their frequency (or inversely proportional to their wavelength). This momentum can be transferred when a photon collides with matter."

I think there's an equation to calculate the momentum of photon, but I have forgotten.

11. Dec 22, 2005

### Ratzinger

try this $$E^2=(pc)^2+(mc^2)^2$$

edit: and E=hf

Last edited: Dec 22, 2005
12. Dec 22, 2005

### f91jsw

From an experimental point of view a photon is what causes a "blip" on a photon counting detector as you shine (low intensity) light on it.

Johannes

13. Dec 22, 2005

### Staff: Mentor

There is refraction, scattering, etc. in the air: That's why the sky is blue and the sun turns red as it gets closer to the horizon.

14. Dec 22, 2005

### Bob3141592

I don't think there is any difference, since it's not the air molecules that the photon makes oscillate. The oscillation is of the electric field transforming into the magnetic field, and then the magnetic field transforming into the electric field. There is no medium required for light, and the medium does not affect the transit of the photon.

I also think that for a photon to disperse, or to change direction in any way, normally requires the absorption of one photon and re-emission of another one. Remember that prisms work because of the index of refraction, which is the ratio of the speed of light in that medium to the speed of light in a vacuum. But light always travels at c, always (don't confuse phase and group velocity, as in recent experiment said to "slow" or "stop" light - those deal with pulses and not individual, persistent photons). The retardation of light in glass is due to the time delay in the absorption/re-emmission process.

That being said, there is one detail where it is possible to refract light in the neighborhood of a very high local gradient in the electromagnetic field. Photons can scatter off photons if their energies are high enough, l but these are very high energy photons. Again, think of how the electric field transforms to the magnetic field. If there is already a very strong field present, it's gradient will distort the photon's oscillating field enough to "lean" it over a bit.

15. Dec 27, 2005

### FeynmanMH42

Momentum = mv
Photon energy = hf
E = mc^2, therefore mass = energy / c^2
Therefore
momentum = E/c^2 * v
For a photon travelling at C momentum = (hf/c^2) * c
therefore momentum p = chf / c^3
so p = hf / c^2

16. Dec 27, 2005

### Vixus

Isn't there always wavelength = h/p?
This links in the idea of wave-particle duality.

17. Dec 27, 2005

### FeynmanMH42

Ah, yes, it's fun messing with formulas to see what equals what...:tongue2:

18. Dec 31, 2005

### matrixx333

ok, I understand all the way up to:
For a photon travelling at C, momentum = (hf/c^2) * c,

but wouldnt (hf/c^2) * c = hfc/c^2? Which in turn would equate to hf/c, right?
But from your math what I am seeing is (hf/c^2) * c/c, which would equal p = hf / c^2.

Also in the equation momentum = E/c^2 * v, doesnt v stand for a vector quantity, and if that is true, is c a scalar quantity or a vector quantity?

J

Last edited: Dec 31, 2005
19. Dec 31, 2005

### FeynmanMH42

AAAAARGH!Yes! Oops Thank you for pointing out my simple error, momentum is hf/c...
I've always seen "c" substituted into equations for light instead of "v", but
c is always the speed of light rather than the velocity...
Light cannot always travel at the same velocity, because it wouldn't be able to change direction... but the speed has to be constant...
I'm confused now. :p I'm probably one of the youngest on here...

20. Dec 31, 2005

### Vixus

At least you're posting something useful. I still need to start thinking in terms of equations. But anyway, what was the question again? :P