What are the properties and behavior of photons?

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In summary: 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_iSo 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
  • #1
debeng
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what exactly is photon? chemical , physical properties? how could they survivr insppeds of light?
 
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  • #2
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
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.
:biggrin:
 
  • #4
How could they survive at light speeds?
Photons are light, that's how they survive. Which is the easy explanation.

Dr.Brain said:
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
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
thanks a lot
 
  • #6
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
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
This is something that has kept me puzzles for long. How can photons that have 0 mass have mommentum-Bob
 
  • #9
Bob3141592 said:
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?


is there any difference between their oscillaiton in the vacuum and in the air ? when they pass through a prism, why would they disperse? and why not disperse in the air?
 
  • #10
vaishakh said:
This is something that has kept me puzzles for long. How can photons that have 0 mass have mommentum-Bob
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
try this [tex]E^2=(pc)^2+(mc^2)^2[/tex]

edit: and E=hf
 
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  • #12
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
debeng said:
is there any difference between their oscillaiton in the vacuum and in the air ? when they pass through a prism, why would they disperse? and why not disperse in the air?
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
debeng said:
is there any difference between their oscillaiton in the vacuum and in the air ? when they pass through a prism, why would they disperse? and why not disperse in the air?

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
Momentum = mv
Photon energy = hf
E = mc^2, therefore mass = energy / c^2
Therefore
momentum = E/c^2 * v
For a photon traveling at C momentum = (hf/c^2) * c
therefore momentum p = chf / c^3
so p = hf / c^2
 
  • #16
Isn't there always wavelength = h/p?
This links in the idea of wave-particle duality.
 
  • #17
Vixus said:
Isn't there always wavelength = h/p?
This links in the idea of wave-particle duality.

Ah, yes, it's fun messing with formulas to see what equals what...:tongue2:
 
  • #18
FeynmanMH42 said:
Momentum = mv
Photon energy = hf
E = mc^2, therefore mass = energy / c^2
Therefore
momentum = E/c^2 * v
For a photon traveling at C, momentum = (hf/c^2) * c
therefore momentum p = chf / c^3
so p = hf / c^2
ok, I understand all the way up to:
For a photon traveling at C, momentum = (hf/c^2) * c,

but wouldn't (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, doesn't v stand for a vector quantity, and if that is true, is c a scalar quantity or a vector quantity?

J
 
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  • #19
matrixx333 said:
ok, I understand all the way up to:
For a photon traveling at C, momentum = (hf/c^2) * c,
but wouldn't (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, doesn't v stand for a vector quantity, and if that is true, is c a scalar quantity or a vector quantity?
J

AAAAARGH!Yes! Oops :blushing: 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
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
 
  • #21
FeynmanMH42 did a good job in clarifying my question.

In the equation momentum = E/c^2 * v, I assumed that v was referring to a vector quantity, which to my knowledge has to specifiy a direction.

But can the scalar quantity (which does not specifiy a direction), be input into the equation, and if so, is it still correct?
 
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  • #22
c is a scaler because it is the speed of light, not the velocity of light. Speed implies simply the magnitude, loosing any information about direction. Wikipedia comes at this questions from a different angle:
In special relativity, the energy-momentum relation is an equation which relates energy, rest mass, and momentum of an object together:
E2 = m2c4 + p2c2,
where c is the speed of light, E is total energy, m is rest mass, and p is momentum.
If the momentum is zero (the object is in its inertial rest frame) then the energy-momentum relation simplifies to
E = mc2.
If the object is massless then the energy momentum relation reduces to
E = pc
as is the case for a photon.
Thus, p = E/c, just as Feynman### deducted.
 
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  • #23
It seems to me we've been talking about nothing since russ's last post.
 
  • #24
silverdiesel said:
c is a scaler because it is the speed of light, not the velocity of light. Speed implies simply the magnitude, loosing any information about direction. Wikipedia comes at this questions from a different angle:
In special relativity, the energy-momentum relation is an equation which relates energy, rest mass, and momentum of an object together:
E2 = m2c4 + p2c2,
where c is the speed of light, E is total energy, m is rest mass, and p is momentum.
If the momentum is zero (the object is in its inertial rest frame) then the energy-momentum relation simplifies to
E = mc2.
If the object is massless then the energy momentum relation reduces to
E = pc
as is the case for a photon.
Thus, p = E/c, just as Feynman### deducted.

Thank you for your reply Silverdiesel, I appreciate it!
 
  • #25
If A Photon Travels With A Certain Vibration Frequency ... And Will The Freq Change The Color... Will There Be Any Change In Frequency When It Changes Medium...
 
  • #26
It can lose energy going through a medium, therefore, it will change color.

This doesn't seem right, am I right? I think I'm forgetting something
 
  • #27
As matrixx333 points out, FeynmanMH42's derivation is not correct, as momentum is a vector. In fact his derivation does give p=E/c, but it is not rigorous, using a mixture of concepts from Newtonian dynamics (p=mv), and relativistic (E=mc^2). The fundamental equation has been given, it is
[tex]E^2=p^2c^2-m^2c^4[/tex]
For light, m=0 so we have the result directly.
This equation is fundamental because by Hamilton's equations, we can derive all the dynamics from derivatives of Energy (called the Hamiltonian H). Thus, for light,
[tex]H=\sqrt{p_x^2+p_y^2+p_z^2}\ c[/tex]
Hamilton's equations:
[tex]{dx\over dt}={\partial H\over\partial p_x}[/tex]
and same for y and z.
We find:
[tex] {dx\over dt}={p_x\over p}\,c,\ {dy\over dt}={p_y\over p}\,c,\ {dz\over dt}={p_z\over p}\,c[/tex]
Now we are in position to find the speed v:
[tex]v=\sqrt{ \left({dx\over dt}\right)^2+\left({dy\over dt}\right)^2+\left({dz\over dt}\right)^2}=c[/tex]
 
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  • #28
Photons are a model to describe the particle nature of light.

They are energy quanta that light carries and since it was found in various experiments like photoelectric effect etc. that they can behave in a manner similar to 'normal' particles, the model was developed.

http://www.geocities.com/physics_all"
 
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1. What is a photon?

A photon is a fundamental particle of light that carries energy and has no mass. It is the smallest unit of light and is responsible for all electromagnetic radiation, including visible light, radio waves, and X-rays.

2. What are the properties of photons?

The properties of photons include their energy, frequency, wavelength, and polarization. They also have a constant speed of light and can travel through a vacuum. Photons are also known to exhibit both wave-like and particle-like behavior.

3. How do photons behave?

Photons behave according to the principles of quantum mechanics. They can act as both particles and waves and can interact with matter through absorption, emission, and scattering. They also have a fixed speed of light and can travel in a straight line or be redirected by a medium.

4. How are photons produced?

Photons are produced by accelerating charged particles or by the emission of energy from atoms and molecules. They can also be created through the annihilation of a particle and its corresponding antiparticle. In some cases, photons can also be produced artificially, such as in lasers.

5. What is the role of photons in the universe?

Photons are essential to the existence of the universe. They play a crucial role in energy transfer and the interaction of matter. Without photons, there would be no light, and the universe would be dark. Photons also contribute to the formation of stars, galaxies, and other celestial bodies.

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