Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: hubble_bubble on 04/09/2012 06:36:23

If we consider time dilation and length contraction using special relativity does the photon therefore have infinite length contraction and time dilation? If so is that why we see it as a massless particle?

having no mass, per se, is the main reason a photon can travel at the speed of light, and thereby establish that rule.
if a photon had mass, it could not achieve the speed of light....
I believe....lol

If we consider time dilation and length contraction using special relativity does the photon therefore have infinite length contraction and time dilation? If so is that why we see it as a massless particle?
Special relativity relies on the fact that physical laws remain the same for all inertial frames. One of the products of this axiom is that no inertial frame of reference can travel at light speed. In order to make the assumptions you are making you need to posit an inertial frame travelling at light speed  once you do that you can no longer use the equations of SR. We just do not know what happens to objects, observers, clocks, rulers etc within an inertial frame that travels at the speed of light  and if our physics is correct, we can never know because it cannot happen

A slight clarification about the difference between Mass and Rest Mass:
 The photon, traveling at the speed of light does have mass, and is deflected by a gravitational well like the Sun  in 1919 this was one demonstration of Special Relativity http://en.wikipedia.org/wiki/Tests_of_general_relativity#Deflection_of_light_by_the_Sun
 This Mass is derived from the photon's energy by the famous E=mc^{2}
 In turn, the photon's energy is derived from it's frequency v by E=hv http://en.wikipedia.org/wiki/Photon#Physical_properties
 This is an example of the prediction of relativity that objects get infinitely more massive as they approach the speed of light. This effect is demonstrated daily in particle accelerators.
 Photons have Zero Rest Mass. This can be roughly described mathematically as: (Finite Mass)/Infinity = 0. If it were possible to slow photons down below the speed of light (in whatever medium they were traveling), they would have no energy, and would cease to be observable.

Light has been slowed and observed.
http://en.wikipedia.org/wiki/Slow_light
The question is why light travels at c in the first place. I can see the point particle principle but if mass increases as we reach light speed then why is the photon not massive? Maybe it is but because of it's other properties this mass can not be detected.

Slow light is through a medium  not in vacuo; completely different.
[you can think of slow light in terms of the easy heuristic time taken is time of absorption and reemission plus actual travelling time at speed of light. this is not really correct  but is a very useful model. the actual reasons are hideously complicated and really tend towards light not being viewed as travelling through as substance as photons.] what slow light isnt is photons going less than c!
see above  SR cannot cannot be used to predict properties of a particle travelling at speed of light. inertial frames dealing with massive objects do not smoothly change into new frames dealing with massless objects; there is a disconnect, and laws for one frame do not apply to the other.

There is the problem of infinity implicit in any equations connected with c. How can we resolve infinite energy/mass/time dilation? Something is obviously missing. Like the quanta at microscopic scale there must be an equivalent concept for the macro scale. If we look at e=mc2 with e at infinity then m must still be finite as it is always less than e. Infinity multiplied by a value makes no sense.

The photon, traveling at the speed of light does have mass,
Sorry, it's wrong. There is only one mass, the one you call "rest mass" but that name is not quite good because it's impossible to find a photon at rest, so with your definition it would be impossible to define a mass for a photon (or a gluon), even a zero mass.
If you really want to give it a name, call it "invariant mass", it reminds you that it doesn't depend on the frame of reference (and this is the reason physicists use this as definition).

Light has been slowed and observed.
http://en.wikipedia.org/wiki/Slow_light
The question is why light travels at c in the first place.
c is just a number, because it depends on the units of measure; it is 1 or it is 0.000001 or any number, in other units. The question is another: why light speed is finite?
I can see the point particle principle
What is this principle?
but if mass increases as we reach light speed then why is the photon not massive?
On the contrary! As another has written, if it had mass, it couldn't reach light speed.

Light has been slowed and observed.
http://en.wikipedia.org/wiki/Slow_light
The question is why light travels at c in the first place.
c is just a number, because it depends on the units of measure; it is 1 or it is 0.000001 or any number, in other units. The question is another: why light speed is finite?
I can see the point particle principle
What is this principle?
but if mass increases as we reach light speed then why is the photon not massive?
On the contrary! As another has written, if it had mass, it couldn't reach light speed.
Yes we can define c as 1 for convenience in equations. Point particles are idealized particles used in equations. There is a dichotomy inherent in what we consider mass. The fermions which make up matter and a boson like the photon behave in distinct ways at velocity. The Fermions can never act like a photon as instead of its mass decreasing with increasing velocity it increases so will never reach a zero point mass. It is going in the wrong direction. The photon is immediately in the correct state when emitted.

Yes we can define c as 1 for convenience in equations. Point particles are idealized particles used in equations. There is a dichotomy inherent in what we consider mass. The fermions which make up matter and a boson like the photon behave in distinct ways at velocity. The Fermions can never act like a photon as instead of its mass decreasing with increasing velocity it increases so will never reach a zero point mass. It is going in the wrong direction. The photon is immediately in the correct state when emitted.
Yes, all this is true because it matches our observations of the universe. So what's the problem? Why should we be troubled that special relativity, which is designed to cover the reference frames only of massive particles, doesn't include reference frames for massless particles? It still covers all the cases we can measure, since we have mass and our detectors have mass, and produces very accurate results.

Yes, all this is true because it matches our observations of the universe. So what's the problem? Why should we be troubled that special relativity, which is designed to cover the reference frames only of massive particles, doesn't include reference frames for massless particles? It still covers all the cases we can measure, since we have mass and our detectors have mass, and produces very accurate results.
What should trouble us as you say is the connection between the speed of light c, gravity and acceleration. Time dilation slows things down at speeds approaching light speed but also inside a gravitational field, which can be considered an accelerating frame. Under extreme conditions inside the Shwarzschild radius of a black hole the constancy of light's forward motion is breached and escape velocity exceeds the speed of light. If we had a mathematical model of massless particles then we would be on our way to a unified field theory incorporating gravity. We would also finally have an understanding of the mechanics of electromagnetism and gravity.

But our models do account for massless particles. We know exactly how they'll act when viewed by any detector we can build! Special relativity works exceedingly well for predicing how massless particles behave. The only catch is that we don't know how to describe what a massless detector would see... but building a massless detector doesn't even make sense (since all its particles would fly away from each other at the speed of light!)
Would a theory that includes the reference frame of a photon include any testable predictions? If not, then while it might seem more complete, it's more a matter of metaphysics than physics.

I don't think that we would detect anything in the reference frame of a photon as theoretically the time dilation would be infinite. This would effectively stop time in that frame. Looking at it this way if we ever could reach light speed we would experience nothing until we impacted some other object. Even then the impact would preclude any interaction with the universe as our matter would vaporize.
The important point is the effect of gravity on light. Gravity affects mass, which suggests that photons must have mass. If not then there is some component missing from the laws of physics. This is fundamental. Photons are also said to have neutral charge, however, this also may be wrong. If the spacetime frame of a photon has infinite dilation then it follows that we would not be able to detect any mass, charge or other reactive properties. Whereas state properties such as luminosity and wavelength are still detectable. This implies a separation between state and reaction. This should also hold for discrete particles of matter. This can be tested under general relativity for ordinary matter and applied to light.

There is a dichotomy inherent in what we consider mass. The fermions which make up matter and a boson like the photon behave in distinct ways at velocity. The Fermions can never act like a photon as instead of its mass decreasing with increasing velocity
Mass decreasing with velocity? Even if you talked of "relativistic mass" (which is not "mass") I can't understand how it could decrease with increasing velocity.
so will never reach a zero point mass.
What is a "zero point mass"?
It is going in the wrong direction.
What does it mean?
The photon is immediately in the correct state when emitted.
What does it mean?

having no mass, per se, is the main reason a photon can travel at the speed of light, and thereby establish that rule.
if a photon had mass, it could not achieve the speed of light....
I believe....lol
http://en.wikipedia.org/wiki/Photon
Physical properties
See also: Special relativity
The photon is massless,[Note 2] has no electric charge,[12] and is stable.
HA !! I was right, no mass......

The only catch is that we don't know how to describe what a massless detector would see... but building a massless detector doesn't even make sense (since all its particles would fly away from each other at the speed of light!)
Infact a massless detector would be, e.g., another photon. Since Even if photonphoton scattering hasn't been observed yet (as far as I know), because of its exceedingly low cross section, it shouldn't be anything particularly bizarre.

There is a dichotomy inherent in what we consider mass. The fermions which make up matter and a boson like the photon behave in distinct ways at velocity. The Fermions can never act like a photon as instead of its mass decreasing with increasing velocity
Mass decreasing with velocity? Even if you talked of "relativistic mass" (which is not "mass") I can't understand how it could decrease with increasing velocity.
so will never reach a zero point mass.
What is a "zero point mass"?
It is going in the wrong direction.
What does it mean?
The photon is immediately in the correct state when emitted.
What does it mean?
What I was saying is that mass INCREASES with velocity so it will never be in a massless state. Going in the wrong direction as in increase of mass with velocity instead of a decrease. The mass is getting less like a photon with velocity. The photon is massless when emitted and doesn't lose mass over time.

Preface: I see that current practice prefers to talk in terms of a particle's momentum, rather than it having a relativistic mass which increases with its speed (http://en.wikipedia.org/wiki/Mass_in_special_relativity#Relativistic_mass). However, I am a bit oldschool, so apologies in advance...
Let's ignore the photon for a while, and take a simpler example of the difference between Relativistic Mass and Rest Mass: the electron neutrino.
 Mass and the Energy are equivalent, due to E=mc^{2}
 Particle physicists quote the Mass of a Particle by its Energy, in electron Volts (eV) or Megaelectron Volts (MeV).
 A neutrino has a RestMass, thought to be around 0.04eV to 2.2eV (but it's very hard to measure: http://en.wikipedia.org/wiki/Neutrino#Mass)
 This means that if you could ever capture a neutrino, stop it moving, and turn it entirely into energy, you would get somewhere around 1 eV (give or take a decimal place)
 However, neutrinos emitted from a nuclear fission reactor have energies of up to 10MeV. (http://en.wikipedia.org/wiki/Neutrino#Artificial)
 This means that if a neutrino from a nuclear reactor interacts with a hydrogen nucleus (eg in http://en.wikipedia.org/wiki/KamLAND), a neutrino from a nuclear reactor will produce particles with an energy millions of times larger than the energy of a stationary neutrino
 This increase in energy (equivalent to an increase in momentum, and an increased relativistic mass) is because they are traveling at very close to the speed of light; a rough calculation suggests that a 4MeV neutrino will be traveling somewhere around 99.999999999997% of the speed of light
 So their momentum and energy is much higher than you would expect from their velocity (effectively c) and their rest mass (1eV).
 No matter how much energy you give the classical neutrino (ie how much you increase its momentum), it cannot reach the speed of light, as this would imply infinite momentum (or infinite relativistic mass, in older terminology).
So, for a 4MeV neutrino:
 It has a small rest mass (= a small energy) of somewhere around 1 eV
 It has a nonzero and finite relativistic mass (which may be millions of times larger)
 It can never reach the speed of light in a vacuum
 Time is slowed down significantly, but it has not stopped, so we can still see phenomena like neutrino oscillations (http://en.wikipedia.org/wiki/Neutrino#Flavor_oscillations)
 The neutrino will have severe length contraction  but not to zero
Heading back to the original question: if we contrast this with a violet photon
 It has a small relativistic mass of around 0.5 eV (= a small energy), similar in magnitude to a stationary neutrino
 The photon always travels at exactly the speed of light in a vacuum.
 Time is stopped, so the photon does not change (unless it interacts with something)
 Length contraction will also apply to photons, but the wave nature of light as a transverse electromagnetic wave means that the size will be nonzero at rightangles to the direction of travel.
 If you could ever slow down a photon in a vacuum, and turn it entirely into energy, you would detect 0eV, ie the photon's restmass would be zero.
PS: If you find an error in my rough calculations, please feel free to point it out!

The only catch is that we don't know how to describe what a massless detector would see... but building a massless detector doesn't even make sense (since all its particles would fly away from each other at the speed of light!)
Infact a massless detector would be, e.g., another photon. Since Even if photonphoton scattering hasn't been observed yet (as far as I know), because of its exceedingly low cross section, it shouldn't be anything particularly bizarre.
There's pair production, but that has to happen in the presence of matter, and there are also people looking at the idea of vacuum nonlinearity (which would be photonphoton interactions) at high energy densities. These effects still don't require a theory of the photon's reference frame, though, since they can be adequately described in terms of any experiment we know how to do, in terms of special relativity's reference frames.

I don't think that we would detect anything in the reference frame of a photon as theoretically the time dilation would be infinite. This would effectively stop time in that frame. Looking at it this way if we ever could reach light speed we would experience nothing until we impacted some other object. Even then the impact would preclude any interaction with the universe as our matter would vaporize.
This is the point that's tripping you up, hubble_bubble. Check what Imatfaal said above. There's no such thing as a photon reference frame. The claims that time stops for photons or that if we traveled at a photon's speed, we'd see a static universe are science fiction, not science. The problem is that special relativity, which is used for this claim, is designed specifically to only work in reference frames in which you, as the observer, see light (in vacuum) always moving at the speed of light. If you were to somehow ride along on a photon at its speed, it couldn't be moving at the speed of light with respect from you, so all of special relativity would break down.
The problem is that if you simply plug in the speed of light for the observer's velocity, you get a lot of infinities in the equations of special relativity (which is a sign that they've broken down). A lot of time these infinities get misinterpreted as the idea that time stops for a photon.
A photon reference frame might be interesting if we could only explain certain phenomena by introducing it, but it doesn't appear to be useful to try to add photon reference frames into special relativity. There is no problem (at least that I know of) that we need them for. And there is now way (again, that I know of) to measure them if they do exist.
The important point is the effect of gravity on light. Gravity affects mass, which suggests that photons must have mass. If not then there is some component missing from the laws of physics. This is fundamental.
On this point, gravity does affect mass, but it does not only affect mass. If you look at what general relativity says, gravity alters trajectories through spacetime. Even massless particles have trajectories. In fact, general relativity predicts very precisely how the gravity massive objects should affect massless photons, and this has been confirmed by observation.
As for creating gravity, anything that has energy or momentum can create gravity. Mass has a lot of energy (E=mc^{2}), so mass creates gravity. But photons, which are massless, have energy and momentum, so they can also create gravity.

Evan, you're not the only one thinking of it in terms of 'relativistic mass', myself I think of in terms of 'invariant mass or as they say restmass, but depending on terminology relativistic mass also works, as I understands it. Although, if we are going to discuss the definition of 'invariant mass' then it is that it has to be the same from all frames measured as I understands it, no matter if on a event horizon or in 'flat space' without gravitation.
How do we know what that mass is?
'Energy' right :)
And what the he* is 'energy'?

If E=mc^2
then
y not the mass stays same and the energy amount changes dependant on the speed of which the photon moves.
I know this is not the way you guys see either stationary or moving at 'c' it but if you had a method to slow the photon down incrementilly then surely the energy it has
would be inverse to the speed at which it travells
i know very little on this subject but is good to read your conversations

yes stu.
EMC2
energy of object is equal to the mass of said object x the speed of light squared.
8.98755179×10^16 m^2/s^2 I believe is speed of light squared.
so this is the rule as it applies to photons.
photons do not have mass, but they do have momentum. The proper, general equation to use is E2 = m2c4 + p2c2 So in the case of a photon, m=0 so E = pc or p = E/c. On the other hand, for a particle with mass m at rest (i.e., p = 0), you get back the famous E = mc2.
This equation often enters theoretical work in Xray and Gammaray astrophysics, for example in Compton scattering where photons are treated as particles colliding with electrons.
http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/960731.html

A slight clarification about the difference between Mass and Rest Mass:
 The photon, traveling at the speed of light does have mass,and is deflected by a gravitational well like the Sun
photons do not have any "mass'
mass
A measure of the total amount of material in a body, defined either by the inertial properties of the body or by its gravitational influence on other bodies.
a Photon does not affect other particles gravitationally, cause it has zero mass....
No, photons do not have mass, but they do have momentum. The proper, general equation to use is E2 = m2c4 + p2c2 So in the case of a photon, m=0 so E = pc or p = E/c. On the other hand, for a particle with mass m at rest (i.e., p = 0), you get back the famous E = mc2.
http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/960731.html
http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html
http://www.weburbia.com/physics/photon_mass.html
http://home.fnal.gov/~pompos/light/light_page31.html
The photon is massless,[Note 2] has no electric charge,[12] and is stable
http://en.wikipedia.org/wiki/Photon

A slight clarification about the difference between Mass and Rest Mass:
 The photon, traveling at the speed of light does have mass,and is deflected by a gravitational well like the Sun
photons do not have any "mass'
mass
A measure of the total amount of material in a body, defined either by the inertial properties of the body or by its gravitational influence on other bodies.
a Photon does not affect other particles gravitationally, cause it has zero mass....
No, photons do not have mass, but they do have momentum. The proper, general equation to use is E2 = m2c4 + p2c2 So in the case of a photon, m=0 so E = pc or p = E/c. On the other hand, for a particle with mass m at rest (i.e., p = 0), you get back the famous E = mc2.
http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/960731.html
http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html
http://www.weburbia.com/physics/photon_mass.html
http://home.fnal.gov/~pompos/light/light_page31.html
The photon is massless,[Note 2] has no electric charge,[12] and is stable
http://en.wikipedia.org/wiki/Photon
I am not sure about your assertion that a photon cannot have a gravitational influence. General Relativity is much more complicated that just saying that mass is required for gravity. Photon have energy  energy and energy flux are parts of the calculations to determine curvature. Now I cannot believe we could possibly measure that gravitational influence at present  but I am not sure that it is zero.
However  one fly in ointment, a photon is a exemplary quantum mechanical object, our theory of gravity is GR which is not quantum.

A slight clarification about the difference between Mass and Rest Mass:
 The photon, traveling at the speed of light does have mass,and is deflected by a gravitational well like the Sun
photons do not have any "mass'
mass
A measure of the total amount of material in a body, defined either by the inertial properties of the body or by its gravitational influence on other bodies.
a Photon does not affect other particles gravitationally, cause it has zero mass....
No, photons do not have mass, but they do have momentum. The proper, general equation to use is E2 = m2c4 + p2c2 So in the case of a photon, m=0 so E = pc or p = E/c. On the other hand, for a particle with mass m at rest (i.e., p = 0), you get back the famous E = mc2.
http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/960731.html
http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html
http://www.weburbia.com/physics/photon_mass.html
http://home.fnal.gov/~pompos/light/light_page31.html
The photon is massless,[Note 2] has no electric charge,[12] and is stable
http://en.wikipedia.org/wiki/Photon
I am not familiar with this equation. For E2 = m2c4 + p2c2 what is p? I am also assuming m^2c^4 + p^2c^2.

Surely this equation could be rewritten E = mc^2 + pc. Unless these are not exponential functions.

A slight clarification about the difference between Mass and Rest Mass:
 The photon, traveling at the speed of light does have mass,and is deflected by a gravitational well like the Sun
photons do not have any "mass'
mass
A measure of the total amount of material in a body, defined either by the inertial properties of the body or by its gravitational influence on other bodies.
a Photon does not affect other particles gravitationally, cause it has zero mass....
No, photons do not have mass, but they do have momentum. The proper, general equation to use is E2 = m2c4 + p2c2 So in the case of a photon, m=0 so E = pc or p = E/c. On the other hand, for a particle with mass m at rest (i.e., p = 0), you get back the famous E = mc2.
http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/960731.html
http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html
http://www.weburbia.com/physics/photon_mass.html
http://home.fnal.gov/~pompos/light/light_page31.html
The photon is massless,[Note 2] has no electric charge,[12] and is stable
http://en.wikipedia.org/wiki/Photon
I am not familiar with this equation. For E2 = m2c4 + p2c2 what is p? I am also assuming m^2c^4 + p^2c^2.
they say it better then I do, hope this helps hubble.
Does the photon have mass? After all, it has energy and energy is equivalent to mass.
Photons are traditionally said to be massless. This is a figure of speech that physicists use to describe something about how a photon's particlelike properties are described by the language of special relativity.
The logic can be constructed in many ways, and the following is one such. Take an isolated system (called a "particle") and accelerate it to some velocity v (a vector). Newton defined the "momentum" p of this particle (also a vector), such that p behaves in a simple way when the particle is accelerated, or when it's involved in a collision. For this simple behaviour to hold, it turns out that p must be proportional to v. The proportionality constant is called the particle's "mass" m, so that p = mv.
In special relativity, it turns out that we are still able to define a particle's momentum p such that it behaves in welldefined ways that are an extension of the newtonian case. Although p and v still point in the same direction, it turns out that they are no longer proportional; the best we can do is relate them via the particle's "relativistic mass" mrel. Thus
p = mrelv .
When the particle is at rest, its relativistic mass has a minimum value called the "rest mass" mrest. The rest mass is always the same for the same type of particle. For example, all protons, electrons, and neutrons have the same rest mass; it's something that can be looked up in a table. As the particle is accelerated to ever higher speeds, its relativistic mass increases without limit.
It also turns out that in special relativity, we are able to define the concept of "energy" E, such that E has simple and welldefined properties just like those it has in newtonian mechanics. When a particle has been accelerated so that it has some momentum p (the length of the vector p) and relativistic mass mrel, then its energy E turns out to be given by
E = mrelc2 , and also E2 = p2c2 + m2restc4 . (1)
There are two interesting cases of this last equation:
If the particle is at rest, then p = 0, and E = mrestc2.
If we set the rest mass equal to zero (regardless of whether or not that's a reasonable thing to do), then E = pc.
In classical electromagnetic theory, light turns out to have energy E and momentum p, and these happen to be related by E = pc. Quantum mechanics introduces the idea that light can be viewed as a collection of "particles": photons. Even though these photons cannot be brought to rest, and so the idea of rest mass doesn't really apply to them, we can certainly bring these "particles" of light into the fold of equation (1) by just considering them to have no rest mass. That way, equation (1) gives the correct expression for light, E = pc, and no harm has been done. Equation (1) is now able to be applied to particles of matter and "particles" of light. It can now be used as a fully general equation, and that makes it very useful.
Is there any experimental evidence that the photon has zero rest mass?
Alternative theories of the photon include a term that behaves like a mass, and this gives rise to the very advanced idea of a "massive photon". If the rest mass of the photon were nonzero, the theory of quantum electrodynamics would be "in trouble" primarily through loss of gauge invariance, which would make it nonrenormalisable; also, charge conservation would no longer be absolutely guaranteed, as it is if photons have zero rest mass. But regardless of what any theory might predict, it is still necessary to check this prediction by doing an experiment.
It is almost certainly impossible to do any experiment that would establish the photon rest mass to be exactly zero. The best we can hope to do is place limits on it. A nonzero rest mass would introduce a small damping factor in the inverse square Coulomb law of electrostatic forces. That means the electrostatic force would be weaker over very large distances.
Likewise, the behavior of static magnetic fields would be modified. An upper limit to the photon mass can be inferred through satellite measurements of planetary magnetic fields. The Charge Composition Explorer spacecraft was used to derive an upper limit of 6 × 10−16 eV with high certainty. This was slightly improved in 1998 by Roderic Lakes in a laboratory experiment that looked for anomalous forces on a Cavendish balance. The new limit is 7 × 10−17 eV. Studies of galactic magnetic fields suggest a much better limit of less than 3 × 10−27 eV, but there is some doubt about the validity of this method.

If the photon has no mass. How can it react on surfaces whith force. In the case of vanes of mill under a vacuum (crookes radiometer)or in the case of this link
http://www.newscientist.com/mobile/article/dn16158photonforceharnessedtodosomelightwork.html
and second part of question
after reacting on these mechanisims what is energy is used or diminished from the photon?

If the photon has no mass. How can it react on surfaces whith force. In the case of vanes of mill under a vacuum (crookes radiometer)or in the case of this link
http://www.newscientist.com/mobile/article/dn16158photonforceharnessedtodosomelightwork.html
and second part of question
after reacting on these mechanisims what is energy is used or diminished from the photon?
http://en.wikipedia.org/wiki/Optical_tweezers
Optical tweezers are capable of manipulating nanometer and micrometersized dielectric particles by exerting extremely small forces via a highly focused laser beam. The beam is typically focused by sending it through a microscope objective. The narrowest point of the focused beam, known as the beam waist, contains a very strong electric field gradient. It turns out that dielectric particles are attracted along the gradient to the region of strongest electric field, which is the center of the beam. The laser light also tends to apply a force on particles in the beam along the direction of beam propagation. It is easy to understand why if one considers conservation of momentum. Photons that are absorbed or scattered by the tiny dielectric particle in its path impart momentum to the dielectric particle. This is known as the scattering force and results in the particle being displaced slightly downstream from the exact position of the beam waist, as seen in the figure.

A slight clarification about the difference between Mass and Rest Mass:
 The photon, traveling at the speed of light does have mass,and is deflected by a gravitational well like the Sun
photons do not have any "mass'
mass
A measure of the total amount of material in a body, defined either by the inertial properties of the body or by its gravitational influence on other bodies.
a Photon does not affect other particles gravitationally, cause it has zero mass....
No, photons do not have mass, but they do have momentum. The proper, general equation to use is E2 = m2c4 + p2c2 So in the case of a photon, m=0 so E = pc or p = E/c. On the other hand, for a particle with mass m at rest (i.e., p = 0), you get back the famous E = mc2.
http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/960731.html
http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html
http://www.weburbia.com/physics/photon_mass.html
http://home.fnal.gov/~pompos/light/light_page31.html
The photon is massless,[Note 2] has no electric charge,[12] and is stable
http://en.wikipedia.org/wiki/Photon
I am not sure about your assertion that a photon cannot have a gravitational influence. General Relativity is much more complicated that just saying that mass is required for gravity. Photon have energy  energy and energy flux are parts of the calculations to determine curvature. Now I cannot believe we could possibly measure that gravitational influence at present  but I am not sure that it is zero.
However  one fly in ointment, a photon is a exemplary quantum mechanical object, our theory of gravity is GR which is not quantum.
I got the definition of mass from this link.
http://imagine.gsfc.nasa.gov/docs/dict_jp.html#mass ( the goddard space centers site )
mass
A measure of the total amount of material in a body, defined either by the inertial properties of the body or by its gravitational influence on other bodies.

A clarification, please Emc2:
a Photon does not affect other particles gravitationally, cause it has zero mass....
... photons do not have mass, but they do have momentum.
Eddington showed that photons are diverted by the Sun's gravity.
This produces a change in the direction of their velocity, which represents a change in the direction of their momentum.
Because momentum is conserved in interactions, a change in the momentum of the photon must be balanced by a change in the momentum of the Sun (as small as that may be).
Or you could look at it in a Newtonian sense  the photon feels a gravitational force towards the Sun, so the Sun feels an equal and opposite gravitational force towards the photon...
This suggests that the photon does affect other objects gravitationally, does it not?
Some of the advantages quoted for the newer momentumbased relativistic notation is that momentum is conserved in all inertial reference frames, and applies equally to all particles regardless of whether they travel at the speed of light or not...

A clarification, please Emc2:
a Photon does not affect other particles gravitationally, cause it has zero mass....
... photons do not have mass, but they do have momentum.
http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/961102.html ( content from link next )
We also knew that photons are affected by gravitational fields not because photons have mass, but because gravitational fields (in particular, strong gravitational fields) change the shape of spacetime
In 1915 Albert Einstein proposed the theory of general relativity. General relativity explained, in a consistent way, how gravity affects light. We now knew that while photons have no mass, they do possess momentum (so your statement about light not affecting matter is incorrect). We also knew that photons are affected by gravitational fields not because photons have mass, but because gravitational fields (in particular, strong gravitational fields) change the shape of spacetime. The photons are responding to the curvature in spacetime, not directly to the gravitational field. Spacetime is the fourdimensional "space" we live in  there are 3 spatial dimensions (think of X,Y, and Z) and one time dimension.
Let us relate this to light traveling near a star. The strong gravitational field of the star changes the paths of light rays in spacetime from what they would have been had the star not been present. Specifically, the path of the light is bent slightly inward toward the surface of the star. We see this effect all the time when we observe distant stars in our Universe. As a star contracts, the gravitational field at its surface gets stronger, thus bending the light more. This makes it more and more difficult for light from the star to escape, thus it appears to us that the star is dimmer. Eventually, if the star shrinks to a certain critical radius, the gravitational field at the surface becomes so strong that the path of the light is bent so severely inward so that it returns to the star itself. The light can no longer escape. According to the theory of relativity, nothing can travel faster than light. Thus, if light cannot escape, neither can anything else. Everything is dragged back by the gravitational field. We call the region of space for which this condition is true a "black hole" (a term first coined by American scientist John Wheeler in 1969).
Now, being scientists, we do not just accept theories like general relativity or conclusions like photons have no mass. We constantly test them, trying to definitively prove or disprove. So far, general relativity has withstood every test. And try as we might, we can measure no mass for the photon. We can just put upper limits on what mass it can have. These upper limits are determined by the sensitivity of the experiment we are using to try to "weigh the photon". The last number I saw was that a photon, if it has any mass at all, must be less than 4 x 1048 grams. For comparison, the electron has a mass of 9 x 1028 grams.

Surely this equation could be rewritten E = mc^2 + pc. Unless these are not exponential functions.
Then, the equation 5^2 = 4^2 + 3^2 could be rewritten as 5 = 4 + 3?

Other discussions about photons mass or mass in relativity:
http://www.thenakedscientists.com/forum/index.php?topic=21363.0
http://www.thenakedscientists.com/forum/index.php?topic=19617.0
http://www.thenakedscientists.com/forum/index.php?topic=16789.0
http://www.thenakedscientists.com/forum/index.php?topic=8298.0

Light has been slowed and observed.
http://en.wikipedia.org/wiki/Slow_light
The question is why light travels at c in the first place. I can see the point particle principle but if mass increases as we reach light speed then why is the photon not massive? Maybe it is but because of it's other properties this mass can not be detected.
This (green text). Interesting question, I still no have a clue in that point.
Also, how can 'born' a photon without any mass @ 300.000 km/s and without any acceleration as well?

Light has been slowed and observed.
http://en.wikipedia.org/wiki/Slow_light
The question is why light travels at c in the first place. I can see the point particle principle but if mass increases as we reach light speed then why is the photon not massive? Maybe it is but because of it's other properties this mass can not be detected.
This (green text). Interesting question, I still no have a clue in that point.
Also, how can 'born' a photon without any mass @ 300.000 km/s and without any acceleration as well?
I don't think anyone's able to answer that scientifically. Right now, we know that this is how the universe appears to work as a result of our measurements, but no one's been able to actually show a reason why the universe works this way.

Light has been slowed and observed.
http://en.wikipedia.org/wiki/Slow_light
The question is why light travels at c in the first place. I can see the point particle principle but if mass increases as we reach light speed then why is the photon not massive? Maybe it is but because of it's other properties this mass can not be detected.
This (green text). Interesting question, I still no have a clue in that point.
Also, how can 'born' a photon without any mass @ 300.000 km/s and without any acceleration as well?
Are you assuming a corpuscolar behaviour for the photon in this situation? That is, are you assuming that the photon behaves as a little bullett, which starts from the source and moves towards the target being spatially localized in between?

a photon is basically pure energy, that it why it can go from particle to wave and visa versa..
(red text) Hi, can you provide an example?

a photon is basically pure energy, that it why it can go from particle to wave and visa versa..
(red text) Hi, can you provide an example?
http://en.wikipedia.org/wiki/Doubleslit_experiment
This result establishes the principle known as wave–particle duality. Additionally, the detection of individual photons is observed to be inherently probabilistic, which is inexplicable using classical mechanics.[3]

There might be one way more too look at it?
'Gravity' is the metric of Space defining it. I become flabbergasted trying to imagining all those 'geodesics', paths of smallest or no 'resistance that becomes the easiest choice for all uniformly moving things. To see the point here you have to imagine all those objects, all able to 'distort' space and all moving relative each other, in different directions simultaneously. It must be very dynamically balanced system SpaceTime. And the photon is just one of those 'objects moving'. What they all comes down too, from the photon to matter, is that they all can be tracked back to 'energy'. So the geodesics gives 'energy' the paths, and they all must change with relative position, simultaneously for all involved. The alternative, as we never see a photon 'move', only when it 'comes to be' as in the recoil from the system it 'leaves' and in its annihilation, is to assume some logic allowing those two to exist without it needing for anything to be there in between. But if so, it still behaves exactly as all other 'moving objects.
Anyway, photonsmatter, expressions of 'energy'.

It is interesting to note that at the Planck scale, if we consider 1 Planck length to be an absolute, that light travels this discrete distance in 1 Planck time. If we don't worry about the macroscopic scale this all makes sense. Consider a shrunken universe contained within say 10 cubic planck lengths. Each cube will contain shrunken particles, of which only photons can jump to the next cube instantaneously. Every other particle is trapped in a particular cube until its momentum moves it to an edge. In this scenario all particles within each cube will see light at the same speed. To move faster necessarily contracts the distance traveled due to dilation. The distance is experienced as less from an external observer.
If a photon starts 10 cubes away, it will reach each cube within a set interval and all particles within a particular cube will 'experience' it at the same instant. This can be scaled up and still works. So it doesn't matter if we measure in Planck scale or metres.
What if the length contraction also applies to the photon moving through this frame? As if spacetime in and around a moving object does contract physically. This could relate the effects of momentum to those of gravity. Ultimately a black hole contracts all spacetime into a singularity (theoretically).

Does anyone know of any work comparing time dilation and length contraction whilst leaving a gravitational field to those approaching light speed?

Surely this equation could be rewritten E = mc^2 + pc. Unless these are not exponential functions.
Then, the equation 5^2 = 4^2 + 3^2 could be rewritten as 5 = 4 + 3?
Yes sorry I had my stupid head on.

EMC2 assertions concerning the claim that the photon is pure energy have been moved to New Theories
http://www.thenakedscientists.com/forum/index.php?topic=45584.0

Does anyone know of any work comparing time dilation and length contraction whilst leaving a gravitational field to those approaching light speed?
Sorry HB  just seen this question. Firstly  I am pretty sure that you only get length contraction in high relative velocity cases. Secondly  I guess you are looking for qualitative differences between the time dilation caused in the two different circumstances; I just don't know. Frankly I cannot think how time a/o time dilation could be different (aprt from quantitatively). And I do know that you can add the two effects to get a single result  this is done all the time in the GPS satellites; high speed makes the clocks slower, and higher grav potential makes the clocks tick faster, but you can just add the two effects and get the actual correction.
At present I cannot imagine how there would be two types of effect (rather than two causes) and in the one case I know of where they both apply they are simply added  so I would guess there is no difference.

A pretty nice question that one, and yes, they differ. If we imagine Earth as 'moving' at one gravity, the time dilation expressed by that 'gravity' will differ from that of a spaceship moving at a constant one gravity, the faster it moves relative, for example. lights blue shift (measured locally in that frame), or Earth. The equivalence found with gravity is in its 'intrinsic properties' expressed, using a constant uniform acceleration, not about 'time'. And the same goes for length contractions.

A pretty nice question that one, and yes, they differ. If we imagine Earth as 'moving' at one gravity, the time dilation expressed by that 'gravity' will differ from that of a spaceship moving at a constant one gravity, the faster it moves relative, for example. lights blue shift (measured locally in that frame), or Earth. The equivalence found with gravity is in its 'intrinsic properties' expressed, using a constant uniform acceleration, not about 'time'. And the same goes for length contractions.
Also a photon does not always theoretically travel at c. It is assumed that light cannot escape an event horizon. Can this be still considered a vacuum? If not then what is inside the event horizon may not be a singularity. As we know light travels slower through a medium. Does this imply that intense gravity acts like a physical medium?

Also a photon does not always theoretically travel at c. It is assumed that light cannot escape an event horizon.
It doesn't matter if it can't escape: light in the void always travels at c by definition.

hubble_bubble, a photon does always travel at c in a vacuum, by definition. In general relativity, this means with respect to the patch of space immediately around the photon. It's a bit like running on a treadmill. If you run at a constant speed with respect to the conveyor belt under your feet and someone slowly turns the speed up, you'll move at different speeds with respect to someone who's not on the treadmill. You'll eventually start going backwards even though you're gong a constant speed with respect to the belt beneath your feet.
This happens because light always moves at c over regions of flat spacetime. Gravity makes spacetime curved, but over any tiny portion of a curve, it looks flat (much like the earth's surface looks flat to us even though it's not). So over a very small neighborhood, light moves at c. Over a large neighborhood, you have to account for curvature, such as when comparing what two observers at different points in curved spacetime see.