Photons can't accelerate, can they?

[gib65]

When a photon is emitted from an electron, is it instantly traveling at c? Doesn't this contradict F=ma*? Or, in other words, doesn't an object have to accelerate if it is to go from one speed (such as rest) to another (such as c)?

 

*I know photons can be considered massless, but then why are they pulled down by gravity?

[swansont]

A photon is never at rest. Light being affected by gravity is a geometrical argument — space is curved and light follows the null geodesic (Ask yourself: what is a straight line in a non-Euclidean geometry?)

[gib65]

So then is the problem solved by saying m=0? That would mean F=0, which would mean it takes no force to get a photon moving, which means that a could be anything (even infinit), which means it could go from 0 to c in no time.

 

Is it right to think of a photon as at rest before it is emitted by an electron (or at least going at the same speed as the electron)? Is the photon 'in' the electron?

[DrRocket]

So then is the problem solved by saying m=0? That would mean F=0, which would mean it takes no force to get a photon moving, which means that a could be anything (even infinit), which means it could go from 0 to c in no time.

 

Is it right to think of a photon as at rest before it is emitted by an electron (or at least going at the same speed as the electron)? Is the photon 'in' the electron?

 

What "problem" ? What "force" ?

 

Photons are never at rest. The expected speed of a photon is always c. Classically the speed is always c.

 

An electron is an elementary particle. So far as is known it contains no other particles. Photons are not "in" the electron, except insofar as the energy of an emitted photon is reflected in a change of the energy of the emitting electron.

[gib65]

What "problem" ? What "force" ?

 

Photons are never at rest. The expected speed of a photon is always c. Classically the speed is always c.

 

An electron is an elementary particle. So far as is known it contains no other particles. Photons are not "in" the electron, except insofar as the energy of an emitted photon is reflected in a change of the energy of the emitting electron.

 

So then what happens when the electron emits the photon? Is the photon created ex-nihilo?

[DrRocket]

So then what happens when the electron emits the photon? Is the photon created ex-nihilo?

 

The electron changes eneergy states and a photon of the frequency corresponding to the energy difference is emitted.

[gib65]

The electron changes eneergy states and a photon of the frequency corresponding to the energy difference is emitted.

 

Yes, but where was that photon before it was emitted? Was it created at the moment of emission?

[mississippichem]

Yes, but where was that photon before it was emitted? Was it created at the moment of emission?

 

One could say so. There is no law against not conserving the number of bosons that I'm aware of.

[DrRocket]

Yes, but where was that photon before it was emitted? Was it created at the moment of emission?

 

Of course.

[gib65]

That makes sense then. The photon doesn't have to start at some speed less than c because it is going at c the moment it is created.

[J.C.MacSwell]

That makes sense then. The photon doesn't have to start at some speed less than c because it is going at c the moment it is created.

 

There is an impulse/reaction when the photon is created. So there is, presumably, a force for some duration.

[swansont]

There is an impulse/reaction when the photon is created. So there is, presumably, a force for some duration.

Momentum will be conserved — the atom recoils. Works in reverse, too. The force is tougher to nail down, because you don't know the duration.

[DrRocket]

There is an impulse/reaction when the photon is created. So there is, presumably, a force for some duration.

 

Not necessarily. There is a momentum change, but dp/dt need not exist is the usual sense of calculus, since the photon does not exist until it is created, and does not accelerate. So p is a step function and dp/dt exists only in the sense of a Schwartz distribution -- the "Dirac delta".

 

The quantum world is not Newtonian. Neither the electron nor the photon are "little marbles".

[between3and26characterslon]

what about a reflected or refracted photon, in both cases a change of direction has occurred and therefore an acceleration.

[Klaynos]

what about a reflected or refracted photon, in both cases a change of direction has occurred and therefore an acceleration.

 

The photon is absorbed, and then re-emitted.

[Widdekind]

Not necessarily. There is a momentum change, but dp/dt need not exist is the usual sense of calculus, since the photon does not exist until it is created, and does not accelerate. So p is a step function and dp/dt exists only in the sense of a Schwartz distribution -- the "Dirac delta".

 

The quantum world is not Newtonian. Neither the electron nor the photon are "little marbles".

 

May I please request a clarification -- physical phenomena are not actually exactly instantaneous, but occur over some "spread" or "uncertainty" in time, [math]\Delta t[/math]. Then, per the HUP, [math]\Delta E \Delta t \ge \hbar[/math] -- where also, for photons, [math]\Delta E = c \Delta p[/math] -- the finite-and-non-zero "generation time" implies that even individual photons are not pure-and-mono-chromatic. To wit, a photon "wave packet", being localized in space, will have some spread in momentum-and-energy.

 

Moreover, all of the other quantum particles involved, in the photon-emitting process, will also be "wave packets", presumably partially localized in space, and so also having their own momentum spreads. Thus, momentum conservation, for the system, will involve some "slop" or "blur", [math]\left( p_1 + \Delta p_1 \right) + \left( p_2 + \Delta p_2 \right) \rightarrow \left( p_1' + \Delta p_1' \right) + \left( p_2' + \Delta p_2' \right) + \left( p_{\nu} + \Delta p_{\nu} \right)[/math].

 

Is this not so ? Were it the case, that [math]\Delta p_{\nu} \ll p_{\nu}[/math], then would not [math]F_{recoil} = dp/dt \approx p_{\nu}/\Delta t_{\nu} \approx E_{\nu}/c \times \Delta E_{\nu} / \hbar[/math], so that [math]F \propto E_{\nu} \Delta E_{\nu}[/math] ? Physically, such would seemingly say, that higher energy, 'harder' photons obviously 'kick' more; and, more mono-chromatic, 'pure' photons, with lower [math]\Delta E_{\nu}[/math], taking longer to generate or 'erect', would smooth out and so slow down that 'kick' ("photons slowly spun off"); and, more multi-chromatic, 'colorful', photons, with higher [math]\Delta E_{\nu}[/math], which were more quickly created -- and which, being more spread in momentum, would be more localized in space -- would concentrate, compress, and so increase the 'kick' ("gun-slung generated & fired from the hip like a BB-sized bullet").

 

The photon doesn't have to start at some speed less than c because it is going at c the moment it is created.

 

Cannot photons 'accelerate', "in frequency", red-shifting up-and-out-grav-well, or blue-shifting down-and-in-grav-well ?

[bombus]

Some have suggested that the path of every photon in the universe (and every particle for that matter) could be set for all time and it is only our movement through time that gives photons/particles the impression of 'moving' through space.

[alpha2cen]

When light moves close to Black Hole the speed of the light might be affected. Does the speed of light change when light moves direct to the Black Hole?

The speed of the light = C + C_BH

where

C_BH; the speed increment cased by Black Hole attraction

 

Position : ---------- near Black Hole (high gravity)----------------------------->far away from the Black Hole(no gravity)

photon property : high speed(C+C_BH) --------------------------------------------the speed of light( C ), shorter wave length.

 

We can detect short wave electromagnetic waves(gamma waves) near Black Hole through the space telescope.

Edited July 31, 2011 by alpha2cen

[swansont]

When light moves close to Black Hole the speed of the light might be affected. Does the speed of light change when light moves direct to the Black Hole?

The speed of light changes because the frame of reference is no longer locally flat.

[khaled]

So the reason why we can observe free electrons, is that when they move from their energy-level in their atoms to a level that is not gravitationally impulsed by the atom, then the energy it lose is producing photons that we observe as light-blue bolts ...

 

What is amazing is that theoretical physicists think that, energy of any primitive particle is not based on the quarks inside them themselves, but it's based on the empty space between those quarks ...

 

That belief implies that gravitational force is an important aspect to energy of primitive particles in an atom, what do you think ?

Edited September 2, 2011 by khaled

[swansont]

So the reason why we can observe free electrons, is that when they move from their energy-level in their atoms to a level that is not gravitationally impulsed by the atom, then the energy it lose is producing photons that we observe as light-blue bolts ...

 

What is amazing is that theoretical physicists think that, energy of any primitive particle is not based on the quarks inside them themselves, but it's based on the empty space between those quarks ...

 

That belief implies that gravitational force is an important aspect to energy of primitive particles in an atom, what do you think ?

 

Huh? No part of QM contends that gravity is responsible for photon emission or any other interaction at this scale.

[MigL]

I assume Khaled means binding energy since he has mentioned the 'forces in the space between quarks' in other threads. He mistakenly assumes the force to be gravitational.

[questionposter]

Wait, an acceleration is a change in direction too, but doesn't changing a direction require a consistant increase in force or energy? But when a photon goes to a lower gravity potential due to distortion in the fabric of space, how is it's potential energy changing, since I thought it was constant? How doesn't a photon deliver more or less force or energy or momentum or w/e based on it's gravity potential?

Edited September 4, 2011 by questionposter

[swansont]

Wait, an acceleration is a change in direction too, but doesn't changing a direction require a consistant increase in force or energy? But when a photon goes to a lower gravity potential due to distortion in the fabric of space, how is it's potential energy changing, since I thought it was constant? How doesn't a photon deliver more or less force or energy or momentum or w/e based on it's gravity potential?

 

It will be red- or blue-shifted if its gravitational potential changes.

[questionposter]

It will be red- or blue-shifted if its gravitational potential changes.

 

Well where does all the extra energy go? If a gamma-ray goes to a lower gravitational potential, it then becomes a less energetic photon? And if so, where did that extra energy go?