# Do Photons Accelerate Upon Creation? [duplicate]

Of course the speed of light is constant relative to everything. Its frequency can change, but its speed is constant. Photons may be absorbed by atoms or release by atoms as energy, but as soon as it is released, does it have infinite acceleration to 300,000 m/s instantaneously as soon as it is released (assuming it is released basically at rest)?

## marked as duplicate by ACuriousMind♦, Kyle Kanos, David Z♦Sep 28 '14 at 2:49

Yours is a subtle question with a rather subtle answer. From the way you ask the question, you seem to be thinking of the photon as a little billiard ball. It is not. It is an excitation of a quantum field, which is described very differently from a "particle" in the classical sense. Even so, the short answer, in some very subtle ways, is that indeed there is a sense wherein the photon can be thought of as accelerating over a nonzero time.

You must think of a whole system which makes a measurement to probe your question. So lets think of a system comprising (1) an excited atom, just after photon absorption, enclosed in and at the centre of a (2) detector sphere, which has a huge number of photon detectors on it so that it will "catch" the photon however it might be emitted. You begin timing and wait for detection. The only meaningful measurable concept of the photon's speed is as $R / \Delta t$, where $R$ is the sphere's radius and $\Delta t$ the time until detection. You would do this experiment many times and with spheres of different radiusses. If your spheres were very big, you would get the answer $c$ as the photon's speed. You would, however, indeed see some weird subtleties if you could do this experiment with very small spheres.

First of all, there is always a nonzero emission lifetime. My main daytime paying gig is as an optical physicist and I work a great deal with fluorophores. So I work a great deal with fluorescence lifetimes, which are much longer (nanoseconds) than the lifetime for emissions you likely have in mind. They are all the same concept for the purposes of this question. The atom emits the photon at a random time, so if you did the experiment with a sphere of 30cm or so with a fluorophore like FITC ($\tau=4.5{\rm ns}$ at standard pH and temperature), you would see a big spread in the $R / \Delta t$ answer you would get from this experiment - and all the answers would be considerably less than $c$. So in this first sense, you can think of the fluorescence (or other, equivalent) lifetime as the acceleration time for the photon. It is simply not meaningful to ask what really happens to a photon between measurable events. In quantum mechanics and quantum field theory in general, this denial of meaning before measurement is very important for keeping QM and QFT consistent with special relativity. The jargon for this is that modern physics rejects counterfactual reality. ("counterfactual reality" is the mouthful standing for the notion that an experiment's outcome has a meaning and existence before the experiment is done).

But I'm guessing you might be a little unsatisfied with a denial and stipulation of fluorescence lifetime as an acceleration time for your answer. But we can still squeeze some more out of our thought experiment: it shows that the fluorescence lifetime can indeed be thought of as the acceleration time in a sense that is near to what you are imagining. I want you to imagine now that your sphere is a few atoms in diameter. Things begin to get very weird indeed now: let's explore what theory has to say a bit more.

A lone, first quantised photon propagates following Maxwell's equations. I explain this in my answer here. We can get a pretty good analogy for our excited atom as a tiny, subatomic sized ring of classical current, with stored energy in the magnetic field through the ring. The circulating charge radiates, thus quenching the current, and the classical, Maxwell equation described radiation field can be reinterpreted as a probability amplitude to absorb the photon at any point as follows: you take the system's whole energy (initially stored as a magnetostatic, inductor energy) and you normalise it to unity. The propagating electromagnetic energy density $\frac{1}{2}\epsilon_0 |\vec{E}(\vec{r},\,t)|^2 + \frac{1}{2}\mu_0|\vec{H}(\vec{r},t)|^2$ becomes the probability density to absorb the photon at position $\vec{r}$ at time $t$. The inductor's (current ring's) current decays exponentially with time constant $L/R_{rad}$, where $R_{rad}$ is the radiation resistance of the ring thought of as a magnetic dipole antenna: this $L/R_{rad}$ is the fluorescence (or generalised emission) lifetime.

The radiation's fastest moving wavefront can indeed leave the ring at speed $c$. That is the nature of a nondispersive wave: it is not made of "stuff" with rest mass and there is truly no acceleration time. Moreover, as in any antenna problem, you calculate the electric and vector magnetic potentials as retarded potentials: at any point $\vec{r}$ and time they are given by the currents in the ring at a time $|\vec{r}|/c$ beforehand. The disturbance is a wave, and it always moves at $c$, so in this sense there is no acceleration time. But the wavefront is of a probability amplitude field: this is very different saying the delay, calculated from the actual absorption time, is $|\vec{r}|/c$. The absorption will be sometime later: the minimum delay is $|\vec{r}|/c$, but it is most likely to be quite a long time after $|\vec{r}|/c$. The nearer you are to the ring, the bigger the discrepancy. When your separation becomes comparable to the ring's dimensions, the variations can be enormous. If you really could get your detectors this near to the ring, your experimental results really would seem to show a nonzero acceleration time for the photon. However, now the matter of your detectors would begin to disturb the boundary conditions for this Maxwellian problem: your detectors themselves would influence the fluorescence lifetime!

You may know that any antenna solution to Maxwells equations in the freespace around the antenna comprises two components:

1. A "farfield": this is a superposition of infinite plane waves, which all move at exactly $c$, without an acceleration time. As I said, that is the nature of a nondispersive wave. Sound doesn't "accelerate", for instance: the disturbance simply passes from one part of the medium to the next as fast as the constituents can play Chinese Whispers;
2. A "nearfield" or "evanescent field": this is the nonpropagating part and is "bound" to the antenna. it changes over lengthscales that are small compared to the radiation wavelength. Its "velocity" is a pretty meaningless concept: the phase and group velocities are all over the place if you try to calculate them. I say more about evanescent waves here, here and here.

So the photon, in leaving the atom, can be thought to partly comprise an analogy of the antenna near, evanescent field. It is this evanescent field which would make our thought experiment seem to say that the photon is accelerating.

"But don't things moving at less than $c$ always have a nonzero rest mass?", I hear you ask. (recall $E^2=p^2 c^2 + m^2 c^4$)

Quite right. The "farfield" part of the probability amplitude field is indeed massless and always moves at exactly $c$. There is no acceleration time for this part of the field. But this is not the whole picture. The evanescent field is bound to the antenna, and confined energy always has a rest mass, as I discussed in my answer yesterday. It is this part of the photon field that begets the weird, "nonzero acceleration time" results to our thought experiment.

In the Feynman diagram world, my "farfield" becomes "photons" and "nearfield" becomes "virtual photons". When physicists say the word "photon", they most often mean "free photon" or "on-shell photon". These always travel at speed $c$. On the other hand, a virtual photon i.e. one whose whole history begins and ends within a Feynman diagram, as opposed to leaving the diagram, is not observable and is off-shell, which means it does not fulfill the fundamental relationship $E^2=p^2 c^2 + m^2 c^4$. In effect, it can be thought of as moving at any speed, and a proper evaluation of the Feynman diagram coherently sums up all the amplitudes for it to be moving at all speeds.

• Wow! Thank you for that incredibly lengthy and detailed answer! What a guy! I certainly seem to have some misconceptions about subatomic particles, though I did understand that photons are not like billiard balls. Thanks, again! – Goodies Sep 29 '14 at 3:18
• No problems. I see your question has already been answered in a slightly different way: there are some good answers and I generally agree with even the ones that say there is no acceleration time: as in my answer, the field propagates at $c$ with zero acceleration time (like a water wave), but the observable phenomenon really does have an acceleration time, manifested through emission lifetimes. – WetSavannaAnimal Sep 29 '14 at 3:59