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Can photons travel faster than $c$? (Feynman Lectures)

Physics Asked by bnosnehpets on April 2, 2021

I apologise for the very non-technical nature of this question. I am new to QED and perhaps am interpreting things in the wrong way, but I’ll ask anyway, and hopefully someone can provide a non-technical response.

There are lots of questions on here about virtual particles travelling faster than the standard speed of light such as this one. However, in Feynman’s book QED, The Srange Theory of Light and Matter, it seems that Feynman is not saying that virtual photons can travel faster than light (which is what these questions are asking about), but that there is a probabilty that (real) photons will travel faster (or slower) than $c$ but that these probabilities cancel out over longer distances. (I have added quotes at the bottom to support this).

Is this, like the virtual photons, just a mathematical construction and not to be taken as reality? From reading the rest of the book I would guess not since Feynman uses words like appear frequently when describing what light appears to do.

As a secondary question, Feynman also seems to suggest that photons do not only travel in a straight line. Instead, they can take all paths, but the probabilities of these are very low and once again cancel out.

Is Feynman describing this in a different way from usual? Or am I misinterpreting what he is trying to say? Or is it really true that over short distances photons can travel faster than light (and seemingly violate relativity)?

Edit:

Here is a quote from Feynman’s book (p89):

“…there is also an amplitude for light to go fsster (or slower) than the conventional speed of light. You found out in the last lecture that light doesn’t only go in straight lines; now, you find out that it doesn’t only go at the speed of light!”

Later he goes on to say:

“The amplitudes for these possibilities are very small compared to the contribution from speed c; in fact, they cancel out when light travels over long distances.”

4 Answers

Technically no, because if something went faster than c it would be classified as a Tachyon. If they did however, they would only be a Tachyon by definition, physically they would still be Photons.

Answered by DragonSlayer3 on April 2, 2021

I just saw a related question in another forum, and a commenter there noted that the non-classical paths "aren't actually taken." But Feynman addressed this too, in his QED lectures that are available on YouTube. In particular, he described an experiment with a mirror, and showed that the answer was dominated by the classically reflecting bit of the mirror, and the contributions from the rest of the mirror cancelled out. However, he then proceeded to turn the mirror into a diffraction grating by removing the pieces well away from the classical path that contributed "negative phase." And in that case the mirror does reflect at an odd angle.

However, if you actuall did this experiment with a diffraction grating designed to make the entire surface reflect (i.e., the grating gradually became a normal mirror in the classically reflecting region), and used a continuous light source, then I'd think the photons that travelled longer path would leave earlier. So your total response at the sensor would be composed of components that were from different cycles of the source.

Answered by Kip Ingram on April 2, 2021

It is important to see a direct quote of where and how Feynman uses the word "light" and where and how "photon".

The quote you give speaks of light.

Light emerges in a complicated quantum mechanical superposition from zillions of photons. Photons are not light, though they are marked by the frequency that light built up by them will display, by E=h.nu. Photons are zero mass particles with spin + or - 1 to their direction of motion. Light in the superposition of the photon wave functions displays all the wave properties of classical Maxwell equations.

I have found that this plot gives an intuition of how this happens:

enter image description here

Even though the photons follow straight paths with just + or -spin, light displays polarization, a complicated function in space, here seen with the electric field vector of the classical electromagnetic wave. The connection to the quantum comes in the spin orientation of the photon.

How the quantum formalism of QFT handles this needs mathematics, and it is outlined here .

My guess, as I do not have the book, is that when Feynman is talking of cancellations he is talking of the classical light field functions built up by the photons.Collective wave behavior with group velocity and phase velocity create complications in light propagation not relevant for photon behavior, which , in my books, always travels at c.

For light:

In vacuum, the phase velocity is c = 299 792 458 m/s, independent of the optical frequency, and equals the group velocity. In a medium, the phase velocity is typically smaller by a factor n, called the refractive index, which is frequency-dependent (→ chromatic dispersion). In the visible spectral region, typical transparent crystals and glasses have refractive indices between 1.4 and 2.8. Semiconductors normally have higher values.

I would be interested in a direct Feynman quote where it says photons may travel faster than c.

Answered by anna v on April 2, 2021

Yes, parts of a wave function can travel faster than light, but from my understanding, much of it has to do with the uncertainty of the position of the particle the wavefunction represented in the first place.

For example, there is active research into how to interpret the results of quantum tunneling experiments that indicate "superluminal tunneling." This recent article from Quanta Magazine‘ explains that research area well. There are several competing definitions on the tunneling time, because time duration is not an quantum observable.

It is thought, but as far as I know not proven, that attempting to use this vanishingly small superluminal part of the wavefunction to send information will always be less efficient than sending the light directly, because for a large barrier, nearly all of the wavefunction is reflected.

(I don't know about how to reason about the case Feynman described, because not enough context about the quote is given.)

Answered by Jonathan Jeffrey on April 2, 2021

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