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Is the photon's frequency its theoretical frequency if time could pass for the photon? Essentially a photon is travelling at the speed of light which according to the Lorenz factor means no time at all will pass for the photon as it travels. It is a hard concept for me to understand, but basically this is my understanding so far.

A photon has no mass whatsoever, and always travels at the speed of light and when colliding with nuclei or electrons it changes its frequency to change its energy without changing its speed. This change in frequency is the change in energy allowing the photon to lose no kinetic energy and this change in frequency is theoretical because the photon never changes in time. The frequency being a measure of the amount of vibrations that would occur in 1 second if 1 second could pass for the photon.

Sorry if what I said didn't make sense or was super obvious and I am uninformed. Just hard to conceive something without mass or something that time stands still for.

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  • $\begingroup$ "The frequency being a measure of the amount of vibrations that would occur in 1 second if 1 second could pass for the photon." -- Photon frequency is not defined in the photon's rest frame (which does not exist). It's measured by an observer and depends on the observer's frame. Note that neither frequency nor energy are invariant quantities. $\endgroup$
    – Sten
    Commented Aug 6, 2023 at 13:00
  • $\begingroup$ The problem is that you're trying to think of things from the photon's rest frame, but this frame isn't well-defined mathematically. It doesn't really make sense to say "time doesn't pass for a photon" because there is no "for a photon". See this thread for details: physics.stackexchange.com/questions/774174/… $\endgroup$
    – Mikayla Eckel Cifrese
    Commented Aug 6, 2023 at 19:47
  • $\begingroup$ A change in wave frequency doesn't require a change in velocity. For light, the relevant equation is $v=\lambda f$ where $\lambda$ is wavelength. Hence, the frequency can change while the velocity remains constant, so long as the wavelength changes inversely. Additionally, as I touched on in my other comment, it's not true that photons never change in time. Their speed doesn't change, but in any defined reference frame, they move through time just like anything else and so changes in frequency aren't just theoretical but physical. $\endgroup$
    – Mikayla Eckel Cifrese
    Commented Aug 6, 2023 at 19:54

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The real meaning of a sentence like

...no time will pass for the photon as it travels.

Is simply that there is no reference frame where the photon is at rest. It may sound weird for people used to a non-relativistic point of view, but Nature is not obliged to follow our limited daily experience with non-relativistic speeds. The statement that no time passes for the photon is a nice pop-sci idea, contradicting the sound physical statement that there is no photon reference system. Better to ignore it if one would like to build sound concepts.

Photons are described by quantum field theory (QFT). They are not classical particles, and in particular, their interaction with other particles is better described as an event where a photon with a given energy disappears, and another photon with the same or different energy appears. From the QFT point of view, there is no vibration of a single photon. Its frequency should be interpreted as synonymous with energy. The connection with frequency is in the fact that a macroscopic superposition of such photons can be interpreted at the macroscopic level as an electromagnetic wave of a frequency related to the single photon energy by the Planck relation $\varepsilon=h \nu$.

The two comments above should clarify the point of view based on the known Physics, preempting your question. Unfortunately, pop science tends to make spectacular statements to capture the attention, even if they are not always rooted in sound Physics.

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Classically, a photon (or plane wave) can be described, in a reference frame, by a 4-wave vector:

$$ k_{\mu} = ({\omega/c, \vec k}) $$

with ofc, $\omega = c|\vec k|$.

Note that neither $\omega$ nor $\vec k$ are Lorentz invariants...they depend on the reference frame in which the photon/wave is observed.

So there is no inherent frequency or wavenumber associated with a photon, you have to pick a reference frame. If it's from a well defined atomic/nuclear transition, it's natural to pick the reference frame of the source, but, really that doesn't mean anything to the lone photon flying through space.

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