# Frequency of a photon

• B
akaSmith
TL;DR Summary
How can a photon have a frequency? Anything moving at the speed of light is predicted to have a zero time rate, e.g. the frequency of a ticking clock would be zero. So no aspect of the light should change along its path - in the same way that no aspect of the moving clock would change.

If special relativity doesn’t apply to light, i.e. a Lorentz transformation is not applied to give light zero frequency, then it cannot be used to multiply a photon's zero rest mass by infinity to give it mass. So photons would have no energy or frequency or even existence. Does the theory apply to light or not?

• weirdoguy

Staff Emeritus
The value of the frequency of a photon, like it's energy, is frame dependent. In fact, quantum mechanics tells us that the energy is Planck's constant times the frequency. So if you have just a photon, you don't know it's energy, not until you specify a particular frame of reference.

I would say it's generally incomplete to talk about the energy of a photon as if it were just a property of the photon for this reason. A possible exception might be the 4-frequencey, https://en.wikipedia.org/wiki/Four-frequency, but I'm assuming that's not the sort of frequency you were talking about.

akaSmith
OK, let's talk about a light beam whose frequency has been measured. Does SR apply to the light or not?

2022 Award
OK, let's talk about a light beam whose frequency has been measured. Does SR apply to the light or not?
Of course. However, attempting to transform into the "rest frame of the photon", as you are implicitly doing, is self-contradictory and leads to the kind of nonsense you are finding. That doesn't mean SR doesn't apply, just that it doesn't work the way you think it does.

akaSmith
I'm not transferring into the photon's rest frame. I'm using SR's predicted time rate, as seen by an observer in their rest frame, not the photon's frame. Does the prediction work or not?

2022 Award
I'm using SR's predicted time rate, as seen by an observer in their rest frame
That's just the inverse transform of the one I mentioned. It's equally self-contradictory. So the predictions you are making do not work, but that's not a problem with SR, that's you applying formulas to situations where they are not valid.

akaSmith
Where does Einstein say in his 1905 paper that SR does not apply to light?

2022 Award
Where does Einstein say in his 1905 paper that SR does not apply to light?
He doesn't. Why would he?

The problem with what you are doing is that you are using maths that relates quantities measured in different inertial frames. However, there are no inertial frames moving at light speed, so you cannot use this formula to relate your measurements to "light's measurements" since there is no coherent way to define the latter.

akaSmith
An inertial frame is a mathematical construct that moves at a constant velocity v. This can take any real value including c. You can't ban inertial frames that you don't like. Nor can you apply the part of SR you like and ignore the part that's nonsense.

Mentor
An inertial frame is a mathematical construct that moves at a constant velocity v. This can take any real value including c. You can't ban inertial frames that you don't like. Nor can you apply the part of SR you like and ignore the part that's nonsense.
An reference frame is a tetrad which consists of one timelike and three spacelike basis vectors. A photon cannot have a reference frame where it is at rest because its tangent vector is null, not timelike. You cannot just make up your own definitions, particularly when they are known to not work.

• etotheipi
Gold Member
An inertial frame is a mathematical construct that moves at a constant velocity v. This can take any real value including c.
No, it can't. If it could a particle with rest mass could move with c.

2022 Award
An inertial frame is a mathematical construct that moves at a constant velocity v. This can take any real value including c.
One of the postulates of special relativity is that the speed of light is the same in all inertial frames of reference. Frames of reference moving at ##c## would involve light moving at ##c## and being stationary at the same time. As such, they are ruled out in relativity. If you ignore this and attempt to construct such a frame anyway, it leads to you requiring two perpendicular vectors that are also parallel.
Nor can you apply the part of SR you like and ignore the part that's nonsense.
The nonsense follows from you implicitly defining a frame that is disallowed by the postulates of the theory.

• etotheipi
akaSmith
So you can't apply the Lorentz factor to give photons mass and they have no energy?

2022 Award
Photons have zero mass. They have non-zero energy. No, this does not follow from the Lorentz transforms.

• etotheipi
Gold Member
2022 Award
I'm not transferring into the photon's rest frame. I'm using SR's predicted time rate, as seen by an observer in their rest frame, not the photon's frame. Does the prediction work or not?
First of all you should not think about photons yet. First you should learn about classical electrodynamics, which is a relativistic field theory and that's why Einstein discovered special relativity while thinking about electrodynamics of moving bodies or more formally about how to transform from one inertial frame of reference to another such that Maxwell's equations are form-invariant, as it must be for any theory.

At the same time the electromagnetic field is somewhat special, because it is a massless vector field. This means that the free electromagnetic field, i.e., a field in a vacuum is a wave field with a frequency ##\omega## and a wave number ##\vec{k}## related by ##\omega=c |\vec{k}|##, where ##c## is the phase velocity, which is the speed of light. Any free massless field has this phase velocity, and it's a special velocity in the theory of relativity, because it's the "limiting speed", i.e., two inertial reference frames can have at most the relative speed ##c##.

Particularly there are no "photons" at rest. It's anyway wrong to think about photons as particles. They are far from being particles already by the fact that there's no proper definition of its position. That's another specialty of the masslessness of the electromagnetic field. It's better think about photons as representing an electromagnetic field with the smallest possible intensity possible for a given frequency. Technically speaking it is a socalled Fock state of the quantized electromagnetic field. The energy of a photon is given by ##\hbar \omega## and its momentum ##\vec{q}=\hbar \vec{k}##.

If you take the analogy with massless particles, it's clear that there cannot be a rest frame of a massless particle, because of the energy-momentum relation ##E^2/c^2=\vec{p}^2##. That means that the speed is ##|\vec{p} c^2/E=c## wrt. to the inertial reference frame we observe the particle in (i.e., our rest frame). Now there's no frame of reference, which moves with a velocity ##c## or larger, i.e., we cannot transform into any frame, where the particle is at rest. It moves with ##c## in any reference frame.

What changes, however, is the frequency as well as the wave vector (or the energy and the momentum of the corresponding photon) when looking at the em. wave from a different frame of reference. A natural specification of the frequency and wave vector is the rest frame of the source of this wave, but you can of course transform to any frame of reference, where the light source moves and the frequency and wave vector change when looking at the wave from a frame, where the light source moves. That's the Doppler effect (i.e., if the light source moves towards you, the frequency is large than when you look at it frome the frame where the light source is at rest). But the change is always such that ##\omega^2-c^2 \vec{k}^2=0## in all frames (or equivalently for the corresponding photon ##E^2-c^2 \vec{p}^2=0##, which tells you that the mass (more precisely the invariant mass, which is the only mass we use in physics nowadays) is always 0. This again implies that in any frame the phase velocity of an em. wave in vacuo is ##c##, the universal limiting speed of special relativity.

• etotheipi
Mentor
So you can't apply the Lorentz factor to give photons mass and they have no energy?
Mass is an invariant so you do not need to transform to the non-existent light pulse's rest frame to determine that a pulse of light has 0 mass and non-zero energy.

akaSmith
Photons have zero mass. They have non-zero energy. No, this does not follow from the Lorentz transforms.

Physicists are forced to conclude photons have zero rest mass because SR multiplies their mass by infinity using the Lorentz factor.

• Mentor
Physicists are forced to conclude photons have zero rest mass because SR multiplies their mass by infinity using the Lorentz factor.
No. Mass is an invariant given by ##m^2 c^2 = E^2/c^2 - p^2##. From this we can get the 0 mass in any frame.

• etotheipi and vanhees71
akaSmith
Mass is an invariant so you do not need to transform to the non-existent light pulse's rest frame to determine that a pulse of light has 0 mass and non-zero energy.

You can't ban a set of moving axes and I'm staying in the observer's frame.

2022 Award
Physicists are forced to conclude photons have zero rest mass because SR multiplies their mass by infinity using the Lorentz factor.
The Lorentz gamma factor is irrelevant to this as we just explained. So your chain of reasoning is wrong. I think you need to learn relativity properly, and then relativistic quantum field theory if you want to talk about photons.

• etotheipi and vanhees71
2022 Award
You can't ban a set of moving axes and I'm staying in the observer's frame.
Are you just going to keep repeating false statements even after we've explained why they are false?

• Mentor
You can't ban a set of moving axes and I'm staying in the observer's frame.
You cannot use a null axis where you need a timelike axis. You don't get to make up your own rules.

• etotheipi
Staff Emeritus
You can't ban a set of moving axes

But there is something we can ban... Seriously, @Ibix is right. Why are you here? Repeating false statements is not going to help anyone. If you have a question, formulate it carefully. If you think relativity is wrong, please read a) the forum rules and b) a textbook.

• • vanhees71, berkeman, weirdoguy and 1 other person
Mentor
Since the OP has been given a vacation, this thread is closed.