Why does light have multiple frequencies?

Question

The wavelength of visible light ranges from 750 - 400 nm, and so do the corresponding frequencies. However, a photon only has one frequency, given by $E =h\nu$, at a given time, and it can’t be changed unless the photon gets energy from somewhere, which isn’t possible in the vacuum; i.e. once it leaves the source it can’t possibly get enough packets of energy to become excited or even lose the energy to some other particle, as space is empty.

So, why does light come with so many wavelengths and frequencies if a single photon can only have one frequency at a time and are emitted from the same source?

A single photon goes with its individual oscillating electric and magnetic fields, right? So multiple photons mean multiple fields. Won’t these different fields affect the adjacent fields in any way and change their properties?

Answer 1

Light comes with so many wavelengths because it is made of so many photons. A typical lightbulb puts out something on the order of $1\ \mathrm{W}$ of power in the visible spectrum, while individual photons in the visible spectrum each have an energy on the order of $10^{-19}\ \mathrm{J}$. Therefore, a typical lightbulb will produce something on the order of $10^{19}$ photons per second. While each photon has one specific frequency, there are so many of them that the ensemble of photons appears to have a continuous range of frequencies.

Answer 2

The question, as I understand it, is: why does a single body emit light of many frequencies?

The problem lies in the question itself. The sources we are dealing with, such as stars or light bulbs, are single macroscopic sources. They are composed of countless particles, each producing light because of its thermal motion.

We know from thermodynamics that a body of a certain temperature T is composed of particles whose energy is not uniform. I have to admit it's been a while since the last time I had to do with thermodynamics, and I don't want to give you wrong information, but I think a look into Boltzmann distribution may be... illuminating.

Once you see that what you call "single source" is actually a group of sources with different energies, it should be easy to see why the body emits photons of different energies.

As for your last question, photons are electromagnetic particles, that's true, but at the same time they have no electric charge, so they can't change the properties of other photons. Only a charged particle can interact with the electromagnetic field.

Answer 3

Light is made up of individual photons and like you said photons get their energy at the source. Depending on the amount of energy it receives, a photon can have any frequency.

Answer 4

Once it is emitted, the wavelength of a photon does not change (if we ignore cosmological redshift due to the expansion of space). But different photons can be emitted with different wavelengths.

Answer 5

Planck's law predicts the spectral energy density $u$ of a radiating blackbody and is given by $$ u(\omega) = \frac{\omega^2}{\pi^2 c^3} \hbar\omega \frac{1}{e^{\frac{\hbar\omega}{k_B T}}-1} $$ A typical derivation involves the canonical partition function from statistical mechanics to derive the average energy of the system. In statistical mechanics every quantity has a certain distribution and isn't exact anymore, because while we could know the exact position and momentum of every constituent this is an unreasonable ask. To get around this problem we discard a lot of the information and ask for simpler quantities such as the pressure or energy of the system instead of asking about every single position. As such we can only ascribe a probability to the system to be in a specific state and this leads to a continuous distribution of the various quantities of interest.

Another thing to note is that as ChiralAnomaly points out in the comments is that on a quantum mechanical level we need to consider the Heisenberg Uncertainty principle. Despite time not being an operator we can derive $$ \Delta E \Delta t \geq \frac{\hbar}{2} $$ So if you consider an atom in an excited state it will necessarily decay into a lower energy state with a certain half life $\Delta t$ and as such the emitted photon won't have an exact energy. If you're doing spectroscopy on real gases it's even worse because the lines you observe will be broadened by the Doppler Effect due to the thermal motion of the atoms and also pressure broadened due to collisions between the atoms.