[yor_on]

And a very nice point there "  Fourier tells us that we should be able to detect harmonics of the fundamental frequency. associated with, say, a LED or similar single spectral line. We can't."

I know of people making what you state there a career Alan. And looking at a 'photon' as a wave function is a tricky part of physics, although making theoretical sense.

[alancalverd]

Indeed. But the wave function of a photon is very easy to use and understand.

And just to go back to the original question, you can't "validate" E = hf because it is the definition of h. You can't validate " a cow is a female bovine" because that is the definition of a cow. What you can do is to make lots of measurements that convince you that h is indeed a universal constant, and cows moo at one end, poo at the other, and make milk in the middle..

[Bill S]

....cows moo at one end, poo at the other, and make milk in the middle.

Surely, every physicist knows that a cow is a sphere, so, which end is which?. 

[Bored chemist]

Fourier tells us that we should be able to detect harmonics of the fundamental frequency. associated with, say, a LED or similar single spectral line. We can't.

Practically every "green laser" you see- and they aren't rare- is actually the second harmonic of an infrared laser beam.

[Bored chemist]

you can't "validate" E = hf because it is the definition of h.

Actually, you can. There are other ways of measuring h
https://en.wikipedia.org/wiki/Brownian_motion#Einstein's_theory
https://en.wikipedia.org/wiki/Noise_temperature

And other ways of measuring the energy- you can count the photons in a laser beam and then see how well the laser warms something up
https://en.wikipedia.org/wiki/Bolometer
And you can measure the wavelength with a ruler.
http://ipl.physics.harvard.edu/wp-uploads/2013/03/15c_s07_5.pdf

And e=hf is not the definition of h anyway.
The definition comes from Planck's work on black body radiation
https://en.wikipedia.org/wiki/Planck%27s_law#The_law

And, even is it was, then we could still validate it.

Imagine that I put forward the idea that the energy of a photon is proportional to the square of the wavelength (it's not , but that's the point).
I could do the same sort of experiments on energies and wavelengths and every time I tried to calculate my measured value for the "constant", it would change.
That observation would invalidate my theory.

Well, by contrast, the fact that every way you measure h gives the same answer does validate the idea that e =hf

[alancalverd]

Fourier tells us that we should be able to detect harmonics of the fundamental frequency. associated with, say, a LED or similar single spectral line. We can't.

Practically every "green laser" you see- and they aren't rare- is actually the second harmonic of an infrared laser beam.

A LED is not a laser. And the individual photons of a laser do not exhibit harmonics. 

[alancalverd]

In 1900, Max Planck empirically derived a formula for the observed spectrum by assuming that a hypothetical electrically charged oscillator in a cavity that contained black body radiation could only change its energy in a minimal increment, E, that was proportional to the frequency of its associated electromagnetic wave.

That is, by assuming E = hf as an axiom (i.e. defining a constant, h) Planck derived an equation that exactly describes the black body spectrum and thus, I admit, validated his axiom.

So the original questions, posed as a tautology, in fact weren't. The validation of E = hf had nothing to do with measuring the energy of an individual photon, but with describing the gross energy/wavelength curve for zillions of photons, that you can measure with a spectrobolometer. Not unreasonably, however, we find that individual photon energies do indeed fit exactly to Planck's assumption.

[alancalverd]

PS I like the cheek of the Harvard paper on "measuring the wavelength of light with a ruler". When you have worked your way through the geometrical optics theory, you get to the experimental setup where the critical element is a 1/64 inch ruler, and the everyday bit of kit you just happen to have in the kitchen drawer is a coherent monochromatic laser.

My next paper will be entitled "crossing the Irish Sea with a magnetised sewing needle". You float the needle carefully on the meniscus of a glass of water (it's a good trick - one method is to float it initially on a cigarette paper, which gradually sinks as it gets wet) then start up your airplane and just fly perpendicular to the needle.... Try this at home, with any steel needle.....

[mxplxxx (OP)]

[

Not unreasonably, however, we find that individual photon energies do indeed fit exactly to Planck's assumption.

A reference to an experiment that corroborates your statement would be nice. If I had a photon generator and a photon meter, the Visual Basic code below would do the trick.

const h = 6.62607004 × 10-34

for each photon in RandomSetOfPhotons
    E = PhotonMeter.MeasureEnergy(photon)
 
    If E = h * photon.frequency then
        MessageBox("photon " & photon.id & " is ok with E = " & E & " for frequency " & photon.frequency
    else
        MessageBox("photon " & photon.id & " not ok - E is " & E " for frequency " & photon.frequency & " - should be " h * photon.frequency
    end if

end for

[mxplxxx (OP)]

I have other, related, questions.

1. Can I generate a photon from anything else bar an electron?
2. For an electron to "leap" to another level, presumably it must have absorbed a photon of the necessary energy. Where does this photon come from?

[chiralSPO]

1. Yes, it is possible to generate photons without the involvement of electrons, but moving electrons (in atoms, or other sources of electric/magnetic fields) is one of the easiest ways.

Photons can be generated by accelerating charged particles (including electrons, or protons, or muons, or alpha particles etc.)

Photons (high energy) can be produced by the annihilation of particles with their antiparticles.

Photons (also typically high energy) can also be produced from nuclear reactions.

Electrons don't necessarily have to change energy levels--simply changing vibrational states in molecules or crystals can lead to emission (or absorbtion) of photons in the infrared region of the spectrum.

2. Electrons can "leap" to higher energy levels for many reasons, including colliding with other electrons or particles. Or high energy vibrations can be converted into excited electronic states with very little vibration. A good example of both of these phenomena is the blackbody radiation of hot objects. Thermal energy is essentially kinetic energy of vibrations and random motions. If there is enough thermal energy (if the sample is hot enough) then some of this kinetic energy is converted to light--hence hot objects glow, and the hotter they are, the more shorter wavelengths can be emitted (blue is hotter than yellow is hotter than red). Another example of collisions creating light is "triboluminescence"

There are also chemical reactions that produce light directly (like in fireflies and glowsticks).

[chiralSPO]

Of course, it is also possible for electrons to be promoted by light. This is how pigments appear to be colored, and how fluorescent dyes glow.

[mxplxxx (OP)]

Photons can be generated by accelerating charged particles (including electrons, or protons, or muons, or alpha particles etc.)

This is counter-intuitive. I would expect the charged particle to be absorbing photons.

[mxplxxx (OP)]

Of course, it is also possible for electrons to be promoted by light. This is how pigments appear to be colored, and how fluorescent dyes glow.

Presumably light of the exact wavelength needed to make the electron "leap". Is this a common event? How does it come about?

[mxplxxx (OP)]

Yet another question:) What is the wavelength relationship (if any) between an emitted photon and the emitting particle?

[mxplxxx (OP)]

Photons can be generated by accelerating charged particles (including electrons, or protons, or muons, or alpha particles etc.)

What of two electrons that are moving at the same speed and are then accelerated at the same rate (i.e. they are stationary with respect to one another), do they emit photons? It seems to me emitting of photons of an accelerating particle contradicts relativity.

[Bored chemist]

PS I like the cheek of the Harvard paper on "measuring the wavelength of light with a ruler". When you have worked your way through the geometrical optics theory, you get to the experimental setup where the critical element is a 1/64 inch ruler, and the everyday bit of kit you just happen to have in the kitchen drawer is a coherent monochromatic laser.

My next paper will be entitled "crossing the Irish Sea with a magnetised sewing needle". You float the needle carefully on the meniscus of a glass of water (it's a good trick - one method is to float it initially on a cigarette paper, which gradually sinks as it gets wet) then start up your airplane and just fly perpendicular to the needle.... Try this at home, with any steel needle.....

Yes, that's right.
This sort of everyday item you may have on a keyring, or in the kitchen drawer.
https://fetch.co.uk/petface-laser-chaser-cat-toy-317684011?gclid=EAIaIQobChMIko-SufGF4AIVFyjTCh2_UAliEAQYASABEgKj3_D_BwE&gclsrc=aw.ds

A LED is not a laser

Nobody said it was, but you said

a LED or similar single spectral line.

And LEDs don't give single spectral lines, but lasers (nearly) do.

[evan_au]

I would expect (accelerating) charged particles to be absorbing photons.

You can accelerate a charged particle by placing it in a static electric field (eg 1 million volts from a Van der Graaf generator).
From the frame of reference of the lab, the accelerating charged particle will radiate photons.

I am no expert in relativity, but I imagine that in the (accelerating) frame of reference of the charged particle, the electrodes producing the electric field are radiating photons(?) - some advice from someone who knows more than I would be appreciated!

In practice, to accelerate charged particles to really high energies (corresponding to billions of volts or more), scientists tend to use:
-  a circular track: which means the charged particles are continually accelerated around the circle, so they are continually radiating due to the acceleration towards the center
- oscillating electric fields, which generate photons of their own...
- This is one reason nobody is allowed in the LHC tunnel while the beam is operating (the other reason is that if a steering magnet failed, the beam could pierce the pipe - and anyone in the tunnel...)

See: https://en.wikipedia.org/wiki/Cyclotron_radiation

Presumably light of the exact wavelength needed to make the electron "leap".

An element in a diffuse gas or plasma has some clearly defined wavelengths which it will absorb or emit, over a very narrow range of wavelengths.

However, when the atoms are "in contact" in a liquid or solid, the Pauli Exclusion principle requires that the electrons take up slightly different energy levels, and this means that the spectrum covers a much broader. range of wavelengths, depending on which levels it jumps from and to.

What is the wavelength relationship (if any) between an emitted photon and the emitting particle?

An electron can be considered to have a certain wavelength, which becomes slightly smaller at higher energies (as in an electron microscope). But an electron in an atom can generate photons of many different frequencies, from microwave through visible to X-Rays.

A proton in an atomic nucleus can also be considered to have a certain wavelength, which is much shorter than the wavelength of an electron. The radiation produced by a proton changing energy levels tends to be in the gamma-ray range, ie much shorter than the wavelengths produced by electrons in atoms.

On the other hand, neutrons (no electric charge) have a similar wavelengths to protons (positive electric charge), but don't tend emit electromagnetic energy in the same way as protons.

So any relationship has a lot of caveats...
See: https://en.wikipedia.org/wiki/Matter_wave

It seems to me emitting of photons of an accelerating particle contradicts relativity.

Einstein's relativity is a more general form of Maxwell's equations, and incorporates Maxwell's equations.

Maxwell's equations describe how accelerating the electrons in the antenna of your smartphone radiates electromagnetic waves, allowing you to connect to WiFi.

Therefore, Einstein's relativity is WiFi-compatible!

[alancalverd]

Photons can be generated by accelerating charged particles (including electrons, or protons, or muons, or alpha particles etc.)

What of two electrons that are moving at the same speed and are then accelerated at the same rate (i.e. they are stationary with respect to one another), do they emit photons? It seems to me emitting of photons of an accelerating particle contradicts relativity.

As with Newtonian physics, relativity distinguishes between constant speed and acceleration.

[alancalverd]

Quote from: alancalverd on Yesterday at 13:58:26 a LED or similar single spectral line. And LEDs don't give single spectral lines, but lasers (nearly) do.
                           
                           

You should consider the whole paragraph, which concerns the "chirp" nature of a single photon.


I just find the "keyring laser" entertaining. Within living memory (mine) we progressed from neolithic oil lamps and candles, and postmen with bags of paper, through biplanes, incandescent lamps, gas mantles and telegrams, to interplanetary flight, hand-held instant worldwide communications, and kids' toys that invoke subtle applications of solid state quantum physics. How sad that politics, economics and religion have remained at their prehistoric level of sophistication. Or indeed that despite all this, some people still need convincing that E = hf, which does rather seem to be the basis of a lot of stuff we take for granted - like the screen you are looking at right now!.