Fun With Photons!

[Xamonas Chegwé]

OK. Here's some weird sciency stuff that has been running through my head for a few days and that I decided to write down somewhere - so why not here!

Feel free to fuck off if equations hurt your brainz - and feel free to tell me I'm a twat if it turns out I am spouting bollocks - but I don't think I am...



So, photons. They are the carrier particles for the electromagnetic force - or - if you prefer - particles of light. So let's assume that we have a photon emitter (could be a lightbulb, an infra-red lamp, or an X-ray machine - all emit photons of different wavelengths - as do radios, microwave ovens, UV bulbs and gamma ray sources such as Uranium 235.) And let's assume that it is firing photons at a stationary (relative to the emitter) receiver somewhere a light-minute away (about 11 million miles) through vacuumy space.

Now, the maths starts...

1. According to Einstein's famous mass/energy equivalency equation, E = mc2, which means that (by simply rearranging the equation) m = E/c2.

2. And, momentum (p) is defined as mass (m) times velocity (v) so p = mv.

3. We can substitute the value for mass in 1 into 2, giving p = Ev/c2

4. And the velocity of a photon is ALWAYS the speed of light (c), so this equation simplifies to p = Ec/c2 ==> p = E/c (Momentum = the energy of the photon divided by the speed of light.)

5. The Planck Relation, a fundamental equation of quantum physics, states that the energy of a photon is equal to its frequency (f) time Planck's constant (h) - E = hf. (Planck's Constant is about 6.6 × 10−34 Joules/s)

6. Also, the wavelength (λ) and frequency (f) of a photon are linked to the speed of light by the equation c = λf.

7. Substituting 5 and 6 for E and c in 4 gives - p = hf/λf ==> p = h/λ.

8. So, the momentum of a photon is simply Planck's constant divided by its wavelength. This means that the momentum of any photon is precisely known and is simply a function of its wavelength.


OK, now let's introduce Planck's old mate, Heisenberg. Heisenberg is most famous for his uncertainty principle - which is exactly what we are going to apply here.

Heisenberg's equation states that the uncertainty in the momentum of a particle, times the uncertainty in its position, can NEVER be less than h/2 (where h is the Reduced Planck Constant = h/2π)

Put algebraically, it states that ∆p.∆x ≥ h/2, which can be rearranged to state that ∆x ≥ h/2.∆p. BUT, in 8 we have shown that there is NO uncertainty in the momentum of any photon! SO ∆x = h/0 - in other words, the uncertainty in the position of a photon is equal to the Reduced Planck Constant divided by zero!!1!!

So what the fuck does this mean? Surely, you can't divide by zero, can you? Well no, not really. But, in this case, the equation boils down to the fact that the uncertainty in the position of the photon is INFINITE! IE. The photon can, quite literally, be ANYWHERE in space!




Which brings us to ask exactly what this means in terms that make sense to the average primate noggin. And for this, we need a little bit of quantum weirdness, along with some more of Einstein's relativity.

A key tenet of QED (Quantum Electrodynamics - the quantum theory of electromagnetism (ie. light) - as proposed by Richard Feynmann) is that quantum particles do not travel in definite paths from one place to another. What actually happens is that the particle takes EVERY POSSIBLE PATH between two points! This may sound insane, but it is precisely what the maths above shows! At any given time, the particle can be absolutely ANYWHERE - its uncertainty of position is 100%! And a little basic algebra on a few standard equations backs this up - which is nice.

The relativistic approach is a little harder - but also backs up the maths.

If we consider an imaginary observer, sitting on our photon, he would (by the principle of relativity) be perfectly entitled to claim that he was at rest. However, since the photon is traveling at the speed of light relative to its emitter and receiver and the space in between, from our observer's POV, this space would be moving, relative to him, at the speed of light. A consequence of this is that time would slow infinitely in that space and space would compress infinitely in the direction of travel. So, according to our observer, there is NO distance to be traveled and it occurs in NO time at all! So the photon is simultaneously at the beginning and at the end of its journey AND EVERYWHERE IN-BETWEEN!

But, if we consider another observer that is stationary relative to the emitter/receiver, he would see something completely different - from his POV, the photon would be infinitely compressed and so become a point and time would stop for the photon. So, from his POV, a point particle would take a minute to travel to the receiver.

The way to reconcile these two, seemingly contradictory, viewpoints is to recognise that relativity absolutely rules out the simultaneity of time between two relatively moving observers. No observer can ever judge that cause comes before effect, yet observers can be chosen that will claim either that two events are simultaneous, or that they happen at some finite time apart!

And this is another description of the maths above. From one viewpoint, emission and reception are simultaneous and the photon has no need to move. From the other viewpoint, a point photon takes every position in between the two extremes! It is everywhere and nowhere (baby) - that's where it's at. I would lay odds that it wears a hippy hat.

[FBM]

I'm going to need more drugs.

[Tero]

I like electrons. You put one into a copper wire and another comes out the other end. No problem with this when how stuff.

[JimC]

I like electrons. You put one into a copper wire and another comes out the other end. No problem with this when how stuff.

But they too have wave-like properties, and quantum weirdness occurs in that one can build double slit interference patterns, even when firing one electron at a time...

The Sum over Paths approach is one way to make at least a little sense of such behaviour...

The other involves dropping tabs of acid...

[MiM]

8. So, the momentum of a photon is simply Planck's constant divided by its wavelength. This means that the momentum of any photon is precisely known and is simply a function of its wavelength.

But how do you know the exact wavelength of any photon

[FBM]

XC, you're slipping, buddy. Thread title should be "Phun with Photons."

[Clinton Huxley]

Photons? It'll be phlogiston next. Modish nonsense.

[mistermack]

I haven't read right through, because I was worried that my head might explode.

So this might be a daft comment. But doesn't E = mc2, only in the frame where a particle is at rest?

And as a photon is never at rest in any frame, but always travelling at C, doesn't that invalidate the equation for photons?

Like I said, probably daft.

[FBM]

mm, photons have been slowed down, trapped, maybe even stopped altogether in the lab. They don't always travel at c. Unless my memory is totally shot, which I can't rule out.

[MiM]

I haven't read right through, because I was worried that my head might explode.

So this might be a daft comment. But doesn't E = mc2, only in the frame where a particle is at rest?

And as a photon is never at rest in any frame, but always travelling at C, doesn't that invalidate the equation for photons?

Like I said, probably daft.

Not daft at all, mister. On the contrary, quite to the point. Interestingly enough XC still manages to derive the correct equation for the momentum (p = h/λ) of the photon from that. The correct derivation would have started from the relativistic equation E2 - p2c2 = mo2c4. When mo=0 that yelds E=pc or p=E/c as XC has it in step 4.

But the real thought mistake in this is still that the momentum (or wavelength) of the photon would be precisely known.

[Gawdzilla Sama]

This is where we use our photon torpedoes and laser blasters, right?

[JimC]

I haven't read right through, because I was worried that my head might explode.

So this might be a daft comment. But doesn't E = mc2, only in the frame where a particle is at rest?

And as a photon is never at rest in any frame, but always travelling at C, doesn't that invalidate the equation for photons?

Like I said, probably daft.

Not daft at all, mister. On the contrary, quite to the point. Interestingly enough XC still manages to derive the correct equation for the momentum (p = h/λ) of the photon from that. The correct derivation would have started from the relativistic equation E2 - p2c2 = mo2c4. When mo=0 that yelds E=pc or p=E/c as XC has it in step 4.

But the real thought mistake in this is still that the momentum (or wavelength) of the photon would be precisely known.

If we obtain our photons from a precisely known jump in energy levels by an electron, or the vibration of a Caesium atom (like atomic clocks), then, because the source frequency is known with great precision, then so should the wavelength of the emitted photon.

[Xamonas Chegwé]

I haven't read right through, because I was worried that my head might explode.

So this might be a daft comment. But doesn't E = mc2, only in the frame where a particle is at rest?

And as a photon is never at rest in any frame, but always travelling at C, doesn't that invalidate the equation for photons?

Like I said, probably daft.

Not daft at all, mister. On the contrary, quite to the point. Interestingly enough XC still manages to derive the correct equation for the momentum (p = h/λ) of the photon from that. The correct derivation would have started from the relativistic equation E2 - p2c2 = mo2c4. When mo=0 that yelds E=pc or p=E/c as XC has it in step 4.

But the real thought mistake in this is still that the momentum (or wavelength) of the photon would be precisely known.

If we obtain our photons from a precisely known jump in energy levels by an electron, or the vibration of a Caesium atom (like atomic clocks), then, because the source frequency is known with great precision, then so should the wavelength of the emitted photon.

It is possible to produce photons of an absolutely known wavelength - another example would be gamma decay - every radionuclide emits a gamma photon of a specific energy/wavelength.

On the subject of E = mc2 only applying to particles at rest - this is usually the case. However, it does apply to photons in a vacuum. I skipped any attempt at justification as I was trying to present a simplified, easy to follow argument.

mm, photons have been slowed down, trapped, maybe even stopped altogether in the lab. They don't always travel at c. Unless my memory is totally shot, which I can't rule out.

I am not sure that a photon has ever been stopped. I have a feeling that that would require either absolute zero temperature or infinite energy - both of which only GOD is licensed to handle! But, if this has been claimed, I would love to see the reference.

But, of course they travel slower than c in any medium other than a vacuum... which is why I specified a vacuum in my opening paragraph.

...and (obviously) there is no such thing as a true vacuum. And there is pair-production and such shit going on, and other complications, and yadayadayada....

[MiM]

In a discussion like this, great precision is something completely different from exactly. Look at any table of gamma energies, and you will see that there are uncertainties there, often quite large. These are partly due to our ability to measure them, but not completely, because even the linewidths of the energy lines within the nucleus (eg the original energy available for the emitted photon) have an uncertainty, which is determined by, surprise, surprise, the Heisenberg uncertainty principle. Other effects, such as recoil of the emitting nucleus, doppler broadening etc adds to the uncertainty with which we can know the exact frequency of a single photon.
---
About stopping photons: http://cerncourier.com/cws/article/cern/28403

[JimC]

In a discussion like this, great precision is something completely different from exactly. Look at any table of gamma energies, and you will see that there are uncertainties there, often quite large. These are partly due to our ability to measure them, but not completely, because even the linewidths of the energy lines within the nucleus (eg the original energy available for the emitted photon) have an uncertainty, which is determined by, surprise, surprise, the Heisenberg uncertainty principle. Other effects, such as recoil of the emitting nucleus, doppler broadening etc adds to the uncertainty with which we can know the exact frequency of a single photon.
---
About stopping photons: http://cerncourier.com/cws/article/cern/28403

So, the uncertainty about the wavelength is not zero, but it can be extremely tiny. This implies a correspondingly tiny uncertainty in the momentum of the photon, and therefore a very large (but not infinite) uncertainty in its position.

So, within the length of a finite photon path from emission to detection, the position uncertainty could be equal to or greater than this length, which has the same overall effect of XC's original picture...