Photons Debate: Is There an Authentic Special Case?

[zoobyshoe]

Last night an EE I was talking to here in San Diego got noticeably upset with me when I asserted that all visible light consists of photons, including light from such sources as the sun, an electric spark, and a lit match. I found myself being treated like a crackpot for asserting what I took to be the most trivial and accepted tenet of Quantum Physics, that all visible light consists of discrete bundles of energy called "photons". His claim was that all the light sources I mentioned emitted "radiant energy", which did not consist of photons. "True" photons, he angrily insisted, were only emitted in the special case where electrons were bouncing in and out of "holes" such as you find in a light emitting diode. Scowling and growling at me, he advised me to google "photonic energy", to be enlightened.

He's an EE and I'm not, so is there really only an authentic special case "true" photon, as he asserts?

 

[Dale]

There are some quantum states (e.g. coherent states) where the number of photons is uncertain. But even so I am with you on this. If you don't like the idea of virtual photons then you may not want to consider a near-field antenna like a microwave to be photons, but it is certainly accepted by the QED community.

 

[stevenb]

Last night an EE I was talking to here in San Diego got noticeably upset with me when I asserted that all visible light consists of photons, including light from such sources as the sun, an electric spark, and a lit match. I found myself being treated like a crackpot for asserting what I took to be the most trivial and accepted tenet of Quantum Physics, that all visible light consists of discrete bundles of energy called "photons". His claim was that all the light sources I mentioned emitted "radiant energy", which did not consist of photons. "True" photons, he angrily insisted, were only emitted in the special case where electrons were bouncing in and out of "holes" such as you find in a light emitting diode. Scowling and growling at me, he advised me to google "photonic energy", to be enlightened.

He's an EE and I'm not, so is there really only an authentic special case "true" photon, as he asserts?

Your story makes me ashamed to be a EE. I apologize for his ignorance.

Of course, the sun emits more than just photons, but I don't know why he would say that the radiation does not consist of photons at all.

By the way, Googling "photonic energy" did not enlighten me at all. How about you?

 

[K^2]

All light is EM field. But all linear fields can be quantized. So while I'd leave alone debate on what really the light is made up of to philosophers and holy men, any light can be properly represented and described physically as "bundles of energy" we affectionately call 'photons'.

To me, as a physicist, that's all the truth I need. If it can always be described as photons, it may as well always be photons, or at any rate, something physically indistinguishable.

The rest is theology.

 

[stevenb]

... "True" photons, he angrily insisted, were only emitted in the special case where electrons were bouncing in and out of "holes" such as you find in a light emitting diode. ...

Well, here is where you can easily make him see his error. Since he thinks photons only come from the combination of hole/electron pairs in a semiconductor, how does he explain the reverse process in which solar panels and photodiodes convert visible light to hole/electron pairs, thus producing usable current and electrical power?

 

[zoobyshoe]

There are some quantum states (e.g. coherent states) where the number of photons is uncertain. But even so I am with you on this. If you don't like the idea of virtual photons then you may not want to consider a near-field antenna like a microwave to be photons, but it is certainly accepted by the QED community.

We weren't even discussing anything like RF or microwaves. He sneered at my assertion the sun itself was emitting photons. He flicked his cigarette lighter on and said "You're trying to tell me that's photons? No."

 

[K^2]

I think you had a moral, if not legal, right to hit him over the head with something heavy for being so smug about being an idiot.

 

[zoobyshoe]

Your story makes me ashamed to be a EE. I apologize for his ignorance.

Of course, the sun emits more than just photons, but I don't know why he would say that the radiation does not consist of photons at all.

I don't know either. It seems to me Huygens, Newton, Young, Planck and Einstein were working to explain light before the invention of the LED, and that the ultimate explanation; that light had to be discrete bundles of energy, was made to explain all the conventional, pre-technological sources of light.

 

[DaveC426913]

The guy's a dope.

 

[zoobyshoe]

I think you had a moral, if not legal, right to hit him over the head with something heavy for being so smug about being an idiot.

Like I said, he's an EE and I'm not. For all I know there could be some technical reasoning I'm not aware of whereby he's right.

 

[DaveC426913]

Like I said, he's an EE and I'm not. For all I know there could be some technical reasoning I'm not aware of whereby he's right.

At best, he has an industry-specific understanding of the subject. It may be the case in his world that that's how they look at it, but you would still be well within your rights to correct him.

 

[zoobyshoe]

The guy's a dope.

At best, he has an industry-specific understanding of the subject. It may be the case in his world that that's how they look at it, but you would still be well within your rights to correct him.

That's what I'm thinking. His notion is some misapplication of an EE method of thinking to general physics.

 

[Antiphon]

As an EE, let me set the record strait- its simply a misunderstanding on his part.

EE's don't need to take quantum effects into account unless:
1) Wavelengths are extremely short
2) Very low noise is important
3) Doing anything in solid state electronics

You get exposed to these in later years so it would be easy for a budding EE to think that quantization was limited to semiconductors.

 

[zoobyshoe]

As an EE, let me set the record strait- its simply a misunderstanding on his part.

EE's don't need to take quantum effects into account unless:
1) Wavelengths are extremely short
2) Very low noise is important
3) Doing anything in solid state electronics

You get exposed to these in later years so it would be easy for a budding EE to think that quantization was limited to semiconductors.

Unfortunately this guy is in his late 40's-early 50's.

 

[Dickfore]

The classical electromagnetic field is described by Maxwell's equations:

$$\begin{array}{rcl} \nabla \cdot \mathbf{E} & = & 4 \pi \, k_{0} \, \rho \\ \nabla \cdot \mathbf{H} & = & 0 \\ \nabla \times \mathbf{E} & = & -\frac{k_{1}}{c} \, \frac{\partial \mathbf{H}}{\partial t} \\ \nabla \times \mathbf{H} & = & \frac{1}{c \, k_{1}} \, \left( \frac{\partial \mathbf{E}}{\partial t} + 4\pi \, k_{0} \, \mathbf{J} \right)$$

Here $\rho$ is the volume density of electric charges and $\mathbf{J}$ is the surface density of electric current.

The second equation implies that the magnetic field $\mathbf{H}$, being solenoidal (divergenceless) can always be represented as a curl of another vector field:

$$\mathbf{H} = \nabla \times \mathbf{A}$$

where $\mathbf{A}$ is called the vector potential of the field. Since the curl of the gradient of any scalar function is zero, we see that this potential is determined up to a gradient of an arbitrary scalar function:

$$\mathbf{A} = \mathbf{A}' + \nabla \Lambda$$

Substituting the expression for $\mathbf{H}$ in terms of $\mathbf{A}$ in the third Maxwell equation, changing the order of derivatives w.r.t. time and coordinates and rearranging, we get:

$$\nabla \times \left( \mathbf{E} + \frac{k_{1}}{c} \, \frac{\partial \mathbf{A}}{\partial t}\right) = 0$$

The term in the parentheses is potential (irrotational) and we can write:

$$\mathbf{E} = -\frac{k_{1}}{c} \, \frac{\partial \mathbf{A}}{\partial t} - \nabla \Phi$$

Here $\Phi$ is called the scalar electromagnetic potential. The minus sign is for traditional reasons. If we change the vector potential according to the above rule, we need to simultaneously change the scalar potential according to:

$$\Phi = \Phi' - \frac{k_{1}}{c} \, \frac{\partial \Lambda}{\partial t}$$

for the electric field $\mathbf{E}$ to remain unchanged.

We may impose an additional condition to the potentials using the arbitrariness of the gauge function $\Lambda$. We will choose it to simplify the remaining to equations as much as possible. Substituting the expressions for the fields in terms of the potentials in the first and fourth Maxwell equations and performing some vector calculus, we get:

$$\begin{array}{l} -\nabla^{2} \Phi - \frac{k_{1}}{c} \, \frac{\partial}{\partial t}\left( \nabla \cdot \mathbf{A} \right) = 4\pi \, k_{0} \, \rho \\ \frac{1}{c^{2}} \, \frac{\partial^{2} \mathbf{A}}{\partial t^{2}} - \nabla^{2} \mathbf{A} = -\nabla\left((\nabla \cdot \mathbf{A}) + \frac{1}{c \, k_{1}} \, \frac{\partial \Phi}{\partial t}\right) + \frac{4\pi \, k_{0}}{c \, k_{1}} \, \mathbf{J} \end{array}$$

to be continued...

 

[arunma]

Wow...I understood all the math in Dickfore's post, but I'm not sure what he was getting at. I eagerly wait the continuation!

Anyway zoobyshoe, you were correct. All light consists of photons. The whole idea of quantum electrodynamics is that you can quantize the EM field as photons. And I can't think of any electromagnetic wave which couldn't be represented as photons. There's no such thing as "radiant energy" in the sense that he's describing. It's just a term that some people (probably in biology and chemistry) fuse for the power transmitted by EM waves.

 

[mingomania]

wth? photon is a unit measurement of light and sunlight is a source of light. spark itself may not be photon but since it the spark probably excites the gases and then the gases stabilize themselves by emitting light or photon. that's all there is to it.

 

[alxm]

You get exposed to these in later years so it would be easy for a budding EE to think that quantization was limited to semiconductors.

And in a funny role-reversal, most chemists are perfectly comfortable with quantized states, but may easily be unaware of where those strange continuous "bands" come from..

 

[zoobyshoe]

Wow...I understood all the math in Dickfore's post, but I'm not sure what he was getting at. I eagerly wait the continuation!

Anyway zoobyshoe, you were correct. All light consists of photons. The whole idea of quantum electrodynamics is that you can quantize the EM field as photons. And I can't think of any electromagnetic wave which couldn't be represented as photons. There's no such thing as "radiant energy" in the sense that he's describing. It's just a term that some people (probably in biology and chemistry) fuse for the power transmitted by EM waves.

It's been my understanding, not just that it is possible to represent an EM wave as quantized, but that Planck demonstrated that light HAS to be quantized, physically, literally, to avoid the "Violet Catasrophe" that would result if light energy were transmitted in a continuous flow. I have not, myself, bothered to dig into Planck's reasoning in any detail, but this dilemma he faced is cited in every introduction to quantum physics.

wth? photon is a unit measurement of light and sunlight is a source of light. spark itself may not be photon but since it the spark probably excites the gases and then the gases stabilize themselves by emitting light or photon. that's all there is to it.

Yes, it has been my understanding the light emitted by a spark is actually an epiphenomenon of the hot, ionized gases the spark causes. Regardless, it's been my understanding the light emitted in this situation consists of photons, like all light.

 

[berkeman]

Zooby, will you see this gentleman again? Hand him a Post-It note with the URL of this thread on it...

 

[DaveC426913]

Regardless, it's been my understanding the light emitted in this situation consists of photons, like all light.

Yep.

 

[arunma]

It's been my understanding, not just that it is possible to represent an EM wave as quantized, but that Planck demonstrated that light HAS to be quantized, physically, literally, to avoid the "Violet Catasrophe" that would result if light energy were transmitted in a continuous flow. I have not, myself, bothered to dig into Planck's reasoning in any detail, but this dilemma he faced is cited in every introduction to quantum physics.

Yup, that's exactly correct (except I think you mean "ultraviolet catastrophe," but whatever). Long story short: you can treat a blackbody as a box with a bunch of waves in it. If you assume that there is a continuous distribution of possible waves in the box and integrate, you reproduce the correct energy density at low frequencies, but it goes to infinity at high energies. But if you assume the waves are discrete and do a summation instead of an integral, the spectrum goes to zero at high frequency. That's the motivation for assuming that light is quantized. However, it turns out to have other applications. For example, the photoelectric effect is explained by light quantization.

 

[Glen Bartusch]

If dude is clueless about photons=EM radiation AT ANY AGE, then he ain't got no bizzness being an Engineer.

 

[Phrak]

Photons are little balls of electricity, right?

 

[K^2]

More like little points of light.

 

[Phrak]

More like little points of light.

So they're really tiny little balls. Is that it?

 

[K^2]

A "ball" implies radius. A photon is a point-particle. All gauge bosons are.

 

[Borg]

Scowling and growling at me, he advised me to google "photonic energy", to be enlightened.

I googled "photonic energy" and first page of hits were for a Danish company with that name. If I type Wikipedia photonic energy, the first hit is for http://en.wikipedia.org/wiki/Captain_America%27s_shield" . To paraphrase Will Smith in Men In Black - Boy, Captain America over here! "Best of the best of the best, sir!" "With honors." Yeah, he's just really excited and he has no clue...

BTW, sorry to pile on your friend Zooby but he is wrong.

 

[arunma]

I googled "photonic energy" and first page of hits were for a Danish company with that name. If I type Wikipedia photonic energy, the first hit is for http://en.wikipedia.org/wiki/Captain_America%27s_shield" . To paraphrase Will Smith in Men In Black - Boy, Captain America over here! "Best of the best of the best, sir!" "With honors." Yeah, he's just really excited and he has no clue...

"Photonic energy" sounds like a term from Star Trek used to describe the composition of holodeck matter. How cool would it be if it were also a meaningful term in physics!?

 

[DaveC426913]

So they're really tiny little balls. Is that it?

A "ball" implies radius. A photon is a point-particle. All gauge bosons are.

K^2, do you feel that tug on your leg?

 

[Phrak]

A "ball" implies radius. A photon is a point-particle. All gauge bosons are.

This sounds like elephants all the way down. Suppose you have a little ball, with light in it, and you shrink it to a point. There's no where left for the electricity to fit!

 

[K^2]

This sounds like elephants all the way down. Suppose you have a little ball, with light in it, and you shrink it to a point. There's no where left for the electricity to fit!

Are you joking? Please tell me you are joking.

 

[DaveC426913]

Are you joking? Please tell me you are joking.

Phrak: pull his other leg; it's got bells on it!

 

[Phrak]

The following is one method we might examine to establish a lower bound, to first order, for the size of an isolated photon.

$$r \approx> 3GM/c^2 \ ,$$​

where r is the radius of the photon sphere of the Schwarzschild solution for mass M. Using

[tex]M = E/c^2 \ \ \ \ \ \ E = h \nu \ \ \ \ \ \ $$\nu = c/\lambda$$​

we get

$$r\approx>3Gh/\lambda c^3$$​

G = 6.67 × 10^-11 m³ kg^-1 s^-2
h = 6.63 × 10^-34 m² kg / s
c = 3.00 × 10⁸ m s^-1

After grinding the numbers,

$$r\approx>5 \times 10^{-69} / \lambda$$​

For red light lambda is about 700 nm, and Bob's your uncle.

 

[sophiecentaur]

But what if there's NO mass?