#362 Survey question: Are photons wave quanta or arbitrary mathematical concepts?
Another way to ask this question: Does a photon have a fixed energy value across all frequencies? Theory seems muddled in this regard. There is the wave quantum given by Planck’s constant h, and there is this account of a photon. They are not the same thing. So what is a photon and is it a real physical entity or an arbitrary theoretical concept?
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In conventional theory, a photon is an “arbitrary theoretical concept”, of course, full of contradictions, and they don’t know what it actually, physically, causally is. And, as usually, they don’t even want to know it, it’s only a set of abstract formulas that matters.
@budrap00 It depends on what you mean by the words “real”, “arbitrary”, “physical”, “theoretical”, “entity”, “wave”, “quanta”, and “concept”.
The way I understand these words (but maybe it’s just me), a “photon” is a real physical-theoretical concept described with the formal language of mathematics. In the lab, a “photon” can also be a causal interaction between an emitter and a receiver, where matter is ionized to produce a free electron.
The $E=h\nu$ photon is boring like a single note in music. By combining many photons together, other interesting phenomena can be described - just like double-stopping on a violin makes music much more interesting.
@RedshiftDrift Not being able to distinguish between a real physical entity and a theoretical concept is a hallmark of mathematicism. A real physical entity is directly observable, measurable, or detectable and exists independently of any theory. A theoretical concept only exists in a theory. It’s hard to do physics if you can’t keep that straight. An electron is a real physical entity, dark matter is a theoretical concept. Maybe you can explain what the term real physical-theoretical concept refers to but to me it sounds like deliberately obfuscatory, oxymoronic mush.
As far as $E=h\nu$ goes, since $h$ is $6.62x10^{-34} J Hz^{-1}$ and $\nu$ is any electromagnetic frequency in $Hz$, the formula gives you the energy contained in a one second interval for a single frequency of electromagnetic radiation. If you are calling that a photon, that’s fine but it doesn’t have any particular physical significance because a second is an arbitrary temporal interval.
A photon is not a quantum of electromagnetic radiation, it has a different value for every frequency, and it is only related to the quantum nature of electromagnetic radiation by the fact that across the electromagnetic spectrum every frequency has the same single cycle (quantum) energy value given by $h$ - the quantum of electromagnetic radiation. A photon is also not a particle in any physically meaningful sense except in the la-la-land of Modern Theoretical Physics.
That’s right, I was asking to make sure you were able to make that distinction.
That is an incorrect theoretical explanation for $E=h\nu$, and it has nothing to do with what a real photon is.
@RedshiftDrift >That is an incorrect theoretical explanation for $E=hν$, and it has nothing to do with what a real photon is.
That wasn’t meant to be a theoretical explanation of the formula but a straightforward logical elaboration of what the math of the formula represents in physical terms. If you have a more lucid theoretical explanation you’d like to share, please do. I’ll be happy entertain it.
As to what a real photon is, that is the question at hand and my contention is that the theoretical meaning of the concept is now so garbled as to defy any precise definition of what the term refers to. For instance in your first comment you said:
So there are boring photons, and not-so-boring photons, and maybe even some really-exciting-photons but that does not address the basic question, What the hell is a photon? The answer appears to be whatever a theorist decides to mumble at the time the question is asked - in other words a photon is an arbitrary mathematical concept. But hey, that’s just my opinion, prove me wrong - put up a concise, logically coherent and physically meaningful definition for your concept of a photon.
A slight improvement to the definition I wrote above: “A photon is a causal interaction between an emitter and a receiver in which the photon’s energy is transferred to a charge.”
A causal interaction between an emitter and a receiver is a well known behavior characteristic of electromagnetic radiation. So now you are defining a photon as that causal interaction while also invoking a photon in that definition of a photon. A couple of days ago you referred to $E=hv$ as a photon. It seems you are trying to prove my point about the photon being an arbitrary (and I would now add, slippery) mathematical concept.
I did not refer to $E=h\nu$ as a photon but as the property of a type of photon of frequency $\nu$. The formal name for a $E=h\nu$ photon is a Fock state. Other types of photons are the mixed state, the thermal state (sunlight), the coherent state (laser), the pure quantum state with minimum uncertainty (used in interferometers), etc.
They all share the property $E = h \int s(\nu) \ \nu {\rm d}\nu$ (where $s(\nu)$ is their spectrum), but that’s not what defines a photon.
Then we agree, but the definition of a photon says more than that. The quantization of the photon is provided by defining the photon in terms of its effect on a charge. These interactions are experimentally observed to be quantized. The two words ‘photon’ refer to two aspects of a photon, not a self-reference.
If your point is that you don’t understand the definition of a photon because you don’t know which experiments led to this definition, then I agree with you.
I don’t know Louis, sometimes I think you’re just playing devil’s advocate but other times I’m afraid you might actually believe this nonsense.
Last week’s version: >The E = hν photon is boring like a single note in music.
Today’s version: >I did not refer to $E=hv$ as a photon but as the property of a type of photon of frequency ν. The formal name for a $E=hv$ photon is a Fock state.
Following that flagrant self-contradiction you list four other types of photon which, you say, all share the same property: >$E = h \int s(\nu) \ \nu {\rm d}\nu$ (where $s(\nu)$ is their spectrum), but that’s not what defines a photon.
Ok, but what then actually defines a photon? How about this trial balloon: >The quantization of the photon is provided by defining the photon in terms of its effect on a charge. These interactions are experimentally observed to be quantized. The two words ‘photon’ refer to two aspects of a photon, not a self-reference.
So the photon is now an “effect” on a charged particle but an “effect” arises when distinct things interact. If you define the photon in terms of its effect without saying what a photon is, you’re just engaging in the usual quantum babble, the purpose of which is to cover up the essential fact that all the math describes is a wave and quantum physicists don’t understand how an electromagnetic wave can produce a particle-like effect on a charged particle so they just mumble photon and run away. (BTW, that last sentence would get you kicked out of a course in basic logic.)
Given that you have offered multiple elliptical or circular definitions, based on allusions to experiments you don’t specify, that amount to an incoherent and illogical performance of quantum babble, my point remains that “photon” is a slippery, abstract mathematical concept that has no identifiable physical correlate. It is of a piece with the rest of Modern Theoretical Physics though. I’ll give you that.
Christiaan Huygens (1690) described light as a wave. Around 1704 Newton described light as ‘corpuscles’ or particles. Thomas Young’s double-slit experiment (1801) affirmed light to behave as a wave. In 1905 Einstein described light as a particle and ‘proved’ this with the photo-electric effect. Ring-a-ring-of-roses …. around we go.
While Einstein was the flavour of the day in the early 1900s, he got tossed under the bus later by Bohr and the likes when QED became really weird. Hence, we are stuck with the wave-particle duality of light, and pretty much: dualities and weirdness for everything else.
My take is that the photon as a particle is king. The photo-electric effect (and some other electron experiments as mentioned by @RedshiftDrift ) cannot be satisfactorily explained by wave behaviour, while on the other hand the double-slit can easily be explained by particle behaviour (even for electron ‘interference’), e.g. De Broglie pilot wave and also by other interpretations as ‘a particle in a wave-like environment’. My interpretation thus is not that the photon is a wave, but it must be riding on something like waves. I know not what they are but these are the waves everyone stopped looking for after M&M claimed they ‘disproved’ the aether, SR didn’t need an aether (because of a misleading postulate), Maxwell’s aether had weird proposed vortices, Lorentz aether was accurate but more complicated than SR, and De Broglie could only match some QED findings, so they all joined Einstein under the bus.
Back to the photon. E=hf=hc/λ says that each photon has a quantity hc per wavelength. It does not say it is spread over the wavelength; it merely states that within each λ there is an ‘hc’ to be detected. So ‘hc’ is discrete. There is nothing saying ‘wave’ here. If you take a quantity h and send it off regularly at frequency f, that is exactly what you will measure .. E=hf. If you take a quantity h and you send the next packet at speed c when the first is a distance λ away, you will measure it as E=hc/λ. It’s the same thing. But it doesn’t mean the photon carries the energy E=hc/λ. No, one photon only carries ‘h’ and it is the repetition of emission that is measured as energy.
Yes they can and I’ve already described that explanation to you - twice. If you prefer your own explanation that’s fine but turning around and claiming there is no alternate explanation just because you don’t like the one offered is either an illogical construct or a disingenuous argument, take your pick.
So we’re back at the central disagreement, is electromagnetic radiation fundamentally a wave or particle phenomenon? I say wave, you say particle. The difference, in my opinion is that my explanation covers all EMR behavior while yours only explains the behavior at emission and absorption.
In order to recover the wave behavior that prevails at all other instances than emission and absorption, the particle model has to invoke subsidiary concepts like superposition of state or, as you seem to prefer, an aether. The problem there is that both of those concepts are metaphysical in nature - they are not an observed physical behavior or physical component of empirical reality; they can only be said to exist in the minds of those who believe in them. To borrow from a t-shirt aphorism, Science doesn’t care what you believe.
This is a good example of how you have to contort your reasoning to maintain the particle model. What does that mean, a photon has energy that cannot be measured unless a second photon is emitted? That doesn’t make physical sense.
You are right, and I can see from this why you abhor the maths, because the photon maths tell us different than we have been taught in electrodynamics and what we should by now firmly ‘believe’. What we perceive as an EM wave must then be a series of regular emissions. This challenges the idea that individual photons are ‘packets of energy’.
The contrary is true in your very sentence an EM wave cannot be explained by one discrete emission. You want light to be a wave without acknowledging that Maxwell’s wave requires an aether.
In answer to your question above; my testable prediction stands: One photon cannot transfer energy (or momentum) because one photon does not have a frequency and is thus not a wave.
The discreet emission and absorption of an EM wave is not trivial. If the energy of the wave was spread over the wavelength, there had to be an effect on the electron like a balloon inflating or deflating, but it is not there. For all frequencies the electron orbital transition is observed as ‘instant’ (up or down) or at least too fast in all cases so as not to be causally linked to the frequency or wavelength of the particular emission. So a discrete ‘packet’ is emitted, which is not energy, but we only perceive it as energy if it comes in a series. That does not nullify the event nor my explanation but challenges our definition of energy as ‘discrete’.
Science doesn’t care what you believe. Your belief is based on circular logic derived from your education that told you photons are particles. In trying to rationalize that claim you have reasoned yourself into an illogical, unphysical corner where:
So your discrete ‘packet’ has neither mass nor energy and that can mean only one thing - it is a metaphysical concept not a scientific one. You justify this unscientific claim with a mathematicist argument that the maths tells us exactly that, but the maths says no such thing.
The maths (Planck’s constant) is defined in terms of a frequency which is a wave measure. Starting from your a priori belief that the photon is a particle you reason yourself into the preposterous conclusion that your discrete ‘packet’ of electromagnetic radiation has no discernible physical characteristics. This is of a piece with your previous assertion that an experiment demonstrating the wave behavior of light somehow supported your belief that light is composed of particles. You are simply fitting the facts (mathematical and physical) to your beliefs.
At any rate your ‘packet’ is not a valid scientific concept, not only because metaphysical assertions that have no physical correlates have no scientific meaning, but also because circular reasoning is a logical error. Mathematics is not the problem. It is the misinterpretation of the mathematics to fit your preconceived belief that EMR is fundamentally a “particle” of some sort that is the source of your incoherent physics.
That premise and the balloon analogy are incorrect; the energy of an electromagnetic wave is not spread out over the wavelength any more than the minute hand of an analog clock is spread out over the face of a clock when tracing a one hour cycle. An electromagnetic wave is a dynamical process involving two in-phase, orthogonally-oriented, transverse waves.
An electron emits a photon when it has sufficient energy $h$ to produce a wave of one complete cycle in the surrounding electromagnetic field. The energy is initially concentrated at the point-like location of the electron as clearly depicted in the graphical illustration previously provided.
When it is emitted, the point-like energy begins expanding out from the electron location into the surrounding electromagnetic medium, beginning the cycle in which the energy sequentially expands and contracts twice to complete one cycle, with the second convergence point being the beginning of the next cycle.
The energy, whether spread out or contracted is always at a definite temporal location at the forward edge of the wave - it is not spread out along the wavelength which only describes the cycle that the energy traces. It is only at the periodic divergence-convergence points that the photon energy is emitted or absorbed by an electron.
The metaphysics are entirely yours, @budrap00 You cannot explain to me how an instantaneous electron orbital drop is translated into an exact wavelength, as if the outgoing wave knows when the next drop will be, and how long it should be stretching itself. How does one ‘wave’ know to spread over nanoseconds while from another electron it might be milliseconds. And what is this reservoir that would hold the energy to release it over time, if it is not the electron deflating (which it is definitely not). No, you are the one believing in waves. If your belief included an aether we might search for common ground, but you want Maxwell’s wave to somehow wave into nothing.
I am assuming you just made a correctible error here because this is just wrong. There is no frequency in Planck’s constant, but there is a frequency in energy through E=hf.
It does not need only a wave to make a frequency though. You can measure many events on a per-second basis which would be considered a frequency in Hertz. E.g. you can put a tacho on your bicycle wheel and maybe measure 1Hz. You can translate this to 60rpm for the wheel, but the 1Hz does not mean something is waving, merely that a sensor is passing the measurement once per second. Now why is it so hard for you to understand that discrete photons passing at a rate can also be interpreted as a frequency?
Then, back to the photon energy, the equation E=hf translates to E = (#)h/s with # as some number of waves, events, cars, light-flashes, cycles, or - as per my reasoning - discrete photons.
I made an error earlier:
which should read: The energy (or momentum) of one single photon cannot be discerned from another single photon.
Why? Because it takes a number of photons per second (frequency) to measure energy. My argument is then that we measure energy as a statistical average of multiple events, and not from a single photon. One photon does not have a frequency.
Maybe you should pause and think but for you write that kind of nonsense. We’ve been over Planck’s constant and its units several times. Here it is again:
$h=6.62x10^{-34} JHz^{-1}$
As has already been discussed $Hz$ is a measure of frequency in units of cycles per second. Therefore my original statement that Planck’s constant is defined in terms of a wave measure, frequency, is correct. Your comment is simply at odds with the facts of the matter; it is inarguably wrong.
Your first paragraph is a clear demonstration that you are not arguing in good faith. In my previous comment I laid out in specific detail why your framing of the wave as having its energy spread out over its wavelength is incorrect. You did not bother to acknowledge that the argument presented even exists let alone attempt to engage with it. You simply ignored it.
Maybe that’s because you did not understand the argument but in that case you should have asked for clarification. You didn’t because fundamentalist believers like you don’t care about other views than their own.
Fundamentalist believers engage in circular arguments based on a priori beliefs and will not engage with any argument that does not acknowledge the primacy of their fundamental beliefs. That’s all you’ve been doing in this conservation. You don’t engage with arguments raised, either against your model or in favor of another. You just ignore anything that doesn’t fit with your model and prattle on, only varying the terminology but not the substance of your circular arguments. In other words sir, you do not argue in good faith.
According to @budrap00 ’s reasoning, a body moving with a constant speed of 1 m/s is a wave with a frequency of 1 Hz.
@HanDeBruijn Maybe in your imagination but before I take it seriously, demonstrate - if you can - the reasoning you used to arrive at your conclusion.
Bud, it’s the one-liner I gave you. Even shorter: $s^{-1} = Hz$ .
Wrong. Photons appear to interact with matter like ‘particles’, but when that analogy is pushed too far one gets into trouble. The idea of a “point particle” is the main cause of all misunderstandings in quantum physics (and other fields of physics too). Every physical “particle” has a physical extent in space, there is no such thing as a point particle.
The photo-electric effect does not prove that photons are particles. The photo-electric effect is described with light waves interacting with electron waves, e.g. https://arxiv.org/pdf/1603.02102 The interaction is at its strongest when these waves overlap, i.e. when their total momentum and energy match.
@budrap00 physics doesn’t say what things “are”, this is the domain of philosophy. Physics defines experiments and finds patterns in the results of those experiments.
There is a ‘bottom’ to the reductionist picture we have of the world. Electrons and photons are currently at the bottom of that hierarchy and what lies underneath remains to be discovered.
Wrong. Shine a faint light on a metal surface and analyze a single electron that comes off the surface (photoelectric effect) with a precise electrostatic electron spectrometer. The spectrometer produces a deviation of the electron (Lorentz force). Measure the electron’s position with a position sensitive semiconductor which produces an electric pulse when it detects an electron. From the deviation calculate the velocity of the electron, and from this its energy. Adding the metal work-function gives the energy of the single photon that produced the electron.
The words themselves specify the experiments. “Particle”, “electron”, “photon”, “frequency”, “spectrum”, “effect”, “de Broglie wave”, “charge”, “aether”, “interaction”, “work-function”, etc. specify experiments. That’s what physicists learn in school so that entire experiments don’t have to be described each time they appear in a discussion.
Every mathematician knows the relations $F=ma$, $F=GMm/r^2$, and $E=h\nu$. The physicist knows how to design an experiment to confirm them.
Not knowing the language of physics is no excuse to claim that physics language is babble.
@HanDeBruijn Oh, so now you want to play symbology games. Ok, for the record $s^{-1}$ by itself does not have any inherent physical meaning, $s^{-1}=Hz$ is a kind of lazy mathematicist convention used to indicate cycles per second because there is no SI unit for cycle. That however has nothing to do with your fatuous claim that somehow my reasoning led you to conclude that meters per second equals cycles per second.
My suspicion at this point is that there is no reasoning at all on your part, only a lame attempt at trolling. Good luck with that. BTW good logical arguments aren’t reducible to bumper sticker aphorisms or one-liners.
@RedshiftDrift >physics doesn’t say what things “are”, this is the domain of philosophy.
Physics uniquely and clearly specifies what an electron is; there is no comparable specification for a photon. As your comments continue to demonstrate, the term photon does not refer to a specific “thing” like an electron. Instead it is a vague mathematicist concept that seems to refer to nothing more than how ever much electromagnetic energy is required to produce the observed out come of any given experiment. It also has nothing to do with Planck’s wave quantum.
When I took physics we were required to write clearly. I wasn’t suggesting you describe the experiments you referred to in detail, just specify them so that the logic of your argument is easier to follow. You do want your arguments to be understandable, don’t you? Not writing clearly is only reasonable if faux-physics babble is your intent.
@RedshiftDrift
A faint light is not provably a single photon. It is still a beam, faint as it may be. A single photon is a single photon and yes, its energy is known to be E-hf or E=hc/λ and this is where our discussion is circling.
My argument is that a single lone photon cannot have a frequency and is thus not a measure of energy, meaning E=hf is not the energy of ‘1 photon’. @budrap00 insists I am not reading his arguments, but his argument presents a photon as a Maxwell wave. Maxwell’s wave requires Maxwell’s aether, which he has yet to acknowledge.
Nobody here said ‘point particle’. We know of the problems that came with that, but someone has yet to explain to me how a photon exists as the length of a wavelength, and I need a satisfactory physical explanation from the instant it exits the electron. Your argument is that one ‘wave’ can knock out an electron, but a wavelength requires a trough and a crest, which I would also like a proposal for how the wave is formed - from the electron exit point.
Then my argument - and it is not an aether based argument - and it is not even an argument for photon particles vs. waves - is that the measurement of energy E=hf requires more than one photon. I argue that the ‘f’ relates to photon emission or arrival rate, for which ‘wavelength’ or ‘frequency’ only makes sense if there is more than one in a beam.
In a way similar to where temperature and pressure is a statistical average and not a measurement of a single particle. If I could describe an analogy it’s something like the response to a button-mashing game. It requires a fast response to reach the target. So the electron doesn’t get flicked by one electron, it gets flicked by E=hf, which is more than one photon - it needs button mashing to escape, and if the button is not mashed fast enough, it stays. Edit: analogy2 like a double-click of a mouse. Click too slow and it doesn’t register as a double-click. If nobody gets this then I will let go of this now.
@budrap00 . Serious now. The $s^{-1}$ or as you write $(1/Hz^{-1})$ in the dimensionality of Planck’s constant has nothing to do with frequency. Consider Planck length $l_P=\sqrt{\hbar G/c^3} = 1.616255(18)\times 10^{-35}\mbox{ }m$ Planck mass $m_P=\sqrt{\hbar c/G} = 2.176434(24)\times10^{−8}\mbox{ }kg$ Planck time $t_P=\sqrt{\hbar G/c^5} = 5.391247(60)\times10^{−44}\mbox{ }s$ Planck temperature $T_P=\sqrt{\hbar c^5/(G k_B^2)} = 1.416784(16)\times10^{32}\mbox{ }K$ .
@Francois-Zinserling >… his argument presents a photon as a Maxwell wave. Maxwell’s wave requires Maxwell’s aether, which he has yet to acknowledge.
First of all, I do not like to use the term photon because it has become so garbled by theorists that it has no clear and distinct meaning; that’s why we’re having this discussion. I prefer to describe $h$ as a wave quantum which is in keeping with its derivation from a wave analysis of blackbody radiation.
Your comment tries to deny my complaint - that you don’t engage my arguments in good faith - only to demonstrate its validity. According to your statement I have to accept a certain belief of yours in order for you to consider my argument. Since I am unwilling to accept your belief and since my argument is based on that non-acceptance you dismiss the argument without considering it on its own terms - because it runs contrary to your belief.
In this case the belief is “Maxwell’s wave requires Maxwell’s aether”. That statement is false. Maxwell’s equations do not contain an aether term and therefore they do not require a physical aether anymore than General Relativity requires a physical spacetime. The reification of human concepts is a common trope of mathematicists used to cover the inconsistencies of their preferred models with empirical reality.
Your comment also seems to contradict your oft stated position that there is a physical aether, This leaves you tied in an illogical knot. If you insist there is an aether, why would you dismiss Maxwell’s wave model because (according to you) it has an aether? Sounds like you’re just making things up without any attempt to be logically coherent. You can’t do science like that.
@HanDeBruijn Your “explanation” demonstrates how often math, and especially mathematical convenience relies on dumbed-down physics. Dropping the defining physical term “cycles” from $Hz$ may allow you to derive Planck units but that is of no physical consequence; it does not mean that the units of $Hz$ are $s^{-1}$ in the context of electromagnetic radiation - because that definition has no physical meaning - in that physical context.
Under discussion here is the physical meaning of the term photon. Mathematical hijinks involving the derivation of Planck’s constant have nothing to do with that subject. Try to stay on topic. Seriously.
@budrap00. >Ok, for the record $s^{-1}$ by itself does not have any inherent physical meaning, $s^{-1} = Hz$ is a kind of lazy mathematicist convention used to indicate cycles per second because there is no SI unit for cycle.
Yes there is one. Find the differences here.
The first statement dumbs down the physics so that the value of the numerator of the final term does not reference anything specific. In the other two statements the meaning of the numerator is clearly specified.
By themselves the numbers 50, 0.3, and 6 don’t have any physical meaning. Adding inverse units doesn’t change that fact - $50s^{-1}$, $0.3m^{-1}$, and $6kg^{-1}$ don’t have any physical meaning unless you specify what the numbers refer to. In your example the physical meaning in the latter two cases is clearly specified as electrons and potatoes. In the first case there is only a kind of circular presentation in which you dilute the meaning of Hertz in electromagnetic theory (cycles per second) to a generic frequency, $s^{-1}$ which has no specific physical meaning. In math terms it means, $Xs^{-1}$, where $X$ is an unspecified variable.
So in the first case $50s^{-1}$ is just a physically meaningless mathematicism. In light of your previous post illustrating that the derivation of Planck units depends on just that dumbed down mathematicism, it raises the question of whether those units have any real physical significance.
As far as I can tell none of this has anything to do with the topic of this post - the meaning of the commonly used term photon. But maybe you’re trying to make a cogent point and not just trying divert a physics discussion to off topic mathematical ruminations. Your intent, however, is not clear from your remarks. If you have a relevant point to make why not just state it.
@budrap00 . Ah, now I see that I am not the first one who tries to persuade you about this. In a reply by @Francois-Zinserling it reads: >Also, f with units of 1/s is not meaningless. If the denominator contains only a number then it translates to events per second. or just number of events per second #/s.
Or number of electrons per cubic metre or number of potatoes per kilogram or .. cycles per second.
But okay, let’s talk again about your definition of the Planck constant as $h = 6.62\times 10^{-34} J Hz^{-1}$ which according to you has to be interpreted as in this reply : >the formula gives you the energy contained in a one second interval for a single frequency of electromagnetic radiation.
With other words: the energy for one cycle? But there are so many different cycles of electromagnetic radiation. Would it mean that we need another Planck constant for each frequency?
Please take a look at the Wikipedia page about Compton scattering - very much experimentally confirmed - and the main formula in there:
$$ \lambda’-\lambda = \frac{h}{m_e c}(1-\cos\mbox{ }\theta) $$
How does that match with the idea that $h$ is an energy per cycle and $Hz \ne s^{-1}$ ?
@HanDeBruijn Back to your old mischaracterization of cherry-picked quotes trick I see. Here’s the full sentence that you extracted a quote from with the quoted section highlighted:
As far as $E=hν$ goes, since $h$ is $6.62x10^{−34} JHz^{-1}$ and ν is any electromagnetic frequency in $Hz$, the formula gives you the energy contained in a one second interval for a single frequency of electromagnetic radiation.
Anyone with even moderate reading comprehension skills can identify that there is only one formula in the sentence and it is $E=hν$. Your comment got it wrong so you could mischaracterize what I said:
(quoted text).
First of all $h$ is a defined scientific constant; it is not my definition. The formula in my sentence was $E=hv$. Maybe you can’t tell the difference between Planck’s constant $h$ and a formula that contains $h$ in the same way you can’t tell the difference between $ms^{-1}$ and $s^{-1}$. I can’t help you with that. To paraphrase an old American saying, You can lead a mathematicist to physics but you can’t make them think.
@budrap00 . >the formula gives you the energy contained in a one second interval for a single frequency of electromagnetic radiation.
That would be energy per second (= Power) and so for the dimensionality : $[h] = J/s = J.Hz$ , instead of the required energy times second $[h] = J\mbox{ }s = J\mbox{ }Hz^{-1}$ . The latter is called an Action in physics. You are mixing up division and multiplication here.
Anyway, you have demonstrated once again that you have a problem with dimensions in physics. As discussed before.
Math isn’t physics and physics isn’t math. Mathematicism is the pseudo-philosophical stance of those who are not good at either. This is a discussion of a physics question you seem incapable of understanding or don’t want to discuss. Your flagrantly disingenuous rants have nothing to do with the subject under consideration.
@budrap00 . You are partially correct. Therefore I have removed some of my flagrantly disingenuous rants in the above reply.
@HanDeBruijn Thank you, I partially accept your partial apology. You are also partially correct. The formula I was referring to is $E=hv$ and I was incorrect to say it gives the energy per second - it only outputs energy - in Joules. To get the energy per second, the convention is to multiply by $s^{-1}$. My argument with that is the convention leaves out the important physical information that the power in electromagnetic radiation comes in integer-number, cyclic units - a Joule.cycle. Since the Joule is a unit of energy, a Joule.cycle is an energy-cycle. This is obvious when you multiply the $E$ output by $Cy\mbox{.} s^{-1}$ instead of just $s^{-1}$ - you get $J\mbox{.}Cy\mbox{.}s^{-1}$.
This doesn’t change the math but it subtly alters the physical description that is implied by the math. This is simply a matter of physical clarity. If Planck’s constant gives the energy of a single electromagnetic energy-cycle then we can describe electromagnetic radiation as consisting of energy cycles. Because the energy-cycle of EMR is based on a defined constant, then by $E=hv$ all frequencies consist of multiples of that energy-cycle and that single energy-cycle is the quantum of electromagnetic radiation.
This brings us back to the the topic of this post - the meaning of the word photon as it is commonly used in theoretical physics. As @RedshiftDrift has repeatedly demonstrated, in common usage the meaning is entirely context dependent. There is an $E=hv$ photon which, since $v$ is a variable, means that there is a different photon for every frequency - and the electromagnetic spectrum contains $10^{26}$ frequencies! There are other photons which are the minimum energy required to produce the observed outputs of various experiments. At any rate the term photon no longer has any connection, to the extent it ever did, to Planck’s wave quantum of one energy-cycle.
I agree. The physics determines the units, but, the units do not determine the physics.
No it doesn’t because the frequency $\nu$ can take any value that is not an integer. The units do not determine the physics.
No ‘physical description’ is implied by math. That’s a “mathematicist” point of view :wink: It is the ‘physical description’ that implies the mathematical formulation.
@RedshiftDrift >The physics determines the units, but, the units do not determine the physics.
Of course not, but a good mathematical formulation describes the physics using units that correspond to physically measurable relationships between objects or events. When that connection is lost the math becomes opaque.
For instance the units meters per second are easily understood as referring to a velocity. The units Joule.second, called an Action by mathematicians, has no clear physical meaning. You can tease the meaning out (in the context of $h$) by stating it as Joules per Hz which makes physical sense but seems to confuse the math-minded. For instance, in response to my statement:
@RedshiftDrift >No it doesn’t because the frequency ν can take any value that is not an integer.
You are mathematically correct but, of course, wrong about the physics. You can describe a frequency of $10.5 Hz$ but in a one second interval the energy detected will always be an integer number times $h$, in this case either 10 or 11 joules. That’s what $h$ means in physics - the energy comes in discrete units given by Planck’s constant as $6.62x10^{-34} J Hz^{-1}$ - that is the minimum electromagnetic energy value. $h$ is the energy of a single cycle of an electromagnetic wave that holds across the entire spectrum - for all frequency-wavelengths. How often do we have to go over this?
I was however wrong about the energy cycle units - they are $J H^{-1}$ not J.cycles as stated.
In an aether model all photons have the same energy. Hear me out … Hopefully this rings true with someone because it seems ‘off the wall’
It takes a stream of photons to make a ‘frequency’ and the equation E=hf only makes sense if there is more than one photon. I have wrestled with this for a long time because from conventional understanding; each photon has a frequency f, and a stream of these same photons will arrive at frequency f. To me this seems a fudge and I have to conclude: One photon does not have a frequency.
A single photon may then be indistinguishable from another, but it is theoretically detectable. In the equation $E=hc/λ$ it can be seen that each photon has an ‘action*speed’ of hc per wavelength. Since there is arguably one photon in a wavelength, it translates to $E=hc/photon$.
In Planckian terms $hc=2π(ℏc)=2π(E_P⋅l_P)$ That is the fundamental photon scale factor. So if one of these arrive at a detector it is indistinguishable, such as trying to discern colours or even shapes in a dark room, but if a series arrive at a frequency f, then it can be detected as E=hf e.g. a unique colour.
Hopefully not too mathy for @budrap00 but if equations can give us an insight then it becomes a useful tool.
In answer to the OP question a photon is definitely real because we can see its result in many meaningful ways. You could argue we also see the effect of dark matter in meaningful ways and yet that doesn’t prove dark matter (thankfully). However, if I have the tools to repeatably recreate a stream of photons e.g. LEDs, and I have the tools to detect it e.g. human eyes, then we have stepped out of the realm of speculation. In response to @RedshiftDrift example - the photo-electric effect - I would then predict a single photon cannot do the job, but only a stream at ~UV rate or higher. Falsifiable prediction.
@Francois-Zinserling You basically have the right idea even if you got there in a typically mathy sort of way, complete with irrelevant math babble (action*speed). Aether theory isn’t required to determine that Planck’s wave quantum $h$ is the same for all frequencies, that comes right out of the math. $h$ is defined as the energy of a one cycle per second ($1 Hz$) electromagnetic wave which means, given $E=hv$, that every single cycle of any frequency has exactly that value - $h$ and you have to multiply it by the total number of cycles (frequency) to get the total energy of a one second radiation interval.
All Planck’s constant does then is define a wave quantum which happens to ultimately mean that a single cycle is a wave minimum - waves can be infinitely subdivided mathematically but there is no physical meaning to a wave that is less than one full cycle. That’s the mundane physical fact that Planck backed his way into mathematically.
The problem with using the term photon as you do is that it has been driven so far from its wave quanta roots that it muddles how you think about things even though you have the right basic idea:
In that statement you are back to treating the photon in the ambiguous way of quantum physicists. If the wave quantum photon is given by Planck’s constant then it has a frequency - of one cycle per second. What you seem to be falling back on is the photon as particle concept that has no real basis in the math. So you end up with a circular argument:
Biological eyes developed to interact with light long before quantum theory came along to muddy the scene. The fact that quantum math is used for engineering LEDs does not justify the interpretation that wave quanta (photons as you and I understand them) are particulate in nature. A wave is by nature a distributed form of energy whereas a particle is a small unit of matter - which is by nature a locally compact form of energy. Their relationship is given by $E=mc^2$. The wave equation $E=hv$ does not contain a mass term because electromagnetic radiation does not have, as is well known, mass - which is a property of matter.
Therefore the misbegotten idea that a wave quanta is in some way a “particle” is a categorical error. And that error has lead to the unnecessary and illogical math based conceit that “particles” are in a superposition of states when not observed. That’s why Feynman had to come to come up with the “path integrals” of QED - to mathematically redistribute the energy that was axiomatically assumed to be particulate in nature. Does the math “work”? Sure it does but it is as physically misleading and unnecessarily complex as Ptolemy’s geocentrism. It also leads directly to the “superposition of states” nonsense.
But if a wave quanta is distributed energy why does it appear to be compact in nature at the point of emission or absorption by a particle? The answer to that comes straight out of the nature of Maxwell’s equations which describe electromagnetic radiation as consisting of transverse, in-phase, orthogonally-oriented electric and magnetic waves:
https://en.wikipedia.org/wiki/Electromagnetic_radiation#/media/File:Onde_electromagnetique.svg
In the illustration the origin (emission) point is on the left where all the energy of the wave is initially at the 3D emitter’s location. The waves then spread out before reconverging to a 3D location at mid-cycle, diverging again and then reconverging at the completion of the cycle where the divergence-convergence cycle begins again. It is at those points of divergence-convergence that the full energy of the wave is available for absorption by a 3D particle like an electron. At no point is the wave a particle - it is always a wave.
The short version: The modern version of a photon that varies with frequency is a purely mathematical construct unrelated to Planck’s wave quanta that was the original basis for the photon concept.
@budrap00
A wave is what you observe when you are looking for waves. The wave-particle duality is not some mysterious property of the photon, it is just a result of observation.
Let’s say a stream of cars are passing you by. All the same speed so the sound is monotone. Now you get to choose one: Look or listen. From the Doppler effect, you will experience cars waving by if you are only listening. Even one car will have a Doppler blue-then-red and one car appears like a wave of sound has just passed. When looking at the cars you can discretely identify each car. If your ears were blocked at this moment it takes effort to see this as a wave, and not as a stream of cars. There is nothing waving now.
The Maxwell visual you gave is misleading too. That wave is e.g. the result of a continuous dipole reversal in an antenna, measured as a streaming wave. You cannot ‘see a photon coming’ so how does the rising sine wave happen, other than maybe instrument lag? If one photon passed you there can only be one cycle. Like one car passing. And there is no way you can position yourselves in the cosmos to measure one photon to look like the continuous Maxwell wave. To then predict that one photon is continuously doing that, when there is no way of measuring it, is a fallacy. It takes a stream of photons to observe the Maxwell wave as you’ve shown.
There is a mathematical error here. Sorry, couldn’t resist! :) The unit of frequency f is 1/s, agreed, but with f as a variable then it is not one cycle per second, but one cycle per fraction of second, inversely proportional to f. The same goes for the $E=hc/λ$ relation I used. λ is variable, but there is always a measure of ‘hc’ per wavelength, iow ‘hc’ per photon. I think we are agreed here though; the photon has a fixed ‘size’ in space and time. My argument is not though that the photon length is stretched to the wavelength. My argument is one photon passing (like the car) is perceived as a wave. I argue the actual photon must be a Planck energy in a Planck length. $ℏc=E_P⋅l_P)$
This is why I deduct the photon is discrete, and the waving is an alternate observation.
I agree that some QED interpretations make it hard to understand. E.g. ‘a photon is a spherical wave emission into the cosmos that collapses onto one point at time of observation’. I call BS. The photon must be discrete and measurable along its entire path, despite the murky maths and QED interpretations. Take a double-slit experiment and blow some smoke through the path after the slits and you will see. There’s no mystery there. D-S with smoke As expected from the scientist, the experimenter’s conclusion was that light is a wave, when a more plausible explanation must be ‘the discrete photon is set on its path at the moment of interaction’.
@Francois-Zinserling >There is a mathematical error here. Sorry, couldn’t resist! :) The unit of frequency f is 1/s, agreed, but with f as a variable then it is not one cycle per second, but one cycle per fraction of second, inversely proportional to f.
Sorry, the error is a logical one and it is all yours. My statement explicitly referred to Planck’s constant which is defined as the energy of a single cycle of an electromagnetic wave. Here it is again:
To be fair that is a little unclear, it is the energy of Planck’s constant that is constant across all frequencies. The one cycle per second frequency is in the definition of Planck’s constant but for waves of higher frequencies the wavelength-frequency of a single cycle varies accordingly while the energy of a single cycle remains constant. And here again is Planck’s constant:
$h=6.63x10^{-34} JHz^{-1}$
$Hz$ is a unit of frequency - in cycles per second. $1Hz$ is one cycle per second. Therefore, Planck’s constant gives the energy of a single cycle of an electromagnetic wave and that energy remains constant at all frequencies according to $E=hv$ - as the frequency increases the wavelength varies inversely, but the energy of a single cycle of an electromagnetic wave of any frequency will always be equal to Planck’s constant.
Defining the frequency unit as 1/s just because SI does not recognize a cycle as a unit of measure results in a physically incoherent mathematical description. 1/s does not have a physical meaning unless you put something physically meaningful in the numerator - in the case of a velocity that is a meter (unit of length). In the case of a frequency that is a cycle.
The subsequent discussion of the wavelength version of $E=hv$, $E=hc/λ$ quickly gets garbled. First you wind up equating a photon with wavelength: “…there is always a measure of ‘hc’ per wavelength, iow ‘hc’ per photon”, only to immediately contradict yourself:
The logic here eludes me.
That is what mathematicism looks like - it happily contradicts actual physical observations with unobserved theoretical constructs. First of all, a single cycle of a wave is as discrete as a particle - but it is not a particle it is a wave quantum (or minimum) and therefore it has a discrete wavelength/frequency. Your discrete photon as you’ve described it here seems to be point-like.
That the experimentally demonstrated wave behavior of light can be most plausibly explained by your discrete photon seems like a leap of faith on your part or just circular logic.. Somehow you have to account for the observed wavelike behavior - which is what Feynman did with his path integrals - it was mathematically possible but has physically implausible consequences like superposition of states and wave particle duality. Can you do that with the aether but without the implausibility? Maybe mathematically, but empirically speaking there is no more evidence for the aether than there is for dark matter.
Sorry, @budrap00 your reasoning still does not compute. Earlier you said:
and
Per second and per Hertz are inverses of each other - not the same thing. Js is not energy per second. J/s would be energy per second but that is not the units of h.
Edit: wait, I think I just realised you mean h is defined as the energy of a 1 second cycle (for f=1, h is measured in J). Not quite how I read it initially but I think I get it now.
Also, f with units of 1/s is not meaningless. If the denominator contains only a number then it translates to events per second. or just number of events per second #/s.
Correct … more correct would be ‘Planck sized’
Still, I think we are aligned that there is a ‘fixed quantity or energy’ in each photon. From what I ascertain you think it is spread across the wavelength, and my opinion is that it is point-like. Like the cars driving past, and the wavelength is just the distance between them. My further argument is that an electron emitting a signal e.g. the hydrogen 21 cm line, which corresponds to a frequency of approximately 1.42 GHz, does not mean it takes 1/(1.42 G) seconds for the electron to drop an orbital. No, it appears to be instant for all frequencies, and the emitted photon must therefore be point-like. A repetition results in wavelength or frequency measurement between the events.
We’ve ended in a mathematical argument which must frustrate you endlessly. Your original question was ‘what is a photon’. I have given you my opinion with heuristic arguments. Maybe you didn’t want to see this solution, or maybe you will still tell us what your version of a photon looks like or maybe you were just raging against photons - and that’s also ok. I’ll join you in that any time.
No, there is no proof of aether, just as there is no proof against it. But there is a terribly weak search for it because Michelson and Moreley falsely claimed that they disproved the aether. At the same time Einstein’s SR provided a simpler solution to ‘current problems’ than Lorentz’s which had an aether base. So the science world went ‘that way’ and lost sight of an important aspect - aether - in my opinion. A Lorentzian aether does not just impose SR with a postulate - it gives full intuition.
Then if the photon is only a wave and nothing else there are more questions than answers. Waves are continuous emitters of energy, or alternately they need continuous stimulation to keep waving. How do you suppose a photon does this without an aether? And what is waving, really? If your answer is ‘the electromagnetic field’, then we are on our way to establish some of the properties for the aether.
So it seems that we share a common viewpoint with regard to the idea that the word photon should only be used to refer to an entity with the energy given by Planck’s constant. Our main disagreement revolves around whether that entity is a wave or particle. My argument in that regard is:
Your argument seems to be:
I already did, I even provided you with a picture based on Maxwell’s equations which you summarily dismissed without apparently giving any real thought to the argument being raised regarding a particular characteristic of the depicted single cycle wave - that its energy periodically diverges from and converges to point-like locations. In other words you dismissed the argument being presented by ignoring it. Give it another try with Planck’s constant in mind.
I specifically said 1/s is physically meaningless because the unitless numerator does does not refer to anything physical. “Translating” the unitless numerator to an “event” unit still does not mean anything physical until you identify what event you’re talking about - maybe it’s just the number of angels arriving on your doorstep. Speaking of which:
The same can be said about angels and devils which is why they are not proper subjects for scientific study.
The electromagnetic field is what’s waving and it is a well and closely studied physical phenomenon. The aether is a metaphysical conjecture that lacks any empirical evidence for its physical existence. The electromagnetic field has no more need of an aether than it does angels and devils.