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| Photon/Energy Quanta Corrupters of Modern Physics |
Modern Physics identified by quantum mechanics/atom mechanics as a revolution of classical non-atomistic continuum wave mechanics, was initiated by Planck in 1900 with his mathematical derivation of the spectrum of blackbody radiation based on a concept of energy quanta $hf$ (Joule) as discrete packets of energy with $h=6,62607015·10^{-34}$ Planck's constant and a $f$ a natural number (1,2,3,...) representing a frequency.
Planck described his long struggle to motivate a high-frequency cut-off needed to avoid an ultra-violet catastrophe with energies tending to infinity from frequencies without upper bound, as follows:
- the whole procedure was an act of despair because a theoretical interpretation had to be found at any price, no matter how high that might be...
Frequencies can range from $10^{12}$ for infrared light to $10^{19}$ for gamma rays with corresponding energy quanta $hf$ ranging from $10^{-20}$ to $10^{-14}$ Joule, thus macroscopically very small. Planck did not view his energy quanta to represent real physics, because atom physics was not yet born, and then only as a mathematical trick to achieve high-frequency cut-off from a statistical argument.
The next step towards quantum mechanics was taken in 1905 by the young Einstein in his "heuristic derivation" of the law of photoelectricity (already formulated by Hertz in 1887 on the basis of experiments), where Einstein picked up the idea of energy quanta $hf$ from Planck, to motivate why shining light on a metal surface releases electrons from the surface only if the light frequency is large enough, as if an energy quanta $hf$ of sufficient strength is needed to release one electron. Einstein's basic "heuristic idea" was thus that exactly one energy quantum later named photon ejects exactly one electron.
Einstein thus suggested to view light as a stream of photons/energy quanta each one if large enough capable of ejecting one electron. But this was only "heuristics" without real physics as admitted by Einstein in 1951:
- All these 50 years of conscious brooding have brought me no nearer to the answer to the question, "What are light quanta"? Nowadays every Tom, Dick and Harry thinks he knows it, but he is mistaken.
A decisive step towards quantum mechanics was taken by the Nobel Prize Committee awarding the 1918 Nobel Prize in Physics to Planck for "his discovery of energy quanta" (in his derivation of black body spectrum), and the 1921 Nobel Prize in Physics to Einstein for "his discovery of the law of the photoelectric effect", thus sending the World a message of light as a stream of particles/photons/energy quanta. This work by Planck and Einstein is still the main "evidence" presented for the "particle nature of light", while all real physics evidence shows "wave nature" expressed in Maxwell's equations for electromagnetics and Schrödinger's equation for quantum mechanics.
This is where modern physics stands today 100 years later coming to expression as the firm belief of a physicist (whether Nobel Laureate or not) that every material body as real physics is emitting an unstoppable shower of light particles/photons depending on its temperature, but not on the medium surrounding the body. This connects to the discussion with Will Happer still without conclusion.
Planck and Einstein viewed energy quanta/photons/light particles as a "heuristic" concept, which could be useful in certain types of theoretical arguments (statistics or cut-off), but which lacked real physics. Unfortunately this is forgotten by physicists of today, who do not object to an alarm of CO2 Warming caused by Downwelling Long Wave Radiation as stream/shower of energy quanta/photons emitted by the atmosphere and being absorbed by the Earth surface.
It is important to distinguish between real physics and phantom physics. In phantom physics you are allowed to use concepts without physical meaning if it serves your objectives. In phantom physics you can view the radiative exchange of heat energy between two bodies as a two-stream flow of photon particles transferring massive heat energy back-and-forth even if the bodies have the same temperature.
It is the same as believing your bank account to be connected to all other bank accounts with a massive one billion dollar transfer back and forth every moment. Or that you are connected to all other people on the web with a constant exchange back and forth of the same one Gbyte message every moment.
This type of belief lacks real physics because it involves massive transfer back-and-forth, which is unstable and so cannot persist over time. To rely on unstable processes is dangerous and will result in misfortune.
Corruption involves massive back-and-forth transfer of services/commodities. Corrupted physics involves massive back-and-forth transfer of heat energy.
For a derivation of the laws of black body radiation and photoelectricity based on real physics carried by electromagnetic waves, see Mathematical Physics of Blackbody Radiation. For quantum mechanics without particles, see Real Quantum Mechanics.
Inlägg
Claes, I have a question for you.
SvaraRaderaI've set up a circuit simulator in the past to analogize thermodynamics to electrical theory... temperature is analogized to voltage via σT^4, emissivity is analogized to resistivity, a diode is analogized to the fact that emissivity only applies to objects which are emitting (not absorbing), shunting current to ground is analogized to emission to 0 K, capacitance is analogized to heat capacity, etc.
I was able to use it to arrive at the same result as the S-B equation to very high precision (I ran the circuit simulator with the display showing 10 decimal places, so at least 10 decimal places). That circuit simulator used DC current. I did it because people can more easily think in terms of voltage and current flow than they can in terms of energy density and radiant exitance.
Here's the circuit simulator, if you want to play around with it.
https://tinyurl.com/yzo8hak9
Ok, so I attempted to do the same with A/C current, based upon your concept of radiative transfer being a resonant phenomenon.
At first, I used a transmission line (which the impedance set to the characteristic impedance of free space, 376.73 Ohms)... one can clearly see the standing waves in the case of a 288 K voltage source connected to each end of the transmission line, and one can clearly see the traveling waves in the case of one 288 K voltage source and one 255 K voltage source. But the implementation of the transmission line is screwy in the circuit simulator, and I couldn't use it for anything but the standing and traveling waves.
Anyway, I calculated the voltage via σT^4 for a voltage source representing a 288 K object, and for a voltage source representing a 255 K object.
I halved that voltage, then offset the voltage output of each voltage source by that half-voltage, so the output voltage would still range to the calculated value, but not go below zero.
I calculated the Wien Displacement peak wavelength and from that the frequency for 288 K and 255 K, which I used for each voltage source.
I hung a resistor on the output of each voltage source while shunting current to ground, then adjusted the resistance until the power output equaled the radiant exitance of a 288 K and 255 K object emitting to 0 K with emissivity of 1. Then I got rid of the grounds.
I hung capacitors on the resistors, analogous to the heat capacity of the objects. The capacitors allow us to see which voltage source is 'winning' in pushing current against the other.
I then connected the two voltage sources... this is akin to introducing the energy density gradient, to allowing the objects to interact via the EM field.
So it was:
288 K voltage source -- resistor -- capacitor -- {free space (a wire)} -- capacitor -- resistor -- 255 K voltage source
Then I set it running. The power flow was nearly exactly what one would expect from the S-B equation, but as the capacitors build charge (the one on the 288 K voltage source going more negative, the one on the 255 K voltage source going more positive), it reaches a point where capacitor voltage will no longer change.
Now, one would think that this means no energy is flowing, but the graphs clearly show a still-significant energy flow (and a negative energy flow in those instances where the destructive interference of the two waves cancels and allows that capacitor voltage to backflow).
Now, if I change the 255 K voltage source to represent a 288 K voltage source, (with capacitor voltage equal to zero volts) literally no energy flows... this is what we would expect of thermodynamic equilibrium... a quiescent state.
So clearly capacitors aren't perfect analogies to heat capacity for an A/C circuit... they're essentially putting out DC, a constant 'push' whereas the voltage sources are A/C, and the destructive interference of the two A/C sources allows that DC to backflow.
So my question is: What is better-used to represent in electrical terms the heat capacity of an object?
Claes, does this pass the sniff test? If not, can you tell me where I went wrong?
SvaraRaderaWe know that pressure and temperature are intricately intertwined, and we know that temperature and energy density are intricately intertwined... to such an extent that we can directly convert between pressure and energy density:
1 Pa = 1 J m-3
Now, we can calculate R_specific (the specific gas constant) of any given gas (directly based upon the gas parameters, rather than relying upon the ideal gas equation approximation), and determine that at 101325 Pa pressure and any given temperature, a gas will expand or contract until it has 101325 J m-3 kinetic energy density.
And given the Equipartition Theorem, that means the gas (if it has any energetically accessible quantum states) will have rotational mode and vibrational mode quantum state energy roughly equal to the kinetic energy density... this is why, at 273 K, H2 has no vibrational mode quantum states which are excited (the first vibrational mode of H2 doesn't unfreeze until 3392 K), and the energy density is so low that most of the rotational mode quantum state energy will only excite the J1 state, with very few J2, very very few J3 and almost no higher rotational mode quantum states being excited... I calculate the quantum state (rotational mode) energy density to be 3071.3719 J mol-1 on average at 273 K, whereas translational mode energy density is 3404.7724 J mol-1.
H2 at 273 K:
Molar kinetic energy density:
(3/2) * 1.380649e−23 J K−1 * 273 K = 5.653757655e-21 J particle-1 * 6.02214076e23 particle mol−1 = 3404.77244213375178 J mol-1
Checking our math:
1/2 m v^2 = (1/2) * 0.0020156799993094936060614464 kg mol-1 * (1838.0116092566284089214803816125227469883595064233 m sec-1)^2 = 3404.77244213375178 J mol-1
M = 2.01568 amu = 1.6605390666050e−27 kg amu-1 * 2.01568 amu * 6.02214076e23 particle mol−1 = 0.0020156799993094936060614464 kg mol-1
We can calculate the actual R_specific (rather than an approximation based upon an ideal gas):
R_specific = 3404.77244213375178 J mol-1 / 273 K = 12.47169392722986 J mol-1 K-1 / 0.0020156799993094936060614464 kg mol-1 = 6187.3382339965950012080922381039 J kg-1 K-1
p = (101325*(0.0020156799993094936060614464))/([12.47169392722986]*273)= 0.05998602825921580387937586617338 kg m^3
V_m = 1/p = 16.670548609731758552477761470539 m^3 kg-1 * 0.0020156799993094936060614464 kg mol-1 = 0.03360249141015299067357512953368 m^3 mol-1
And the translational mode energy density would be 3404.77244213375178 J mol-1 / 0.03360249141015299067357512953368 m^3 mol-1 = 101325 J m-3.
We can do the same for any given temperature and for any given gas and for any given pressure... if the gas is unconstrained and allowed to freely expand or contract (as it would be in our atmosphere), it will expand or contract until the energy density in J m-3 equals pressure in Pa. This is what drives convection in our atmosphere.
Thus CO2 would have nearly no effect upon temperature except to increase heat capacity of any given parcel of air for a higher CO2 concentration (the molar heat capacity of CO2 is higher than the molar heat capacity of bulk air), thus that parcel of higher-concentration CO2 air would convectively transit more energy from the surface, which would have the effect of attempting to decrease temperature differential with altitude (ie: a reduction of the lapse rate), while at the same time the higher concentration of CO2 would radiatively cool the upper atmosphere faster than it can convectively warm it... IOW, a higher concentration of CO2 would cause more cooling, not warming.
Think of the negative feedback mechanism here... the planet cools, air becomes denser and thus pressure (in Pa) rises, and given that 1 Pa = 1 J m-3, this attempts to increase air temperature. Conversely, the planet warms, air becomes less dense and thus pressure (in Pa) falls, and given that 1 Pa = 1 J m-3, this attempts to decrease air temperature.
Or, to put it as simply as possible, our atmosphere always attempts to equilibrate, on average, such that the volumetric energy density of air (units: J m-3) is equal to atmospheric pressure (units: Pa), because 1 J m-3 = 1 Pa.
SvaraRaderaIf a parcel of air has higher volumetric energy density in J m-3 than the pressure in Pa, it will convect. If a parcel of air has lower volumetric energy density in J m-3 than the pressure in Pa, it will descend in altitude.
This is what drives convection. This is why temperature drops as altitude increases (and thus air pressure drops). CO2 has next to nothing to do with atmospheric temperature (and thus next to nothing to do with surface temperature)... and what it does have to do with it, it's a net atmospheric radiative coolant, just as water vapor is (except water acts as a literal refrigerant (in the strict 'refrigeration cycle' sense) below the tropopause, transiting energy via latent heat capacity, whereas CO2 only has molar heat capacity).
That's why water vapor is the predominant atmospheric radiative coolant below the tropopause, and CO2 is the predominant atmospheric radiative coolant above the tropopause (and in fact is a net atmospheric radiative coolant at all altitudes except at the tropopause, where it absorbs a greater proportion of cloud-reflected solar IR insolation and radiation from cloud condensation).
Our atmosphere is trading one form of energy density for another... volumetric energy density for pressure, pressure for work/volume in a constant-pressure process, and given that volume = mass / air density, that affects air density, which affects buoyancy, which affects convection.
Energy Density: [M1 L-1 T-2] ... is:
Pressure: [M1 L-1 T-2] ... which translates to:
Force: [M1 L1 T-2] ... over:
Area: [M0 L2 T0] ... which translates to:
Mass: [M1 L0 T0] ... times:
Acceleration: [M0 L1 T-2] ... over:
Area: [M0 L2 T0] ... which translates to:
Work: [M1 L2 T-2] ... over:
Volume: [M0 L3 T0]
Also...
Work: [M1 L2 T-2] ... over:
Pressure: [M1 L-1 T-2] ... (a constant pressure) changes:
Volume: [M0 L3 T0] ... which is:
Mass: [M1 L0 T0] ... over:
Air Density: [M1 L-3 T0] ... which causes:
Mass: [M1 L0 T0] ... to flow:
Rate Of Flow: [M0 L3 T-1] ... over:
Time: [M0 L0 T1]
A true black-body (BB) consists of numerous atoms having different internal energies. If one of them absorbs a photon, it does not matter where it came from. Hence, a BB of higher temperature can absorb photons from a BB of lower temperature; BUT the net energy transfers is - with high probability and for a sufficiently large BB - always from hot to cold - in accordance with the 2nd law of thermodynamics.
SvaraRaderaI know it intuitively sounds like it might happen that way, but it doesn't, John... mainly because idealized blackbody objects are provable contradictions which do not and cannot exist.
SvaraRaderaWe're not talking about idealized blackbody objects, which assume emission to 0 K, which have their emissivity and absorptivity pinned to 1 at all times, which must absorb all radiation incident upon them (ie: it is assumed that all quantum states are in unexcited states at all times and thus available to absorb radiation), and which must emit all radiation they absorb (ie: they have no thermal capacity).
The above is why idealized blackbody objects do not and cannot exist, they're idealizations. The closest we can come is laboratory blackbodies (which are just really good absorbers and emitters), but even they do not absorb all radiation incident upon them, do not emit all radiation they absorb (ie: they have thermal capacity), do not have emissivity and absorptivity pinned to 1 at all times, do not assume emission to 0 K, etc.
We're talking about real-world graybody objects, which assume emission to greater than 0 K, emissivity less than 1, emissivity and absorptivity which tends to zero as energy density gradient tends to zero (and which is zero at thermodynamic equilibrium... real-world graybody objects become perfect reflectors at thermodynamic equilibrium... see below).
https://i.imgur.com/QErszYW.gif
A warmer object will have higher radiation energy density at all wavelengths than a cooler object.
https://i.imgur.com/kS20QG1.png
Thus what you claim, John, cannot happen. Energy does not and cannot spontaneously flow from a lower radiation energy density to a higher radiation energy density.
The definition of emissivity: The ratio of the total emissive power of a body to the total emissive power of a perfectly black body at that temperature.
RaderaThe definition of absorptivity: The ratio of the absorbed to the incident radiant power.
While an idealized blackbody object emits when its temperature is > 0 K, a graybody object emits when its temperature is > 0 K above its ambient. This is plainly evident in the animated version of the S-B graphic above.
Do remember that temperature (T) is a measure of radiation energy density (e), equal to the fourth root of radiation energy density divided by Stefan’s Constant.
As Δe → 0, ΔT → 0, q → 0. As q → 0, the ratio of graybody object total emissive power to idealized blackbody object total emissive power → 0. In other words, emissivity → 0. At thermodynamic equilibrium for a graybody object, there is no radiation energy density gradient and thus no impetus for photon generation.
As Δe → 0, ΔT → 0, photon chemical potential → 0, photon Free Energy → 0. At zero chemical potential, zero Free Energy, the photon can do no work, so there is no impetus for the photon to be absorbed. The ratio of the absorbed to the incident radiant power → 0. In other words, absorptivity → 0.
α = absorptivity = absorbed / incident radiant power
ρ = reflectivity = reflected / incident radiant power
τ = transmissivity = transmitted / incident radiant power
α + ρ + τ = 100%
For opaque surfaces τ = 0% ∴ α + ρ = 100%
If α = 0%, 0% + ρ = 100% ∴ ρ = 100% … all incident photons are reflected at thermodynamic equilibrium for graybody objects.
This coincides with standard cavity theory… applying cavity theory outside a cavity, for two graybody objects at thermodynamic equilibrium, no absorption nor emission takes place. The photons remaining in the intervening space set up a standing wave, with the wavemode nodes at the object surfaces by dint of the boundary constraints. Nodes being a zero-crossing point (and anti-nodes being the positive and negative peaks), no energy can be transferred into or out of the objects. Photon chemical potential is zero, they can do no work, photon Free Energy is zero, they can do no work. Should one object change temperature, the standing wave becomes a traveling wave with the group velocity proportional to the radiation energy density gradient and in the direction of the cooler object.
We start with Energy: [M1 L2 T−2] and we subtract...
RaderaForce: [M1 L1 T-2] *
times Length: [M0 L1 T0] = [M0 L0 T0]
We are left with nothing on the 'transmitting' end... [M0 L0 T0]. In other words, that Energy is used to apply a Force along a Length.
It’s obvious then, that if an equal and opposing Force were applied along that Length, no energy can flow… this is just as true radiatively as it is mechanically.
That Force applied along a Length gives us (on the 'receiving' end):
Force: [M1 L1 T-2] *
Length: [M0 L1 T0] =
Work: [M1 L2 T-2]
You'll note that Energy and Work have the same units:
Work: [M1 L2 T-2] = Energy: [M1 L2 T−2]
For those who want to put it in terms of Momentum:
Momentum: [M1 L1 T−1] *
Velocity: [M0 L1 T-1] =
Work: [M1 L2 T−2]
That means Energy Expended = Force * Length = Momentum * Velocity = Work
There's a reason for that. Free Energy is defined as that energy capable of performing work.
The above is why photons are considered the force carrying gauge bosons of the EM interaction.
Radera