I rounded off Part 3 of this series by suggesting the following:
“Next up: How do you heat a planetary surface, then? If not by the Earth’s own thermal radiation, a result of its temperature rather than a cause of it … How does the atmosphere insulate the surface?”
Not so. This will have to wait a bit still. Next post, perhaps. I will rather try to clarify my stance on the whole ‘bidirectional flow’ concept thing, seeing how this topic has a tendency of stirring up both emotions and misconceptions.
There is quite a bit of confusion surrounding the whole issue of electromagnetic radiation, the Stefan-Boltzmann Law and the thermodynamic concept of ‘energy transfer’.
I will try to explain why there can be no such thing as a bidirectional energy transfer between two objects radiating at each other. Yes, they are radiating at each other! Radiation goes in all directions.
But, radiation in itself does not universally constitute a thermodynamic ‘energy transfer’. For instance, a cool object cannot and will not transfer any of its energy via radiation to a warmer object. This should go without saying, since this would be a direct violation of the 2nd Law of Thermodynamics.
Recall that a thermodynamic ‘energy transfer’ is always unidirectional, always comes in the form of either ‘heat’ [Q] or ‘work’ [W], and always changes the ‘internal energy’ [U] (and therefore, normally, the temperature) of the two objects/regions involved in the transfer.
The confusion simply arises from not being able to distinguish between a potential ‘energy transfer’ and an actual (real) ‘energy transfer’.
A cool object would always potentially be able to transfer energy to any other object, simply from the basic fact that it radiates energy in all directions by virtue of it having a temperature above absolute zero.
However, if a nearby object happens to be warmer than our first one, then this potential will not be realised in that particular direction. Yup, it still radiates in the direction of the warmer object, but it does not and cannot transfer energy to it. In this case, its radiation is a potential ‘energy transfer’ only.
Sounds like a contradiction in terms? Counter-intuitive? Yes. I agree. But this subject is a strange one. It is quantum theory stuff. Things we cannot really see, only imagine.
The reason why radiation doesn’t at all times necessarily translate into an energy transfer, however, is not that hard to explain.
What you have to realise is that the ‘bidirectional flow’ concept of radiative heat transfer is all about mental images. It is a mental model constructed solely for the purpose of trying to grasp with our human mind what is actually going on.
So we mentally and hypothetically split up an observed physical process into separate, individual parts. As if these parts or pieces were independent from one another, somehow working all by themselves within the process examined, the process in our case of course being a radiative heat transfer. We look first at the one piece, then at the other, and then finally we put the theorised individual effects of these two pieces into one ‘net effect’. Working backwards, so to say. After all, we already know the ‘net effect’. It is what we actually observe and detect in the real world. We simply imagine this to be the result (the ‘sum’) of two opposing constituent effects.
And in a way, this is true. But in saying so, one should not forget that the two individual effects making up the ‘net’ aren’t actually real (in the sense of being realised) effects. They are merely potential effects. The only real (realised) effect is the one we call the ‘net’, the one we actually experience.
It is absolutely essential to understand the distinction here: Real/actual (observed) vs. potential (hypothetical) effects.
In reality, the potential effect of our two imagined individual processes would only become real if the opposite process didn’t occur at the same time, occupying the same space.
Relating this to the Stefan-Boltzmann Law, it is like pointing out that the cool object would have transferred its energy by radiation, based solely on its own temperature and emissivity, to the warm object if (and only if) the warm object were rather an object (or surroundings) at absolute zero. So that it didn’t radiate anything back. This is in fact precisely what the radiative heat transfer equation is saying, in going from this form:
q = σT4
q = σ(Th4 – Tc4)
We realise that (in the latter version) the q term on the lefthand side is the actual (observed, detected) ‘energy transfer’, while the T terms on the righthand side are just potential (hypothetical) energy transfers. They are not individually realised any of them. A T term on the righthand side is only ever realised as an actual energy transfer if it stands alone, like in the upper equation.
So we call the actual ‘energy transfer’ (the heat transfer), the q, a ‘net result’. But is it really? Well, yes, it is hypothetically and mathematically the net of two opposing potentials. It is, however, not the net of two real, separately working physical processes. There is only the one process realised – the heat transfer.
This might still not make much sense. The cooler object is after all radiating energy in the direction of the warmer one. So how could it not transfer any energy to it?
First of all, because it simply doesn’t happen in nature. It is prohibited by the fundamental laws of thermodynamics. You cannot have a situation where energy from a cooler object is transferred to a warmer one and have that energy, directly and all by itself, increase the ‘internal energy’ and thus the temperature of the warmer one in absolute terms. A situation like with the ‘Steel Greenhouse’, or with the postulated rGHE for Earth. As explained here.
This would be plainly at odds with the 2nd Law. There is no way around it.
So we have to find another way of explaining what happens in a radiative heat transfer. There simply is no and cannot be any ‘bidirectional transfer of energy’ between two radiating objects, because a ‘transfer of energy’ in thermodynamics has a very specific meaning; it is a very specific process yielding a very specific effect. One which can only go one way, in a heat transfer situation, from hot to cold.
Secondly, it is because of the fundamental nature of light. Light moves like a propagating wave, a set of wavefronts at a particular wavelength and amplitude, through space. Energy transfers with the waves. So if the wavetrain moves in one direction, the energy moves in that same direction.
When two waves moving in the opposite direction meet, however, they interfere. If they have the same wavelength and amplitude (containing the same amount of energy), they create a so-called standing wave between them. The standing wave is the ‘net result’ of the two opposing waves:
The oppositely moving green and red waves portrayed here would not be observed in the real world. They basically do not exist as physical entities. They exist separately and independently only up to the point where they meet. After this point, their individual physical existence is lost. Their potentials carry on, though. But only the ‘net result’ of the two, the blue standing wave pattern, is to be observed. Only that is realised.
This is not a trivial point. It is THE point to make.
Once the standing wave pattern is up, there is no further transfer of energy between the two sources. Because the standing waves between them are not moving in either direction. The energy was travelling freely through space until the two opposing wavetrains met. Then it stopped. You can observe what happens here:
The two point sources appear to be of equal strength. Watch what happens to the waves they send out toward each other. They meet halfway and then there is no further horizontal movement. No energy from the one source ever reaches the other.
In a heat transfer situation, this would represent a thermal equilibrium. The two objects have reached the same temperature and so no more energy (heat) can be transferred between them.
You would have the same thing with perfect reflection. The reflected energy from the source does not move back to the source. It is ‘caught’ in a standing wave in the space between. No energy transfer.
So what happens in a radiative heat transfer where a warm object heats a cooler one, like say from an initial temperature of 0 K to a final temperature equal to that of the warm object?
We can opt to explain this in two ways:
- A wavelength-specific amplitude surplus resulting in a partial standing wave.
- A wavelength-specific ‘photon’ surplus resulting in a certain ratio between standing waves cancelling and travelling waves transferring.
Now, bear in mind that both of these explanations/descriptions are most decidedly merely conceptual models – simplifications of reality – just as much as the ‘bidirectional flow’ concept of radiative heat transfer. The difference is, these two have the advantage of not ending up violating the 2nd Law of Thermodynamics 😉
1. An amplitude surplus (‘The block approach’)
If two facing planar blackbody surfaces at different temperatures radiate at each other, we can represent each specific wavelength band of their respective EM spectrums as one ‘total’ wave. The energy transferred by this ‘total’ wave is determined not by its wavelength (which, after all, is already set by its spectral band), but rather by its amplitude. This would make it equivalent to an ocean wave or a sound wave, but not really to an individual electromagnetic wave, whose energy is derived specifically from its wavelength: E = hc/λ.
The one ‘total’ wave is thus basically rather an abstract representation of the sum of all the individual EM waves emitted in the given spectral band, and therefore of the total energy transferred from that band, for the blackbody in question, determined only by its temperature.
Since a warmer blackbody would emit more energy at every wavelength than what a cooler blackbody would, we can know that the amplitude of the ‘total’ wave at each particular wavelength emitted by it would be larger than the amplitude of the equivalent ‘total’ wave emitted by the cooler blackbody.
We would then end up with a set of ‘partial standing waves’ moving unidirectionally from the warmer BB to the cooler BB.* The result of which would be the actual ‘energy transfer’, the radiative heat.
*A full standing wave corresponds to the case where the amplitudes of the forward and backward-travelling waves (of equal wavelength) are equal. Conversely, a regular travelling wave corresponds to the case where one amplitude is zero.
We therefore end up going from a set of full, regular travelling waves from the warmer object to the cooler one at ti (Tw >> Tc (0 K)) to a set of full, regular standing waves between the two at tf (Tw = Tc). From full transfer to no transfer of energy.
2. A ‘photon’ surplus (‘The swarm approach’)
This explanation/description is basically the same as the first one, only via a different route. Instead of adding up all the individual EM waves emitted in a given spectral (wavelength) band into one ‘total’ wave of a certain cumulative amplitude before letting it face the opposing wave, one could rather ‘count’ the total number of individual waves (or ‘photons’) emitted in that spectral band and stack each of them up against the individual waves/’photons’ emitted by the opposing object of the same band and see who comes out on top.
Since all individual EM waves within a particular wavelength band naturally have the same wavelength and hence carry the same amount of energy, there will be a multitude of standing waves between the two opposing ‘wave streams’. The warmer object of the two, however, will emit a higher total number of ‘photons’ (a higher energy density) toward the cooler than the cooler emits toward the warmer, so even as all the ‘photons’ from the cooler object is tied up in standing waves with equal energy ‘photons’ from the warmer object, the warmer object will still have ‘photons’ to spare. And these ‘photons’, then, will meet no opposition at all. They will transfer their energy like regular travelling waves to the cooler object.
This ‘photon’ surplus (like the amplitude surplus of the ‘block approach’ above) will become smaller and smaller the closer the temperatures of the two objects get, in the end reaching nil.
The end result of this conceptual model description will be exactly the same as in the first one – because they both simply describe (and attempt to explain) the actual (observable) ‘energy transfer’, the radiative heat.
This superposition of waves works at all points in time and space throughout the integrated radiation field between the two planar BBs, from the one surface to the other. When the ‘net’ wave pattern is set up (happens spontaneously and instantaneously as soon as the two objects are brought into thermal contact), then its movement through the field is what ‘pulls’ the energy from the warmer object and ‘deposits’ it at the cooler.
‘Photons’ versus EM waves
At this point it feels as if a brief discussion on ‘photons’ is warranted.
What is a ‘photon’? Is it something other than an EM wave? Well, its energy tends to be defined in a different (albeit mathematically interchangeable) way from that of a single EM wave, by its frequency: E = hν; an EM wave’s energy would rather normally be defined by its wavelength: E = hc/λ. It would make little sense defining the energy of a single EM wave by frequency. Why? Well, can a single wave be said to possess a frequency? What is ‘frequency’ in the first place? It is the passage of a certain number of waves through a specified interval over a particular period of time. So you can basically only measure the frequency of light from a fixed point outside the actual light beam. If you travel with an individual EM wave flying forward at the speed of light (duh!), then you could only ever assess its length; the wave itself would not come with a frequency. Mathematically, we can of course determine a hypothetical frequency as the speed of light (c) divided by its wavelength (λ), but it doesn’t really mean anything physical. Not for that single wave. It just means that, since all EM waves travel at the speed of light (they are light, after all), we can know for sure that it is their wavelength alone that will determine in the end how many of them will pass through our designated interval. Hence the simple relation: c/λ = ν.
So you see the artificial (and ultimately unnecessary) distinction being made this way between an EM wave and a ‘photon’ …
Defining the energy of the ‘photon’ by its frequency seems to set it apart as a separate entity from the EM wave. As if light is the combined product of two different things put together, a wave and a particle. In most people’s minds, I’m sure, a ‘photon’ is a particle as good as any other elementary particle, a concrete ball of substance flying independently around. In this ‘wave-particle duality’ perspective on the nature of light, one gets the distinct feeling that the ‘photon’, the particle, is meant to sort of be riding on the back of the waves; not on the back of a single wave, but along the rollercoaster course set up by the wavetrain in its entirety. One gets the sense that the wavetrain is somehow frozen in time and space, functioning only as the cresty and troughy highway along which the ‘photons’ fly. In this perspective, the ‘photon’ is the sole carrier of light’s energy, while the waves are just there to provide it with a means to transfer this energy from one place to the next.
This clearly seems to be the corollary of the “‘photon’ gets its energy from its frequency” approach. I’m sure the old masters never meant for it to be this way. But it is. The energy of the ‘photon’ is ostensibly determined by how many individual (and – one must assume, then – unmoving) waves it can pass within a certain amount of time, usually a second. Since presumably it always travels at the speed of light, the length of the waves it passes underway, then, determines its energy.
This all comes off as weird. For instance, if a ‘photon’ travels along the path of an undulating wavetrain, how could it ever manage to move forward in space at the speed of light (c)? Wouldn’t it then have to move considerably faster than the speed of light, along its actual path, in order to cover the distance from wave crest to wave crest at the speed of light:
Such a view is of course absurd.
No, the ‘photon’ is simply an individual light wave, one distinct EM wavefront. It is the wavefront itself that moves forward at the speed of light, not something moving along its length.
The wave and the particle are one and the same. The ‘photon’ simply represents a particular property of EM waves, namely how they come in discrete packets rather than in a continuous stream like mechanical (macroscopic) waves do.
Light is a strange thing. No point in making it even stranger …
What this entails is the following: Rather than being ‘unstoppable’ particles of light, travelling forward no matter what happens, no matter what the waves might do, ‘photons’ simply represent the quanta of light energy moving with and in each single EM wave in the train. So if the waves stop moving forward, like in a standing wave pattern, then the ‘photons’ (and thus the energy) naturally stop moving forward also. The ‘photon’ is not a separate thing from the EM wave.
This situation is in fact summed up quite well by Wikipedia:
“Eventually the modern theory of quantum mechanics came to picture light as (in some sense) both a particle and a wave, and (in another sense), as a phenomenon which is neither a particle nor a wave (which actually are macroscopic phenomena, such as baseballs or ocean waves). Instead, modern physics sees light as something that can be described sometimes with mathematics appropriate to one type of macroscopic metaphor (particles), and sometimes another macroscopic metaphor (water waves), but is actually something that cannot be fully imagined.”
Yes, it’s both and neither, in one.
So, finally, if we had two opposing blackbody spectrums (Figure 2 below), what would be the ‘net’ result of them coming together? Quite simply the ‘energy transfer’ actually observed between the objects emitting them (Figure 3 below). The ‘hot’ spectrum minus the ‘cold’ spectrum. At each wavelength. The result: the radiative heat from hot to cold. There is no double transfer anywhere. Only two potential transfers resulting in one real (as in ‘realised’) transfer: q, the heat.
Everything but the heat is purely conceptual. It doesn’t really exist, doesn’t really occur. Only on paper. And in our minds. Nowhere in the real world.
Figure 3. Both of these figures, I will freely admit, are borrowed (for their clarity) from Joe Postma’s blog.
The black curve of the resulting actual ‘energy transfer’ – the radiative heat, q – between the hot and the cold object, does not in itself represent a blackbody emission spectrum like the other two (the potentials), even though it mathematically emulates the shape of one. The energy moving between two warm objects as heat, after all, is not emitted as radiation by any one of the two objects (nor by a third and unknown ‘heat emitter’). That doesn’t make any difference. The heat is still the only real flux (transfer) of energy around.
How can you tell, then, that the black curve in Figure 3 does not itself represent a third BB spectrum? Apart from the fact that two spectrums could never magically generate a third one between them. Because how would that work …?
You can tell it straight from the graph. The peak wavelength of the ‘heat curve’ is shifted further to the left than the peak of the ‘hot’ spectrum. Which makes no physical sense. Mathematically, of course, this is just the result of subtracting the T1 curve from the T2 curve. But physically it has no meaning. As if the heat flux from the hot to the cold object were emitted by an object even hotter than the hot one, only putting out much less total energy (the area below the curve), as if it were somehow hotter, but much downscaled (like the solar flux at 1AU).