Light Polarizer and the Second Law of Thermodynamics

WetSavannaAnimal aka Rod Vance 12/10/2014. 4 answers, 303 views
thermodynamics optics entropy polarization blackbody

I have stumped myself with a thought experiment of my own devising.

Suppose I take a beam of wholly depolarised, but otherwise plane wave light. Its von Neumann entropy per photon is $\log(2)$ nats or one bit per photon - each photon's state takes one bit to describe.

Now I pass the beam through a polariser, knocking out one of the polarisation states. There are now photons in only one polarisation eigenstate left and the entropy per photon is now nought.

In doing this, half the beam's power $P$ has been absorbed by the polariser. Consider a sample of the beam propagating in time $\mathrm{d}t$. This beam sample's entropy before polarisation is $\frac{P}{h\,\nu}\,k_B\,\log 2\,\mathrm{d}t$, where $\nu$ is the light's frequency. After polarisation, the remaining light's entropy is nought. Assuming that the polariser is in thermodynamic equilibrium with its surroundings at temperature $T$, in time $\mathrm{d}t$ the polariser absorbs energy $\frac{1}{2}\,P\,\mathrm{d}t$ and so its entropy increase, by definition of the thermodynamic temperature, is $\frac{1}{2}\,\frac{P}{T}\,\mathrm{d}t$. The total entropy change wrought by the polarisation process is therefore:

$$\mathrm{d}S = \left(\frac{1}{2\,T}-\frac{1}{h\,\nu}\,k_B\,\log 2\right)\,P\,\mathrm{d}t\tag{1}$$

and so the second law is "safe" as long as:

$$T\leq\frac{h\,\nu}{2\,k_B\,\log2}\tag{2}$$

So far so good. Work this out for $600{\rm THz}$ visible light (500nm) and we get $T<21000K$. We have doubts about our polariser working as planned if the temperature is thrice that of the Sun's surface, no surpises there, so no Nobel prizes for working that one out!

However, what about a microwave polariser? - a grid of wires in one direction that lets the microwaves pass if their $\vec{E}$s are orthogonal to the wires. Working at $6{\rm GHz}$. Now our temperature bound is $T<0.2K$. Obviously, the polariser will be imperfect and may re-radiate in complicated ways, but if we imagine the whole scenario in deep space where everything around the thought experiment is at $T=2.7K$, then I can't see how polariser cannot re-radiate light with more entropy than what is lost by the polarisation of light. If you radiate heat ${\rm d} Q$ into the CBMR then the entropy gain of the thermalised EM field is ${\rm d} Q/T_{CMBR}$, so we still would seem to hit up against the limit in (1) even after we account for re-radiation.

What (1) seems roughly, intuitively to be saying is that if you polarise light, then you must somehow be able to re-radiate at least as many photons as you absorb, so that there is an overall entropy increase. (The number of photons per unit heat radiated is roughly $\propto\frac{1}{T}$). And this sounds reasonable to me.

And yet, microwave polarisers do work!

What am I missing here?

It seems such a simple thought experiment, I'd be surprised if there weren't a paper on it somewhere, so I'd also appreciate a reference as an answer.

4 Comments
1 higgsss 12/10/2014
My immediate guess is that the polarizer probably works less perfectly as the temperature is raised.
WetSavannaAnimal aka Rod Vance 12/11/2014
@higgsss The more I think about this, the more I think your explanation is exactly what my thought experiment means. See my answer and see what you think.
WetSavannaAnimal aka Rod Vance 12/11/2014
@PiotrMigdal Indeed. I believe my answer is in keeping with this.

4 Answers


By Symmetry 12/10/2014.

As always the answer is a simple thing. You calculated the change in entropy using the definition of entropy \begin{equation} \mathrm{d}S = \frac{\mathrm{d}Q_\mathrm{rev}}{T} \end{equation} Note that this applies to heat transferred reversibly.

More generally we must use Clausius theorem \begin{equation} \mathrm{d}S \ge \frac{\mathrm{d}Q}{T}\end{equation} Where the $\mathrm{d}Q$ now may be none reversible.

Now a process is thermodynamically reversible if it can be reversed by an infinitesimal change in the conditions, which certainly is not the case here, so all we can conclude about the change in entropy (without doing a more detailed analysis) is that it must be greater than the quantity you calculated, so there never was a problem.

1 comments
1 WetSavannaAnimal aka Rod Vance 12/11/2014
Thanks for answering. I actually don't think this helps here, because we're calculating $\mathrm{d} S = \mathrm{d} P/T$ and, as you know, $\mathrm{d} P\geq \delta Q$ (sorry for my sloppy symbols), which is where the Clausius theorem comes from - the inequality as opposed to equality is essentially the notion of "wasting work" to produce entropy or getting more entropy that calculated by heat absorption alone. Anyhow, I know that's not a full counter argument (which your answer definitely deserves), so I am answering in detail in the next day. Please be patient and then we can discuss some more

Firstly, the OP is forgetting that the classic microwave polariser experiment is done with EM radiation in a pure state, not a mixture. We simply have polarised light from, say, a Gunn diode and this pure quantum superposition is forced into a polarisation eigenstate by the polariser. So we begin with near to zero entropy light, absorb some of it (adding entropy to the polariser as it heats), and the leftover light is also of near zero entropy. No problem. So this easily explains how the OP claims to remember microwave polarisation experiments shown to him when he was 17 and that they work perfectly at 300K.

But what about depolarised microwaves? In this case, the second law of thermodynamics puts a limit on how well a polariser can work at a given temperature.

Let's imagine a finite-length, depolarised beam of microwaves moving through deep space. All quantum state variables - direction, spin, frequency - aside from polarisation set to known states so the light is fully describable by a $2\times 2$ density matrix written with respect to the polarisation eigenstate basis. The photons don't need to be in frequency (energy) eigenstates and direction (momentum) eigenstates, but instead can be a quantum superposition of such eigenstates (not a mixture), so that they form a pulse can be finite in space and time. The photon ensemble is moving through deep space separated from its source so that we can think of the whole initial system as:

  1. The microwave photon ensemble, described by a classical mixture of two pure states (quantum superposition of energy and momentum eigenstates) that differ only by being in orthogonal polarisation states;
  2. The freespace quantum light field, in thermodynamic equilibrium at the CMBR temperature of $T_{CMBR}=2.7K$, so it is a mixture of thermalised photons described by the Planckian BB spectrum at $T_{CMBR}$;
  3. The polariser itself, also initially at thermodynamic equilibrium at $T_{CMBR}$.

It should be pointed out that depolarised microwaves whose other quantum properties are otherwise perfectly known are exotic creatures and I cannot think of how one might produce such things experimentally, contrasting with the situation for light where such mixtures are much more plausible. Nonetheless, there doesn't seem to be any in principle reason why such mixtures can't exist for microwaves if they do for visible light.

We consider the pulse sundered from its source, so we have already accounted for the entropy increase in the production of the light, which, as Wolphram jonny's Answer points out, will produce a great deal of entropy. There is no problem with the second law here. But we now wish to know what the entropy changes are for the system that begins as incoming microwave energy, polariser and CMBR.

As By Symmetry's Answer discusses, initially the polariser is not at equilibrium and heating is irreversible. It will begin to heat up. But it must eventually reach a steady state: it may have temperature gradients in it with a "hot spot" where the beam is absorbed, but it will eventually be described as a temperature distribution $T(\vec{x})$. This steady state is reached when the sum of the incoming power from the beam together with the heat absorbed by the polariser from the CMBR equals the heat re-radiated to the CMBR radiation field in unit time. I sketch these ideas below.

Steady State System

At steady state, the polariser's macrostate is unchanging with time, therefore its total entropy must be steady. We can therefore think conceptually of the conversion happening as the light is absorbed by the polariser and, later, the equal nett power is radiated to the CMBR field as drawn in my sketch below:

Conversion of Microwaves

At steady state, therefore, we need to account for:

  1. The entropy lost from the beam through any polarisation: the entropy of the partially polarised beam output less that of the input beam;
  2. The entropy increase of the radiation field as the energy input to the polariser from the beam is re-radiated.

Now the maximum entropy that the radiation field can "soak up" in absorbing power $\mathrm{d}\,P$ is $\frac{\mathrm{d}P}{T_{CMBR}}$; this is because:

  1. The proportional error between the total state information content of a system of particles (here thermalised EM field quantum oscillators) and the information calculated assuming that the system is in its maximum entropy, thermodynamic equilibrium state approaches nought as the number of particles increases without bound, as discussed further in my answer here to the question "Why are the laws of thermodynamics “supreme among the laws of Nature”?" and
  2. The definition of thermodynamic temperature is $\frac{1}{T} = \frac{\partial\,S}{\partial\,U}$ where $U$ is the internal energy of a system and $S$ its entropy.

Therefore, the entropy increase of the radiation field must exceed he entropy lost from the beam through any polarisation in all cases, and so:

$$(1-\alpha)\,\frac{\mathrm{d}P}{T_{CMBR}} \geq \frac{\mathrm{d}P}{h\,\nu}\,k_B\,\left(-\mathrm{tr}\left(\rho_{in}\,\log\,\rho_{in}\right)+\alpha\,\mathrm{tr}\left(\rho_{out}\,\log\,\rho_{out}\right)\right)$$

and this is the statement of the second law of thermodynamics manifesting itself as a limit on how much a polariser can actually polarise light, where $\alpha$ is the fraction of the input beam's that is transmitted by the polariser, $\rho_{in}\approx\left(\begin{array}{cc}\frac{1}{2}&0\\0&\frac{1}{2}\end{array}\right)$ is the density matrix describing the mixed state of each photon input to the polariser and $\rho_{out}$ is the density matrix describing the mixed state of each photon directly transmitted by the polariser. Therefore, the second law limits the quality of polarisation possible and:

$$-\mathrm{tr}\left(\rho_{out}\,\log\,\rho_{out}\right) \geq \frac{\log\,2}{\alpha}-\frac{1-\alpha}{\alpha}\,\frac{h\,\nu}{k_B\,T}$$

defines the maximum "quality" of polarisation allowed by the second law of thermodynamics, where the quantity $-\mathrm{tr}\left(\rho_{out}\,\log\,\rho_{out}\right)$ on the left hand side is positive, gets smaller with increasing quality of polarisation and has a value of nought nats when the polarisation is perfect. Here $T$ is the effective temperature of the ambient radiation field, and it must be greater than or equal to $T_{CMBR}$. For a given quality of polarisation to happen, a necessary condition is then:

$$T \leq \frac{(1-\alpha)\,h\,\nu}{k_B\,\left(-\mathrm{tr}\left(\rho_{in}\,\log\,\rho_{in}\right)+\alpha\,\mathrm{tr}\left(\rho_{out}\,\log\,\rho_{out}\right)\right)}$$

which reduces to the OPs formula in the case of fully depolarised input light and perfectly polarised output light.


By Symmetry 12/10/2014.

I think the factor you are ignoring is that the polariser will emit thermal radiation. If we continue with the ideal polariser, then it should only emit the polarisation which it absorbs (ideal components are weird). This means that there will still be a component of the absorbed polarisation in the beam after the polariser and so there will always be some entropy in the final beam. This entropy will increase with temperature and decrease with frequency.

1 comments
1 By Symmetry 12/10/2014
Having thought about this more carefully, the emitted thermal radiation does not scale with the power of the input beam, so I don't think this can be right

Wolphram jonny 12/10/2014.

I believe the error is in assuming that the polarized beam is a pure state of zero entropy. If you characterize it in terms of polarization only, then the characterization is not complete. You need a complete set of commuting observables to charactherize a pure state. The macroscopic polarized beam is still compatible with many different quantum microstates (for instance, the spin orientations on the polarization plane will still be random). For the same reason the assumption that the entropy per photon will be log(2) is incorrect.

But let us assume a sligthly different but equivalent situation in which the original light source is prepared in such a way that that you know the total state of the system, that is, you are preparing it with all sets of quantum numbers other than polarization perfectly known. The mistake now, I believe is that you only considered the increase in entropy on the polarizer, but not the increase in lost the non-polirizad light itself. During the absorption you will loose a lot of information about the non polarized light (technically will not be lost but mixed with the rest of the polarizer. You didn not include that term. You only included the amount gained by the reservoit due top heay transfer, but that doesnt include all the other information lost because half or your system is no longer in its original state.

3 comments
By Symmetry 12/10/2014
We have the photon's direction, frequency and spin. what other variables do we need for a complete commuting set?
Wolphram jonny 12/10/2014
You are correct, I think in that you can assume a sligthly different but equivalent situation in which the original light source is prepared in such a way that that you know the total state of the system, that is, you are preparing it with all sets of quantum numbers other than polarization perfectly known. Let me think what would happend then.
WetSavannaAnimal aka Rod Vance 12/11/2014
Yes you will lose information and it does become mixed with the polariser and, later, the radiation field as it is reradiated. But that's the point of the $\mathrm{d}P/T$ term: this is the information mixing as you say. But do I misunderstand? Anyhow, I am writing up a detailed answer, in the next day, so lets talk about it then. Your points definitely contribute to my understanding, so +1 and thanks: it's much appreciated.

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