Say we have a square metal plate $A$, 1 $m^2$ on each side, of negligible thickness, in a vacuum with no other objects nearby and no other heat coming into the system. Say its temperature is at an equilibrium of 20°C.

Using the Stefan Boltzmann Law we calculate that $A$ will be radiating 418.8 W/$m^2$ or 837.6 W total -- or in other words, its internal heat source is constantly outputting 837.6 W in order to keep it at 20°C.

Now say we take another metal plate $B$ of identical size and shape, this one with no internal heat source. It starts off at just above 0K. We approach $B$ and place it right next to $A$ but with a little space. The only heat transfer is via radiation, and $A$ will start heating $B$. How will the system end up at equilibrium?

I can see one of two possibilities:

  1. If we model energy in = energy out for $A$, $B$, and the system as a whole, with $A$ emitting all it receives in two directions with $a$ in each direction, and similarly for $B$ with $b$, then we have the following system of equations:

    • $837.6 + b = 2a$ (energy into $A$ equals energy out of $A$)
    • $a = 2b$ (energy into $B$ equals energy out of $B$)
    • $837.6 = a + b$ (total energy into the system equals total energy out of the system)

    Some simple math results in $a = 558.4$ W and $b = 279.2$ W.

    In other words, $A$ will increase in temperature to 41.87°C while $B$ will end up with a temperature of -8.2°C.

  2. Taking it from a different angle, $B$ will gradually heat up until we eventually reach a new equilibrium where both $A$ and $B$ reach 20°C, similarly as if they were touching and thus transferring heat by conduction. As both sides will be radiating 418.8 W out to space, total energy in still equals total energy out.

I cannot argue with the math and logic of #1, but #2 seems more intuitive and sensible in a "common-sense" type of reasoning, though I cannot explain it fully with equations (perhaps because it is wrong).

Further if #1 is the case, it seems strange that if we bring the two plates to touch into $AB$ we will end up $AB$ itself having 20°C (assuming good conductivity). Then if we separate them a little, $A$ heats up again. Then if we connect them again, $A$ cools back down. It seems a little odd, unfamiliar and non-intuitive, and as we could separate and re-attach the two plates with extremely minimal work, couldn't we take advantage of the new temperature gradient to do more work than that, thus violating some thermodynamic law?

My questions are:

  1. What is the case -- will the equilibrium be as in #1 or as in #2, or something else? And why?
  2. Have any experiments been done that actually show what will happen? For me this is the ultimate evidence.
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    $\begingroup$ I have no idea how you came up with scenario 1. $\endgroup$ Commented Apr 6, 2023 at 21:10
  • $\begingroup$ My thought was that $B$ now has a temperature too, so it must be radiating some amount onto $A$ which contributes to $A$'s temperature... very keen to hear why it's wrong, if it is, in detail! $\endgroup$
    – Cloudyman
    Commented Apr 6, 2023 at 21:48
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    $\begingroup$ Intuitively #2 makes sense if they are in perfect thermal contact (probably touching, with conduction). If they are not, there is no reason to think they should have the same temperature. Indeed, a series of alternating air pockets and sheets of identical material is how thermal insulation foam works! $\endgroup$ Commented Apr 6, 2023 at 21:56
  • $\begingroup$ @user253751 is the answer #1 then? Or something else, where $A$ remains at 20°C and $B$ ends up a lower temperature? $\endgroup$
    – Cloudyman
    Commented Apr 6, 2023 at 22:04
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    $\begingroup$ Isn't the situation with A and B exactly analogous to the Sun and the Earth, respectively? If so, I don't find it surprising that they end up at the different temperatures. And the increase in A's temperature due to the presence of B is (very roughly) what happens in the greenhouse effect. $\endgroup$ Commented Apr 7, 2023 at 11:57

2 Answers 2


Wow—I'm quite surprised that a more detailed quantitative analysis supports the conclusion that blocks A and B (assumed to be excellent conductors) do indeed converge to steady-state temperatures of 41.85°C and -8.24°C as they are brought toward each other, switching to a uniform steady-state temperature of 20°C when they come in contact to become a monolithic block. My original answer is below, but I conclude key parts are incorrect. I pick up again in the bold section below.

I cannot argue with the math and logic of #1

I can. You disrupted the symmetry when you put something to the side of A (and the same argument holds for B), so the radiative heat transfer is no longer symmetric. Thus, you cannot split left-and-right heat fluxes evenly to give $2\times a$ and $2\times b$.

The new equilibrium temperature of A is $20^\circ\text{C}-\delta T_1$, and the equilibrium temperature of B is $20^\circ\text{C}-\delta T_1-\delta T_2$, where $\delta T_1$ and $\delta T_2$ are small positive temperatures that decrease with increasing aspect ratio of A and B and decreasing spacing between A and B. They cannot be zero for real objects with nonzero spacing and nonzero thickness, as some radiative heat transfer occurs from the sides (with more surface area now that B is present) and view factor fraction that doesn't include the adjacent plate.

The paradox essentially emerges from multiplication of $\sim$$\infty$ by $\sim$$ 0$: A and B are so nearly the same object that it's tempting to idealize them as being the same temperature, but only A is internally heated, so B must get its energy from somewhere. To account for this plate-to-plate heating, we must multiply a very large number (the side surface area, relative to the thickness and spacing) by a very small number (the temperature difference $\delta T_2$). It would be incorrect to assume that the product is also necessarily very small.

A precise calculation of $\delta T_1$ and $\delta T_2$ will depend on the actual geometry, but the apparent paradox is resolved nonetheless.

Cloudyman prompted me to justify this conclusion with a heat transfer analysis. It indicates that the presence of unheated block B does indeed make heated block A warmer. My answer above that block A is always cooler than 20°C is thus incorrect. I was relying on a smooth temperature transition from not touching to touching, but this transition could be very sharp if the block material is a very good thermal conductor, corresponding to a very low Biot number.

The energy balances for plates with side length $L$, thickness $t$, spacing $S$ and emissivity 1 (for simplicity) surrounded by ambient temperature $T_\infty$ are



where $\sigma$ is the Stefan–Boltzmann constant and where the view factor between equal square plates is $$F=\frac{1}{\pi x^2}\left(\ln\frac{y^4}{1+2x^2}\right)+4xz,$$

with $x\equiv\frac{L}{S}$, $y\equiv\sqrt{1+x^2}$, and $z\equiv y\arctan\frac{x}{y}-\arctan{x}$. (Experimentally, I find that $F\approx 1-\frac{2S}{L}$ for small $S$.)

Now let $T_\infty=0$ and $t\to 0$ for simplicity and solve equations (1) and (2) simultaneously for $T_A$ and $T_B$. We obtain

$$T_A^4=\left(\frac{P}{\sigma L^2}\right)\left(\frac{2}{(2-F)(2+F)}\right);$$

$$T_B^4=\left(\frac{P}{\sigma L^2}\right)\left(\frac{F}{(2-F)(2+F)}\right).$$

When plate B is distant ($F\to 0$), $T_A^4=\frac{P}{2\sigma L^2}$ and $T_B=0$, as expected. When the plates are nearly touching ($F\to 1$), $T_A^4=\frac{2P}{3\sigma L^2}$ and $T_B^4=\frac{P}{3\sigma L^2}$, with block A achieving a steady-state temperature $2^{1/4}$ hotter than block B on an absolute temperature scale. Fascinating!

Now, since an object doesn't exist with zero thickness and infinite thermal conductivity, in practice, we'll see some radiative heat transfer from the edges and some internal spatial variation in the temperature. However, I no longer think that this has as big an impact on the qualitative results as the earlier version of this answer asserted.

  • $\begingroup$ I am utterly intrigued. Would it be possible for you to work through an example with say 1cm thick plates spaced 1cm apart? Or direct me to a resource where I could learn the math, I should be able to learn it. $\endgroup$
    – Cloudyman
    Commented Apr 6, 2023 at 22:34
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    $\begingroup$ @Cloudyman If you space them apart, then some of the radiation escapes at the edge. How much it is depends on the reflectivity of the metal surfaces. Since the problem is now inhomogeneous, we have to consider the full radiation (and heat) transport problem. $\endgroup$ Commented Apr 6, 2023 at 22:39
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    $\begingroup$ The net heat flow between A and B is not almost zero. You've again multiplied a large number (the area) by a small number (the temperature difference) and concluded that the product is small, but it's not. A needs to heat B by nearly exactly the same amount that it heated the void. I recommend working on understanding the equilibrium case thoroughly before moving on to the transient case. $\endgroup$ Commented Apr 6, 2023 at 23:14
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    $\begingroup$ Yes, thanks! Please see the edited answer. $\endgroup$ Commented Apr 7, 2023 at 1:04
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    $\begingroup$ You've convinced me! I've changed my answer considerably based on our discussion. $\endgroup$ Commented Apr 7, 2023 at 4:25

A few remarks (too long for comments and related to the discussion in the other thread):

  1. If the plate is being internally heated, there is an energy pumped into the system, so we never reach a thermal equilibrium, and we shouldn't apply equilibrium concepts to discuss this problem.
  2. If the plates are in open vacuum, there will be energy leaving (in the form of radiation), so the best that we can expect is a steady state, where the energy losses are equal to the energy add by the internal source in the plate.
  3. If the plates are big, so that their internal energy is much bigger than the energy lost/acquired via radiative transfer, we still can consider them to be in thermal equilibrium internally and attribute to them temperature (but not in thermal equilibrium with the environment and the other plate.)
  4. The temperature of the second plate in the steady state achieved does not have to be the same as the temperature of the first plate.
  5. If the second plate radiates some of the radiation back or reflects some of the radiation, the temperature of the first plate might be higher than it would be, if the second plate were not there.
  6. Much depends on the assumptions about the heating regime: is this a constant energy flow? Or is it aimed at maintaining constant temperature? Something else?
  • $\begingroup$ Thanks for the comments! 1+2: What is the distinction of “steady state” vs “thermal equilibrium”? 6: constant energy flow. 5: that is the conclusion, which many find odd! As approaching a “colder surface” to a “hotter” one results in the hotter one being even hotter. I can’t argue with the math or logic of it though! $\endgroup$
    – Cloudyman
    Commented Apr 7, 2023 at 11:55
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    $\begingroup$ In a steady state nothing changes in time, but there are still non-zero fluxes (of energy, matter, etc.) In equilibrium these fluxes are zero. As a fluid dynamics analogy: A river is in a steady state, but not in equilibrium (water flows.) A lake is in a steady state, even though in mechanics we would probably refer to it as a static rather than equilibrium state. $\endgroup$
    – Roger V.
    Commented Apr 7, 2023 at 11:59
  • $\begingroup$ That's a very nice distinction actually, I think it explains some of the confusion surrounding this topic generally. Re 3: then, wouldn't the plate itself also never be in thermal equilibrium, since it always have non-zero flux (energy in from internal source, energy out from radiative heat loss). Then we have the surface temperature as proportional to whatever energy is radiated out from the surface. $\endgroup$
    – Cloudyman
    Commented Apr 7, 2023 at 12:27
  • $\begingroup$ I would call it a quasi-equilibrium, since it is approximate (as described above). This however allows applying many equilibrium results - like Planck's formula, proportionality of radiation to temperature, etc. See physics.stackexchange.com/questions/645671/… and physics.stackexchange.com/q/644253/247642 $\endgroup$
    – Roger V.
    Commented Apr 7, 2023 at 12:44

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