A previous question asked why the road sometimes appears wet on hot days.

The reason is that when there's a temperature gradient in the air, it causes a gradient in the index of refraction, causing the light to bend. This diagram (from Lagerbaer's answer to the previous question) is

enter image description here

I have two questions about this.

First, Wikipedia says that in order for the effect to appear, there must be a temperature gradient on the order of a few degrees per hundred meters above the asphalt.

To investigate this claim, I assumed the air near the surface of the asphalt is at constant pressure, so the density is inverse-proportional to the temperature (i.e. I assumed the density variation due to the weight of the air is small compared to the density variation induced by temperature gradient. Otherwise we would see mirages all the time.)

Then I assumed the difference of index of refraction of air from one is proportional to the density, i.e. $n = 1 + a\frac{\rho}{\rho_0}$ with $n$ the index of refraction of air, $a$ some dimensionless constant, and $\rho_0$ some reference density. Looking up the index of refraction of air online, I guessed $a = 0.0003$.

Then I assumed the temperature above the Earth is modeled by $T = T_0 + gy$ with $y$ the height and $g$ a temperature gradient in ${}^\circ\mathrm C/\mathrm m$.

This completes the model. Fermat's principle gives a variational problem to solve for the path of the light. However, the resulting differential equation was hard to work with, so to first order I approximated that the light starts at a horizontal distance $x = -L$ at height $y =h$, slopes down towards the ground with some slope $m$ until it gets to $x=0$, then slopes back up at the same slope $m$ until it gets to $x = L$ and $y = h$ again. Then I chose $m$ to minimize the travel time of this path.

To first order for small $h$, I found

$$m = \frac{agL}{2T_0}$$

The problem is that when I plug in $g = 5\ \mathrm{^\circ C}/100\ \mathrm m$, $a = 0.0003$, $T_0 = 300\ \mathrm K$, and $L = 100\ \mathrm m$ I get $m = 2.5\times10^{-6}$, which is too small to explain the mirage. It would only allow light to dip a quarter millimeter in the path from a car $200\ \mathrm m$ away to me. In the Wikipedia image of a mirage, the light clearly dips down at least 1000 times that much (check out the blue car).

enter image description here

So my first question is: what's wrong? Why are there mirages when this model predicts that there are not?

My second is: why do we see them only when it's hot? This model depends only weakly on absolute temperature, and hot absolute temperatures actually decrease the effect. Evidently, there are only high temperature gradients on hot days, by why should that be? A temperature gradient over the asphalt comes from sun heating the asphalt directly more than it heats the air. Shouldn't that happen on cold days as well as on hot days?

  • $\begingroup$ Using a thermal gradient g of 5 C/100 m would mean that the air next to the road is only 0.05 C hotter than the air at waist level. If you put your hand down near a sun-baked road, you can feel the air is perceptibly warmer (e.g. 5 C hotter) than the air higher up. So multiplying g (therefore m) by a hundred seems warranted. $\endgroup$
    – DMPalmer
    Feb 22, 2018 at 22:06

2 Answers 2


I have not done the math but would expect that the radiation from the asphalt as $T^4$ will favor larger gradients for higher temperatures. I have the impression that air goes something like $T^6$, so even conduction energy transferred will have larger gradients the hotter it is. Your g is temperature dependent I guess.

Edit in response to edit of question.

Shouldn't that happen on cold days as well as on hot days?

I am copying from the comments below:

if you do the calculation of black body, asphalt at $40^\circ \mathrm{C}$ radiates $547\mathrm{watts}/\mathrm{m}^2$. At $41^\circ\mathrm{C}$ 551, i.e. difference $4\mathrm{watts}/\mathrm{m}^2$, whereas at $50^\circ\mathrm{C}$ 617 and at $51^\circ\mathrm{C}$ 628, i.e. $11\mathrm{Watts}/\mathrm{m}^2 \Delta T$. that is what I mean by g is temperature dependent. If there is convection the asphalt remains cooler and the equilibrium between heating from incoming radiation and cooling from black body is lower.

On a cold day even if windless with no convection the equilibrium temperature will be much lower since it starts heating from a much lower ground temperature. You can fry eggs in the summer in Greece on the asphalt, not in the winter.

In addition $5^\circ\mathrm{C}$ per meter is a low number. The asphalt may be at $50^\circ\mathrm{C}$ or $60^\circ \mathrm{C}$ but at 1 meter it will not be more than 35, in Greece, summer. In northern latitudes maybe $25^\circ\mathrm{C}$?

  • $\begingroup$ First problem is, that those mirages do not appear on warm days, but on sunny days with low or no wind. The asphalt will easily reach 50 °C or more, and not radiate much! So a thin layer of air above the asphalt is heated to rather high temperatures (by conduction!), and acts as a "mirror". What do You mean with this "air T⁶"? $\endgroup$
    – Georg
    Nov 4, 2011 at 15:10
  • $\begingroup$ The black asphalt radiates as T^4 (black body formula) Air is a bad conductor. On sunny windless days, without convection mixing the air, it will be radiation which will be heating the immediate layer of air, the infrared part of the spectrum that will jiggle the molecules in passing. The derivative of T^4 is T^3, the higher the asphalt temperature the higher the increment. The air layers also radiate, but as T^6 ( I would have to search for a link), so each layer heating the next layer has even more incremental temperature dependence. $\endgroup$
    – anna v
    Nov 4, 2011 at 15:29
  • $\begingroup$ @georg the above for you $\endgroup$
    – anna v
    Nov 4, 2011 at 15:54
  • $\begingroup$ @ anna, nitrogen and oxygen do not absrb any IR! The absorption by the trace CO2 is negligible in this dimensions, and last not least, at 50 to maybe 100 °C radiation of the asphalt is negligible! $\endgroup$
    – Georg
    Nov 4, 2011 at 21:01
  • $\begingroup$ @ Georg if you look at the absorption spectrum astronomy.ohio-state.edu/~pogge/Ast161/Unit5/atmos.html you will see that they do absorb, not as much as H2O, ( which btw is never non existent in air on earth) but they do. Otherwise how would rooms be heated by radiant heaters? Air is a bad conductor, used for insulation. And the radiation from asphalt is not negligible. $\endgroup$
    – anna v
    Nov 5, 2011 at 5:39

The illusion is generated by your brain. The light is only a quarter of a millimeter lower, at your eye. So you detect it as being feet lower at a distance of 200M. Your brain interprets the light as if it took a straight path to your eye, so only a small deviation in angle is required to make the mirage possible.

If the object is small enough (or the heat gradient is large enough)it can be completely obscured by the mirage.


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