I have read an article about glass (zerodur) with low thermal expansion coefficient. It is mentioned that large casts of such glass are covered with reflective layer of Aluminium and used as mirrors in space observatories.

Low CTE is so important in this glass, because changes in size of it would distort the picture taken with the telescope. But what about said layer of aluminium? It is metal and it's CTE is much larger than the one of the glass.

So how does it happen that the thin layer of aluminium doesn't distort the picture? Doesn't it expand/shrink?

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    $\begingroup$ Good question, I've used aluminium on glass mirrors without stopping to think about problems with thermal expansion. I would guess that the thinner the film the lower the force pulling it away from the glass substrate, and since the films in question are typically a few 100nm thick the forces due to thermal expansion aren't big enough to make the film buckle off the glass. If would be nice to see a quantitative description of this if anyone feels up to it ... $\endgroup$ Commented Apr 30, 2015 at 17:49
  • $\begingroup$ John is basically correct - The net stress applied by a thin film on a thick substrate is quite small (and the equation is pretty easy to find in the thin film literature). On top of that, the metal is bot fairly elastic, and will gracefully deform plastically should the yield stress be exceeded. If the situation were reversed (thick metal, thin glass), than cracking or delamination of the brittle glass would be likely. $\endgroup$
    – Jon Custer
    Commented Apr 30, 2015 at 18:47
  • $\begingroup$ @JonCuster - if you made a 100 nm coating of glass on metal, the metal would not bend under heating (unless you introduced a thermal gradient across the thickness) and the thin glass coating would not crack. Dissimilar coatings are used all the time, and the thinner they are the less likely they are to crack / delaminate. $\endgroup$
    – Floris
    Commented Apr 30, 2015 at 19:53
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    $\begingroup$ @JohnRennie - I have tried a "back of the envelope" quantitative estimate of the effect. Would be interested in your critique of my method. $\endgroup$
    – Floris
    Commented Apr 30, 2015 at 19:53
  • $\begingroup$ @JohnRennie: Than thicker the Al coating than less it has mirror qualities. Such mirrors are used in industrial imaging where thermal expansion or weight does not matter. $\endgroup$ Commented Apr 30, 2015 at 19:56

1 Answer 1


Let us take the example of the Hubble primary mirror. It has a diameter of 2.4 m and a mass of 828 kg. It is actually made in a sandwich structure - glass-honeycomb-glass - making it about 30 cm thick (for stiffness) but light.

The mirror is coated with an aluminum coating of thickness t = 65 nm, with a 25 nm MgF2 protective coating on top.

Coefficient of thermal expansion of aluminum is $2.2 \cdot 10^{-5} \mathrm{m/m\cdot K}$ and the Young's modulus is 69 GPa. If you constrain the coating to be of constant size, then you increase the strain by $2.2 \cdot 10^{-5}$ per °C, and for a square sheet of aluminum with sides $L$ and thickness $t$, the force this creates would be

$$F = \sigma E A = 2.2\cdot 10^{-5} \cdot 69 \cdot 10^9 \cdot 2.4 \cdot 10^{-9} \approx 0.4 N $$

A force of 0.4 N across a 30 cm thick mirror gives a bending moment of about 0.06 Nm; the radius of curvature can be approximated by

$$R = \frac{EI}{M}$$

Again, we will approximate the mirror as a square with 1 D deflection, in which case the second moment of area is given by approximately (t = thickness of glass = 3.5 cm, s = spacing = 25 cm)

$$I = \frac{Lt}{2 s^2} \approx 0.001 m^4$$

Thus we find for the (additional) curvature of the mirror:

$$R = \frac{70\cdot 10^9 \cdot 0.001}{0.4}=3\cdot 10^9 m$$

This results in a deflection across the 2.4 m diameter of

$$\Delta h = \frac{r^2}{2R} = \frac{1.44}{2\cdot 3\cdot 10^9} \approx 0.3 \mathrm{nm}$$

For comparison it is worth noting that the specification of the mirror calls for the shape to be accurate to within 10 nm, and the error that led to the initial blurry images was on the order of 2 µm. In other words, while the coating might have an effect, it is quite a bit smaller than the specification of the mirror.

And there is another thing - the mirror has a heater on it which keeps the temperature constant within half a degree (to maintain shape). Originally this temperature was 21 degrees, but that makes it work less well in the near IR. Later the temperature was dialed down to 15°C (source - as above).

So yes - thermal expansion of coatings can affect the shape of the mirror; but no, it is not significant.

  • $\begingroup$ That's great and elaborate answer! Thanks! :) But what about forces that try to tear apart such thin layer of Aluminium? Length of such layer is quite big and layer is very thin. So why layer doesn't fragment? Is it so powerfully fixed to the glass, or is it just too durable? $\endgroup$
    – user46147
    Commented May 1, 2015 at 16:37
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    $\begingroup$ It is no different than if you had a thick layer of aluminium - the strain (relative change in length) is just a few parts in 100,000 which is not enough to tear the coating. $\endgroup$
    – Floris
    Commented May 1, 2015 at 17:12

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