# Heating and Recooling of an Object

Consider a piece of metal of length $L$ and linear thermal expansion coefficient $\alpha$. We eat the metal $\Delta T$ degrees, causing the metal to increase to length $$L' = L + L \alpha \Delta T$$ Now, cool the object back to the original temperature. This causes the metal to decrease in length to $$L'' = L' - L'\alpha\Delta T \\ = L + L \alpha \Delta T - (L + L \alpha \Delta T)\alpha \Delta T \\ =L (1 - \alpha^2 \Delta T^2)$$

My intuition would lead me to believe that heating up and then recooling an object would cause it to return to the same size it began at (if it did not, then bridges which repeatedly warmed and cooled would continually shrink) but this is not true according to my mathematics, which indicates that warming and recooling an object leaves it slightly smaller than it was before.

Do objects really not return to their original size when recooled? If so, then why do objects which warm and cool on a daily basis not slowly shrink?

I suspect that this may be related to $\alpha$ varying over the range of temperatures, but I didn't think that was a significant effect for solids.

Your intuition is right, an object (at least of the ideal sort these formulas are used to describe) returns to the same size when it is brought back to the same temperature. The reason your math isn't giving that result is that $L' = L + L\alpha\Delta T$ is a first-order approximation, valid for small values of $\alpha\Delta T$. This means, among other things, that when doing calculations with it, your results only have to agree to first order - in other words, if you have two quantities that differ only by a multiple of $(\alpha\Delta T)^2$ (or any higher power), they can be considered the same. By this reasoning, $L'' = L$ to first order; or, as it would be written in symbols, $L'' = L + \mathcal{O}\bigl((\alpha\Delta T)^2\bigr)$.

If you want to be a little more rigorous about this whole business, you can derive the full, non-approximate formula for thermal expansion like this. For a small value of $\Delta T$, the corresponding change in $L$ is

$$\Delta L = \frac{\mathrm{d}L}{\mathrm{d}T}\Delta T$$

From the approximate thermal expansion equation, you know that $\Delta L = L\alpha\Delta T$, which gives

$$L\alpha = \frac{\mathrm{d}L}{\mathrm{d}T}$$

Integrating this gives

$$L' = Le^{\alpha(T' - T)}$$

which is the real expression (or at least a mathematically consistent better approximation). You can verify that $L' = L\alpha\Delta T$ is the first-order approximation to this.

Another way to do this derivation is to think about heating a material from temperature $T$ to $T' = T + \Delta T$, which takes it to length

$$L' = L(1 + \alpha\Delta T)$$

then to a slightly higher temperature $T'' = T' + \Delta T$, which takes it to length

$$L'' = L'(1 + \alpha\Delta T) = L(1 + \alpha\Delta T)^2$$

then to a slightly higher temperature $T'''$, which takes it to

$$L''' = L''(1 + \alpha\Delta T) = L(1 + \alpha\Delta T)^3$$

and so on. If your desired final temperature is $T_f = T + n\Delta T$, then the final length will be

$$L_f = L(1 + \alpha\Delta T)^n = L(1 + \alpha\Delta T)^{(T_f - T)/\Delta T}$$

In the limit as $\Delta T\to 0$, this becomes

$$L_f = L e^{\alpha\Delta T}$$

Isn't this a disguised version of the "markup-discount" problem? If a retailer starts with an item with a price P, marks the price up by, say, 10 percent, and then discounts that amount by 10 percent, the result is not going to be P.

The thermal expansion rule is applied to a specific reference length $L_0$. If the temperature is changed by $\Delta T$, its length changes by a proportion given by $\alpha \Delta T$. You are choosing the altered length as a new reference length, so naturally a different proportion is required in order to achieve the original reference length.

The distinction between relative and absolute length changes is deceptive here. The object of length $L$ requires the same temperature change to expand to $L'$ and to return to the original length. An object of initial length $L'$ is a different object and will require a slightly different temperature change to "shrink" to $L$ and then be restored to $L'$.