Stand on a railway platform in Rajasthan and look along an older track. Every thirteen metres two rails meet at a joint held by a steel plate, and a small gap is left between them. In a January dawn the gap is a few millimetres wide. On a May afternoon it has almost closed. The wheels click at every gap.
A lopsided pull
A solid is a crowd of atoms held in place by forces between neighbours. Push two atoms closer and they repel strongly. Pull them apart and they attract, but the pull weakens slowly with distance. The potential energy of a pair against their separation is therefore a lopsided well. It is steep on the near side and gentle on the far side.
Each atom vibrates in this well. You met such wells in the chapter on energy. A particle with energy \(E\) swings between the two points where the curve meets the line of height \(E\). Heating the solid gives each atom more energy, so the line rises. On the steep side the turning point hardly moves. On the gentle side it moves far out.
In Fig 4.1 the middle of the swing sits at \(1.03\,r_0\) when cold and moves to \(1.30\,r_0\) when hot. So the average spacing of neighbours grows with temperature. Every pair of neighbours moves apart a little. Added over billions of atoms in a row, the solid becomes measurably longer.
Picture a child on a swing hung next to a wall. The wall stops the swing sharply on one side, while on the other side the swing rises freely. Push harder and the swing goes much further on the open side than towards the wall. The middle of its motion moves away from the wall.
Suppose the force between two atoms were exactly like an ideal spring, so the energy well was symmetric. What would happen when the solid is heated?
Linear expansion
A rod of length \(L\) is warmed by a small amount \(\Delta T\). Each small piece of it grows by the same fraction, so the rod grows in proportion to its length. The growth is also in proportion to the rise in temperature. So \(\Delta L = L\alpha\,\Delta T\). The constant \(\alpha\) belongs to the material and is its coefficient of linear expansion.
The value of \(\alpha\) is small. For steel it is \(1.2 \times 10^{-5}\,\text{K}^{-1}\), for brass \(1.9 \times 10^{-5}\), for aluminium \(2.4 \times 10^{-5}\) and for ordinary glass \(9 \times 10^{-6}\). The alloy invar has only about \(1 \times 10^{-6}\). A steel rod warmed by \(45\,\text{K}\) grows by \(1.2 \times 10^{-5} \times 45 = 5.4 \times 10^{-4}\) of its length, which is \(0.054\,\%\). A long rod and a short rod grow by the same percentage.
The rail and its gap
Now the rails of the phenomenon. Each steel rail is \(13\,\text{m}\) long, with \(\alpha = 1.2 \times 10^{-5}\,\text{K}^{-1}\). From the cold morning when the track was laid to a Rajasthan summer afternoon, the rails may warm by \(45\,\text{K}\). Then each rail grows by \(\Delta L = 13 \times 1.2 \times 10^{-5} \times 45\,\text{m}\). That is \(7.0 \times 10^{-3}\,\text{m}\), or \(7.0\,\text{mm}\).
Watch one gap in the sim above. The rail on its left pushes its right end in by half its growth, \(3.5\,\text{mm}\) at \(45\,\text{K}\). The rail on its right pushes its left end in by the same amount. So the gap closes by the growth of one whole rail, and a gap of \(7.0\,\text{mm}\) is just enough. Lesson 6 shows what happens when no gap is left.
A steel span is \(457\,\text{m}\) long, the length of the main span of the Howrah Bridge in Kolkata. Between a winter night and a summer afternoon its temperature rises by \(25\,\text{K}\). By how much does it lengthen? Take \(\alpha = 1.2 \times 10^{-5}\,\text{K}^{-1}\).
(a) \(\Delta L = L\alpha\,\Delta T = 457 \times 1.2 \times 10^{-5} \times 25 = 0.137\,\text{m}\).
(b) The span lengthens by about \(14\,\text{cm}\) every summer day and shortens by as much at night. Its joints and bearings are built to slide by this amount.
An aluminium rod is \(2.0\,\text{m}\) long. It is warmed by \(50\,\text{K}\). Take \(\alpha = 2.4 \times 10^{-5}\,\text{K}^{-1}\). By how many millimetres does it lengthen?
A steel rod (\(\alpha = 1.2 \times 10^{-5}\,\text{K}^{-1}\)) is longer than an aluminium rod (\(\alpha = 2.4 \times 10^{-5}\,\text{K}^{-1}\)) by \(12\,\text{cm}\) at every temperature. How long is the steel rod, in cm?
A copper wire \(1\,\text{m}\) long and a copper wire \(10\,\text{m}\) long are warmed by the same \(40\,\text{K}\). Which statement is correct?
Area and volume
Most solids are isotropic: they expand by the same fraction in every direction. Take a square plate of side \(a\) and warm it by \(\Delta T\). Each side becomes \(a(1 + \alpha\,\Delta T)\). The area grows by a fraction \(\beta\,\Delta T\), where \(\beta\) is the coefficient of area expansion.
\[ A' = a^2(1 + \alpha\,\Delta T)^2 = a^2\left(1 + 2\alpha\,\Delta T + \alpha^2\,\Delta T^2\right) \]
Why. each side becomes \(a(1 + \alpha\,\Delta T)\), and the area is the side squared. In Fig 4.2 the three terms are the old square, the two strips and the corner.
\[ A' \approx A(1 + 2\alpha\,\Delta T) \]
Why. \(\alpha\,\Delta T\) is about \(10^{-3}\), so its square, about \(10^{-6}\), is far too small to matter. The corner is dropped.
\[ \beta = 2\alpha \]
Why. compare with the definition \(A' = A(1 + \beta\,\Delta T)\).
\[ V' = a^3(1 + \alpha\,\Delta T)^3 \approx V(1 + 3\alpha\,\Delta T), \quad \text{so} \quad \gamma = 3\alpha \]
Read it. a cube has three sides that each grow by the same fraction; again drop the squares and cubes of \(\alpha\,\Delta T\).
The coefficient of volume expansion \(\gamma\) is the one that matters most for liquids and gases. A liquid has no shape of its own, so it has no length to expand. It has only a volume, and only \(\gamma\) describes it.
A crystal such as calcite is not isotropic: it expands by different fractions along its three axes, with coefficients \(\alpha_x\), \(\alpha_y\) and \(\alpha_z\). Multiply the three new edges of a box and drop the products of small terms. The result is \(\gamma = \alpha_x + \alpha_y + \alpha_z\), and a face in the \(xy\) plane has \(\beta = \alpha_x + \alpha_y\). For an isotropic solid all three are \(\alpha\), and \(\gamma = 3\alpha\) again.
(a) A steel sheet (\(\alpha = 1.2 \times 10^{-5}\,\text{K}^{-1}\)) has an area of \(2.0\,\text{m}^2\). Find the increase in its area, in cm², when it warms by \(50\,\text{K}\). (b) A copper cube (\(\alpha = 1.7 \times 10^{-5}\,\text{K}^{-1}\)) has edges of \(10\,\text{cm}\). Find the increase in its volume when it is heated by \(100\,\text{K}\).
(a) \(\beta = 2\alpha = 2.4 \times 10^{-5}\,\text{K}^{-1}\). \(\Delta A = 2.0 \times 2.4 \times 10^{-5} \times 50 = 2.4 \times 10^{-3}\,\text{m}^2 = 24\,\text{cm}^2\).
(b) \(\gamma = 3\alpha = 5.1 \times 10^{-5}\,\text{K}^{-1}\) and \(V = 1000\,\text{cm}^3\). \(\Delta V = 1000 \times 5.1 \times 10^{-5} \times 100 = 5.1\,\text{cm}^3\). The exact cube of the new edge gives \(5.11\,\text{cm}^3\), so dropping the small terms costs almost nothing.
A metal ball is heated, and its diameter grows by \(0.1\,\%\). By what percentage does its volume grow?
A steel plate has a round hole in the middle. The plate is heated evenly. What happens to the hole?
The chips show why. Heating makes every distance between atoms grow by the same fraction, as in a photographic enlargement. The distance across the hole is one of those distances. Imagine the hole filled with a disc of the same steel. The disc would grow with the plate and still fit exactly. So the hole grows as if it were made of the material around it.
The same holds for a cavity. The inside of a glass flask grows as if the cavity were solid glass. A flask that holds \(1000\,\text{cm}^3\) at \(20\,°\text{C}\) holds \(1000 \times 3 \times 9 \times 10^{-6} \times 100 = 2.7\,\text{cm}^3\) more at \(120\,°\text{C}\). Lesson 5 needs this result.
Shrink-fitting
A village blacksmith fits an iron tyre on a wooden cart wheel by using a hole that grows. The tyre is made slightly smaller than the wheel. He heats it in a fire until its inner width grows past that of the wheel, drops it on and pours water over it. The tyre shrinks as it cools and grips the wheel with great force. Steel rims of railway wheels are fitted the same way.
An iron tyre has an inner diameter of \(100.00\,\text{cm}\) at \(25\,°\text{C}\). It must slide onto a wheel of diameter \(100.12\,\text{cm}\). To what temperature must it be heated? Take \(\alpha = 1.2 \times 10^{-5}\,\text{K}^{-1}\).
The inner diameter behaves like a length of iron. It must grow by \(\Delta d = 0.12\,\text{cm}\).
\(\Delta d = d\alpha\,\Delta T\), so \(\Delta T = \dfrac{\Delta d}{d\alpha}\).
\(\Delta T = \dfrac{0.12}{100.00 \times 1.2 \times 10^{-5}} = 100\,\text{K}\).
The tyre must be heated to at least \(25 + 100 = 125\,°\text{C}\). In practice it is heated well beyond, so that it slides on easily.
A steel plate has a hole \(4.0\,\text{cm}\) across. The plate is heated by \(200\,\text{K}\). Take \(\alpha = 1.2 \times 10^{-5}\,\text{K}^{-1}\). By how many millimetres does the diameter of the hole change?
A glass jar (\(\alpha = 9 \times 10^{-6}\,\text{K}^{-1}\)) has a tight aluminium lid (\(\alpha = 2.4 \times 10^{-5}\,\text{K}^{-1}\)). The top of the jar is held under hot water. What happens to the grip of the lid?
In 1731 the Dutch professor Pieter van Musschenbroek built an instrument to measure how much a metal rod lengthens when heated. One end of the rod was fixed, and its free end turned a train of levers and gears that moved a pointer across a dial. A growth far thinner than a hair became a visible swing of the pointer. He called the instrument a pyrometer, and with it he compared how much iron, copper and tin expand.
Drive over a long flyover in Mumbai or Delhi and the car thumps across a strip of interlocking steel teeth every few hundred metres. These are expansion joints. The concrete deck lengthens on a hot afternoon and shortens at night, and the teeth slide into each other to take up the change. Without them the deck would push against its supports and crack.
- Using \(\beta = \alpha\) or \(\gamma = \alpha\) for an area or a volume.For an isotropic solid \(\beta = 2\alpha\) and \(\gamma = 3\alpha\): an area has two lengths, a volume three.
- Thinking a hole shrinks when the plate around it is heated.Every distance grows by the same fraction. The hole grows as if it were filled with the same material.
- Leaving \(L\) in centimetres and \(\Delta L\) in metres.\(\Delta L/L = \alpha\,\Delta T\) is a pure number, so \(\Delta L\) comes out in the unit of \(L\).
- Making a rail gap half the growth of a rail.Each gap is closed from both sides, so it must equal the full growth of one rail.