Why does a solid and hollow sphere go through the same expansion in volume when raised to the same temperature?

Please explain it at the molecular level with an example and not via a formula.

  • $\begingroup$ It's the same material so why wouldn't it? I would intuitively expect a big thick bar to increase in length by the same amount as a thin one. $\endgroup$ – DKNguyen Jun 12 at 4:29
  • $\begingroup$ Is there a particular formula? If so, what is it? For example, does it have a name (or not)? $\endgroup$ – Peter Mortensen Jun 12 at 14:46
  • $\begingroup$ It is Sunday... $\endgroup$ – Peter Mortensen Jun 12 at 14:47
  • 5
    $\begingroup$ How exactly would someone explain it quantitatively, but without any math (which i assume you mean by "formula")? Are you sure you don't mean qualitatively? $\endgroup$ – Hearth Jun 12 at 17:02
  • $\begingroup$ does it though? has someone experimentally measured them? $\endgroup$ – lineage Jun 12 at 19:50

If you aren't interested in the dilemma your question sent me into, skip straight to the highlighted What all of this points to below

To me the shell based explanation did not make intuitive sense. e.g. Consider the $1$D analog of a rod positioned in the interval $[0,L']$, clamped at $0$. Lets chop off its end so its described as $[0,L]\cup[L,L+\Delta L]$.

  1. If this rod is heated what would be its new length? It would be the sum of the expansions of the two parts - the entire mass expands. It would therefore be $L(1+\alpha\Delta T)+ \Delta L(1+\alpha\Delta T)$ exactly the same as if the rod hadn't been chopped. This made sense. This is analogous to the solid sphere case.$^1$

  2. Now consider removing $[0,L]$. Also clamp the chopped piece at $L$. Now heat. If I ask you, "What is the new length of the rod ?", would you answer $L'(1+\alpha\Delta T)$? No, that would be silly since the only thing expanding is that chopped piece - which expands only $\Delta L (1+\alpha\Delta T)$.

It is this second case that put a knot in my stomach. You see, the 'hollowed' rod doesn't expand as much as the solid rod. As it shouldn't. When we heat a material, the net expansion is the sum of the expansion of all the material.$^2$ Remove material and you will remove expansion.

Yet the second case is analogous to the hollowed out sphere.$^3$ Without all that material inside how could the outer shell expand as much as the solid sphere? And yet the volume expansion is the same. How do we reconcile this?

I often find that when faced with a conceptual dilemma, reducing the dimensionality of the problem helps. Lets go back to the $1D$ case.

What is the change in length of the rod in the second case? We concluded earlier that it's $\Delta L (1+\alpha\Delta T)$. This was since most of the rod had been removed, its length was now considered changed. However for the sphere, even though most of the material is removed, its volume remains the same. And therein lies the discrepancy. If we define the length of the rod to always be from the origin, the thermal expansion would become same in both cases. While this weird '$1D$ volumetric expansion' makes no sense, it does in $3D$.

What all of this points to

is that the two scenario's undergo different physics

  1. In case of the solid sphere, the expansion occurs in the bulk of the solid. The net volumetric expansion is the sum of the expansion of all regions of the solid.

  2. In case of the hollow sphere, the expansion occurs in the surface of the solid. This surface, having expanded, must lead to a change in the volume of the solid according to its geometry. So even though there is hardly any material, volumetric expansion will occur.

Why these two volumetric changes should be same in general isn't obvious to me. It is certainly not true for larger temperature ranges, or anisotropic materials or even real world materials for which $\gamma$ and $1.5 \beta$ may not be exactly same $^4$. Also the fact that they must be so, independent of geometry, for small changes, though mathematically easy to show, isn't satisfying. In some ways, the mathematical equivalence illumines me more to the nature of geometry rather than the physics.

So why is the math telling us that the volume change caused directly and that induced via surface expansion must be same ? What physical constraint have we baked into out linear expansion model that causes this?

Well it is the condition of small change. That is what allows us to model thermal expansion linearly. Since change is measured relative to the original dimensions, this is saying that the geometry shouldn't distort too much during thermal expansion.

And therein lies the key observation - same initial geometries when slightly expanded, expand the same volumetrically regardless of what fills them$^5$. The spherical geometry could have been solid, a sphere, a perforated sphere like a spherical sieve, a collection of diameters, a sprinkling of random points or a sponge. As long as the solid and hollowed object had the same initial geometry, for small changes the volumetric change would be same.

$^1$ Note that there is internal stress and change in relative geometry when the rod expands. It is to relieve exactly this stress that the outer part moves outwards, till there is none.

$^2$ This is reflected in the mathematical modelling of the phenomenon too wherein we integrate to sum the expansions of all differential elements.

$^3$ Yes a sphere is three dimensional while a rod is one dimensional. But a sphere can be thought to be made up of thinner and thinner spokes jutting radially outwards from its center. Then the above discussion would apply exactly.

$^4$ $\frac{\Delta L}{L \Delta T}=\alpha$,$\frac{\Delta A}{A \Delta T}=\beta$,$\frac{\Delta V}{V \Delta T}=\gamma$

$^5$ of the same material


One geometric way to look at it is as follows:

  1. Imagine that the solid sphere is in fact a collection of concentric shells (like a spherical Russian doll).

  2. The hollow sphere on the other hand obviously is exactly the same thing, except that it has only one shell, namely the outermost one.

  3. Let’s ignore the contribution of the inner shells in a first approach. The dilation then is the same in both cases, namely it is only a function of temperature (and of a constant which is the same in both cases).

  4. The only question remaining is: do the inner shells not contribute more to the overall expansion in the case of a solid sphere? The answer is no. Why? Well simply because since all the inner shells expand linearly with the temperature, their relative geometrical positions respectively to each other remain the the same as initially at all times during the dilation process. In other words as the whole mass of the solid sphere expands the respective (ideal) shells it is made of do not « push outwards » on the next shell from the innermost shell to the outermost shells (again: because they expand in the same geometric proportions). In other words, the « picture before and after expansion is indistinguishable from a zoom effect on a Russian doll.

  • $\begingroup$ @user253751 : done. Thanks. $\endgroup$ – Serge Hulne Jun 12 at 17:49

There is only one equation, but fairly straightforward and I have included an explanation at the atomic level also.

A uniform mass distribution when heated to a certain temperature will expand to a certain volume according to $$\Delta V =V_0 \alpha \Delta T$$

where $\Delta V$ is the change in volume of the mass, $\alpha$ is the coefficient of thermal expansion and $\Delta T$ is the change in the temperature of the mass.

Note how this equation does not care about how the mass is distributed. Sphere or spherical shell, the amount of volume increase is the same for the same increase in temperature, meaning it does not matter if the mass is a sphere or a shell (hollow sphere).

What is happening at an atomic level:

An important point about volumetric thermal expansion is that, in general, all homogeneous substances expand (and contract when cooled) when their temperature increases, and the expansion (or contraction) occurs isotropically, or the same in all directions (substances that expand this way in all directions are themselves also called isotropic).

The atoms that make up the substance, are a certain distance away from each other, and this distance stays the same if the temperature (and other variables like pressure, volume etc) stays the same. The atoms are vibrating, though their average distance apart remains constant.

When heated, the kinetic energy of the atoms that make up the material will increase. Because of this, the vibrational energy of each atom increases, and so the average distance between each atom also increases. Hence the volume increases.

Again (and using the same reasoning as above), the distribution of the mass is irrelevant. A spherical shell of one material will increase in volume by the same amount as a sphere of the same material.

  • $\begingroup$ The anharmonicity of the pair potential well should be mentioned as a key component of the consensus understanding of thermal expansion in most materials. Without this asymmetry, the atoms would just vibrate more around the same equilibrium spacing, causing no net expansion. $\endgroup$ – Chemomechanics Jun 12 at 17:20

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