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I found this elegant physics question in the Q/A section of the research gate. My own judgment is that this is true except, and only except, for complete spherical shapes. . In other words, it applies well to both hemispheres and to all shapes other than solid spheres. I tried to test the accuracy of this question to no avail. The exception of the sphere versus the cube puzzles me. The reason may be the physical component of the question: the cube and the sphere both have a degree of freedom! I don't have a strict explanation. Why?

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    This seems to be a math question rather than a physics question. – David Hammen Nov 03 '22 at 15:12
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    No, this is quite false. Take two arbitrary shapes which are not scaled versions of each other, say a cube and a ball. There's a unique way to scale one of them to have the same volume-to-area ratio as the other, and there's no reason this should imply they have the same volume and area. – Qiaochu Yuan Nov 03 '22 at 16:56
  • Probably you mean "of the same shape". Then it is true but trivial. – user Nov 03 '22 at 16:58
  • Qiaochu Yuan Thank you for your helpful response. The question is not entirely mathematical, it has a physical component. Also, the question itself states that "except for full spherical shapes, although it applies to hemispheres". where the difficulty arises, I have no rigorous explanation for this behaviour. Thanks again. –  Nov 03 '22 at 17:08
  • user Of course not !!. I mean literally what is written. Thanks. –  Nov 03 '22 at 17:12
  • A more specific pointer might help people see what you're asking about (like a URL and maybe some other information if one needs to navigate within a document). As written here, there's no apparent physics component (except for the word "physics"). – David K Nov 03 '22 at 17:27
  • If you increase the radius of a sphere by a factor $k$, the area increases by $k^2$ and the volume increases by $k^3$, so the volume/area ratio increase by $k^3/k^2 = k$. Why do you say "except for complete spherical shapes"? – David K Nov 03 '22 at 17:30
  • David K I tried to test the accuracy of this question to no avail. The exception of the sphere versus the cube puzzles me. The reason may be the physical component of the question: the cube and the sphere both have a degree of freedom! –  Nov 03 '22 at 17:38
  • The stated question--at least, the part in the title--has been unequivocally and definitively answered. The sphere vs. cube is not an exception; it is merely an example of the general rule. Any two shapes can be scaled to have the same volume to area ratio. Perhaps if you were to actually show some step-by-step calculations in the question, it would enable people to find the source of misunderstanding. The ability to actually calculate results is essential not only to mathematics but also to physics and engineering; if you can't calculate, you're not doing physics. – David K Nov 04 '22 at 02:46
  • i-The spherical shape is the exception that affirms the general rule. ii-Me (+80 published research and +120 citations) not doing physics, hope you know what you're talking about. –  Nov 04 '22 at 08:16

2 Answers2

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This is obviously false as @QiaochuYuan says. Take any two shapes from this Wikipedia page under the heading Mathematical examples. They have the SA/V ratio parametrized by the shape's characteristic length.

For instance:

SA/V for a tetrahedron is $\rho_{\text{tetrahedron}} = {6\sqrt {6} \over {a}}$ where $a$ is the edge length.

SA/V for a octahedron is $\rho_{\text{octahedron}} = {3\sqrt {6} \over {b}}$ where $b$ is the edge length.

These may be scaled so that ${\rho_{\text{tetrahedron}} \over \rho_{\text{octahedron}}} = 1$.

The shapes have different volumes, surface areas, but their SA/V ratios are equal.

vvg
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  • I thank you for your helpful answer. This question of extreme importance belongs to physics as well as to mathematics as it seems at first sight.from WIKIPEDIA- Surface-to-volume ratio, also called surface-to-volume ratio and variously denoted sa/vol or SA:V, is the amount of surface area per unit volume of an object or set of objects.

    SA:V is an important concept in science and engineering. It is used to explain the relationship between structure and function in processes occurring through surface and volume. Good examples of such processes are the processes governed by the heat equation

     –  Nov 03 '22 at 19:57
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If I got you question right, I think that your assertion is false. Take an object 1 with Volume/Area ratio equal to $\frac{V_1}{A_1} = R_1$ and a second object with $\frac{V_1}{A_1} = R_2$. Two remarks before going on:

  • in general $R_1 \ne R_2$;

  • the Volume/Area ratio of a family of scaled objects is proportional to a characteristic dimension $\ell_1$ of the object of the family,

    $\frac{V_i}{A_i} = R_i(\ell_i) = K_i \ell_i$,

    where the constant $K_i$ depends on the shape of the objects.

- as the dimension of the object goes to zero $\ell \rightarrow 0$, the Volume/Area ration goes to zero as well,

$\lim_{\ell_i \rightarrow 0} \frac{V_i}{A_i}(\ell_i) = \lim_{\ell_i \rightarrow 0} R(\ell_i) = K_i \lim_{\ell_i \rightarrow 0} \ell_i = 0$.

Comparing the volume to area ratio of the two families of objects to find the ratio of the characteristic length we get the following equation

$\dfrac{V_1}{A_1} = \dfrac{V_2}{A_2}$$ \qquad \rightarrow \qquad$$K_1 \ell_1 = K_2 \ell_2$$ \qquad \rightarrow \qquad$$\dfrac{\ell_2}{\ell_1} = \dfrac{K_1}{K_2}$.

basics
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