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While reading Thorne's 'The Science of Interstellar', I came across this piece of information:

'Now, the Sun’s equatorial plane divides space into two identical halves, that above the plane and that below. Nonetheless, Figure 4.4 shows the equatorial plane as warped like the surface of a bowl. It bends downward inside and near the Sun, so that diameters of circles around the Sun, when multiplied by π (3.14159 . . . ), are larger than circumferences—larger, in the case of the Sun, by roughly 100 kilometers.'

Has anyone that has read this book got an explanation for me, I do not get the d*pi compared to circumference part, why does it differ?

Qmechanic
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Walter Nap
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2 Answers2

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I suggest you to read the wikipedia article about.

However, note well: in general, the Schwarzschild radial coordinate does not accurately represent radial distances, i.e. distances taken along the spacelike geodesic congruence which arise as the integral curves of $\partial_r$.

Annibale
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The circumference of a circle is only $\pi$ times its diameter in flat space.

The space around and inside the Sun is curved by its mass-energy. As a result, the proper distance between two radial points is larger than the difference in their radial coordinates. However, the circumference (a path of constant radial coordinate) is still $2\pi$ times that radial coordinate.

The way to imagine this is a circle drawn on the curved surface of a sphere. The radius you would (hypothetically) measure is the length of the path along the curved surface to the centre. However, the circumference is $\pi$ times the diameter of the circle projected onto a flat surface, so is less than $2\pi$ times the radius you previously measured.

Curved space

See also How much extra distance to an event horizon?, though the maths for calculating the proper radius for an object with a density profile (like the Sun) is more complicated than for a black hole.

ProfRob
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