2

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
  • 201,751
Walter Nap
  • 123
  • 1
  • 6

2 Answers2

2

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
  • 474
  • 3
  • 11
  • To be frank, this complicates things... I would expect it to not be this difficult since Thorne goes into great detail in this book about a lot he claims. With this, he does not say anything so it must be a small detail I missed. – Walter Nap Apr 07 '18 at 17:00
  • 1
    Maybe the wikipedia article is not the best, but every book in general relativity could be good. The problem is that the coordinates you’re using when you search a spherically symmetric static solution to Einstein equation (that is schwarschild one) couldn’t retain the feature that the circumference is long 2pi*r if the space isn’t flat, because this is a property of euclidean space, but a curved (static) 3-space is not euclidean at all. – Annibale Apr 07 '18 at 17:07
2

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
  • 130,455