When a massive object warps the space around it, does the amount of space expand?

I may have watched those physics shows for too long, but the warping of space is usually described as a surface which bends. Whenever that happens, the surface they are showing increases in surface area. (like those weighty spheres on a stretched rubber surface)

I was wondering if this is practically the same in reality. Does space really ... expand when in the presence of a massive object?

Spacetime isn't a thing so it can't expand. The rubber sheet analogy is actually quite good but gives the wrong impression that spacetime is some stretchy material.

But I think we can easily show that, despite my reservations about the terminology, spacetime has expanded near a black hole. Suppose we consider two concentric spheres, one with a radius $r_1$ and another with a radius $r_2$, and we place a tape measure between the two spheres: The distance measured between the two spheres, labeled $d$ on my diagram, is just:

$$d = r_2 - r_1$$

But now suppose we place some large mass $M$ at the centre of the spheres. The presence of the mass warps the spacetime geometry so now the distance between the two spheres is greater than $r_2 - r_1$:

$$d \gt r_2 - r_1$$

Let me emphasis exactly what I mean:

If our scientist standing on the outer sphere lets down a tape measure until it touches the inner sphere the length of the tape measure will be greater than $r_2 - r_1$.

Actually calculating the length of the tape is a little involved. If you're interested I give the details in my answer to How much extra distance to an event horizon? However the conclusion is that due to the curvature of spacetime there is more distance between the spheres than there would be if the mass wasn't there.

For those who want to probe a bit more deeply into this: there is an ambiguity in the above i.e. what exactly do I mean by the radii of the spheres $r_2$ and $r_1$? Isn't the radius just what is measured by a tape measure?

Well there are two ways of measuring the radius of a circle:

1. the distance measured from the circle to its centre

2. the circumference of the circle $c$ divided by $2\pi$ i.e. using the fact that the circumference of a circle is $c = 2\pi r$

And when we draw a circle on a piece of flat paper these two methods for finding the radius $r$ give the same answer. In my discussion above when I talk about the radii $r_1$ and $r_2$ I mean the circumference measured around the sphere divided by $2\pi$ i.e. method (2) above.

The example I've given of the distance between the spheres is a result of the fact that in a curved spacetime the circumference of a circle is generally not equal to $2 \pi r$. In the black hole spacetime I described above the circumference is less than $2\pi r$, but there can be circumstances in which the circumference is greater than $2\pi r$, for example in a universe that is negatively curved.

• Isn't this the explanation for the Shapiro Effect? – JEB Sep 20 '17 at 17:44
• @JEB: sort of – John Rennie Sep 20 '17 at 19:09 If you're talking about something like this, no, the amount of space does not actually increase. They show it like this simply to appeal to our intuition - we think of "down" as "stuff falls into the hole" because we're used to gravity on earth.

It would be better to think of the large ball "sucking in" the grid around it at a certain rate. The rate at which it's sucking in space is just enough to keep the small moon in this picture, which is traveling in straight line with respect to the 'graph lines' of the grid, in orbit.

I see how you mean, but it's actually not space that is stretching, it's spacetime. There is a difference. And a simple way of representing this, without diving into the mathematics, is showing a picture or animation of a cloth being stretched by a massive ball... To illustration at least the gravitational effect of the curved spacetime.

Just a slight hint (have to put kids in bed...)... Whenever space, in spacetime, is stretched or compressed, and you try to measure it with your meter stick, it will show the correct length, since your meter stick is stretched or compressed the same way. And if you would measure a velocity things would end up right since your clock is ticking slower or faster (relative another frame).

Google and you'll find loads of examples of this which will deepen your understanding.. Which in turn will turn your question in to nonsense (sorry, couldn't find any other appropriate word)