Why a static universe cannot work (8.286)

Before Hubble, everyone, Newton included, assumed the universe was static: the stars as a whole were not going anywhere. But Newton's own law of gravity pulls everything toward everything else. How could a universe full of matter stand still? This page follows Guth's account of Newton's answer and why it fails. It is the starting point for the dynamics of the expanding universe.

Newton's argument#

Newton saw the problem clearly. In a letter to the theologian Richard Bentley dated 10 December 1692, he argued that a finite collection of matter would fall together into one great spherical mass. Matter spread evenly through infinite space, he thought, would instead gather into many separate masses scattered through space, which could become the sun and the stars. (He added that he could not explain by natural causes why some matter became shining suns and some opaque planets.)

The modern way to put Newton's intuition, as Guth does, is a symmetry argument: in an infinite, evenly filled universe every particle is pulled equally in every direction, so nothing has a reason to move. Bentley had raised the idea that the infinite pulls on each side of a particle might exactly balance. In his second letter, 17 January 1693, Newton replied that infinities are neither equal nor unequal, so such a balance cannot be taken for granted. The problem of gravity in an infinite Newtonian universe was taken up seriously again only in the 1890s, by Hugo von Seeliger and Carl Neumann, and later by Einstein.

The verdict of Gauss's law#

The cleanest way to settle the question is not to add up infinitely many pulls, but to use a local law. For gravity, the counterpart of Gauss's law in electrostatics says that the flux of the gravitational field $\vec g$ out of any closed surface is fixed by the mass inside:

Gauss's law of gravity

\oint \vec g \cdot d\vec a = -4\pi G\,M_\text{enclosed}

In words

the loop integral of vector g dot d vector a equals minus 4 pi G M sub enclosed

Source: 8.286 Lecture 6 slides (review of Gauss's law, Poisson's equation, 'But what about symmetry?'), slide 1

Now suppose the universe is static and uniform. Then nothing accelerates, so $\vec g = 0$ everywhere, and the left side is zero for every surface. But a uniform density $\rho > 0$ means every surface encloses some mass, so the right side is not zero. Contradiction: a static, uniformly filled universe is impossible.

The same result follows from the differential form, Poisson's equation:

Poisson's equation

\nabla^2\phi = 4\pi G\rho, \qquad \vec g = -\nabla\phi

In words

del squared phi equals 4 pi G rho , vector g equals minus del phi

Source: 8.286 Lecture 6 slides (review of Gauss's law, Poisson's equation, 'But what about symmetry?'), slide 2

If $\vec g = 0$ everywhere, the potential $\phi$ is constant, so $\nabla^2\phi = 0$ , which forces $\rho = 0$ .

Why adding up the pulls goes wrong#

If Gauss's law is so clear, why did the symmetry argument seem convincing? Because the infinite sum of pulls is ambiguous. A simple integral shows the problem. Take $f(x) = +1$ for $x > 0$ and $-1$ for $x < 0$ , and integrate over a window of half-width $L$ :

An order-dependent integral

\lim_{L\to\infty}\int_{-L}^{L} f(x)\,dx = 0, \qquad \lim_{L\to\infty}\int_{a-L}^{a+L} f(x)\,dx = 2a

In words

the limit as L goes to infinity of the integral from minus L to L of f(x) dx equals 0, the limit as L goes to infinity of the integral from a-L to a plus L of f(x) dx equals 2a

Source: 8.286 Lecture 5 transcript, worked on the board

Shift the centre of the window and the "limit" changes: you can get any answer you want. An integral like this, which converges only because positive and negative parts cancel, is called conditionally convergent, and it is dangerous: rearranging the terms changes the result.

The gravitational sum in an infinite universe is exactly like that. Add up the matter in spherical shells centred on a point P: each shell pulls P equally in all directions, so $\vec g = 0$ , as Newton argued. Now add up the same matter in shells centred on a different point Q, a distance $b$ away. The ball of radius $b$ around Q pulls P toward Q, while every shell outside it surrounds P and exerts no net force:

Shells centred on Q

\vec g = \frac{G M}{b^2}\,\hat e_{QP}, \qquad M = \frac{4\pi}{3}\rho\,b^3 \quad\Longrightarrow\quad |\vec g| = \frac{4\pi}{3}G\rho\,b

In words

vector g equals G M over b squared e hat sub QP , M equals 4 pi over 3 rho b cubed which gives | vector g | equals 4 pi over 3 G rho b

Source: 8.286 Lecture 5 slides (Newton–Bentley letters; shells around P and Q), slide 8

Here $\hat e_{QP}$ points from P toward Q. Choose Q anywhere, and the "answer" can have any direction and any size. The matter is the same in both cases; only the bookkeeping changed. Newton's symmetry argument picked one bookkeeping and declared it correct.

Worked example

How strong is the ambiguity?

Take a uniform density $\rho = 2.7\times10^{-27}\ \text{kg/m}^3$ (roughly the average density of matter in today's universe). If we add up shells around a point Q that is $b = 100$ Mpc $= 3.1\times10^{24}$ m from P, what acceleration does the "Q bookkeeping" give at P?

Use the formula

$|\vec g| = \tfrac{4\pi}{3}G\rho\,b = 4.19 \times 6.67\times10^{-11} \times 2.7\times10^{-27} \times 3.1\times10^{24}$ .

Evaluate

$|\vec g| \approx 2.3\times10^{-12}\ \text{m/s}^2$ . Tiny — but it doubles if $b$ doubles, and points wherever Q is. The sum has no single answer.

What decides physically

Not the order of adding but the physics: Gauss's law says $\vec g$ cannot be zero everywhere. The universe must be accelerating toward collapse or decelerating in expansion.

But what about symmetry?#

Newton's intuition was not entirely wrong. His mistake was to assume there is one absolute standard of "not accelerating" — for him, the fixed stars. In a collapsing (or expanding) universe there are no fixed stars. Accelerations, like velocities, are then only relative.

Any observer may regard themselves as unaccelerated and will see every other galaxy accelerating toward them, with acceleration proportional to distance. That looks as if it has a centre — but every observer sees the same pattern, exactly like Hubble's law for velocities. The universe can be homogeneous and gravitating; it just cannot be static. The next page turns this into an equation for $a(t)$ .

Summary#

Newton argued an infinite uniform universe could be static by symmetry.
Gauss's law (or Poisson's equation) shows it cannot: $\vec g = 0$ everywhere would require zero density.
The direct sum of gravitational pulls is conditionally convergent; different groupings (shells around P or Q) give different answers.
In an evolving universe accelerations are relative, so a homogeneous universe can collapse or expand without a centre.

Symbols on this page

Show the 17 symbols used here

$\pi$ pi: The ratio of a circle's circumference to its diameter, π ≈ 3.14159. Say it: “pi”.
$\rho$ rho: Density: mass (or energy divided by c²) per unit volume. ρ_c is the critical density, ρ̄ the mean density. Say it: “rho”.
$\phi$ phi: The gravitational potential: g = −∇φ. Say it: “phi” (sounds like “fye”).
$\infty$ infinity: Infinity: larger than any number. Say it: “infinity”.
$\approx$ approximately equal: Is approximately equal to. Say it: “is approximately”.
$\to$ arrow: Goes to (tends to), or becomes. Say it: “goes to”.
$\Rightarrow$ implies: Implies: what follows is a consequence of what came before. Say it: “which gives”.
$\times$ times: Times (multiplication). Between two vectors: the cross product. Say it: “times”.
$\cdot$ dot: Times (multiplication). Between two vectors: the dot product. Say it: “times”.
$\nabla$ nabla (del): Del: ∇f is the gradient, ∇·v the divergence, ∇×v the curl, ∇² the Laplacian. Say it: “del”.
$\int$ integral: Integral: the sum of infinitely many small pieces, such as the area under a curve. Say it: “the integral of”.
$\oint$ loop integral: An integral around a closed loop. Say it: “the loop integral of”.
$\lim$ limit: Limit: the value approached as the variable written below approaches its target. Say it: “the limit”.
$\vec v$ vector arrow: A vector: a quantity with a size and a direction. Say it: “vector”.
$\hat\imath$ hat: A unit vector: a vector of length 1 that only gives a direction (î, r̂). Say it: “hat”.
$a$ a: The scale factor a(t); in the integral limits a is just a number, and in d a⃗ (with an arrow) a small piece of surface area. Say it: “a”.
$G$ G: Newton's gravitational constant, G ≈ 6.674 × 10⁻¹¹ m³ kg⁻¹ s⁻². Say it: “G”.

Tap or click a symbol in a formula to see the same explanation. All symbols: symbols and notation.

Check your understanding

Why did Newton think an infinite uniform universe could stay at rest?

Gauss's law of gravity says the flux of

\vec g

through a closed surface equals

-4\pi G M_\text{enclosed}

. What does that imply for a static, uniformly filled universe?

Adding up the pull of shells centred on P gives

\vec g = 0

; shells centred on another point Q give a force toward Q. What is the lesson?

In a collapsing uniform universe, which galaxy is 'not accelerating'?

0 of 4 answered

Why a static universe cannot work

You need this first

Newton's argument#

The verdict of Gauss's law#

Why adding up the pulls goes wrong#

How strong is the ambiguity?

Use the formula

Evaluate

What decides physically

But what about symmetry?#

Summary#

Symbols on this page

Where this leads

Pages that build on this

Check your understanding

Flashcards