With how many Newtons of force is the universe expanding?

Question

The universe is expanding at an accelerating rate. They key word here is "accelerating". Meaning that there is an equivalent "force" (in Newtons) that would cause that same acceleration. My question is, around how many Newtons is that force? If mass plays a non-trivial factor, then I'll be just as happy if you can give me the acceleration induced by the expanding universe (e.g. like how $g=9.8$).

To give a more concrete scenario for my question: two neutrons are in free space. At what rate will they accelerate away from each other (due to the expansion of the universe)? Does it depend on the mass of the neutrons, or the distance between them? Feel free to ignore the gravitational force attracting the two neutrons together, or any other forces (like electromagnetic forces, etc.)

Edit: if saying they're "accelerating" away from each other is problematic wording, then feel free to instead answer "what is the second-derivative (w.r.t. time) of the distance between them?"

Answer 1

To give a more concrete scenario for my question: two neutrons are in free space. At what rate will they accelerate away from each other (due to the expansion of the universe)? Does it depend on the mass of the neutrons, or the distance between them? Feel free to ignore the gravitational force attracting the two neutrons together, or any other forces (like electromagnetic forces, etc.)

Using neutrinos is not a good example for this case, so I'll consider two galaxies. The reason is that the universe's expansion does not affect locally bounded systems or sub-atomic particles. It works on much larger scales since we are also considering/using GR.

The dynamics of the LCDM model can be crudely expressed as a modified Newtonian Gravitation Law, given by

$$F = - \frac{GMm}{r^2}+\frac{\Lambda mr}{3}$$

which represents the force $F$ on a test particle of mass $m$ due to a body of mass $M$. Here $\Lambda$ is the cosmological constant.

In this case, the acceleration of the test particle can be written as

$$a = - \frac{GM}{r^2}+\frac{\Lambda r}{3}$$

To give an example, consider these values.

$r = 1\rm{Mpc} \sim 10^{22}\rm{m}$,

$\Lambda \sim 10^{-36}\rm{s^{-2}}$

$M = 10^{12}M_{\odot} \sim 10^{42}\rm{kg}$

$G \sim 10^{-11}\rm{m^{3}kg^{-1}s^{-2}}$

Thus,

$$a \sim -\frac{10^{-11}\rm{m^{3}kg^{-1}s^{-2}} \times 10^{42}\rm{kg}}{10^{44}\rm{m^2}} + \frac{10^{-36}\rm{s^{-2}} \times 10^{22}\rm{m}}{3}$$

or

$$a \sim -10^{-13}\rm{m s^{-2}} + 10^{-14}\rm{m s^{-2}} \sim -10^{-13}\rm{m s^{-2}}$$

You'll see that for these values, after $r \approx 3.5\rm{Mpc}$, $a$ will be positive, which means that the Newtonian acceleration will be repulsive and not attractive.

Note w.r.t @chausies question: As we can see, if we ignore the attractive force of gravity we obtain, $a = \frac{\Lambda r}{3}$. Since $a = \ddot{r}$, the equation becomes

$$\ddot{r} = \frac{\Lambda r}{3}$$

which has a solution in the form of

$$r(t) \propto e^{\sqrt{\Lambda/3}t}$$

and has a similar form compared to the de-Sitter universe where $$a(t) \propto e^{\sqrt{\Lambda}t}$$

Note: You can also look at this question,

Modification of Gravitational force's law and related problem

I have derived the Newtonian Friedmann Equation by using this modified Newtonian Gravitational Force. But this force expression is very crude. It's clear that we cannot apply Newtonian Dynamics to the universe so take it with a grain of salt.

Answer 2

Zero for both proper acceleration and force. There is no proper acceleration associated with expansion. See: wikipedia for the difference between proper acceleration, which corresponds to forces and is absolute; and coordinate acceleration, which does not correspond to forces and depends on choice of coordinate system.

Hubble Factor and Coordinate Acceleration

Distant objects become more distant - without acceleration with respect to one another - at a rate of approximately $70 km/s/Mpc$. One megaparsec ($Mpc$) is approximately 3.3 million light years. This gives us a differential equation in proper time for comoving distance:

$d'(t) = H d(t)$

where the Hubble Factor $H \approx 70 km/s/Mpc$ and $t$ is the time since some reference $t_0$. Solving the ODE while approximating $H$ as a constant$^1$,

$d(t) \approx d_0e^{Ht}$

$d''(t) \approx d_0H^2e^{Ht}$

Although the second derivative of distance is positive and nonzero, this is neither a proper acceleration nor what is referred to as the universe expanding at an accelerating rate.

Proper acceleration is what an accelerometer measures, and both objects, even though the distance between them is increasing at an increasing rate, measure zero acceleration. The space between them is getting bigger, without them having to move.

If you assign a cartesian coordinate system such that one comoving body is fixed at the origin and the other comoving body is free to "move", ignoring gravity, $d''$ is the "moving" body's coordinate acceleration with respect to the first body.

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1: A safe approximation over cosmically short time scales, since the measurement uncertainty in H is large compared to the rate of change of H. However, each time we take the derivative, the potential error that we're introducing by assuming that $H'=0$ gets bigger.

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Scale Factor and Accelerating Expansion

When people say that the expansion rate of the universe is accelerating, they mean that the scale factor $a(t)$ of the universe has a positive second time derivative. (No relation to $\vec a$ for acceleration.) The scale factor is defined such that

$H := a' / a$

$d(t) = a(t) d_0$

$a(0) = 1$

The middle expression gives the distance between two comoving objects, given their starting distance, the time, and a function for the time-varying scale factor of the universe.

Note that $a''$ is not identical to $H'$ and need not even have the same sign.

$H' = \dfrac{a a'' - a'^2}{a^2} = -H^2 (1-\frac{aa''}{a'^2})$

$a$ is always positive, because negative distances are meaningless. $a'$ is positive for an expanding universe. $a''$ is positive for a universe with so-called accelerating expansion. So, for a universe with positive and increasing expansion, $\frac{-a a''}{a'^2}$ is negative. If it's less than $-1$, $H'>0$, and if it's more than $-1$, $H' < 0$. Experimental data indicate $\frac{-a a''}{a'^2}$ has a present-day value of about $-0.55$, corresponding to a negative $H'$. The scale factor is increasing at an increasing rate while the Hubble factor is decreasing.

Answer 3

Space is expanding, and there is no force (which is also why the universe can expand faster than the speed of light).

To give a more concrete scenario for my question: two neutrons are in free space. At what rate will they accelerate away from each other (due to the expansion of the universe)? Does it depend on the mass of the neutrons, or the distance between them? Feel free to ignore the gravitational force attracting the two neutrons together, or any other forces (like electromagnetic forces, etc.)

The rate depends on how far the neutrons are from each other and is broadly given by the Hubble parameter (and some other stuff involving the light-travel time for very faraway objects). There is no force, and the mass is irrelevant. In other words, if you took the Milky Way and some faraway galaxy - far enough for peculiar velocity to be irrelevant - and replaced them with two neutrons, they'd still be receding from each other at the same speed, even though galaxies are much more massive than a neutron.

Answer 4

Expansion force depends on distance between galaxies or objects of interest. For the rough estimate we can use centrifugal force analogy, because Hubble constant has units of $\text{Hz}$ in SI unit system, namely $\to~H_0 = 2.27 \times 10^{−18} ~\text{Hz}$.

Centrifugal force in magnitude is same as centripetal force, so,: $$ F_{exp} = M_{gal}2D\left(\pi H_0\right)^{2}\,, $$ where $M_{gal}$ is mass of typical galaxy, $D$ is observable universe diameter. i.e. maximum possible distance which can separate galaxies, and $H_0$ is present day Hubble constant. Substituting all parameters, gives: $$ F_{exp} = 10^{42} kg × 2 × (8.8 × 10^{26}~m) × (\pi×2.27×10^{−18}~Hz)^2 \approx 10^{34}~N $$

So most distant parts of universe are pushed apart with about $10^{34}~N$ force.

For the reference, Planck force is: $$ F_{_P}={\frac {m_{_{P}}c}{t_{_{P}}}}={\frac {c^{4}}{G}}=1.210295\times 10^{44}{\text{ N}} $$, which is about 10 orders of magnitude higher than this edge-case expansion force $F_{exp}$.