Does gravity slow the expansion of the universe?

I read through the thread http://www.physicsforums.com/showthread.php?t=322633 and I have the same question. I know that the universe is not being stopped by gravity, but is the force of gravity slowing it down in any way? Without the force of gravity, would space expand faster?

Help me formulate this question better if you know what I am asking.

  • $\begingroup$ In GR, gravity is not a force, it is a curvature of spacetime; it is geodesic deviation. So, to formulate your question better, you should start with sharpening your notion of what "without gravity" means. $\endgroup$ – Alfred Centauri Jul 17 '12 at 1:10
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    $\begingroup$ @AlfredCentauri: Does that really matter? "If spacetime isn't curved , would the universe expand faster?" is essentially the same question. $\endgroup$ – MSalters Jul 17 '12 at 11:54
  • $\begingroup$ I'd like to add my personal modification: Is the term "gravity" clearly defined here, i.e. is there a measure of the amount of gravity in spacetime (maybe the action is a valid one)? And how do the expansion equations (Friedmann?) depend on this real parameter. I formulate it that way because it seems invalid to me to ask about the influence on gravity on the expansions of the universe like that (the terminology "slowing down" seems dubious to me), if gravity is what brings the expansion about. If merely energy has negative influence on metric expansion, not gravity is slowing things down. $\endgroup$ – Nikolaj-K Jul 17 '12 at 13:04
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    $\begingroup$ @MSalters, what follows from if spacetime isn't curved? $\endgroup$ – Alfred Centauri Jul 17 '12 at 14:46
  • $\begingroup$ @AlfredCentauri: That's the question here! I don't have a background in astrophysics, so I can't give a good answer. $\endgroup$ – MSalters Jul 19 '12 at 8:46

The answer is that yes gravity does slow the expansion of space (leaving aside dark energy for the moment), but to get a better grasp on what's going on you need to look into this a bit more deeply.

If we make a few simplifying assumptions about the universe, e.g. it's roughly uniform everywhere, we can solve the Einstein equation to give the FLRW metric. This is an equation that tells us how spacetime is expanding, and actually it seems to be a pretty good fit to what we see so we can be reasonably confident it's at least a good approximation to the way the universe behaves.

To reduce gravity you simply reduce the density of matter in the universe because after all it's the matter generating the gravity. At low densities of matter the FLRW metric tells us that the universe expands forever. As you increase the matter density the expansion slows, and for densities above a critical density (known as $\Omega$) the expansion comes to a halt and the universe collapses back again.

So yes, gravity does slow the expansion and the FLRW metric tells us by how much. If you want to pursue this further try Googling for the FLRW metric. The Wikipedia article is very thorough but a bit technical for non GR geeks, but Googling should find you more accessible descriptions.


This answer is intended to address Nick Kidman's reformulation of the question:

is there a measure of the amount of gravity in spacetime (maybe the action is a valid one)? And how do the expansion equations (Friedmann?) depend on this real parameter.

The way that cosmologists answer this is in terms of the energy density of the universe. This energy can come from radiation, matter, a cosmological constant, or any other form of dark energy if it exists.

The rest of the answer is very similar to the discussion found in textbooks such as Ryden. For simplicity, we'll consider the imaginary case where the energy density of the universe is dominated entirely by matter - that is, we'll ignore radiation energy and dark energy. This will allow us to discuss how expansion of the universe depends on a single parameter, the energy density of the matter (I'll just call it 'matter density' from now on). Including the other energies will complicate the picture but not change the fundamental nature of the answer.

The Friedmann equations are second order in time. We'll choose our two integration constants based on the size and rate of expansion we observe now in today's universe (even though today's expansion is dominated by dark energy, this is just a choice of numbers to set a convenient point of reference). Then, we can vary the matter density and solve the Friedmann equations to see how the early and late phases of the universe's expansion would change.

Here is a graph showing three possible scenarios:

Expansion history for varying amounts of matter density

Let's focus on the middle one first. Here, the expansion rate $\dot{a}$ approaches zero asymptotically for $t \rightarrow \infty$. The magnitude of the density in today's universe corresponding to this type of expansion is called the critical density, and we can use it to define a dimensionless measure of density called the density parameter $\Omega$. The middle curve corresponds to $\Omega = 1$.

The lower curve in the plot corresponds to $\Omega > 1$. Here the expansion eventually reverses iteself into a big crunch.

The upper curve corresponds to $\Omega < 1$. In this case the expansion continues to accelerate at late times, leading to a 'big freeze' or 'big rip'.

Closed form analytical solutions to the Friedmann equations in a matter-only universe with arbitrary $\Omega$, such as those used to generate the graph, can be found in many cosmology textbooks including the one I linked to above.

There are other important things that change with $\Omega$, such as the topology and curvature of the universe.

Now for some fine print: In our universe, we actually measure $\Omega$ to be close to 1, meaning that the topology and curvature of the universe appear to match what we expect for $\Omega = 1$. But we also think that the universe will continue to expand in an accelerated matter. This is because of the presence of dark energy, which modifies the the solutions to the Friedmann equations.

  • $\begingroup$ Also, since matter/energy isn't created or destroyed locally, the matter density will change with time, and so will the critical density that is used to normalize $\Omega$. However, you can show that if $\Omega$ starts out equal to 1, it will stay that way. It will also stay greater than 1 or less than 1 if it starts that way. $\endgroup$ – kleingordon Jul 20 '12 at 23:17

The Friedmann equations for the expansion of space are (assuming flat space for simplicity):

$(1)\ (\frac{\dot a}{a})^2 = \frac{8 \pi G \rho + \Lambda}{3}$

$(2)\ \frac{\ddot a}{a}= -\frac{4 \pi G}{3}(\rho + 3P) + \frac{\Lambda}{3}$

where $a$ is the scale factor (roughly, how "expanded" space is), $\dot a$ is the rate of expansion and $\ddot a$ is the acceleration of the expansion.

If, "without the force of gravity", you mean "with $G = 0$", then we have:

$(3)\ (\frac{\dot a}{a})^2 = \frac{\Lambda}{3} \rightarrow a(t) = a(0)e^{\pm t \sqrt{\frac{\Lambda}{3}}}$

$(4)\ \frac{\ddot a}{a}= \frac{\Lambda}{3}$

So, "without gravity" in this particular sense, with $G = 0$, space is either expanding or contracting exponentially with time (for the special case of $\Lambda = 0$ , $\dot a = \ddot a = 0$)

Now, in the context of your question about an expanding universe, by inspection of equation (2), see that introducing "gravity" via giving $G$ a positive value (and, of course, assuming there is a non-zero mass density), this term "opposes" the cosmological constant term and can even reverse the acceleration of the expansion of space by making $\ddot a$ negative thus slowing the expansion.

  • $\begingroup$ Could you elaborate more on the conclusion regarding "$G=0$" from the solution of the second order differential equation? $\endgroup$ – Nikolaj-K Jul 17 '12 at 18:46
  • $\begingroup$ Sigh, is mentioning the decaying solution likely to help the OP's understanding? $\endgroup$ – Alfred Centauri Jul 17 '12 at 19:08
  • $\begingroup$ It's an honest question. You clearly start out with some finite $a$ and then, if you say "space expands exponentially" you will have to justify that you don't drop the constant of integration, which gives growth. Otherwise it's no explaination as the solution $a(t)=a(0)e^{-\sqrt{\Lambda/3}t}$ is a solution too and not a expanding one. $\endgroup$ – Nikolaj-K Jul 17 '12 at 19:11
  • $\begingroup$ @NickKidman, edited to address your concerns. $\endgroup$ – Alfred Centauri Jul 17 '12 at 21:04
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    $\begingroup$ I specifically wrote "in the context of your question about an expanding universe" to indicate we're considering the positive root solution. $\rho$ depends on $a$. What are you intentions? $\endgroup$ – Alfred Centauri Jul 17 '12 at 21:27

For most of the 20th century, it was thought that expansion slowed gravity. Many weighed in on the value of omega, it being thought that the universe was so close to critical balance it could not be determined whether it would expand forever or collapse. That mystery should have evaporated with the discovery that the universe is accelerating - yet science TV shows are still promulgating the mystery as a fine tuned feature of the universe. Fine tuning of conflicting factors that lead to perfect balance indicates something is wrong unless one things of creation in the same vein as a potter working on clay. Some years ago the idea evolved that gravity and expansion are not opposites in conflict, but rather one depends from the other - that they appear to be independent perfectly balanced factors, is due to the fact that neither Newtonian gravity nor GR addresses the cause of the acceleration factor G that needs to be inserted into both formalism to predict the motion of one body in the vicinity of another. Embellishing upon an idea first put forth by Richard Feynman suggesting that gravity might be explained as a pseudo force, one can make a quick calculation using Friedmann's equation to express G as 3H^2/4(pi)p where p is the density of the Hubble sphere and H is the Hubble constant. In an accelerating (q=-1) universe, the value of G then reduces to (c^2)/4(pi)R[S] where S represents the Hubble sphere as an area density of one kg/m^2 and R is the effective Hubble scale.


Short answer: yes, gravity slows the expansion of the universe, in the sense that we'd see even greater expansion if gravity* were (slightly) weaker, and everything else was kept the same.

*more precisely the gravitational constant


What is this gravity thing?

In a metaphysical context, gravity just is, it is fundamental, a primary. Gravity can't be explained in terms of other "more fundamental" things because that would be a contradiction.

In a physics context, we have a mathematical model for gravity, the General Theory of Relativity which, in a nutshell, is:

$\textbf{R} + (\Lambda - \frac{1}{2}R)\textbf{g} = \dfrac{8 \pi G}{c^4}\textbf{T}$

Now, the terms in these equations are geometric objects, tensors, (the Ricci tensor, the metric tensor, and the stress-energy tensor) and this equation relates these tensors. Moreover, the LHS of the equation involves the geometry of spacetime. The RHS involves the mass-energy content of spacetime.

So, I think it is the case that, in this context, gravity is the relation between these tensors, between the geometry of spacetime and the contents of spacetime.

Note that I'm not claiming that gravity is this equation; I'm claiming that gravity is a relation expressed by this equation, a fundamental relationship between spacetime geometry and mass-energy.

So, to tie this in with the OP's interesting question starting with "Without the force of gravity...", let's consider what that actually means.

Stipulating that gravity is a relationship between the geometric objects above, then "without gravity" must mean "without a relationship between these geometric objects".

What would such a world be like?

(more to come)

  • $\begingroup$ Is that an answer of a question? What are the (at least two) geometric objects? Or what is not a geometric object then? $\endgroup$ – Nikolaj-K Jul 19 '12 at 18:57
  • $\begingroup$ A tensor is a geometric object. Now, it also the case that the metric tensor, a geometric object, is also a description of the geometry of spacetime. So the relation that I'm calling gravity relates the a description of the stress-energy in spacetime to the geometry of spacetime. I'll edit my answer for clarity. $\endgroup$ – Alfred Centauri Jul 19 '12 at 20:52
  • $\begingroup$ Mhm, okay. Notice that I'm foremostly interested in the main question. A definition of the expression gravity might be helpful, but only to communicate the answer about the expanding. $\endgroup$ – Nikolaj-K Jul 19 '12 at 21:03
  • $\begingroup$ I'm primarily interested in the quoted question. I find it interesting and think that others might too. I hope to get some constructive comments on my answer and I may pose it as a question myself. ---- I don't mean this in other way than as a simple statement of fact: the question of what you're foremostly interested in never crossed my mind and it isn't likely too. $\endgroup$ – Alfred Centauri Jul 19 '12 at 21:12

I think that Gravity accelerates the expansion of space: Consider two objects falling one after the other, directly into a gravitational well. The closer one feels a stronger pull, so it accelerates away from the object behind. Then consider two objects falling side by side, directly into a gravitational well. Deeper in the well, the path that light takes between them is more greatly bent, and therefore longer. So, those distances are also greater. Any two galaxies in our cosmos are falling into the gravitational well of the cosmos as a whole. So, the distance between them must be increasing due to gravity.

A lot of people get confused between the deflection of an object toward a gravitational well, and the effect of the gravitational well on the distances between objects falling into it.


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