I may know the answer to my question. Mass must curve the universe but the pressure due to energy density may curve the universe in the opposite direction, leading to a flat universe, if they exactly cancel. But this leads to a supplementary question.

It seems to me that in a universe with positive curvature, gravity is a force encouraging the contraction of the universe while the pressure due to the density of energy is a force encouraging the expansion of the universe. But in a universe with negative curvature, the opposite should apply. Gravity should encourage the expansion of the universe while pressure should encourage its contraction. And in a flat universe neither gravity nor pressure should have any effect on the expansion of the universe. So if our universe is flat, dark energy can not be encouraging accelerating expansion. Any thoughts?

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    $\begingroup$ Gravity $\neq$ force in General Relativity. $\endgroup$
    – ProfRob
    Commented Jun 7, 2021 at 14:21
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    $\begingroup$ Are you talking about spatial flatness or about spacetime flatness? They are very different things. We think the universe is (very close to) spatially flat (which it can be without a cosmological constant) but I don't think anyone thinks spacetime is flat. $\endgroup$
    – user38308
    Commented Jun 7, 2021 at 16:05
  • $\begingroup$ @tfb If space is flat time is also flat. So spacetime is also flat. $\endgroup$ Commented Jun 8, 2021 at 8:18
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    $\begingroup$ @Barbierium: that's simply false, sorry. $\endgroup$
    – user38308
    Commented Jun 8, 2021 at 8:56
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    $\begingroup$ @Barbierium: given a metric $ds^2 = dt^2 - \frac{R^2(t)}{c^2}\left( dr^2 + d^2d\theta^2 + r^2\sin^2\theta d\phi^2\right)$ where $R(0) = 0$, & $R(t)$ is (say) monotonic increasing. This is spatially flat. You can either now just calculate the curvature, or instead just consider that all past-going timelike & null curves meet at $t=0$, which can't be the case in a flat spacetime. $\endgroup$
    – user38308
    Commented Jun 8, 2021 at 9:44

3 Answers 3


"Mass must curve the universe but the pressure due to energy density may curve the universe in the opposite direction, leading to a flat universe"

This is not correct. What produces curvature is the total energy density of the universe (adding up the contributions from mass, radiation, and dark energy) relative to the critical density. If the total is $<$ the critical density, then the curvature is negative; if it is $>$ than the critical density, then the curvature is positive (and if it is equal to the critical density, then the curvature is zero and the universe is flat).

Expansion is a completely separate issue. Mass (and radiation) will slow down the expansion, while dark energy will increase it, regardless of what the curvature is.

  • $\begingroup$ Dark energy has the property to give negative curvature (while normal energy gives positive curvature. When dark energy supplants mass energy in the universe the overall structure of spacetime will have negative curvature (as in the early universe). That's the weird thing about dark energy. It works oppositely to normal energy. It's connected with negative mass which makes objects repel from each other (tese masses give spacetime a negative curvature, related to expansion). $\endgroup$ Commented Jun 8, 2021 at 8:16
  • $\begingroup$ @Barbierium you take expansion as dictated by curvature, but it is not like that in the Friedman equation,which has a cosmology constant $\endgroup$
    – Alchimista
    Commented Jun 8, 2021 at 9:17
  • $\begingroup$ @Alchimista A negatively curved spacetime can only expand. Negative pressure (caused by negative, or imaginary energy, or negative mass) causes expansion. As in the early universe (the inflaton field contained a lot of these hypothetical particles). It cannot exist stationary. So, expanding spacetime has a negative curvature (what else would be the difference with a flat spacetime). $\endgroup$ Commented Jun 8, 2021 at 9:32
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    $\begingroup$ @Barbierium It's possible you are confusing spacetime curvature with spatial curvature; it's the latter that is being asked about and is what is meant by "curvature" in cosmological discussions. The spatial curvature is determined by the energy density, and the energy density of dark energy is positive (otherwise, it cannot provide acceleration), so it provides the same kind of contribution to the curvature as ordinary matter. See, e.g., physics.stackexchange.com/questions/96551/… $\endgroup$ Commented Jun 8, 2021 at 9:45
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    $\begingroup$ @Barbierium "Only negatively curved space can expand" -- this is, simply, wrong, both as a matter of GR and observationally (i.e., our universe is expanding, and is spatially flat). $\endgroup$ Commented Jun 8, 2021 at 11:41

There are two separate issues here, which need to be distinguished: one is whether the universe is spatially curved and the other is whether spacetime is flat or curved.

The answers to these questions are different: in any sane cosmological model spacetime is curved. But the universe may also be spatially flat (and it seems that it is very close to spatially flat).

The Robertson-Walker metric

If you assume that the cosmological principle is true – that the universe is, on large scales, homogeneous and isotropic – then the most general metric you can come up with for the universe as a whole is the Robertson-Walker metric:

$$ds^2 = dt^2 - \frac{R^2(t)}{c^2}\left(\frac{dr^2}{1-kr^2} + r^2d\theta^2 + r^2\sin^2\theta d\phi^2\right)$$

In this metric

  • $t$ is the time coordinate;
  • $R(t)$ is a scale factor (with dimensions of length);
  • $r,\theta,\phi$ are 'reduced' spherical polar coordinates, which are all dimensionless (the length dimension is absorbed into $R$);
  • $k$ is a dimensionless constant and $k \in \{-1,0,1\}$.

There are other ways of representing this metric, quite often you will see it in terms of a dimensionless scale factor $a$ instead of my $R$, when $r$ has dimensions of length and $k$ has dimensions of $l^{-2}$.

This metric doesn't have any real physics in it: it's just the most general metric you can come up with that satisfies the cosmological principle. Note, for instance, that if you make $k=0$ and $R$ constant, you get back Minkowski space: that's how general it is.

But note couple of things: Firstly if $\dot{R} \ne 0$ then this metric doesn't describe a flat spacetime. I'm not going to calculate the Ricci tensor for it, but Wikipedia does and you can see that in general the curvature is not zero.

Secondly you can consider 'slices' of this thing for a given $t$, or equivalently a given $R(t)$: you can think about slices though the universe at a given scale factor, in other words. And you can look at the metric of these 3-dimensional slices, which is, at some time $t_0$, and dropping the $1/c^2$:

$$R(t_0)\left(\frac{dr^2}{1-kr^2} + r^2d\theta^2 + r^2\sin^2\theta d\phi^2\right)$$

Now you can compute the 3-dimensional curvature of these slices. And what you find is that it depends on $k$:

  • $k=1$ gives you a spherical curvature;
  • $k=0$ gives you a flat space;
  • $k=-1$ gives you hyperbolic space.

Note again, this is the curvature of the spatial sections through the universe, not the spacetime curvature.

So the next question is: can you try and work out what $k$ might be for the universe?

Adding physics

To say anything useful about this you need to plug in some physics, in the form of the field equations of General Relativity. I'm not going to go that here, but have a look at the Friedmann equations which is what you end up with. There are two of these, and the first one is the interesting one here. First of all define the Hubble parameter

$$H(t) = \frac{\dot{R}(t)}{R(t)}$$

And now we can write the first Friedmann equation as:

$$H^2(t) = \frac{8\pi G}{3}\rho(t) - \frac{kc^2}{R^2} + \frac{\Lambda c^2}{3}\tag{*}$$


  • $\rho(t)$ is the density of the universe;
  • $\Lambda$ is the cosmological constant.

For the moment assume $\Lambda = 0$. We can reorganise this in terms of $\rho(t)$:

$$\rho(t) = \frac{3}{8\pi G}\left[H^2(t) + \frac{k c^2}{R^2}\right]$$

But $k \in \{-1,0,1\}$, so there are three possibilities here:

$$\rho(t) = \begin{cases} \frac{3}{8\pi G}\left[H^2(t) - \frac{c^2}{R^2}\right] &k = -1\\ \frac{3}{8\pi G}H^2(t) &k = 0\\ \frac{3}{8\pi G}\left[H^2(t) + \frac{c^2}{R^2}\right] &k = 1 \end{cases}$$

So, in particular what falls out of this is that there's a critical value for $\rho$. If we call $H(t_0) = H_0$ then

$$\rho_\text{crit} = \frac{3 H_0^2}{8\pi G}$$

I'm not an experimentalist and I get confused about what people can measure. However the point here is that there is a critical density at which the universe is spatially flat. And I believe that, when you go out and measure things, you find that it's extremely close to that density.


One important thing is that, after (*) above I just said let's assume $\Lambda = 0$. What this means is that you don't need a cosmological constant for a spatially-flat universe. $\Lambda$ changes things a little, and it changes the future evolution of the universe (for which you need the other Friedman equation) in such a way that some $k=1$ universes continue expanding for ever, but you don't need it for spatial flatness.


The three conclusions are.

  • Spacetime curvature and spatial curvature are different things, and when cosmologists talk about a 'flat' universe they are usually talking about spatial curvature.
  • No plausible universe has zero spacetime curvature.
  • It is quite plausible to have zero spatial curvature however.
  • You do not need a cosmological constant for a spatially flat universe.


I've bodged this answer together from slightly fading memory and looking a couple of things up which inevitably use different sign conventions etc. I may have made some mistakes. The underlying points are, I think, correct however.


Mass curves spacetime and so does energy density. They both find their place in the time (energy) component of the stress-energy tensor, which functions as the source of the gravity field (like charge for the electromagnetic field). So positive mass and energy curve spacetime in the same way.
The only way to give the universe a negative curvature is therefore to introduce negative energy (dark energy). This makes mass-energy move away from each other. In the inflation era there was supposed to be an inflaton field which made spacetime grow. The field is thought to consist of particles with imaginary mass or imaginary momentum, because $E^2=m^2c^4+p^2$. If the particles are massless then $E^2=p^2$ so the equality holds if the particles have negative momentum.
If you consider these particles to have negative mass and negative momentum $E^2$ will be positive, but this is just as imaginary as imaginary mass and momentum.
The same argument holds for dark matter although this has not a comparable energy as the energy contained in the inflaton field.

So the curvature of spacetime can be negative while there is positive mass-energy in it. Positive mass-energy curvature is canceled by negative mass-energy curvature. Positive pressure can be compensated by negative pressure. The state of the early universe was one of hugely negatively curved spacetime. When real matter appeared this curvature was inversed in a positive curvature, but in the future, the underlying negative curvature will reappear (as it is already now because of the diminishing mass-energy density) and drive matter apart again. Due to the strong negatively curvature that will come to be in the future, there will be a new possibility for virtual particle fields to become real again, so in the future a new big bang will happen in our spacetime (this implies a universe that is infinite in space as well in time).

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    $\begingroup$ Dark energy is not "negative energy" -- it is positive energy density with negative pressure. All forms of energy -- normal matter, radiation, dark energy -- contribute in the same way to the curvature of the universe. The way to get negative curvature is to have the total energy density (of all kinds) be less than the critical density. $\endgroup$ Commented Jun 7, 2021 at 21:23
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    $\begingroup$ There's a lot of confusion and error in this answer: e.g., "imaginary mass", "imaginary momentum", "negative mass", "positive mass-energy is canceled by negative mass-energy", etc. $\endgroup$ Commented Jun 7, 2021 at 21:24
  • $\begingroup$ @PeterErwin The problem is how to make $E^2=m^2 c^4+p^2$ hold if the pressure is negative. Imaginary mass is just another way of saying that it's negative, but the conclusion for energy will be the opposite. $\endgroup$ Commented Jun 7, 2021 at 21:31
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    $\begingroup$ I think you're trying to treat dark energy as though it's made up of individual particles, which is probably not a good idea. $\endgroup$ Commented Jun 7, 2021 at 21:44
  • $\begingroup$ @PeterErwin Isn't the inflaton field a particle field? Or the dark energy field (virtual particles). Or is it spacetime itself that contains the energy (as in quantum loop gravity)? $\endgroup$ Commented Jun 7, 2021 at 21:50

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