# Intuition on the Friedman equation: curvature and expansion

1. Sep 16, 2014

### center o bass

The Friedmann equation states that

$$(\frac{\dot a}{a}) = \frac{8\pi G}{3} \dot \rho + \frac{1}3 \Lambda - \frac{K}{a^2},$$
where $a, \rho, \Lambda, K$ respectively denotes the scale factor, matter density, cosmological constant and curvature.

Now, I'm trying to get at an intuition on why, according to the above equation, space expands slower with a bigger curvature K? Is this somehow intuitive?

2. Sep 16, 2014

### Staff: Mentor

First, a small correction: there should not be a dot over the $\rho$ in the equation.

The way I would look at this, intuitively, is to look at the second Friedmann equation first:

$$\frac{\ddot{a}}{a} = - \frac{4 \pi G}{3} \left( \rho + 3 p \right) + \frac{\Lambda}{3}$$

This is the equation that actually gives the dynamics. Note that it does *not* include the curvature $K$. The first equation is just telling you that, if you take a "snapshot" of the universe at a particular instant of time, the relationship between the expansion rate and the other parameters will depend on the spatial curvature. But the relationship between the *change* in the expansion rate and the other parameters, which is what the second equation gives, will not.

In fact, a more illuminating way of looking at the first equation may be to look at it as telling you that the spatial curvature depends on the relationship, at a given instant, between the expansion rate and the other parameters. In other words, it's actually an equation *for* the spatial curvature.

Last edited: Sep 16, 2014
3. Sep 16, 2014

### ChrisVer

You have to understand that that equation indeed doesn't give any dynamics about the expansion of universe. $\dot{a}/a$ itself gives you the Hubble parameter at some time t and not how it evolves. That's the reason Peter gave the second derivative of this which corresponds to some $\dot{H}$.
Also don't forget that $K \in \left \{ -1, 0, +1 \right \}$ so it's not a part of bigger or smaller. Changing the $K$ parameter you are changing the whole geometry...

4. Sep 16, 2014

### Staff: Mentor

Actually, that depends on the coordinates you choose. Physically, there is in fact a continuous difference between, for example, a universe which is exactly spatially flat, and a universe which is very, very, very slightly spatially curved. In the coordinates in which $K$ is discrete, the continuous difference is all contained in the scale factor $a$; note that the curvature term in the Friedmann equation is not just $K$, but $K / a^2$, so a universe which is exactly spatially flat, with $K = 0$ and therefore a zero curvature term, can still be very, very close to a universe that is very, very slightly spatially curved, with $K = 1$ but $a$ equal to some very, very large number, so that the curvature term $K / a^2$ is very, very close to zero.

5. Sep 16, 2014

### ChrisVer

To that I am quiet skeptic. Although as a sentence it makes perfect sense, I don't understand how can you let $a$ become very very large. It's not uncommon to see that $0 < a <1$ (maybe that's coordinate dependent) where you have set $a_0 =1$. But nevertheless, the value of $a$ is obtained by solving the Friedmann equations so you can't just set it very large -there would have been a time it was very small.

6. Sep 16, 2014

### Staff: Mentor

That's true, mathematically, but it doesn't change what I was saying. The point (of yours) that I was responding to was the implicit assertion that, since $K$ is discrete (it can only be +1, 0, or -1), there can't be a continuous variation between flat universes and curved universes. That's wrong; there can.

Remember that we are looking at spatial slices, so "close" does not refer to two spacetimes--one with flat spatial slices and one with curved spatial slices--it refers to two slices, one flat and one curved (but only very, very slightly curved). The curved slice might have to be a slice from some curved universe at a late enough time in that universe's evolution for the radius of curvature to be very, very large, yes; but that doesn't matter for the comparison being made.

To see the point another way, look at it from the other end, so to speak: here we are, in a universe, and we're trying to figure out whether it's spatially flat or curved. We don't know for sure which it is: but we *do* know that, *if* it is curved, the radius of curvature must be very, very large. That does imply that, if the universe is curved, it must be old enough to have expanded to a very, very large radius of curvature, yes. But that doesn't change the fact that the spatial slice of the universe that we are now in, if it is not perfectly flat, is very, very close to flat, even though, if it is curved, it has a value of $K$ that is discretely different from zero.