# How is the curvature term of the Friedmann Equation calculated with the Newtonian derivation?

I'm trying to develop and intuitive understanding of the Friedmann equation. I'm afraid I get lost with the relativistic derivation as it's just a lot of crank-turning. When I derive it from the Newton iron-sphere concept, I can see that the constant of integration has the physical interpretation of 'total energy at the surface of the sphere'. So far, so good.

I don't understand how the total energy at the surface of the 3-Sphere leads to curvature. I came across this video (which I think is excellent) Astrophysics (Cosmology) 2.4. At the 9:58, they almost connect the dots. They make the jump from: $$U=\frac{1}{2}m\dot{a}^2x^2-\frac{4\pi}{3}G\rho a^2x^2m$$ to multiply both sides by $$\frac{2}{ma^2x^2}$$, and then to this equation: $$\left(\frac{\dot{a}}{a}\right)^2=\frac{8\pi G}{3}\rho-\frac{kc^2}{a^2}$$ $$k=\frac{-2U}{c^2x^2}\tag1$$ But they don't explain how the curvature got into the equation. Can anyone tell me from where Eq. (1) comes? I don't recognize it and it's the last piece I need to understand curvature.

• Questions on this site are supposed to be self-contained and not require watching a video. It is very unclear what $U$, $m$, and $x$ are. I have never seen any of these appear in any derivation of the Friedmann equations. – G. Smith Jul 10 '20 at 17:18
• In my opinion, any intuition about curvature that you get from a non-GR flat-space analog (which is what I assume an “iron sphere” universe is) is probably going to be misleading. – G. Smith Jul 10 '20 at 17:32
• Minor comment: the left hand side of the first equation you quote should be $\left(\frac{\dot{a}}{a}\right)^2$. – Urb Jul 10 '20 at 18:51
• @Urb - Thank you. Corrected. – Gluon Soup Jul 10 '20 at 19:08
• I am writing this as a comment because I would have to watch the video carefully to write a proper answer. You're never going to derive the curvature in a Newtonian derivation, since it happens in flat space. The best you can do is to note that you have some constant; you have to compare with the actual relativistic equation to identify it as the curvature. Also, it makes no sense to me to say that $U$ depends on $x$, because $U$ is the total energy, not a position dependent field. I think the Newtonian "derivation" just can't be pushed too far. – Javier Jul 16 '20 at 2:40

A genuine derivation of the Friedmann equation would go through general relativity. You would start with the spacetime metric, which determines the curvature $$k$$, and then evaluate the Einstein field equations to get the result.

When gravitational effects are weak, general relativity reduces to Newtonian mechanics. Therefore, in some limits, it should be possible to describe the same observable behavior entirely within Newtonian mechanics. In other words, there should be relativistic equations and Newtonian equations that lead to the same $$a(t)$$.

However, these equations aren't going to look exactly the same. The relativistic equation will have a term involving the curvature $$k$$ (which can't be defined in Newtonian mechanics), while the Newtonian equation will have an analogous term involving the total energy $$U$$ (which is tricky to define in general relativity).

The best you can do is say, "for this value of $$k$$ in the relativistic derivation, plugging in this equivalent value of $$U$$ in the Newtonian derivation would give the same solutions for $$a(t)$$". It turns out the equivalence is $$k = - 2 U / c^2 x^2$$, but this can't be derived in Newtonian mechanics, because there's no such thing as curvature in Newtonian mechanics; everything is perfectly flat.

Can anyone tell me from where Eq. (1) comes?

The curvature come from the right-hand side ($$U$$) of your first equation (modified a bit, merged $$a$$ and $$x$$ into a single $$a$$, since $$x$$ in your equation is apparently a fixed constant which can be absorbed into $$a$$ or set to $$x=1$$ in the chosen unit): $$U=\frac{1}{2}m\dot{a}^2-\frac{4\pi}{3}G\rho a^2m$$

In the Newtonian derivation, you can regard $$\frac{1}{2}m\dot{a}^2$$ as kinetic energy, $$\frac{4\pi}{3}G\rho a^2m$$ as potential energy, and $$U$$ as total energy. The equation can be understood as the conservation of the total energy ($$U$$). $$U$$ is therefore a constant.

If you multiply both sides by $$\frac{2}{ma^2}$$, you got: $$\frac{2U}{ma^2}=\frac{\dot{a}^2}{a^2}-\frac{8\pi}{3}G\rho,$$ which yields: $$\frac{\dot{a}^2}{a^2}=\frac{8\pi}{3}G\rho + \frac{2U}{ma^2}.$$ If you define $$k=\frac{-2U}{mc^2}$$ The previous equation reduces to the final Friedman equation: $$\frac{\dot{a}^2}{a^2}=\frac{8\pi}{3}G\rho - \frac{kc^2}{a^2}.$$

The zero total energy $$U=0$$ implies $$k=0$$ which means flat space (not flat space-time! $$k$$ in the FLRW metric is only related to space curvature, not space-time curvature.).

Currently, all observations point to zero $$U$$ or $$k$$. The flatness of the universe is one of the basic assumptions of the standard cosmology $$\Lambda$$CDM model.

The underlying cause for the flatness hasn't been fully settled so far. The inflation hypothesis purportedly proffers an explanation. But is everyone convinced? Probably not.

• Thank you. That's very helpful background, but I need a citation for Eq. (1). I've seen some version of it now in two videos and three papers, without a reference. From where does this relation come? – Gluon Soup Jul 13 '20 at 18:06
• And currently, there's a crisis in LCDM. The $H_0$ constant obtained from SNe Ia is $73.2$ and the $H_0$ from CMB radiation is $67.4$. The error bars on the measurements have shrunk to such a point that there's a profound contradiction. One of the resolutions to the conflict is closed space, but that creates more problems than it fixes. Anyway, I wouldn't bet a lot of money on that 'universe is flat' conclusion just yet. – Gluon Soup Jul 13 '20 at 20:08
• @GluonSoup, yes there are proposals to resolve the Hubble tension with non-flat universe. As you mentioned, it "creates more problems than it fixes". So far, all attempts of naively tinkering with one or more parameters (including $k>0$ closed universe) in the $\Lambda$CDM have failed. – MadMax Jul 13 '20 at 20:11
• Any chance you can help me out with the $k=\frac{-2U}{mc^2}$ reference? Just found it in another paper: people.ast.cam.ac.uk/~pettini/Intro%20Cosmology/Lecture02.pdf but still no reference to tell me from where the relation comes. – Gluon Soup Jul 13 '20 at 21:12
• All you've done here is re-state the problem. You've done nothing to answer the original question: from where does the relation $k=\frac{-2U}{mc^2}$ come? – Gluon Soup Jul 14 '20 at 0:51

I've seen this "derivation" before. You first start with the EOM (Newton's second law) for a particle at the boundary of the sphere

$$m\ddot{r}=-\frac{GMm}{r^2}.$$

If we eliminate $$m$$, substitute $$\ddot{r}=\ddot{a}x$$ and $$M=\displaystyle\frac{4}{3}\pi r^3\rho$$ we get to

$$\ddot{a}x=-\frac{4\pi G}{3}\rho\ r$$ and dividing by $$x$$ to $$\ddot{a}=-\frac{4\pi G}{3}\rho\ a.$$

We can integrate this equation by noticing that $$\displaystyle\ddot{a}=\frac{d\dot{a}}{dt}=\frac{d\dot{a}}{da}\frac{da}{dt}=\frac{d\dot{a}}{da}\dot{a}$$, therefore

$$\int\dot{a}d\dot{a}=-\int\frac{4\pi G}{3}\rho\ ada\tag{1}$$

Now, integrating $$(1)$$ will give you a constant on one side of the equation, let's name it $$C$$:

$$\frac{\dot{a}^2}{2}=-\frac{4\pi G}{3}\int\rho_0\frac{1}{a^2}da$$ $$\frac{\dot{a}^2}{2}=\frac{4\pi G}{3}\rho_0\frac{1}{a}+C$$ $$\frac{\dot{a}^2}{2}=\frac{4\pi G}{3}\rho\ a^2+C.$$

The constant $$C$$ must have the same dimensions as the other two terms in the last equation, i.e. it must have dimensions of $$[T]^{-2}$$. Hence, nothing prevents me from writing the constant $$C$$ as

$$C\equiv \frac{kc^2}{2},$$

where $$c$$ is the speed of light (with dimensions $$[L]/[T]$$)and $$k$$ is a curvature, i.e. it's something that has dimensions of curvature $$([L]^{-2})$$. By making this renaming of the constant of integration, you arrive to the Friedmann equation in the way it is usually stated (as derived from General Relativity).

$$\left(\frac{\dot{a}}{a}\right)^2=\frac{8\pi G}{3}\rho-\frac{kc^2}{a^2}\tag{2}$$

From here, to show the relation between $$k$$ and $$U$$ that you state just multiply both sides by $$a^2x^2/2$$, you get $$\frac{1}{2}\dot{a}^2x^2=\frac{4\pi G}{3}\rho a^2x^2-\frac{1}{2}kc^2x^2.$$ Now since $$v=x\dot{a}$$ and $$\rho=\displaystyle\frac{M}{\displaystyle\frac{4}{3}\pi r^3}$$ $$\frac{1}{2}v^2=\frac{4\pi G}{3}\displaystyle\frac{M}{\displaystyle\frac{4}{3}\pi r^3} r^2-\frac{1}{2}kc^2x^2$$ $$\frac{1}{2}v^2=\frac{GM}{r}-\frac{1}{2}kc^2x^2$$ $$\frac{1}{2}v^2-\frac{GM}{r}=-\frac{1}{2}kc^2x^2.$$ Finally, identify the LHS with the total energy per unit mass $$U/m$$ and solve for $$k$$.

One last bit, the interpretation of $$k$$ is done in the same way as in general relativity. First, we know experimentally that $$\dot{a}>0$$ at the present time, i.e., the Universe is expanding right now. By looking at equation $$(2)$$, we see that if $$k<0$$ the RHS of Eq. $$(2)$$ will be positive at all times, indicating that $$\dot{a}(t)>0$$ for all $$t$$ and the Universe will expand forever. To see this more clearly, rewrite the Friedmann equation as $$\dot{a}^2=\frac{8\pi G}{3}\frac{\rho_0}{a}-kc^2.\tag{3}$$ Notice how for $$a\rightarrow\infty$$, $$\dot{a}^2$$ approaches a constant value $$-kc^2$$. If $$k=0$$, then the RHS of Eq. $$(2)$$ will also be positive forever, but in such a way that the expansion eventually stops at very large $$a$$, cf. Eq. $$(3)$$, $$\dot{a}\rightarrow 0\, \text{ as } a\rightarrow\infty\quad (k=0).$$

Finally, if $$k>0$$, the RHS of Eq. $$(2)$$ will vanish at a value of $$a_{max}=\displaystyle\frac{8\pi G\rho_0}{3kc^2}$$. At that point the expansion will stop and the Universe will start contracting $$(\dot{a}<0)$$. The sign of $$k$$ then, separates Universes that expand forever from Universes that recollapse in the future.

• The steps above start with the equations of motion for gravity, $$F=ma$$ $$\frac{GMm}{r^2}=ma$$ $$\frac{GM}{r^2}=a$$ $$\frac{GM}{r^2}dr=a dr$$ $$\frac{GM}{r} + U = \frac{1}{2}\left(\frac{r}{dt}\right)^2$$ and work towards the last two steps I showed. You are going backwards. You start with the Friedmann equation and work your way backwards to the equations of motion. I want to know where the constant term $\frac{kc^2}{a^2}$ comes from and how it's related to the kinetic and potential energy in the Newton derivation. – Gluon Soup Jul 10 '20 at 19:13
• @GluonSoup I've edited the answer. – Urb Jul 16 '20 at 12:58

NEWTON

the energy is:

$$\frac{1}{2} \dot{a}^2+V(a)=E\tag 1$$

where $$V(a)$$ is the potential energy and $$\frac{1}{2} \dot{a}^2$$ the kinetic energy.

you can describe equation (1) using the Habbel parameter $$H=\frac{\dot a}{a}$$

$$H^2-\frac{2E}{a^2}=\frac{8\pi\,G}{3}\,\epsilon\tag 2$$

where $$\epsilon$$ is the energy density

with equation (1) in (2) you obtain for $$V(a)$$

$$V(a)=-\frac{8\pi\,G}{6}\,a^2\,\epsilon$$

EINSTEIN

The Friedmann equation is:

$$H^2+\frac{k}{a^2}=\frac{8 \pi\,G}{3}\,\epsilon\tag 3$$

compare equation (2) with (3) you obtain for $$k$$

$$k=-2\,E$$

More details you find in v. MUKHANOV book " Physical Foundations of Cosmology"