# How can the Friedman Equation produce negative pressure or density?

As I understand it, the Friendman Equation provides the driving mechanism for Inflation.$$\frac{\ddot a}{a}=-\frac{4\pi G}{3}\left(\rho+\frac{3p}{c^2}\right)$$If $$\rho$$ or p is negative enough, you get a positive acceleration (i.e. expansion). I've tried reading through several papers and text, but they all regress into this techno-babble that sounds like Alice-In-Wonderland logic. Could someone explain in terms of known physics how either the density or pressure of the early universe can be negative, especially during a period of hyper-accelerated expansion?

• Negative pressure is just the same thing as tension. If you pull a rope, that rope carries negative pressure. Negative energy on the other hand is quite exotic. – knzhou Dec 9 '18 at 13:46
• Negative pressure is when an enclosed area has lower pressure than the area around it. How does that apply to a homogeneous universe? – user32023 Dec 9 '18 at 13:49
• It's the same as energy. Outside of relativity, there is no absolute energy, just areas with higher or lower energy. In relativity, energy is absolute, you don't have to compare it with anything, e.g. see $E = m c^2$. And similarly, pressure is also absolute. – knzhou Dec 9 '18 at 13:55
• Well, as I said in the first comment, negative pressure isn't even weird in Newtonian mechanics. It's usually called tension. Do you believe tension can't exist? – knzhou Dec 9 '18 at 14:07
• In Mukhanov's book: Physical foundations of Cosmology, in chapter 5 which is about Inflation, more precisely in sections $5.3$ and $5.4$, called respectively "How can gravity become "repulsive"?" and "How to realize the equation of state $p \approx = - \epsilon$ " There is a good treatment and explanation of this. I could copy an answer from there but it would surely far worse than his explanations. – RenatoRenatoRenato Dec 9 '18 at 15:33

The Friedmann equation doesn't produce negative pressure or energy density; it just tells you that if $$\rho+3p$$ is negative then $$\ddot{a}$$ will be positive, so you'll get accelerating expansion of the universe. We observe accelerating expansion of the universe today, and there is also reasonable evidence that cosmic inflation occurred in the early universe. Thus, we have to consider what physical systems are capable of producing negative pressure or energy density.

Negative energy density is considered extremely exotic, but negative pressure not so much. I'll discuss two situations in which you get negative pressure.

The first one, which is used in many inflation models, is a spin-0 (i.e., scalar) field that fills the universe. Such a field can have negative pressure. We know that such fields exist, because the Higgs boson is a quantum of just such a field that fills the universe. However, the spin-0 field usually assumed to be responsible for inflation is not the Higgs field but another spin-0 field that we haven't detected except, apparently, through its cosmological effects.

I don't know how to give a non-techno-babble explanation of why the pressure of a spin-0 field can be negative. The problem is that spin-0 fields aren't like any of the fields that make up things you have everyday experience with. For example, matter is made from spin-1/2 (i.e., spinor) fields interacting via spin-1 (i.e., vector) fields, and electromagnetic radiation is a spin-1 field. Both of these have positive pressure and energy density. But humans have no intuition about what the pressure and energy density of a spin-0 field are, and have to rely on mathematics to work it out.

Consider a spin-0 field with a self-interaction of the form $$V(\phi)$$. Then the field theory for a spin-0 field says that there are three contributions to its energy density: the first comes from how fast the field changes in time, the second from how fast the field changes in space, and the third from how it interacts with itself. The energy density turns out to be

$$\rho=\frac{1}{2}\dot{\phi}^2 + \frac{1}{2}\frac{(\nabla \phi)^2}{a^2} + V(\phi)$$

and the pressure is

$$p=\frac{1}{2}\dot{\phi}^2 - \frac{1}{6}\frac{(\nabla \phi)^2}{a^2} - V(\phi)$$

(A derivation of these formulas can be found at http://hep.itp.tuwien.ac.at/~wrasetm/files/2017S-GRplusScalar.pdf.)

So, assuming $$V(\phi)$$ is positive-definite, the energy density is always positive, but the pressure can be negative. In a homogeneous and isotropic universe like ours appears to be, the gradient term (the second one) would be zero. If we assume that the field varies slowly enough that its "kinetic" term (one the first one) is small compared with its "potential" term (the third one), then the relationship between pressure and density is

$$p=-\rho.$$

This negative pressure then makes $$\rho+3p$$ equal to the negative value $$-2\rho$$, so $$\ddot{a} > 0$$ and the expansion accelerates.

The second way to get negative pressure is through vacuum energy. Again, humans have no intuition about how vacuum energy works, because it is so tiny in our universe. Math is again the guide. The energy-momentum-stress tensor for the vacuum, if it isn't exactly zero, must be proportional to the metric tensor, because in a vacuum state there are no other tensors available to construct it out of. So

$$T_\text{vacuum}^{\mu\nu}=\Lambda g^{\mu\nu}$$

In terms of components, this looks like the following in flat spacetime:

$$\begin{pmatrix} \rho & 0 & 0 & 0 \\ 0 & p & 0 & 0 \\ 0 & 0 & p & 0 \\ 0 & 0 & 0 & p \end{pmatrix} =\Lambda \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}$$

and once again we get

$$p=-\rho$$

and accelerating expansion.

So there are at least two mechanisms for getting enough negative pressure to explain accelerating expansion of the early universe and of today's universe.

• +1 for successfully reducing "techno-babble" to something a non-physicist can grapple with! – Chappo Hasn't Forgotten Monica Dec 10 '18 at 7:20
• What I fail to understand is this: every equation of state is a function of pressure, volume and temperature. That is f(p,T,V)=0 in one form or another. I can imagine a scenario where the math leads us to the possibility of negative pressure, but, if volume is positive, I can't understand a physical reality where negative pressure doesn't result in negative temperature. – user32023 Dec 10 '18 at 14:18
• Yes, having negative pressure with positive temperature is very unintuitive. I think the problem is that you are expecting the ideal gas law $PV=Nk_B T$ to apply to the contents of the universe, but it doesn’t in the case where strange contents like spin-0 fields or vacuum energy dominate. The statistical mechanics of this stuff is different from that of normal matter, and it allows negative pressure along with positive temperature. – G. Smith Dec 10 '18 at 17:25
• @G.Smith - No, not the Ideal Gas Law, per se, but every Equation of State known to science has the same basic relation: if volume is positive, then the sign of pressure and temperature have to be the same. Aren't you a little bit worried that you've ventured into the domain of mathematical witchcraft? Many of these concepts aren't disprovable, which should make you seriously wonder if it's science at all. – user32023 Dec 10 '18 at 17:54
• @G.Smith - However, I'm up-voting because you have done the best job yet of making these ideas accessible. – user32023 Dec 10 '18 at 17:56