# Linearly expanding vacuum solves cosmological constant problem? [duplicate]

The flat FLRW metric written in terms of conformal time $$\eta$$ is given by $$ds^2=a^2(\eta)(d\eta^2-dx^2-dy^2-dz^2)\tag{1}$$ where the scale factor $$a(0)=1$$. The interval of proper time $$\tau$$ measured by a comoving observer is given by $$d\tau=a(\eta)d\eta.$$ The coordinate system used to describe the metric in Eqn.(1) is that of a stationary observer fixed at the origin $$(\eta,x,y,z)=(0,0,0,0)$$ with four-velocity $$U^\mu=(1,0,0,0).$$ Assume that he observes a particle moving along a nearby path $$x^\mu(\lambda)$$ with four-velocity $$V^\mu=\frac{dx^\mu}{d\lambda}.$$ The four-momentum of the particle is $$P^\mu=mV^\mu.$$ The energy $$\epsilon$$ of the particle as measured by the stationary observer fixed at the origin $$\eta=0$$ is given by $$\epsilon=U_\mu P^\mu.$$ Now let us assume that the path $$x^\mu(\lambda)$$ is a geodesic. Using the metric in Eqn.(1) it can be shown that the covariant derivative of the energy $$\epsilon$$ along $$x^\mu(\lambda)$$ is given by $$\frac{d\epsilon}{d\lambda}=m\frac{\dot a}{a}V^\mu V_\mu.$$ Consider two types of particle; a massless particle and a stationary massive particle. The massless particle has $$V^\mu V_\mu=0$$ therefore $$\begin{eqnarray} \frac{d\epsilon_{\rm massless}}{d\lambda} &=& 0\\ \epsilon_{\rm massless} &=& {\rm constant.} \end{eqnarray}$$ The massive particle has $$V^\mu V_\mu=1$$ therefore $$\begin{eqnarray} \frac{d\epsilon_{\rm massive}}{d\tau} &=& m\frac{da/d\eta}{a}\\ &=& m\frac{da/d\tau\ d\tau/d\eta}{a}\\ &=& m\frac{da}{d\tau}\\ \epsilon_{\rm massive} &=& ma. \end{eqnarray}$$ Thus the constant energy $$\epsilon_{\rm massless}$$ is the energy of a massless particle as it propagates into the future with respect to the stationary observer fixed at the origin at $$\eta=0$$. The energy $$\epsilon_{\rm massive}=ma$$ is the energy of a stationary massive particle as it propagates into the future with respect to the stationary observer fixed at the origin at $$\eta=0$$.

Now, using natural units with $$\hbar=c=1$$, the Friedmann equation for the vacuum is given by $$\Big(\frac{da/d\tau}{a}\Big)^2\sim \frac{1}{M_{\rm pl}^2}\rho_{\rm vac}$$ where $$M_{\rm pl}$$ is the Planck mass. The vacuum density $$\rho_{\rm vac}$$ is given by $$\rho_{\rm vac} \sim P_{\rm pl}^4$$ where $$P_{\rm pl}$$ is the Planck momentum.

According to the stationary observer fixed at the origin $$\eta=0$$ $$\begin{eqnarray} M_{\rm pl} &=& M_{\rm pl}^0\ a(\tau)\\ P_{\rm pl} &=& P_{\rm pl}^0 \end{eqnarray}$$ where $$M_{\rm pl}^0$$ and $$P_{\rm pl}^0$$ are the Planck mass and momentum measured at the origin.

Thus, using Planck units where $$M_{\rm pl}^0=P_{\rm pl}^0=1$$, the Friedmann equation for the vacuum becomes $$\begin{eqnarray} \Big(\frac{da/d\tau}{a}\Big)^2 &\sim& \frac{1}{a^2}\\ a(\tau) &\sim& \tau. \end{eqnarray}$$ Thus the scale factor $$a$$ is a linear function of the proper time $$\tau$$ as measured by a co-moving observer. This is very different from the standard vacuum cosmology which expands exponentially.

In fact we can define the scale factor $$a(\tau)=\tau$$ so that the scale factor is simply the age of the universe in Planck units.

We can also redefine the energy scale so that by definition a co-moving observer always measures a constant Planck mass. In that case the energy density of the vacuum, $$\rho_{\rm vac}$$, must vary as $$\rho_{\rm vac} \sim \frac{1}{a^2} \sim \frac{1}{\tau^2}.$$

The age of the universe, $$\tau$$, is approximately $$10^{60}$$ in Planck units so that the current density of the vacuum $$\rho_{\rm vac}\sim 10^{-120}$$. This is consistent with observation.

• Soooo, ... what was the question again ? – Cham Jun 7 '19 at 17:51
• In GR can an observer fixed at the origin of a coordinate system have a consistent view about the energy of a particle located at a distant spacetime point? – John Eastmond Jun 7 '19 at 18:10
• What about radiation and dust dominated era, which are consistent with observation for most of the cosmological evolution? I don't understand your concept of the stationary massive particle and how it relates to the cosmological constant or the dark energy? What is the meaning of an observer fixed at $\eta=0$? – p6majo Jun 7 '19 at 18:18
• The linear expansion is only a half of the truth. The other half is that the FLRW metric is incorrect from the start, so any meaningful result based on it is only accidental.. – safesphere Jun 8 '19 at 3:59
• I'm assuming that the vacuum has such a high energy density that its effect overrides the radiation and dust to produce a linear expansion at all times. An example of the stationary massive particle is the Planck mass whose mass increases with the scale factor $a$ from the perspective of an observer at the origin at $\eta=0$ or equivalently $\tau=\tau_0$. This implies that Newton's $G \sim 1/a^2$ with respect to the observer fixed at the origin $\eta=0$. – John Eastmond Jun 9 '19 at 16:03