The origin of the cosmological constant in unimodular gravity

Abstract

It is well-known that in unimodular gravity (UG) the cosmological constant is not sourced by a constant energy density, but rather appears as some sort of integration constant. In this work we try to flesh this out by studying in some detail a couple of examples, one from cosmology and the other from gravitational collapse.

1 Introduction

Unimodular Gravity (UG) is a modification of General Relativity (GR) where only unimodular metrics (determinant, $g\equiv \left| det\,g_{\mu \nu }\right| =1$) are considered. Even in the path integral we are instructed to integrate over unimodular metrics only. The equations of motion of UG are not identical to GR Einstein’s ones with the same source. In fact the second Noether theorem (Bianchi identities) allows to recover the lost trace as an integration constant. What happens then is that given a fixed value for the cosmological constant (CC) (for example $\Lambda =0$) any solution of GR is also a solution of UG, but the converse is untrue. If we represent by E the space of classical solutions of the equations of motion, then

$$\begin{aligned} E_{GR (CC=\Lambda )}\subset E_{UG} \end{aligned}$$

(1)

and, somewhat symbolically,

$$\begin{aligned} \sum _{\Lambda \in \mathbb {R}} E_{GR (CC=\Lambda )}\equiv E_{UG} \end{aligned}$$

(2)

From the physical point of view, UG is interesting because the value of the CC is not determined by the constant vacuum energy, which does not weigh at all in UG. The natural question is then, what determines the value of the CC in UG?.

In this paper we want to study this question under two different aspects, the first one dealing with cosmological solutions and the second one with the spherical collapse.

To begin with, in [1] we asked the same question in the framework of standard cosmology. We found there (and we review here) that in vacuum there are two solutions: one in which the scale factor is constant (which corresponds to flat space)

$$\begin{aligned} b(t)=b_0 \end{aligned}$$

(3)

and another where

$$\begin{aligned} b(t)=(3t-t_0)^{4/3} \end{aligned}$$

(4)

In this case at least, the divide between these two solutions lies in whether the initial condition on the derivative of the scale factor $\dot{b}(t_0)$ is different from zero or not. This initial condition on a vacuum solution is admissible only in UG.

Our second subject is the detailed study of gravitational collapse. This is of course a complicated aspect, but it is believed that all $j\ge 2$ multipoles of the matter are eventually radiated away, and that the final state is stationary, with axial symmetry [5]. In this paper we shall consider the simplest models of spherical collapse of a compact matter: the one originally proposed by Schwarzschild [9], where the source is assumed to be an incompressible perfect fluid (that is, with constant density) and another where the source is assumed to be a presureless perfect fluid (a.k.a. dust) [11].

What we find is that in GR the energy density of the collapsing cloud $\rho$ is bound to be related to the curvature by

$$\begin{aligned} \rho =-{R\over \kappa ^2} \end{aligned}$$

(5)

owing to Einstein’s equations. In UG this is not the case, and in fact

$$\begin{aligned} \rho =-{R\over \kappa ^2}+\text {constant} \end{aligned}$$

(6)

(note, however that the derivative $\dot{\rho }$ obeys the same restriction as in GR). In fact this constant is intimately related to the value of the CC.

2 Cosmological solutions

The Friedmann–Lemaître metric in the unimodular gauge of GR (which is the only admissible metric in UG cf. for example [2]) reads

$$\begin{aligned} ds^2=b(t)^{- 3/2}\,dt^2-b(t)^{1/2}\, \delta _{ij}dx^i dx^j \end{aligned}$$

(7)

where the function b depends on time only, $b=b(t)$. The cosmic normalized four velocity vector field, $u^\mu u_\mu =1$, is given explicitly by

$$\begin{aligned} u^{\mu }=\left( b^{3/4},0,0,0\right) \end{aligned}$$

(8)

It is easy to check that this congruence is geodesic

$$\begin{aligned} \dot{u}^\mu \equiv u^\nu \nabla _\nu u^\mu =0 \end{aligned}$$

(9)

and the expansion reads

$$\begin{aligned} \theta \equiv \nabla _\mu u^\mu =\frac{3}{4}b^{-1/4}\frac{db}{dt} \end{aligned}$$

(10)

The equation of motion in UG (the traceless piece of Einstein’s) reads

$$\begin{aligned} R_{\mu \nu }-\dfrac{1}{4}R\,g_{\mu \nu }=\kappa ^2 \left( T_{\mu \nu }-\dfrac{1}{4}\,T\,g_{\mu \nu }\right) \end{aligned}$$

(11)

We assume matter as a perfect fluid, that is

$$\begin{aligned} T_{\mu \nu }\equiv \left( \rho +p\right) u_\mu u_\nu -p g_{\mu \nu } \end{aligned}$$

(12)

the energy-momentum conservation, $\nabla _\nu T^{\mu \nu }=0$ is then equivalent to

$$\begin{aligned} \dot{\rho }+\left( \rho +p\right) \theta =0 \end{aligned}$$

(13)

Using the equation of motion (11) and the property (9), Raychaudhuri’s equation [8] reduces to

$$\begin{aligned}{} & {} \dot{\theta }+\frac{1}{n-1}\theta ^2+\sigma _{\alpha \beta }\sigma ^{\alpha \beta }-\omega _{\alpha \beta } \omega ^{\alpha \beta }\nonumber \\{} & {} \qquad +\frac{1}{n}R+\frac{2(n-1)}{n}\kappa ^2(\rho +p)=0 \end{aligned}$$

(14)

The scalar curvature reads

$$\begin{aligned} R=-2u^\mu \nabla _\mu {\theta }-\frac{4}{3}\theta ^2 \end{aligned}$$

(15)

Assuming as usual for simplicity vanishing shear and rotation, $\sigma _{\alpha \beta }=\omega _{\alpha \beta }=0$, and in the physical dimension $n=4$

$$\begin{aligned} \dot{\theta }+3\kappa ^2(\rho +p)=0 \end{aligned}$$

(16)

It is worth remarking that it is not possible to express R in terms of T.

We can use Ellis’ clever trick [6] to define a length scale through

$$\begin{aligned} \theta =3{\dot{ l}\over l} \end{aligned}$$

(17)

Then

$$\begin{aligned} b\sim l^4. \end{aligned}$$

(18)

Finally we can write Raychaudhuri’s equation (16) as

$$\begin{aligned} u^\mu u^\nu \Big [ l\nabla _\nu \nabla _\mu {l}-\nabla _\mu l\nabla _\nu l\Big ]+\kappa ^2(\rho +p)l^2=0 \end{aligned}$$

(19)

2.1 Vacuum solutions

Vacuum implies $p=\rho =0$ and Raychaudhuri’s equation reduces to

$$\begin{aligned} \dot{\theta }=0 \end{aligned}$$

(20)

In our case

$$\begin{aligned} \dot{\theta }=u^\nu \nabla _\nu \nabla _\mu u^\mu =-\frac{3}{16\sqrt{b}}\left[ \left( \frac{db}{dt}\right) ^2-4b\frac{d^2 b}{dt^2}\right] \end{aligned}$$

(21)

It follows that

$$\begin{aligned} \left( \frac{db}{dt}\right) ^2-4b\frac{d^2 b}{dt^2}=0 \end{aligned}$$

(22)

Its general solution is given by either a constant

$$\begin{aligned} b=H_0^{4 \over 3} \end{aligned}$$

(23)

which corresponds to flat space; or else

$$\begin{aligned} b(t)=H_0^{4 \over 3}\left( 3t-t_0\right) ^{4\over 3} \end{aligned}$$

(24)

which corresponds to de Sitter¹

space; $H_0\equiv 3\theta$ being the constant expansion. In this solution it is arbitrary, because there is no physical scale in the problem that determines it. It is to be emphasized that this solution depends on two parameters, whereas the constant solution depends only on one, being thus less generic.

This could be anticipated, because the vacuum equation of motion in unimodular gravity are just Einstein spaces

$$\begin{aligned} R_{\mu \nu }={1\over 4} R g_{\mu \nu } \end{aligned}$$

(29)

Flat space is just a quite particular solution; constant curvature space-times [12,13,14] are a more generic one.

2.2 The line $p+\rho =C$

Let us examine the inhomogeneous equation of state $\rho +p=C$. This is one of the most interesting results of UG, where physics depends on the value of the constant C only.

Indeed, depending on the value of the constant C, it is possible that both p and $\rho$ are positive. Only in the case $C=0$ is this situation strictly equivalent to a vacuum energy density.The differential equation (19) then reads

$$\begin{aligned} \frac{1}{l}{d^2 l\over dt^2}-\frac{1}{l^2}\left( {dl\over dt}\right) ^2+C\kappa ^2l^{-4}=0 \end{aligned}$$

(30)

whose general solution reads

$$\begin{aligned} l_C(t)=e^{-\sqrt{C_2}(t+C_3)}\Big \{\frac{1}{6C_2}\Big [e^{6\sqrt{C_2}(t+C_3)}-3C\kappa ^2C_2\Big ]\Big \}^{1/3} \nonumber \\ \end{aligned}$$

(31)

Obviously when $C=0$ this solution reduces to the vacuum solution

$$\begin{aligned} l_{\Lambda }(t)=l_0 e^{H_0t} \end{aligned}$$

(25)

with $C_3=0$ and $C_2^2=H_0$.

More interestingly, this solution is an atractor asymptotically when $t\rightarrow \infty$. Any solution tends asymptotically to de Sitter.

For $C C_2 >0$ there is an origin of time $t_0$. For earlier times $t< t_0$ the solution becames unphysical. To be specific,

$$\begin{aligned} t_0={1\over 6 \sqrt{C_2}}\log \,\left( 3 C \kappa ^2 C_2\right) - C_3 \end{aligned}$$

(26)

2.3 Unimodular gravity versus general relativity

The unimodular gauge of General Relativity (GR) is of course fully equivalent to the usual formulation of GR in comoving coordinates [4, 10] where the metric reads

$$\begin{aligned} ds^2= d\tau ^2-a(\tau )^2 \sum \delta _{ij} dx^i dx^j \end{aligned}$$

(32)

with a four velocity

$$\begin{aligned} u^\mu =(1,0,0,0) \end{aligned}$$

(33)

and

$$\begin{aligned} \dot{u}^\mu =0 \end{aligned}$$

(34)

so that in this case

$$\begin{aligned} \theta =3\frac{1}{a}\frac{da}{dt} \end{aligned}$$

(35)

We insist that he only difference between GR and UG stems from the equations of motion. Let us spell this out in some detail.

Now the equation of motion is the usual Einstein one

$$\begin{aligned} R_{\mu \nu }-\dfrac{1}{2}R\,g_{\mu \nu }=\kappa ^2 T_{\mu \nu } \end{aligned}$$

(36)

in this case the scalar of curvature reads

$$\begin{aligned} R=-\frac{6}{a^2}\left[ \left( \frac{da}{dt}\right) ^2+a\frac{d^2a}{dt^2}\right] \end{aligned}$$

(37)

Then Raychaudhuri’s equation in comoving coordinates yields

$$\begin{aligned} -3\left( \frac{da}{dt}\right) ^2+2a^2\kappa ^2\rho =0 \end{aligned}$$

(38)

In this case, the vacuum solution reduces to

$$\begin{aligned} \frac{da}{dt}=0 \end{aligned}$$

(39)

i.e $\theta =0$ which is just flat spacetime, it is a subset of the UG result, $\dot{\theta }=0$ (20), which is obviously a more general equation of motion.

3 Spherical collapse

We shall present two very simplified analysis. The first analysis follows closely the classic approach pioneered by Schwarzschild [9] and later by Oppenheimer and Volkoff [7] for an incompressible fluid (which means constant density). The second one corresponds to what astrophysicists call dust (that is pressureless matter) cf. for example [11].

The main idea is the following. It is assumed the existence of a radial coordinate, r, such that at a given value $r=a$ a matching can be performed between the interior solution sourced by the energy momentum tensor of the perfect fluid, and an exterior solution, which according to Birkoff’s theorem [3] must be Schwarzschild’s.² The first thing to note is that the interior solution is not Ricci flat (in fact the trace of the GR equations of motion implies that the curvature scalar is given by $R=-\kappa ^2 T$, where T is the trace of the energy-momentum tensor. For the equation of state this is just

$$\begin{aligned} R=-\kappa ^2 \left( \rho -3p\right) . \end{aligned}$$

(40)

This is at variance with the exterior solution, which is indeed Ricci flat according to Birkhoff’s theorem. Then in the matching there is necessarily a discontinuity in the second derivatives of the metric.

As we shall see in the next paragraph, this is not so in UG, where the interior metric is allowed to be Schwarzschild-(anti)-de Sitter, so that

$$\begin{aligned} R=-\kappa ^2\rho -4 \Lambda \end{aligned}$$

(41)

where $\Lambda$ is the cosmological constant. What happens is that the UG equations of motion are traceless, which loosens somewhat the constraint on the scalar curvature, allowing it to be modified by a constant term, precisely related to the CC. The main purpose of our work is to understand better the physical meaning of the CC from the collapsing matter viewpoint.

Let us first analyze some general properties of the energy-momentum tensor. For matter which can be modelled as a perfect fluid

$$\begin{aligned} T_{\mu \nu }=(\rho +p )u_\mu u_\nu -p g_{\mu \nu } \end{aligned}$$

(42)

Covariant conservation of the energy-momentum tensor $\nabla _\nu T^{\mu \nu }=0$ is equivalent to

$$\begin{aligned} \nabla ^\lambda p=\left( \dot{\rho }+\dot{p}\right) u^\lambda +\left( \rho +p\right) \left( \theta u^\lambda +\dot{u}^\lambda \right) \end{aligned}$$

(43)

In the synchronous gauge this implies

$$\begin{aligned} \dot{\rho }+(\rho +p)\theta =0 \end{aligned}$$

(44)

3.1 A short review of the GR spherical collapse

• When dealing with an incompressible fluid, where $\rho =\rho _0$, the metric reads [9]

  $$\begin{aligned} ds^2=f_4dx_4^2-f_1dx_1^2-f_2\Big [\frac{dx_2^2}{1-x_2^2}+(1-x_2^2)dx_3^2\Big ] \nonumber \\ \end{aligned}$$

  (45)

  where the functions $f_i$ depend on $x_1$. If we choose to work in the unimodular gauge of GR (as Schwarzschild did) we have to impose

  $$\begin{aligned} f_4f_1f_2^2=1 \end{aligned}$$

  (46)

  The interior solution turns out to be

  $$\begin{aligned} ds^2= & {} \left( \frac{3\cos \chi _a-\cos \chi }{2}\right) ^2dt^2-\frac{3}{\kappa ^2\rho _0+\Lambda }\nonumber \\{} & {} \times \left[ d\chi ^2+\sin ^2\chi \Big [\frac{dx_2^2}{1-x_2^2}+(1-x_2^2)dx_3^2\Big ]\right] \nonumber \\ \end{aligned}$$

  (47)

  whereas the exterior solution reads

  $$\begin{aligned} ds^2= & {} \Big [1-\frac{r_s}{\bar{r}}\Big ]d\bar{t}^2-\frac{d\bar{r}^2}{\Big [1-\frac{r_s}{\bar{r}}\Big ]}\nonumber \\{} & {} -\bar{r}^2\Big [\frac{dx_2^2}{1-x_2^2}+(1-x_2^2)dx_3^2\Big ] \end{aligned}$$

  (48)

  Matching both solutions, at the surface of the sphere $\bar{r}_a$, leads to

  $$\begin{aligned} \cos ^2\chi _a=1-\frac{r_s}{\bar{r}_a} \end{aligned}$$

  (49)

  where

  $$\begin{aligned} \bar{r}_a=\sqrt{\frac{3}{\kappa ^2\rho _0}}\sin \chi _a \end{aligned}$$

  (50)

  The total mass of our sphere will be

  $$\begin{aligned} M=\frac{3}{4\kappa ^2}\sqrt{\frac{3}{\kappa ^2\rho _0}}\left( \chi _a-\frac{1}{2}\sin 2\chi _a\right) \end{aligned}$$

  (51)

• Assume now that the collapsing cloud is made of dust ($p=0$), with synchronous metric [11]

  $$\begin{aligned} ds^2=dt^2-f(r)S^2(t)dr^2-S^2(t)r^2d\Omega _2^2 \end{aligned}$$

  (52)

  We have

  $$\begin{aligned}{} & {} \theta =3 {\dot{S}\over S}\nonumber \\{} & {} \dot{u}^\lambda =0 \end{aligned}$$

  (53)

  Conservation of the energy-momentum tensor implies that

  $$\begin{aligned} \rho (t)=\rho (0)S^{-3}(t) \end{aligned}$$

  (54)

  The interior solution, valid for $r\le a$, reads

  $$\begin{aligned} ds^2=dt^2-S^2(t)\Big [\frac{dr^2}{1-\lambda r^2}+r^2d\Omega ^2_2\Big ] \end{aligned}$$

  (55)

  where S(t) stands for

  $$\begin{aligned}{} & {} S[\psi [t]]=\frac{1}{2}\left( 1+\cos [\psi [t]]\right) \nonumber \\{} & {} t[\psi ]=\left( \frac{\psi +\sin \psi }{2\sqrt{C}}\right) \end{aligned}$$

  (56)

  with $C=\frac{\kappa ^2\rho _0}{3}$ The metric outside the sphere $r\ge a$ to be matched with this interior one must be the usual Schwarzschild metric owing to Birkhoff’s theorem

  $$\begin{aligned} ds^2=\Big [1-\frac{r_s}{\bar{r}}\Big ]d\bar{t}^2-\frac{d\bar{r}^2}{\Big [1-\frac{r_s}{\bar{r}}\Big ]}-\bar{r}^2d\bar{\Omega }^2 \end{aligned}$$

  (57)

  The actual matching of both solutions, in the surface of the sphere $\bar{r}=S(t)a$, leads to

  $$\begin{aligned} M=\rho _0\frac{4\pi }{3}a^3 \end{aligned}$$

  (58)

  This metric can be easily written in the unimodular gauge by using new coordinates x and $\tau$ such that the dependence between new and old coordinates reads r(x) and $t(\tau )$. Then $dr=r'(x)dx$ and $dt=t'(\tau )d\tau$ the metric becomes

  $$\begin{aligned} ds^2= & {} \left[ t'(\tau )\right] ^2d\tau ^2\nonumber \\{} & {} -S^2[t(\tau )]\Big [\frac{\left[ r'(x)\right] ^2}{1-\lambda r^2(x)}dx^2+r^2(x)d\Omega ^2\Big ] \end{aligned}$$

  (59)

  with the unimodular condition

  $$\begin{aligned} 1=\left[ t'(\tau )\right] ^2S^6[t(\tau )]\frac{\left[ r'(x)\right] ^2}{1-\lambda r^2(x)} r^4(x) \end{aligned}$$

  (60)

  we split in two equations

  $$\begin{aligned}{} & {} \left[ t'(\tau )\right] ^2S^6[t(\tau )]=C\nonumber \\{} & {} \frac{\left[ r'(x)\right] ^2}{1-\lambda r^2(x)} r^4(x)=\frac{1}{C} \end{aligned}$$

  (61)

  Therefore the unimodular metric, with $C=1$, is

  $$\begin{aligned} ds^2=\frac{d\tau ^2}{S^6[t(\tau )]}-S^2[t(\tau )]\Big [\frac{dx^2}{r^4(x)}+r^2(x)d\Omega ^2\Big ] \end{aligned}$$

  (62)

  It should be noted that the trace of Einstein’s equations

  $$\begin{aligned} R+\kappa ^2 \rho =0 \end{aligned}$$

  (63)

  is automatically enforced at all points, so the density at the surface (the point where the interior solution should be matched with the exterior solution) is not arbitrary, but determined by the geometry through the trace equation, as we have pointed out already.

3.2 Unimodular presureless collapse

After this short review, let us repeat the analysis in UG using the same simplifying hypothesis on the matter source as before.

First, we assume $\rho (r,t)$ and $p(r,t)=0$ and the unimodular metric to be

$$\begin{aligned} ds^2= & {} A^2(r,t)dt^2-B^2(r,t)dr^2\nonumber \\{} & {} -\frac{1}{A(r,t)B(r,t)}\Big [\frac{d\psi ^2}{1-\psi ^2}+(1-\psi ^2)d\phi ^2\Big ] \end{aligned}$$

(64)

In this case, the conservation of energy-momentum tensor reduces to

$$\begin{aligned} \nabla _\mu T^{\mu \nu }=u^\nu u^\mu \nabla _\mu \rho +\rho \left[ u^\nu \nabla _\mu u^\mu +u^\mu \nabla _\mu u^\nu \right] \end{aligned}$$

(65)

Bianchi identity yields

$$\begin{aligned} R+\kappa ^2\rho =\text {constant} \end{aligned}$$

(66)

Later on, this constant will be identified with $-4\Lambda$.

Assume again a separable solution [10].

$$\begin{aligned}{} & {} A(r,t)=\frac{1}{S^3[t(\tau )]}\nonumber \\{} & {} B(r,t)=\frac{S[t(\tau )]}{b^2(r)} \end{aligned}$$

(67)

that is, the metric is

$$\begin{aligned} ds^2= & {} \frac{d\tau ^2}{S^6[t(\tau )]}-\frac{S^2[t(\tau )]}{b^4(r)}dr^2\nonumber \\{} & {} -b^2(r)S^2[t(\tau )] \Big [\frac{d\psi ^2}{1-\psi ^2}+(1-\psi ^2)d\phi ^2\Big ] \end{aligned}$$

(68)

If we combine the equation of motion and the Bianchi identity, we obtain

$$\begin{aligned}{} & {} S^2[t(\tau )]\Big \{\Lambda +\frac{\kappa ^2\rho _0}{S^3[t(\tau )]}-3S^4[t(\tau )][S'[t(\tau )]]^2 \left[ t'(\tau )\right] ^2\Big \}\nonumber \\{} & {} \quad =\frac{3}{b^2(r)}\left[ 1-b^4(r)[b'(r)]^2\right] \end{aligned}$$

(69)

where we denote $S'[t(\tau )]=\frac{S[t(\tau )]}{dt}$ and the relation

$$\begin{aligned} 1+b^4(r)[b'(r)]^2+b^5(r)b''(r)=0 \end{aligned}$$

(70)

From the unimodular condition (61) it follows

$$\begin{aligned} \Lambda S^2[t]+\frac{\kappa ^2\rho _0}{S[t]}-3[S'[t]]^2=\kappa ^2\rho _0 \end{aligned}$$

(71)

It is plain that in the case $\Lambda =0$, we recover the GR solution.

It is easy to find a formal solution of

$$\begin{aligned} \Lambda S^2[\psi [t]]+\frac{\kappa ^2\rho _0}{S[\psi [t]]}-3[S'[\psi [t]]]^2=\kappa ^2\rho _0 \end{aligned}$$

(72)

namely

$$\begin{aligned} S[\psi [t]]=\frac{1}{2}\left( 1+\cos [\psi [t]]\right) \end{aligned}$$

(73)

and

$$\begin{aligned}{}[\psi '[t]]^2=\frac{\kappa ^2\rho _0}{3}\sec ^4\left[ \frac{\psi [t]}{2}\right] +\frac{\Lambda }{3}\cot ^2\left[ \frac{\psi [t]}{2}\right] \end{aligned}$$

(74)

whose implicit solution reads

$$\begin{aligned} t=\pm \sqrt{3}\int _1^{\psi }dx\frac{\sqrt{2+\cos [x]-2\cos [2x]-\cos [3x]}}{\sqrt{16\kappa ^2\rho _0 \left( 1-\cos [x]\right) +\Lambda \left( 10+15\cos [x]+6 \cos [2x]+\cos [3x]\right) }} \end{aligned}$$

(75)

On the other hand, the proper energy density

$$\begin{aligned} \rho =\frac{\rho _0}{S^3[\psi [t]]} \end{aligned}$$

(76)

Therefore when $\psi =\pi$ the density $\rho$ diverges.

3.2.1 Expansion in ${\Lambda \over \kappa ^2 \rho _0}$

Let us expand in ${\Lambda \over \kappa ^2 \rho _0}$ in order to get a grasp of the physical properties of our UG solution

$$\begin{aligned} \Lambda S^2[t]+\frac{\kappa ^2\rho _0}{S[t]}-3[S'[t]]^2=\kappa ^2\rho _0 \end{aligned}$$

(77)

with

$$\begin{aligned} S[t]=S_0[t]+\Lambda f[t] \end{aligned}$$

(78)

Introducing in (77) the GR solution which is valid at order zero in the expansion,

$$\begin{aligned}{}[S'_0[t]]^2=\frac{\kappa ^2\rho _0}{3}\Big [\frac{1}{S_0[t]}-1\Big ] \end{aligned}$$

(79)

At first order,

$$\begin{aligned} S_0^2[t]-\frac{\kappa ^2\rho _0 f[t]}{S_0^2[t]}-6f'[t]S_0'[t]=0 \end{aligned}$$

(80)

Let us now perform the change of variables

$$\begin{aligned} S_0^2[\psi [t]]-\frac{\kappa ^2\rho _0 f[\psi [t]]}{S_0^2[\psi [t]]}-6f'[\psi [t]]S_0'[\psi [t]][\psi '[t]]^2=0 \nonumber \\ \end{aligned}$$

(81)

in terms of the the GR solution

$$\begin{aligned}{} & {} S_0[\psi [t]]=\frac{1}{2}\left( 1+\cos [\psi [t]]\right) \nonumber \\{} & {} t[\psi ]=\left( \frac{\psi +\sin \psi }{2\sqrt{\frac{\kappa ^{2}\rho _{0}}{3}}}\right) \end{aligned}$$

(82)

Substituting (82) into (81) we get:

$$\begin{aligned} \cos ^8\left[ \frac{\psi [t]}{2}\right] -\kappa ^2\rho _0 \Big (f[\psi [t]]-\sin [\psi [t]]f'[\psi [t]]\Big )=0 \nonumber \\ \end{aligned}$$

(83)

where $\psi [t]$ is given by (82). Then

$$\begin{aligned} f[\psi [t]]= & {} C_1\tan \left( \frac{\psi [t]}{2}\right) +\frac{560 t + 512 \cot (\tfrac{\psi [t]}{2}) + 376 \sin (\psi [t]) + 40 \sin (2 \psi [t]) + \tfrac{8}{3} \sin (3 \psi [t])}{512 \kappa ^2 \rho _0 \cot (\tfrac{\psi [t]}{2})} \end{aligned}$$

(84)

The proper energy density is given by

$$\begin{aligned} \rho= & {} \frac{\rho _0}{S^3(\psi [t])}=\frac{\rho _0}{\left( S_0(\psi [t])+\Lambda f(\psi [t])\right) ^3}\nonumber \\= & {} \frac{\rho _0}{S_0^3(\psi [t])} -\frac{3\rho _0f(\psi [t])}{S_0^4(\psi [t])}\Lambda +\mathcal {O}(\Lambda ^2) \end{aligned}$$

(85)

in such a way that when $\psi =\pi$, using (82)

$$\begin{aligned} T=\frac{\pi }{2}\sqrt{\frac{3}{\kappa ^{2}\rho _{0}}} \end{aligned}$$

(86)

the density $\rho$ diverges.

3.2.2 Matching with the exterior solution

The generalized Birkhoff’s theorem ensures that the metric outside the sphere $r=a$ must be the usual Schwarzschild-(anti)de Sitter

$$\begin{aligned} ds^2=\Big [1-\frac{r_s}{\bar{r}}-\frac{\bar{r}^2\Lambda }{3}\Big ]d\bar{t}^2 -\frac{d\bar{r}^2}{\Big [1-\frac{r_s}{\bar{r}}-\frac{\bar{r}^2\Lambda }{3}\Big ]}-\bar{r}^2d\bar{\Omega }^2 \nonumber \\ \end{aligned}$$

(87)

Inside the sphere $r\le a$ we have

$$\begin{aligned} ds^2= & {} \frac{d\tau ^2}{S^6[t(\tau )]}-\frac{S^2[t(\tau )]}{b^4(r)}dr^2\nonumber \\{} & {} -b^2(r)S^2[t(\tau )]\Big [\frac{d\psi ^2}{1-\psi ^2}+(1-\psi ^2)d\phi ^2\Big ] \end{aligned}$$

(88)

In order to match both solutions at the surface, is necessary to redefine $\bar{r}=S[t(\tau )]b(r)$. After some simple algebra, one gets

$$\begin{aligned} ds^2= & {} \Big [1-\frac{\kappa ^2\rho _0}{3\bar{r}}b^3(a)-\frac{\bar{r}^2\Lambda }{3}\Big ]d\bar{t}^2\nonumber \\{} & {} -\frac{d\bar{r}^2}{\Big [1-\frac{\kappa ^2\rho _0}{3\bar{r}}b^3(a)-\frac{\bar{r}^2\Lambda }{3}\Big ]}-\bar{r}^2d\bar{\Omega }^2 \end{aligned}$$

(89)

where $r_s=2MG$ and the mass is given by

$$\begin{aligned} M=\rho _0\frac{4\pi }{3} b^3(a) \end{aligned}$$

(90)

3.3 UG with incompressible fluid (constant density)

In this final section, we assume that the density is constant, $\rho (r,t)=\rho _0$ whereas the pressure is to be calculated p(r, t). We further assume that the unimodular metric only depends on the radial coordinate

$$\begin{aligned} ds^2= & {} A^2(r) dt^2-B^2(r) dr^2-r^2 \nonumber \\{} & {} -{1\over A(r)B(r)}\Big [\frac{d\psi ^2}{1-\psi ^2}+(1-\psi ^2)d\phi ^2\Big ] \end{aligned}$$

(91)

In this case, the conservation of energy-momentum tensor reduces to

$$\begin{aligned} \nabla _\mu T^{\mu \nu }=u^\nu u^\mu \nabla _\mu p +\left( \rho _0+p\right) u^\mu \nabla _\mu u^\nu -\nabla ^\nu p =0 \end{aligned}$$

(92)

because $\theta =0$. The timelike components is identically satisfied. The radial component implies

$$\begin{aligned} \rho _0+p(r,t)=\frac{\gamma }{A(r)} \end{aligned}$$

(93)

where $\gamma$ is a constant.

The equations of motion read

$$\begin{aligned} \hat{R}_{\mu \nu }-\frac{1}{4}g_{\mu \nu } \hat{R}=\kappa ^2\left( T_{\mu \nu }-\frac{1}{4}g_{\mu \nu }T\right) \end{aligned}$$

(94)

Bianchi identity implies

$$\begin{aligned} \hat{R}+\kappa ^2\left( \rho _0-3p(r)\right) =-C \end{aligned}$$

(95)

Again, later on this constant will be identified with the CC, $C=4\Lambda$.

Assume now (cf. [9])

$$\begin{aligned}{} & {} A^2[r]\equiv \zeta [r]\eta ^{-1/3}[r]\nonumber \\{} & {} B^2[r]\equiv \frac{1}{\zeta [r]\eta [r]} \end{aligned}$$

(96)

Then the equations of motion reduces to

$$\begin{aligned}{} & {} 2\zeta \eta ''=-3\kappa ^2\gamma \zeta ^{-1/2}\eta ^{1/6}\nonumber \\{} & {} \zeta ''=-2\eta ^{-\frac{5}{3}}+3\kappa ^2\gamma \eta ^{-\frac{5}{6}} \eta ^{-\frac{1}{2}}+\frac{5}{3}\zeta \eta ^{-1}\eta '' \end{aligned}$$

(97)

and the Bianchi identity, with the expression of $\zeta ''$ (97)

$$\begin{aligned} 2\zeta \eta ''+\zeta '\eta '=3\eta ^{-2/3}-3\kappa ^2\rho _0-3\Lambda \end{aligned}$$

(98)

To summarize, we can express the interior solution like

$$\begin{aligned} ds^2= & {} \left( \frac{3\cos \chi _a-\cos \chi }{2}\right) ^2dt^2-\frac{3}{\kappa ^2\rho _0+\Lambda }\nonumber \\{} & {} \times \left[ d\chi ^2+\sin ^2\chi \left( \frac{d\psi ^2}{1-\psi ^2}+(1-\psi ^2)d\phi ^2\right) \right] \end{aligned}$$

(99)

where

$$\begin{aligned}{} & {} \eta \equiv \left( \frac{3}{\kappa ^2\rho _0+\Lambda }\right) ^{3/2} \sin ^3\chi \nonumber \\{} & {} \zeta =\sqrt{\frac{3}{\kappa ^2\rho _0+\Lambda }} \sin \chi \left( \frac{3\cos \chi _a-\cos \chi }{2}\right) ^2 \end{aligned}$$

(100)

It is plain that when $\Lambda =0$ we go back to Schwarzschild’s spacetime.

3.3.1 Matching with the exterior solution

The metric outside the sphere must still be the usual Schwarzschild-(anti)de Sitter

$$\begin{aligned} ds^2=\Big [1-\frac{r_s}{\bar{r}}-\frac{\bar{r}^2\Lambda }{3}\Big ]d\bar{t}^2 -\frac{d\bar{r}^2}{\Big [1-\frac{r_s}{\bar{r}}-\frac{\bar{r}^2\Lambda }{3}\Big ]}-\bar{r}^2d\bar{\Omega }^2 \nonumber \\ \end{aligned}$$

(101)

Inside the sphere we have the solution (99), In order to match solutions at the surface, we need to redefine $\bar{r}^2=\frac{3}{\kappa ^2\rho _0+\Lambda }\sin ^2\chi$, so that

$$\begin{aligned} ds^2= & {} \left( \frac{3\cos \chi _a-\cos \chi }{2}\right) ^2dt^2-\frac{1}{\cos ^2\chi }d\bar{r}^2 -\bar{r}^2\nonumber \\{} & {} \times \Big [\frac{d\psi ^2}{1-\psi ^2}+(1-\psi ^2)d\phi ^2\Big ] \end{aligned}$$

(102)

At the surface of the sphere $\bar{r}_a$

$$\begin{aligned} \cos ^2\chi _a=1-\frac{r_s}{\bar{r}_a}-\frac{\bar{r}_a^2\Lambda }{3} \end{aligned}$$

(103)

where

$$\begin{aligned} \bar{r}_a=\sqrt{\frac{3}{\kappa ^2\rho _0+\Lambda }}\sin \chi _a \end{aligned}$$

(104)

4 Conclusions

In this paper we have examined the possible origin of the cosmological constant in the context of Unimodular Gravity (UG). The popular belief that UG is just equivalent to General Relativity (GR) in the unimodular gauge is unfortunately not correct, or at least in need of serious nuances.

The main difference with GR lies in the equations of motion. In GR they read

$$\begin{aligned} R_{\mu \nu }-{1\over 2} R g_{\mu \nu }=\kappa ^2 T_{\mu \nu } \end{aligned}$$

(105)

which in the free case reduces to

$$\begin{aligned} R_{\mu \nu }=0 \end{aligned}$$

(106)

In UG only the trace-free part of the equations of motion holds; that is

$$\begin{aligned} R_{\mu \nu }-{1\over 4} R g_{\mu \nu }=\kappa ^2 \left( T_{\mu \nu }-{1\over 4} T g_{\mu \nu }\right) \end{aligned}$$

(107)

which in the free case reduces to

$$\begin{aligned} R_{\mu \nu }-{1\over 4} R g_{\mu \nu }=0 \end{aligned}$$

(108)

which allows for non-vanishing constant curvature solutions. This clearly shows that UG and GR are not equivalent, even in the unimodular gauge for the latter.

What is true instead, is that given a particular solution of UG there is some value of the cosmological constant (CC) such that this metric is a solution of the GR equations with this particular value of the CC.

But the essential point we would like to make in this paper is that this value is not determined by the constant vacuum energy density (in case there is one such), but by boundary conditions in the equations of motion. Those are the ones we spelled out in this paper, both in the cosmological setting and also in some simplifyied models of spherical collapse.

Our hope is to have explained clearly that the boundary (or initial) conditions in the UG equations of motion that give rise to a corresponding GR solution with nonvanishing CC are peculiar to UG; those particular initial conditions would not be admissible in GR. There we would need to put a CC by hand. For example, the initial condition on the cosmological scale factor

$$\begin{aligned} b(t=0)\ne 0 \end{aligned}$$

(109)

is only admissible in UG. The same thing happens with the initial condition

$$\begin{aligned} R+\kappa ^2\rho \ne 0 \end{aligned}$$

(110)

in spherical gravitational collapse. In GR the second member is only allowed to vanish, whereas in UG it can have any real value.