General Relativity: Homework 4 Solutions - Department of Physics ...

General Relativity: Homework 4 Solutions
1. Carroll Problem 4.1
The Lagrange density for electromagnetism in curved space is
where J µ is the conserved current.
L = √ −g ( − 1 4 F µν F µν + A µ J µ)
a) Derive the energy-momentum tensor by functional differentiation with respect to the metric. You can assume
that the A µ J µ term does not contribute to the energy momentum tensor.
b) Consider adding a new term to the Lagrangian,
L ′ = βR µν g ρσ F µρ F νσ
How are Maxwell’s equations altered in the presence of this term? Einstein’s equation? Is the current still
conserved?
Solution:
a) To start let’s make all of the metric dependence of L explicit. In particular the trace of the field strength
depends on the metric so let’s lower all of the indices and drop the piece associated with the conserved current
because of the assumption stated by the problem.
L = − 1 4√ −g g λµ g ρν F λρ F µν
Now perform the variation of the metric, g µν ↦→ g µν + δg µν . Under such variations we have that the square
root of the determinant varies as
√ √ 1√ −g ↦→ −g − −g gµν δg µν
2
Now make the corresponding changes in the Lagrangian
L + δL = − 1 4 (√ −g − 1 2√ −g gµν δg µν )(g λµ + δg λµ )(g ρν + δg ρν )F λρ F µν
Now expand all of the factors out, keeping terms only linear in the variation
L + δL = − 1 √ ( 1 −g 1 −
4 2 g στ δg στ )( g λµ g ρν + δg λµ g ρν + g λµ δg ρν) F λρ F µν
L + δL = − 1 4√ −g
(
g λµ g ρν + δg λµ g ρν + g λµ δg ρν − 1 2 g στ δg στ g λµ g ρν) F λρ F µν
Now subtract the original Lagrangian from both sides leaving just the variation
δL = − 1 4√ −g
(
δg λµ g ρν + g λµ δg ρν − 1 2 g στ δg στ g λµ g ρν) F λρ F µν
= − 1 √ (
−g δg λµ g ρν F λρ F µν + g λµ δg ρν F λρ F µν − 1 4
2 g στ δg στ g λµ g ρν )
F λρ F µν
1

Now take some care in permuting the indices of F µν in the fist term, remembering that it is antisymmetric,
so we will flip the indices of both factors and raise one of them
δL = − 1 √ (
−g δg λµ F ρλ F ρ µ + δg ρν F µ
4
ρF µν − 1 2 g στ δg στ F µν )
F µν
Since there are no free indices we can relabel any paired indices within a given term. In particular we should
look to match all of the indices of the variation term.
⇒
δL
δg µν = −1 4
δL = − 1 √ (
−g δg µν F ρµ F ρ ν + δg µν F ρ
4
µF ρν − 1 2 g µνδg µν F λρ )
F λρ
√ (
−g Fλµ F λ ν + F λ µF λν − 1 2 g µνF λρ ) 1√ (
F λρ = − −g 2F
λ
4
µ F λν − 1 2 g µνF λρ )
F λρ
So finally we have the energy-momentum tensor
T µν = − 2 √ −g
δL
δg µν = F λ µF λν − 1 4 g µνF λρ F λρ
b) Let’s look at the full Lagrangian (where the field strength tensor has all lower indices) to get an idea of how
the resulting equations of motion might change
L + L ′ = √ −g ( − 1 4 gλµ g ρν F λρ F µν + A µ J µ + βR µν g ρσ F µρ F νσ
)
To see how this new term affects Maxwell’s equations we should focus on the parts containing the field
strength tensor (see Carroll §1.10). Using the freedom to relabel the indices, we will make it so that the field
strength tensors in each term have matching indices, and then factor that term out
L+L ′ = √ −g ( − 1 4 gλµ g ρν F λρ F µν +A µ J µ +βR λµ g ρν F λρ F µν
)
=
√ −g
(
−
1
4 gλµ g ρν +βR λµ g ρν) F λρ F µν + √ −gA µ J µ
Now to derive Maxwell’s equations (i.e. the equations of motion for the field A µ ) we should derive the Euler-
Lagrange equations corresponding to variations of A µ and ∇ ν A µ (demanding covariant equations). However
note that the only term dependent upon ∇ ν A µ is the field strength tensor.
δ
δ(∂ σ A τ ) [F δ
λρF µν ] =
δ(∇ σ A τ ) [(∇ λA ρ − ∇ ρ A λ )(∇ µ A ν − ∇ ν A µ )]
= (δ σ λδ τ ρ − δ σ ρ δ τ λ)(∇ µ A ν − ∇ ν A µ ) + (∇ λ A ρ − ∇ ρ A λ )(δ σ µδ τ ν − δ σ ν δ τ µ)
= (δ σ λδ τ ρ − δ σ ρ δ τ λ)F µν + F λρ (δ σ µδ τ ν − δ σ ν δ τ µ)
Therefore the variational derivative of the Lagrangian is
δ(L + L ′ )
δ(∂ σ A τ ) = √ −g ( − 1 4 gλµ g ρν + βR λµ g ρν) [(δ σ λδ τ ρ − δ σ ρ δ τ λ)F µν + F λρ (δ σ µδ τ ν − δ σ ν δ τ µ)]
= √ −g ( F τσ + 2β(R σµ g τν − R τµ g σν ) √ (
)F µν = −g F τσ + 2βR µ[τ F σ] )
µ
We can now write out Maxwell’s equation
(
∇ µ F νµ + 2βR λ[ν F µ] )
λ = J
ν
A quick check shows that we recover the usual equation when β goes to zero. Because mixed partial derivatives
commute and the new term is antisymmetric in µ and ν, we see that the current is still conserved locally ,
that is ∂ ν J ν = 0, however because covariant derivatives do not commute, the current is not conserved globally ,
that is ∇ ν J ν ≠ 0.
2

Now we must consider variations of the metric to see how this new term affects Einstein’s equation. Let’s
consider the terms that will vary.
L ′ + δL ′ = β( √ −g + δ √ −g)(R µν + δR µν )(g ρσ + δg ρσ )F µρ F νσ
= √ −gβF µρ F νσ (R µν g ρσ + g ρσ δR µν + R µν δg ρσ − 1 2 Rµν g ρσ g λγ δg λγ )
⇒ δL ′ = √ −gβF µρ F νσ (g ρσ δR µν + R µν δg ρσ − 1 2 Rµν g ρσ g λγ δg λγ )
We must now figure out δR µν in terms of the variation of the metric (note that the argument in Carroll that
this variation disappears, fails in this case because we are not contracting with g µν ). Recall that
R µν = g µα g νγ R λ αλγ
and so the variation is
δR µν = g µα g νγ δR λ αλγ
Now let’s pluck the variation of the Riemann tensor and Christoffel symbols from Carroll (eqs. 4.62 and 4.64)
δR λ αλγ = ∇ λ (δΓ λ γα) − ∇ γ (δΓ λ λα)
δΓ λ γα = − 1 2 [g ρα∇ γ (δg ρλ ) + g ργ ∇ α (δg ρλ ) − g ασ g γτ ∇ λ (δg στ )]
Putting this into the expression for the variation of the Riemann tensor gives us
δR λ αλγ = − 1 2 [g ργ∇ λ ∇ α (δg ρλ ) − g ασ g γτ ∇ λ ∇ λ (δg στ ) + g ασ ∇ λ ∇ γ (δg σλ ) − ∇ γ ∇ α (g ρλ δg ρλ )]
δR λ αλγ = − 1 2 [∇ λ∇ α (δg λ γ) − ∇ γ ∇ α (δg λ λ) + ∇ λ ∇ γ (δg λ α) − g ασ g γτ ∇ λ ∇ λ (δg στ )]
⇒ δR µν = g µα g νγ δR λ αλγ = − 1 2 [∇ λ∇ µ (δg λν ) − ∇ ν ∇ µ (δg λ λ) + ∇ λ ∇ ν (δg λµ ) − δ µ σδ ν τ ∇ λ ∇ λ (δg στ )]
δR µν = − 1 2 [∇ λ∇ µ (δg λν ) − ∇ ν ∇ µ (δg λ λ) + ∇ λ ∇ ν (δg λµ ) − ∇ λ ∇ λ (δg µν )]
Now just focus on the variation of the action due to the variation of the Ricci tensor
∫ ∫
√−gβFµρ
δS R = F νσ g ρσ δR µν = − 1 −gβFµρ F νσ g
2√ ρσ [∇ λ ∇ µ (δg λν )−∇ ν ∇ µ (δg λ λ)+∇ λ ∇ ν (δg λµ )−∇ λ ∇ λ (δg µν )]
Now integrate by parts twice, using the fact that the variation and it’s covariant derivative vanish at the
boundary to arrive at
∫
δS R = − 1 √ [
−gβ ∇ µ ∇ λ (F µρ F νσ g ρσ )δg λν −∇ µ ∇ ν (F µρ F νσ g ρσ )δg λ
2
λ+∇ ν ∇ λ (F µρ F νσ g ρσ )δg λµ −∇ λ ∇ λ (F µρ F νσ g ρσ )δg µν]
Now do some relabelling
∫
δS R = − 1 √ [
−gβ ∇ λ ∇ µ (F λρ F νσ g ρσ )−∇ λ ∇ γ (F λρ F γσ g ρσ )g µν +∇ λ ∇ µ (F νρ F λσ g ρσ )−∇ λ ∇ λ (F µρ F νσ g ρσ ) ] δg µν
2
So now the variation of the new term is
δL ′ = √ (
−gβ − 1 [
∇ λ ∇ µ (F λρ F νσ g ρσ ) − ∇ λ ∇ γ (F λρ F γσ g ρσ )g µν + ∇ λ ∇ µ (F νρ F λσ g ρσ ) − ∇ λ ∇ λ (F µρ F νσ g ρσ ) ]
2
+F µρ F νσ R ρσ − 1 2 Rλγ g ρσ F λρ F γσ g µν
)
δg µν
3

and Einstein’s equation now reads
(
R µν − 1 2 Rg µν + β − 1 ∇
2[ λ ∇ µ (F λρ F νσ g ρσ ) − ∇ λ ∇ γ (F λρ F γσ g ρσ )g µν + ∇ λ ∇ µ (F νρ F λσ g ρσ )
−∇ λ ∇ λ (F µρ F νσ g ρσ ) ] + F µρ F νσ R ρσ − 1 2 Rλγ g ρσ F λρ F γσ g µν
)
= T µν
where T µν is the energy momentum tensor associated with the electromagnetic field derived in part a.
2.
The point of this exercise is to compute the first correction to the Newtonian solution caused by a source that
rotates. In Newtonian gravity, the angular momentum of the source does not affect the field: two sources with
identical densities, ρ(x i ), but different angular momenta have the same gravitational field. This is not so in General
Relativity, since all components of T µν generate the field. This is an example of a post-Newtonian effect.
a) Suppose a spherical body of uniform density ρ and radius R rotates rigidly about the x 3 axis with constant
angular velocity ω. Write down the components T 0ν in a Lorentz frame at rest with respect to the center of
mass of the body, assuming ρ, ω, and R are independent of time. For each component, work to the lowest
non-vanishing order in ωR.
b) The general solution to the equation ∇ 2 f = g, that vanishes at infinity, is the generalization of Φ(r) = −Gm/r,
f(x) = − 1 ∫
4π
g(y)
|x − y| d3 y
Use this to solve Einstein’s Equation, keeping only terms linear in the perturbation, for the components of
the trace-reversed perturbation h 00 and h 0i , where h µν = h µν − 1 2 ηµν h, for the source described in part (a).
Obtain solutions only outside the body and only to the lowest non-vanishing order in r −1 , where r is the
distance from the center of the body. Express the result for h 0i in terms of ω. Find the metric tensor within
this approximation and transform it to spherical coordinates.
c) Since the metric is independent of t and the azimuthal angle, φ, particles orbiting this body will have constant
p 0 and p φ along their trajectories. Consider a particle of non-zero rest mass in a circular orbit of radius r in
the equitorial plane. To lowest order in ω, calculate the difference between its orbital period when it’s orbiting
with the direction of the body’s rotation (prograde) and when it’s orbiting against the direction of the body’s
rotation (retrograde). (Define the period to be the coordinate time taken for one orbit of ∆φ = 2π).
Solution:
a) Given that we are doing a small perturbation to the Newtonian solution, we should assume that the rotation
of the body is non-relativistic (ωR

1
from this we can determine the factor γ = √ , where we will use the unperturbed metric η
dx µ dx ν
µν to
gµν
dt dt
calculate.
1
⇒ γ =
1 − ω 2 r 2 ≈ 1 + ω2 r 2
Now we can write out the 4-velocity
U µ = dxµ
dτ
= γ dxµ
dt
= γ(1, −ωr sin ωt sin θ, ωr cos ωt sin θ, 0)
which to first order in ωR is just dxµ
dt
(i.e. γ = 1). Now we can calculate the components T 0ν for the
energy-momentum tensor (for r = √ x 2 + y 2 + z 2 ≤ R):
T 00 = ρU 0 U 0 = ρ
T 01 = ρU 0 U 1 = −ρωy
T 02 = ρU 0 U 2 = ρωx
T 03 = ρU 0 U 3 = 0
b) We will now use Einstein’s equation (with upper indices) to solve for the perturbation to the metric.
R µν − 1 2 Rgµν = 8πGT µν
where
g µν = η µν − h µν
We will be dealing with the case where |h µν |

equations, we are free to choose our coordinates. There are four coordinates, so we can equivalently enforce
four conditions on the solution to our metric. Before we decide what conditions to impose, let’s write the
Einstein tensor in terms of the trace-reversed perturbation h µν = h µν − 1 2 hηµν
G µν = 1 (
∂ µ ∂ λ (h νλ + 1 2
2 hηνλ )+∂ ν ∂ λ (h µλ + 1 2 hηµλ )−∂ λ ∂ λ (h µν + 1 2 hηµν )−∂ µ ∂ ν h+η µν ∂ λ ∂ λ h−η µν ∂ λ ∂ ρ (h λρ + 1 2 hηλρ ) )
which simplifies to
G µν = 1 (
∂ µ ∂ λ h νλ + ∂ ν ∂ λ h µλ − η µν ∂ λ ∂ ρ h λρ − ∂ λ ∂ λ h µν )
2
Now we can decide what conditions we would like to put on the perturbation terms in order to simplify the
calculations. Noting that the four-divergence of the perturbation comes up in three of the four terms, if we
enforced the four equations
∂ λ h µλ = 0
then, after neglecting the time derivatives, we are led to the tidy expression
G µν = − 1 2 ∇2 h µν
and now Einstein’s equation reads
− 1 2 ∇2 h µν = 8πGT µν
The only non-zero terms in the energy-momentum tensor are those with at least one time component. Therefore
we can set h ij = 0 for i, j = 1, 2, 3.
Consider the first equation
Let’s use the general form of the solution to solve
∫
h 00 (r) = 4Gρ
∇ 2 h 00 = −16πGρ
1
|r − r ′ | d3 r ′ ≈ 4Gρ 1 ∫
r
d 3 r ′ = 16πGρR3
3r
= 4GM
r
1
where we have just taken the first order term of the Taylor expansion
|r−r ′ | = 1 r′·ˆr
r
(1 +
r
+ ...) for r >> r ′ .
Now consider
∇ 2 h 01 = −16πGρωy
the solution to which is
∫
h 01 y ′ d 3 r ′
(r, θ, φ) = 4Gρω
|r − r ′ |
1
where the integral is taken over the source matter. Now expand
|r−r ′ |
to the first two orders, this will be
necessary because we will see that the first order term will be zero.
∫
h 01 (r, θ, φ) = 4Gρω
( 1 r + r′ · ˆr
r 2 )y′ d 3 r ′ = 4Gρω
∫ ( 1
r y′ + r′ · ˆr
r 2 y′) d 3 r ′
The source matter has a spherical distribution, and as such the region of integration is symmetric under sign
changes of the coordinates. We may then see that the first term in the integral, being odd in y ′ , will yield no
contribution.
h 01 (r, θ, φ) = 4Gρω 1 ∫
r 2 (r ′ · ˆr)y ′ d 3 r ′ = 4Gρω 1 ∫
r 2 (x ′ cos φ sin θ + y ′ sin φ sin θ + z ′ cos θ)y ′ d 3 r ′
h 01 (r, θ, φ) = 4Gρω sin θ sin φ < y2 >
r 2
6

where we have used the notation < y 2 >= ∫ y ′2 d 3 y ′ , to denote the average of the square of the y-coordinate
on the sphere and that ˆr = (cos φ sin θ, sin φ sin θ, cos θ). We have also used symmetry to deduce that the x ′ y ′
and z ′ y ′ terms in the integrand will not contribute. Similarly we deduce
We also know that because T 03 vanishes, h 03 = 0.
h 02 (r, θ, φ) = − 4Gρω sin θ cos φ < x2 >
r 2
Now let’s translate back to the metric perturbations, which gives us the following equations
h µµ = h µµ − h
h 01 = h 01
h 02 = h 02
h 03 = h 03
h = −h 00 + h 11 + h 22 + h 33
We can see readily that
h 01 = 4Gρω sin θ sin φ < y2 >
r 2
h 02 = − 4Gρω sin θ cos φ < x2 >
r 2
h 03 = 0
Because h ii = 0 for i = 1, 2, 3 we have that h ii = 1 2 h (which implies all hµµ are equal) so we can write the
trace entirely in terms of h 00 h = 2h 00
and we have h 00 = h 00 − 1 2 (2h00 ) which implies
h µµ = 1 2 h00 = 2GM
r
Now let’s transform from Cartesian to spherical coordinates using the transformation law for (0, 2)-tensors.
h 0r = ∂x
∂r h 01 + ∂y
∂r h 02 = (cos φ sin θ) ( −4Gρω sin θ sin φ < y 2 >
r 2 )
+ (sin φ sin θ)
(4Gρω sin θ cos φ < x 2 >
r 2 )
= 0
where we’ve used that by the spherical symmetry of the source distribution < x 2 >=< y 2 >.
h 0φ = ∂x
∂φ h 01 + ∂y
∂φ h 02
h 0φ = (−r sin φ sin θ) ( −4Gρω sin θ sin φ < y 2 > ) (4Gρω sin θ cos φ < x 2 > )
+ (r cos φ sin θ)
r 2 r 2
h 0φ = 4Gρω sin2 θ(< x 2 > + < y 2 >)
r
While they are equal we have kept the distinction between < x 2 > and < y 2 > so that we may make an
observation. In particular the combination ρ(< x 2 > + < y 2 >) = ∫ ρ(x ′2 + y ′2 )d 3 r ′ is equal to the moment
of inertia about the z-axis of the sphere (for a uniform sphere this is I = 2 5 MR2 = 8π 5 ρR5 ). We may then
rewrite this component as
h 0φ = 4G sin2 θL
r
where L = Iω is the angular momentum of the source matter.
Let’s focus on the remaining non-zero components of the metric,
7

h rr = ( ∂x
∂r
) 2h11
+ ( ∂y
∂r
) 2h22
+ ( ∂z ) 2h33
= 8πGρR3
∂r
3r
h φφ = ( ∂x ) 2h11
+ ( ∂y ) 2h22
+ ( ∂z ) 2h33
= 8πGρR3 r sin 2 θ
∂φ ∂φ ∂φ
3
h θθ = ( ∂x) 2h11
+ ( ∂y ) 2h22
+ ( ∂z ) 2h33
= 8πGρR3 r
∂θ ∂θ ∂θ
3
It is easily verified that the remaining off-diagonal terms are zero. Now we can write out the metric as
where Φ = − GM r
ds 2 = −(1 + 2Φ)dt 2 + 4G sin2 θL
(dtdφ + dφdt) + (1 − 2Φ)(dr 2 + r 2 sin 2 θdφ 2 + r 2 dθ 2 )
r
= − 4πGρR3
3r
.
c) A particle in orbit in the equatorial plane will have the trajectory
x µ = (t, r, φ(t), π 2 )
Now consider the r-component of the geodesic equation for such trajectories (i.e. constant r and θ = π/2)
d 2 r
dτ 2 + ( dt ) 2 Γr tt + Γ
r
(dφ) 2 (
dτ φφ + 2Γ
r dt dφ) dτ
φt = 0
dτ dτ
⇒ Γ r ( dt ) 2
tt + Γ
r
(dφ) 2 (
dτ φφ + 2Γ
r dt dφ) dτ
φt = 0
dτ dτ
Because the metric is t and φ independent we see that
Γ r tt = − 1 2 grr ∂ r g tt
Γ r φφ = − 1 2 grr ∂ r g φφ
Γ r φt = − 1 2 grr ∂ r g φt
If we assume that the body is orbiting slowly (so that we may assume t ≈ τ) the geodesic equation may then
be written as
(dφ) 2 (dφ) ∂ r g tt + ∂ r g φφ + 2∂r g φt = 0
dt
dt
Plugging in the metric components we have
−∂ r (1 − 2GM ) + ∂ r (r 2 + 2GMr) ( dφ) 2
+ 2∂r ( 4GL
r
dt
r
)( dφ) = 0
dt
− 2GM
r 2 + 2(r + GM) ( dφ) 2 8GL(dφ) − = 0
dt r 2 dt
which we can rewrite as
r 2 (1 +
r
GM )( dφ) 2 4L(dφ) − − 1 = 0
dt M dt
Recognizing this equation as a quadratic, we solve for dφ
dt
4L
dφ
dt =
M ± √
16L 2
M 2 + 4r 2 (1 + r
2r 2 (1 +
Now we may calculate the difference in the orbital period
∆T = 2π | dφ
dt − | − | dφ
dt + |
| dφ dφ
dt + dt − |
r
GM )
GM )
( 4L )(
M
= 2π
r 2 (1 +
r
GM )
∆T = 8πL
M
= 16π
5 R2 ω
r 2( 1 +
r ) )
GM
Note that when L → 0 the difference in orbital period is zero, as we should expect.
8

3.
In the presence of a cosmological constant Λ, Einstein’s equation is
R µν − 1 2 Rg µν + Λg µν = 8πGT µν
a) Calculate the gravitational potential Φ in the Newtonian limit outside a spherically symmetric point source
with ρ = Mδ 3 (r).
b) A good description of the gravitational forces in our solar system is given by Newton’s potential, Φ =
−GM ⊙ /r, where M ⊙ is the mass of the sun. Use this fact to determine an upper limit for the vacuum energy
density ρ Λ = Λ/8πG, given that the radius of Pluto’s orbit is r P luto ≈ 6 × 10 12 m. Express your answer in
GeV 4 .
Solution:
a) Follow the steps in Carroll §4.2 to derive Poisson’s equation from Einstein’s equation in the Newtonian limit.
∇ 2 Φ = 4πGρ
Now set ρ = Mδ 3 (r) − ρ Λ . The spherical symmetry of the distribution requires that the potential should only
depend on r, the distance from the point source, that is we should look for a solution of the form Φ(r).
Poisson’s equation now reads
∇ 2 Φ = 4πG(Mδ 3 (r) − ρ Λ (r))
Now integrate both sides over the volume of a sphere of radius r ′
∫
∫
∇ 2 Φ dV = 4πG (Mδ 3 (r) − ρ Λ (r))dV
First look at the right hand side and note that we can apply Stokes’ theorem
∫
∫
∫
∇ 2 Φ dV = ∇ · (∇Φ)dV = (∇Φ) · dS
where the integral is now over the surface of the sphere. Due to spherical symmetry ∇Φ must point in the ˆr
direction and be constant on the surface of the sphere, therefore the integral now becomes
∫
∇ 2 Φ dV = 4πr ′2 |∇Φ|
Now we deal with the right hand side:
∫
4πG
(Mδ 3 (r) − ρ Λ (r))dV = 4πGM − 16π2 G
ρ Λ r ′3
3
Putting the two together we have:
∇Φ(r ′ ) = ( GM
r ′2 − 4π 3 Gρ Λr ′)ˆr
Now to solve for Φ we just integrate ∇Φ from the edge of the universe (let’s call it ∞) to r.
Φ(r) = Φ(∞) +
∫ r
∞
∇Φ · dˆr = Φ(∞) +
∫ r
∞
( GM
r ′2 − 4π 3 Gρ Λr ′ )dr ′ = − GM r
− 4π 3 Gρ Λr 2
where we’ve cancelled out the potential at infinity with the evaluation of the anti-derivative at the edge of
the universe.
9

) To find an upper bound on the vacuum energy density, we will find what value is necessary to keep Pluto
bound to the solar system. We can write this as the requirement that the negative gradient of the potential
(i.e. the acceleration) be negative (so that the acceleration is towards the center of the solar system).
Isolating the vacuum energy density we arrive at
−∇Φ(r P luto ) = − GM ⊙
rP 2 + 4π
luto
3 Gρ Λr P luto < 0
ρ Λ < 3
4π
M ⊙
r 3 P luto
= 3 1.989 × 10 30 kg ( 1GeV )(1.97 × 10 −16 m
4π (6 × 10 12 m) 3 1.78 × 10 −27 kg GeV −1 ) 3
ρ Λ < 9.44 × 10 −30 GeV 4 ≈ 10 −29 GeV 4
10