LINEARISED EINTEIN EQUATIONS
Rosemberg Toala Enriquez
We present an approximation method in order to find solutions of Einstein vacuum equations which are close to a given solution Rab = Λgab .
Consider (M, gab (λ)) a one-parameter family of spacetimes, with gab (0) = gab . Let us
assume that the family admits a Taylor expansion in terms of the parameter
(1)
(2)
gab (λ) = gab + λhab + λ2 hab + ...
The first goal is to obtain a similar expansion for the Ricci tensor
(1)
(2)
Rab (λ) = Rab + λRab + λ2 Rab + ...,
(i)
the idea is, provided each Rab satisfies Einstein’s equations for a vacuum then so does Rab (λ).
(i)
(i)
This is also true for a cosmological constant, i.e., we want to solve Rab = Λhab for each i.
In the following we will think of gab as a fixed background metric, we will also fix its associated connection ∇. Recall that the change of the connection induced by gab (λ) is measured
c
by the tensor Cab
(λ) defined by
c
Cab
(λ)ωc = ∇λa ωb − ∇a ωb
or equivalently,
c
Cab
(λ) =
1 cd
g (λ)(∇a gbd (λ) + ∇b gda (λ) − ∇d gab (λ))
2
from these, we can deduce a formula for the Riemann and Ricci tensors
d
e
d
Rabc d (λ) = Rabc d − 2∇[a Cb]c
(λ) + 2Cc[a
(λ)Cb]e
(λ)
(1)
c
2∇[a Cc]b
(λ)
(2)
Rab (λ) = Rab −
+
e
c
2Cb[a
(λ)Cc]e
(λ)
First Order Approximation
Now we proceed to compute explicitly the linear terms, for readability let us write hab =
(1)
hab .
Using the compatibility of the metric with the connection, ∇a gbc = 0, we get the linear
c
terms for Cab
(λ)
1 cd
g (λ)(∇a gbd (λ) + ∇b gda (λ) − ∇d gab (λ))
2
1 cd
g (∇a hbd + ∇b hda − ∇d hab ) + O(λ2 ).
= λ
2
c
(λ) =
Cab
1
c
, hence it
And for the Ricci tensor (note that the last term of equation (2) is quadratic in Cab
contributes to higher order terms) we have
c
(λ) + O(λ2 )
Rab (λ) = Rab − 2∇[a Cc]b
1 cd
= Rab − λ
g (−∇a ∇b hcd + ∇c ∇a hbd + ∇c ∇b hda − ∇c ∇d hab ) + O(λ2 ).
2
If we define h = g cd hcd , the linearised equation can thus be written as
1
(1)
Rab = (−∇a ∇b h + ∇c ∇a hbc + ∇c ∇b hac − ∇c ∇c hab ) = Λhab
2
or equivalently
1
(−∇a ∇b h + ∇a ∇c hbc + ∇b ∇c hac + 2Rc ab d hcd + Ra c hbc + Rb c hac − ∇c ∇c hab )
2
0 = −∇a ∇b h + ∇a ∇c hbc + ∇b ∇c hac + 2Rc ab d hcd − ∇c ∇c hab
Λhab =
where in the last equality we have used the fact that Ra b = Λδa b . Finally, it is worth noticing
that the trace of the last equation is
0 = −∇a ∇a h + ∇a ∇c hac − Rac hac
Λh = −∇a ∇a h + ∇a ∇c hac .
Choice of a gauge
Let ϕλ : M → M be a 1-parameter family of diffeomorphisms generated by the vector field
X , we know that (M, gab (λ)) and (M, ϕ∗λ gab (λ)) are physically equivalent. Moreover
a
ϕ∗λ gab (λ) = ϕ∗λ (gab + λhab + O(λ2 ))
= gab + λ(hab + LX gab ) + O(λ2 )
= gab + λ(hab + ∇a Xb + ∇b Xa ) + O(λ2 ).
Where LX is the Lie derivative. Hence, if we expand
ϕ∗λ gab (λ) = gab + λh0ab + O(λ2 )
We obtain the corresponding expression, h0ab = hab + ∇a Xb + ∇b Xa , relating two physically
equivalent perturbations. It is also convenient to define the bar operator, h̄ab = hab − 21 hgab . It
is straightforward to check
∇b h̄0ab = ∇b h̄ab + ∇b ∇b Xa + Ra b Xb
then, by solving the equation ∇b h̄ab + ∇b ∇b Xa + Ra b Xb = 0 to find Xa , we can set
∇b h̄0ab = 0.
2
(3)
Therefore, the linearised problem is equivalent to solve (after dropping the 0 )
1
∇c ∇c h̄ab − gab ∇c ∇c h̄ − 2Rc ab d h̄cd − Λgab h̄ = 0
2
(4)
together with the gauge condition, ∇b h̄ab = 0. The previous expression can be simplified by
taking the trace
∇c ∇c h̄ − 2∇c ∇c h̄ + 2Rcd h̄cd − 4Λh̄ = 0
∇c ∇c h̄ = 2Λh̄
Hence equation (4) reduces to
∇c ∇c h̄ab = −2(Λgab ∇c ∇c h̄ + Rc ab d h̄cd ).
It is worth noticing that there is still one last gauge freedom given by h0ab = hab + ∇a Xb +
∇b Xa provided that
∇c ∇c Xa + ΛXa = 0
since this condition is necessary and sufficient to preserve equation (3).
Perturbations in Minkowski Spacetime
The simplest case is when the background metric is flat and with some symmetry assumptions we can solve explicitly the linearised Einstein’s equations. In particular we will consider
stationary and axisymmetric perturbations.
Theorem. The metric in spherical coordinates corresponding to a stationary and axisymmetric perturbation of Minkowski spacetime is of the form
ds2 = −(1 − htt )dt2 + (1 + htt )(dr2 + r2 dθ2 + r2 sin2 θdφ2 ) + htφ dtdφ
where
htt =
∞
X
Rl (r)Pl0 (cos θ),
l=0
htφ =
∞
X
r sin θRl (r)Pl1 (cos θ),
l=1
Plm are the Legendre polynomials, Rl (r) = Al rl +
3
Bl
rl+1
and Al , Bl are constants.
Remark. The firsts Legendre polynomials are
P00 (cos θ) = 1
P10 (cos θ) = cos θ
P20 (cos θ) = 3 cos2 θ − 1
P11 (cos θ) = sin θ
P21 (cos θ) = sin θ cos θ
hence the monopole solution, l = 0, for htt represents an approximation to Schwarzschild
spacetime and together with the l = 1 solution for htφ , they correspond to Kerr spacetime.
Proof. Recall that we want to solve the equations
∇c ∇c h̄ab = 0
∇b h̄ab = 0
In Cartesian coordinates the firsts equations decouple and reduce to the wave equation for
each component. Moreover, since they must be time-independent, we get Laplace equation
which we know how to solve in terms of spherical harmonics, hence
h̄µν =
X
Clm Rl (r)eimφ Plm (cos θ)
l,m
for l = 0, 1, 2, ..., −l ≤ m ≤ l and Clm constant. Also, we have hµν = h̄µν − 21 gµν h̄ and for
simplicity we will assume h̄ij = 0 for the spatial components, i, j = x, y, z. Thus, htt = −hii =
1
h̄ and hti = h̄ti , i = x, y, z.
2 tt
Axisymmetry automatically forces m = 0 for the htt component, so we get the desired
expression. Now we proceed to compute the other components in spherical coordinates, we
can check that
htr = sin θ cos φhtx + sin θ sin φhty + cos θhtz
htθ = r cos θ cos φhtx + r cos θ cos φhty − r sin θhtz
htφ = −r sin θ sin φhtx + r sin θ cos φhty
which must be φ−independent. Using this assumption for htφ it can be seen that the only
possibility corresponds to m = ±1 together with a relation between the coefficients. That is,
X
htx =
Al Rl sin φPl1 (cosθ) + Bl Rl (r) cos φPl1 (cosθ)
l
hty =
X
Bl Rl sin φPl1 (cosθ) − Al Rl (r) cos φPl1 (cosθ)
l
4
Plugging in these expressions into the relation for htr (or hrθ ) and using once again the
P∞
l
0
φ−independency we get that htz =
Finally, the gauge condition
l=0 Cl R (r)Pl (cos θ).
∂x htx + ∂y hty + ∂z htz = 0 implies (after a quite lengthy computation) Bl = Cl = 0. Therefore
we obtain the expressions for htt , htφ and htr = htθ = 0. References
[1] Damour, T., The problem of motion in Newtonian and Einsteinian gravity, in Hawking,
S.W. and Israel, W., eds., Three Hundred Years of Gravitation, pp. 128198, Cambridge
University Press, Cambridge, 1987.
[2] Einstein A., Infeld L., Hoffmann B. The gravitational equations and the problem of motion. Annals of Mathematics 39, No. 1, pp. 65-100, January 1938.
[3] Hawking S.W., Israel W. Three Hundred Years of gravitation. Cambridge University
Press, 1989.
[4] Hawking S.W., Israel W. General Relativity: An Einstein Centenary Survey. Cambridge
University Press, 2010.
[5] Misner C.W., Thorne K.S., Wheeler J.A. Gravitation. W.H. Freeman and Company,
1973.
[6] Wald R.M. General Relativity. U. Chicago Press, 1984.
5