Notes from phz 6607, Special and General Relativity
University of Florida, Fall 2004, Detweiler
Notes on Differential Geometry
These notes are not a substitute in any manner for class lectures. Please
let me know if you find errors.
I.
CHRISTOFFEL SYMBOLS
A covariant derivative operator which is compatible with a metric necessarily satisfies
∇a gbc = 0 = ∂a gbc − Γdab gdc − Γdac gbd ,
(1)
so that
∂a gbc = Γdab gdc + Γdac gbd ,
∂b gca = Γdbc gda + Γdba gcd ,
∂c gab = Γdca gdb + Γdcb gad .
(2)
The second two of these follow from the first with cyclic permutations of the indices. Add
the first two and subtract the third to obtain
∂a gbc + ∂b gca − ∂c gab = 2Γdab gdc ,
(3)
where the symmetry of the Christoffel symbols is used. Now, raising the c index results in
1
Γdab = g dc (∂a gbc + ∂b gca − ∂c gab ) .
2
A.
(4)
Identities involving the Christoffel symbols and covariant derivatives
1 ∂ √
Γaab = √
−g
−g ∂xb
1 ∂ √
a
∇a ξ a = √
−gξ
−g ∂xa
∂
ξb]
∂x[a
√
1 ∂
a
ab ∂ψ
∇a ∇ ψ = √
−gg
−g ∂xa
∂xb
1 ∂ √
∇a F [ab] = √
−gF [ab]
a
−g ∂x
∇[a ξb] =
for Aab = Aba :
g bc ∂gad ad
g ba ∂ √
c
∇a Aab = √
−gA
−
A
a
−g ∂xc
2 ∂xc
1
(5)
(6)
(7)
(8)
(9)
(10)
In a generic coordinate system
√
1
∇2 φ = ∇a ∇a φ = g ab ∂a ∂b φ − g ab Γc ab ∂b φ = √ ∂a
−gg ab ∂b φ .
−g
(11)
It is often convenient to define
√
1
Γc ≡ g ab Γc ab = − √ ∂a
−gg ab ,
−g
(12)
∇2 φ = g ab ∂a ∂b φ − Γc ∂c φ.
(13)
so that
II.
THE RICCI IDENTITY AND THE RIEMANN TENSOR
Let ξa be an arbitrary vector field, and consider
∇a ∇b ξc − ∇b ∇a ξc = ∂a ∂b ξc − Γdbc ξd − Γeab ∂e ξc − Γdec ξd − Γeac ∂b ξe − Γdbe ξd
− ∂b ∂a ξc − Γdac ξd + Γeba ∂e ξc − Γdec ξd + Γebc ∂a ξe − Γdae ξd
= −∂a Γdbc ξd − Γeab ∂e ξc − Γdec ξd − Γeac ∂b ξe − Γdbe ξd
+ ∂b Γdac ξd + Γeba ∂e ξc − Γdec ξd + Γebc ∂a ξe − Γdae ξd
= −∂a Γdbc ξd − Γeac ∂b ξe − Γdbe ξd
+ ∂b Γdac ξd + Γebc ∂a ξe − Γdae ξd
= −∂a Γdbc ξd + Γeac Γdbe ξd
+ ∂b Γdac ξd − Γebc Γdae ξd
= −∂a Γdbc ξd + ∂b Γdac ξd − Γebc Γdae ξd + Γeac Γdbe ξd ,
(14)
where the first equality follows from the description of the covariant derivative in terms of the
Christoffel symbols, the second from the commutation of partial derivatives, the third from
the symmetry of the Christoffel symbol, the fourth from the Libnitz rule for differentiation,
and the fifth by rearranging terms. From the Ricci identity we also have
∇a ∇b ξc − ∇b ∇a ξc = Rabc d ξd .
(15)
With ξa being arbitrary, it is necessary that
Rabc d = −∂a Γdbc + ∂b Γdac − Γdea Γebc + Γdeb Γeac ,
(16)
after a rearrangement of terms and indices.
A.
Algebraic identities of the Riemann tensor
Consider ∇[a ∇b ∇c] ψ for an arbitrary scalar field ψ. From the Ricci identity
1
∇[a ∇b] ∇c ψ = Rabc d ∇d ψ
2
2
(17)
Antisymmetrizing over a, b, and c gives
1
∇[a ∇b ∇c] ψ = R[abc] d ∇d ψ
2
(18)
but ∇[b ∇c] ψ = 0 because the connection Γd bc is symmetric. Because ψ is arbitrary, it follows
that
R[abc] d = 0,
(19)
for any Riemann tensor. A second way of stating this same identity relies upon the antisymmetry of the first two indices of the Riemann tensor and is
Rabc d + Rcab d + Rbca d = 0.
(20)
Consider ∇a ∇b (ξ c λc ), for arbitrary vector fields ξ c and λc . It follows that
∇a ∇b (ξ c λc ) = ∇a (λc ∇b ξ c + ξ c ∇b λc )
= λc (∇a ∇b ξ c ) + (∇a λc ) (∇b ξ c ) + ξ c (∇a ∇b λc ) + (∇a ξ c ) (∇b λc ) .
(21)
Antisymmetrize over a and b: the left hand side is zero because the connection Γd ab is
symmetric; the second and fourth terms on the right hand side cancel each other. Use the
Ricci identity on the third term on the right to obtain
0 = λc ∇[a ∇b] ξ c + ξ c Rabc d λd .
(22)
The vector λc is arbitrary, and we conclude that
∇[a ∇b] ξ c = −Rabc d ξ c .
(23)
Officially, this is the Ricci identity for contravariant vectors. Our interest is usually on a
derivative operator which is compatible with the spacetime metric gab , in which case we can
raise and lower indices on either side of a derivative operator. Thus, it also follows that
∇[a ∇b] ξc = −Rab c d ξc .
(24)
From a straight application of the Ricci identity
∇[a ∇b] ξc = Rabd c ξc .
(25)
Rabcd = Rab[cd] .
(26)
Rabcd = R[ab][cd]
(27)
Rabcd + Rcabd + Rbcad = 0.
(28)
We conclude that
At this point we have shown that
and that
We may now conclude that
Rabcd =
=
=
=
=
−Rcabd − Rbcad
Rcadb + Rbcda
−Rdcab − Radcb − Rdbca − Rcdba
Rdcba + Radbc + Rdbac + Rcdab
Rdcba − Rbadc − Rdbac + Rdbac + Rcdab
3
(29)
where we use Eq. (28) in the first equality, interchange the first two indices of every term
to obtain the second equality, use Eq. (28) on both terms to obtain the third equality,
interchange the last two indices of every term to obtain the fourth equality, and use Eq. (28)
on only the second term to obtain the fifth equality. In the fifth line the third and fourth
terms cancel, and we are left with
Rabcd = Rdcba − Rbadc + Rcdab
= Rcdab − Rabcd + Rcdab ;
(30)
the second line follows from interchanging each pair of indices on the right hand side. This
easily simplifies, and we finally have
R[ab][cd] = Rabcd = Rcdab = R[cd][ab] .
(31)
To simplify the understanding of the independence of these algebraic identities, assume
that Eq. (31) holds for a 4-indexed covariant tensor, Rabcd , and see how much additional
information is learned if Eq. (28) also holds. Eq. (31) immediately implies that Eq. (28)
holds if any two of the indices a, b or c are equal. So nothing new is learned from Eq. (28)
unless a, b and c are all different. But, Eq. (31) also immediately implies that Eq. (28) holds
if d is equal to any of a, b or c. We may conclude that Eq. (28) gives new information,
beyond that contained in Eq. (31), only if a, b, c and d are all distinct.
Now we are in position to count up the number of algebraically independent components
of the Riemann tensor of a four dimensional manifold. An antisymmetric pair of indices [ab]
may be chosen in 6 different ways, so there are 6 ways to chose the [ab] in Rabcd and 6 ways to
pick the [cd]. Also, Rabcd = R[cd][ab] , so Rabcd looks like a 6×6 symmetric matrix, which has 21
algebraically independent components. Eq. (28) provides precisely one more algebraically
independent relationship (when all four indices are distinct), and we conclude that the
Riemann tensor has twenty algebraically independent components on a four dimensional
manifold.
B.
Differential identities of the Riemann tensor
The Bianchi identity is a differential identity of the Riemann tensor. Start with
1
∇a (Rbcd m µm )
2
1
1
= µm ∇a Rbcd m + Rbcd m ∇a µm ,
2
2
∇a ∇[b ∇c] µd =
(32)
and
∇[a ∇b] ∇c µd =
1
1
Rabc m ∇m µd + Rabd m ∇c µm .
2
2
(33)
Now, antisymmetrize each of these over [abc]. The righthand sides of the antisymmetrized
versions each of these equations are equal,
µm ∇[a Rbc]d m + R[bc|d| m ∇a] µm = R[abc] m ∇m µd + R[ab|d| m ∇c] µm
4
(34)
The algebraic identities of the Riemann tensor implies that the first term on the right hand
side is zero and that two of the other terms cancel with the result that
µm ∇[a Rbc]d m = −R[bc|d| m ∇a] µm + R[ab|d| m ∇c] µm
= 0.
(35)
Because µm is arbitrary, we conclude that
∇[a Rbc]d m = 0,
(36)
which is the Bianchi identity.
The contracted form of the Bianchi identity follows from first contracting over a and m,
0 = 3∇[a Rbc]d a = ∇a Rbcd a + ∇c Rabd a + ∇b Rcad a
= ∇a Rbcd a − ∇c Rbd + ∇b Rcd .
(37)
Now contract over b and d,
0 = ∇a Rbc ba − ∇c Rb d + ∇b Rc d
= ∇a Rc a − ∇c R + ∇a Rc a ,
(38)
which may be written as
1
(39)
∇a (Rc a − gc a R) = 0.
2
This last result is also referred to as the Bianchi identity, or sometimes the contracted
Bianchi identity.
III.
THE ALTERNATING TENSOR
In four dimensional space-time
√
abcd = [abcd] = (±1, 0) −g
(40)
depending upon whether (a, b, c, d) is an even, odd or no permutation of (t, x, y, z). Also
1
abcd = [abcd] = −(±1, 0) √ .
−g
(41)
The contractions of the product of the two ’s have simple expressions. First, with no
contractions
4! e [f g h]
[f
h]
[f
e]
h]
[e
h]
abcd ef gh = −4!δa[e δbf δcg δd = −
δa δb δc δd − δaf δb δcg δd − δag δb δce δd − δah δb δcg δd . (42)
4
For the second equality, note that the right hand side is explicitly antisymmetric in [e, f, g, h];
the factor of 14 is the required “normalization.” A similar step is performed at the end of
the following equations.
5
For one contraction,
[f
h]
[a
h]
[f
h]
[f
a]
abcd af gh = −3! δaa δb δcg δd − δaf δb δcg δd − δag δb δca δd − δah δb δcg δd
[f
h]
[f
h]
= −3! × (4 − 3) × δb δcg δd = −3! × 1! × δb δcg δd
h]
f]
h]
= −2! × 1! × δbf δc[g δd − δbg δc[f δd − δbh δc[g δd
(43)
For two contractions,
abcd abgh
= −2! × 1! ×
h]
δbb δc[g δd
−
h]
δbg δc[b δd
−
b]
δbh δc[g δd
h]
h]
= −2! × 1! × (4 − 2)δc[g δd = −2! × 2! × δc[g δd = −1! × 2! × δcg δdh − δch δdg .(44)
For three contractions,
abcd abch = −1! × 2! × δcc δdh − δch δdc = −1! × 2! × (4 − 1)δdh = −1! × 3! × δdh .
(45)
Finally for all four pairs of indices contracted,
abcd abcd = −3! × δdd = −4!
(46)
Summarizing these formulae, with different labeling of the indices, we have
h]
abcd ef gh = −4!δa[e δbf δcg δd
(47)
abcd ef gd = −3! × 1! × δa[e δbf δcg]
(48)
f]
abcd ef cd = −2! × 2! × δa[e δb
abcd ebcd
= −1! × 3! ×
(49)
δae .
(50)
and
abcd abcd = −4!.
IV.
(51)
THE PROJECTION OPERATOR
An observer with four-velocity ua may construct a special spatial alternating tensor
√
(52)
abc = [abc] ≡ abcd ud = (±1, 0)(ut or ut ) −g
And an observer may use a projection operator ha b to project tensor indices perpendicular
to his four-velocity,
ha b ≡ g a b + ua ub .
(53)
Note that
ha b ub = (g a b + ua ub )ub = ub − ub = 0,
(54)
b
where the second equality follows from the normalization of the four-velocity, ub u = −1. It
easily follows that
ha b hb c = ha c ,
(55)
as would be expected for a projection operator.
Any tensor index which is perpendicular to ua may be raised or lowered by either gab and
g ab or by hab and hab .
If a normalized, timelike vector field ua is hypersurface orthogonal, then the projection
operator hab also plays the role of the metric of the three dimensional, spatial hyper-surface
which is perpendicular to ua .
6
V.
ELECTRICITY AND MAGNETISM
Maxwell’s equations for the electromagnetic field F ab = F [ab] with a source J a :
∇[a F bc] = 0
(56)
∇b F ab = 4πJ a .
(57)
and
For a given observer with four-velocity ua , the electromagnetic field may be decomposed
into its electric E a and magnetic B a parts by projecting F ab parallel and perpendicular to
ua ,
E a = F ab ub
(58)
and
1
(59)
Ba = abcd F bc ud .
2
It is easy to show that E a ua = 0 and that B a ua = 0 so the electric and magnetic fields are
spatial vectors to the observer, ua .
We may directly write Fab in terms of its components as
Fab = 2u[a Eb] + abcd B c ud ;
(60)
this may be verified by substituting this expression into the above equations for E a and B a .
The force on a charged particle of charge q and mass m moving with four-velocity v a is
the right hand side of
mv b ∇b v a = qF ab vb ,
(61)
which is the equation of motion of a charged particle in “free-fall” through an electromagnetic
field.
VI.
MAXWELL’S EQUATIONS
Maxwell’s equations for the electromagnetic field F ab = F [ab] with a source J a :
∇[a F bc] = 0
(62)
∇b F ab = 4πJ a .
(63)
and
Imagine a cloud of charged dust moving through spacetime with four-velocity v a and
co-moving number density n. Each bit of dust has a mass m and charge q. The conservation
of dust implies that
∇a (nv a ) = 0.
(64)
An observer with four-velocity ua sees a charge density −qnua v a and current density qnha b v b .
To see that Maxwell’s equations require the conservation of charge, evaluate
∇a ∇b Fcd − ∇b ∇a Fcd = Rabc e Fed + Rabd e Fce
7
(65)
from the Ricci identity. Now contract Eq. (65) with g ac and g bd and use the antisymmetry
of Fab to see that the left hand side of the contracted Eq. (65) is
∇a ∇b F ab − ∇b ∇a F ab = 2∇a ∇b F ab .
(66)
The right hand side of the contracted Eq. (65) is
g ac g bd Rabc e Fed + g ac g bd Rabd e Fce = Rbe F eb − Rae F ae = 0
(67)
where the first equality follows from the definition of the Ricci tensor, Rbd ≡ Ra bad , and the
second follows from the symmetry of Rab and the anti-symmetry of F ab . Thus, from Eq. (63)
4π∇a J a = ∇a ∇b F ab = 0,
(68)
and the four-current density must be conserved for the consistency of Maxwell’s equations.
Eq. (62), ∇[a F bc] = 0, is the integrability condition for the local existence of a vector
potential Aa , such that
∇a Ab − ∇b Aa = Fab .
(69)
~ · F~ = 0 then there exists
This is very similar to the Euclidean geometry theorem that if ∇
~
~
~
~
a vector A such that F = ∇ × A.
Let us find the equation governing the vector potential. From Eq. (69)
∇c Fab = ∇c (∇a Ab − ∇b Aa )
= Rcab d Ad + ∇a ∇c Ab − ∇c ∇b Aa ,
(70)
from the Ricci identity. After contraction with g bc , this becomes
∇b Fab = 4πJa
= R a d A d + ∇ a ∇ b A b − ∇ b ∇ b Aa .
(71)
∇b ∇b Aa − ∇a ∇b Ab − Ra d Ad = −4πJa .
(72)
This is equivalent to
In the Lorentz gauge, where ∇b Ab = 0, this simplifies to
∇b ∇b Aa − Ra d Ad = −4πJa .
A.
(73)
Alternative E&M
Generally, when we consider an area of physics which is well understood in special relativity, such as electricity and magnetism, and try to find the generalization of the relevant
equations to curved spacetime, we do as little as possible to the equations and typically just
replace ordinary derivatives with respect to Minkowski coordinates by covariant derivatives
with respect to an arbitrary coordinate system. However, this process is not unambiguous.
With this in mind consider the vector potential Aa defined in terms of the electromagnetic
field by
∇a Ab − ∇b Aa = Fab .
(74)
In flat spacetime the vector potential satisfies
∇b ∇b Aa − ∇a ∇b Ab = −4πJa .
8
(75)
It might appear reasonable to develop curved-spacetime electricity and magnetism, by starting directly with these equations for Aa and Fab while considering the derivatives to be
covariant derivatives of curved spacetime.
Show that this version of curved-spacetime electricity and magnetism has an unpleasant
feature. Hint: Look at the notes to see the accepted version of curved-spacetime electricity
and magnetism.
From Eq. (75), we can evaluate the divergence of J a . Specifically,
−4π∇a J a = ∇a ∇b ∇b Aa − ∇a ∇a ∇b Ab
(76)
Focus on the first term on the right hand side.
∇ a ∇ b ∇ b Aa =
=
=
=
∇b ∇a ∇b Aa + Ra b b d ∇d Aa + Ra ba d ∇b Ad
∇ b ∇ a ∇ b Aa − R a d ∇ d Aa + R b d ∇ b A d
∇ b ∇ a ∇ b Aa
∇b ∇b ∇a Aa + Rab a d Ad
= ∇b ∇b ∇a Aa + ∇b Rbd Ad
(77)
where the first equality follows from the Ricci identity after interchanging the order of the
covariant derivatives, the second follows from the definition of the Ricci tensor in terms of
the Riemann tensor, the third follows from the symmetry of the Ricci tensor, the fourth
follows from again interchanging derivatives and using the Ricci identity, and the fifth from
again using the definition of the Ricci tensor in terms of the Riemann tensor. Now substitute
this final result back into Eq. (76) to obtain
−4π∇a J a = ∇b ∇b ∇a Aa + ∇b Rbd Ad − ∇a ∇a ∇b Ab
= ∇b Rbd Ad .
(78)
Generally, the right hand side of this last expression is not zero, and we see that in this
“other version” of electricity and magnetism in curved spacetime charge is not conserved.
This is considered an “unpleasant feature”.
VII.
STRESS-ENERGY TENSOR
The stress-energy tensor for the electromagnetic field is
1
4πTab = Fa c Fbc − gab F cd Fcd .
4
(79)
The conservation of stress-energy then implies
1
4π∇a Tab = (∇a Fa c )Fbc + F ac ∇a Fbc − F cd ∇b Fcd
2
1
= (∇a Fa c )Fbc + (F ac ∇a Fbc − F ac ∇c Fba − F ac ∇b Fac )
2
1
a
c
= (∇ Fa )Fbc − f ac (∇a Fcb + ∇c Fba + ∇b Fac )
2
= (∇a Fa c )Fbc ,
= −4πJ a Fba = 4πJ a Fab
9
(80)
where the penultimate line follows from Eq. (62) and the last line from Eq. (63).
Express Tab in terms of the vector potential as
1
4πTab = (∇a Ac − ∇c Aa )(∇b Ac − ∇c Ab ) − gab (∇c Ad − ∇d Ac )(∇c Ad − ∇d Ac )
4
1
= (∇a Ac − ∇c Aa )(∇b Ac − ∇c Ab ) − gab (∇c Ad ) (∇c Ad − ∇d Ac ) .
(81)
2
10