Differential Geometry and Tensor analysis With applications to General Relativity
This is an article from my home page: www.olewitthansen.dk
Ole Witt-Hansen 2015
Contents
1. Linear algebra ...... 1 2. Generalized coordinates...... 5 3. Tensors...... 10 4. Covariant derivative...... 12 4.1 Christoffel symbols as expansion coefficients...... 14 4.2 Expressing the Christoffel symbol by the metric tensor...... 15 4.3 Geodesics and covariant derivatives ...... 16 4.4 Directional change under parallel transport of a vector...... 20 4.5 Angular excess and its correspondence to curvature ...... 21 4.7 Generalization of the Riemann tensor to n-dimensional space...... 24 4.8 Symmetries and contractions of the Riemann curvature tensor...... 25
Covariant derivative and geodesics 1
1. Linear algebra The concept of a tensor is much easier to grasp, if you have a solid background in linear algebra. So we start out with, some fundamental issues in that field.
A linear vector function is a vector function which obeys the following condition.
(1.1) f (a b) f (a) f (b) ,
The relation being valid for arbitrary vectors a and b and arbitrary real numbers and . If we have a base on the vector space consisting of mutually orthogonal unit vectors:
e1 ,e2 ,e3 ,.....,en
The linear function is completely determined by the mapping of these vectors, since a vector
x x1e1 x2e2 x3e3 .... xnen
As a consequence of (1.1) will be mapped into
(1.2) f (x) x1 f (e1) x2 f (e2 ) x3 f (e3 ) .... xn f (en )
Each of the vectors f (e1 ), f (e2 ), f (e3 ),...., f (en ) can then be written as a linear combination of the basic vectors.
(1.3) f (ek ) a1k e1 a2k e2 a3k e3 .....an k en
I follows (because the base vectors are mutually orthogonal), that
a j k e j f (ek )
(1.3) is usually written in matrix form a a a . a 11 12 13 1n a21 a22 a23 . a2n (1.3) ( f (e ), f (e ), f (e ),...., f (e )) (e ,e ,e ,.....,e ) a a a . a 1 2 3 n 1 2 3 n 31 32 33 3n . . . . . an1 an 2 an3 . an n
Or we can write the above relation in tensor form using Einstein’s summation convention:
That whenever the same index appears twice, summation is implied.
Covariant derivative and geodesics 2
n (1.4) f (ek ) a j k e j is written as j1 f (ek ) a j k e j (Summation j = 1…n is implied)
If y f (x) we get from (1.2) and (1.3) using tensor notation:
f (x) x1 f (e1 ) x2 f (e2 ) x3 f (e3 ) .... xn f (en ) x j f (e j ) And f (ek ) a1k e1 a2k e2 a3k e3 .....ank en ai k ei
We get: y f (x) ai j x jei
Which implies: yi ai j x j
Or when written in matrix form:
y a a a . a x 1 11 12 13 1n 1 y2 a21 a2 2 a23 . a2 n x2 (1.5) y a a a . a x 3 31 32 33 3n 3 . . . . . . . a a a . a yn n1 n 2 n3 n n xn
And when written in symbolic matrix form.
y = A x
Where we use double underscore to mark a matrix symbol.
The multiplication of two matrices A and B with elements ai j and bi j to give the matrix C with elements ci j goes as follows: One makes the “scalar product” of the i’th row in A and the j’th column in B . Written in tensor notation:
(1.6) ci j ai k bk j (implied summation over k)
The determinant of a matrix A is written det(A) or |A|
The determinant of a 2x2 matrix:
a a det 11 12 a a a a 11 2 2 21 12 a21 a2 2
Whereas the determinant of a 3x3 matrix is:
Covariant derivative and geodesics 3
a a a 11 12 13 (1.7) deta21 a22 a23 a11a22a33 a11a32a23 a21a12a33 a21a32a23 a31a12a23 a31a22a13 a31 a32 a33
The rule is that each factor in the product a1i a2 j a3k belongs to a different row and a different column from the other factors. If [i, j, k] is an even permutation of [1, 2, 3] then the term is signed with a plus, if it is an uneven permutation, the term is signed with a minus.
The determinant, can also be expressed with the help of the Levi-Civitas symbol
1 if i, j, k is an even permutation of 1,2,3
i j k 1 if i, j, k is an uneven permutation of 1,2,3 0 otherwise
The determinant of the matrix shown above can then be written:
det(A) = a1i a2 j a3k i j k
Since there are six permutations of three elements, the number of terms is six.
The extension to higher dimensions including the Levi-Civitas symbol is straightforward.
The unit matrix E has the elements i j , where i j is the Koneke symbol
(1.7) i j = 1 if i = j else 0
Any n x n matrix that has a non zero determinant, is called regular. Any regular matrix A has an inverse matrix A-1, defined by the equations:
(1.8) A A-1 = A-1A = E
If the elements in the inverse matrix are bi j , then (1.7) can be written
ai kbk j i j
1.1 Coordinate transformations First we shall consider an ordinary coordinate system in the plane, which is rotated an angle θ. The base vectors (e ,e ) are rotated by an angle θ 1 2 into (e ',e ') . From the figure we see, that: 1 2 e1 ' e1 cos e2 sin and e2 ' e1 sin e2 cos
When written in matrix form.
Covariant derivative and geodesics 4
cos sin (e1 ',e2 ') (e1 ,e2 ) sin cos
According to (1.3) and (1.5), we get the corresponding transformation for the coordinates (x1 , x2) and (x1’ , x2’).
x1' cos sin x1 (1.9) x2 ' sin cos x2
In 3-dimensional space, this transformation corresponds to a rotation around the z-axis, an angle θ. The transformation matrix then becomes:
x' cos sin 0 x (1.10) y' sin cos 0 y (Rotation about the z-axis) z' 0 0 1 z
For the rotations around the y-axis or the x-axis, we have quite similar expressions
x' cos 0 sin x y' 0 1 0 y (Rotation about the y-axis) z' sin 0 cos z
x' 1 0 0 x y' 0 cos sin y (Rotation about the x-axis) z' 0 sin cos z
To bring the coordinate system into an arbitrary angular position, we need three rotations. They are traditionally chosen as a rotation around the z-axis an angle α, followed by a rotation around the new y-axis (y’) an angle β, and finally an angle γ around the new z-axis z’’. The overall rotation is found by multiplying the three matrices. The angles α, β, γ are called the Euler angles. They are illustrated below:
R(,, ) Rz ''( ) Ry '( ) Rz () cos sin 0cos 0 sin cos sin 0 sin cos 0 0 1 0 sin cos 0 0 0 1 sin 0 cos 0 0 1
cos cos cos sin sin cos sin cos cos sin sin cos cos cos sin sin cos cos sin sin cos cos sin sin sin cos sin sin cos
Covariant derivative and geodesics 5
Showing the Euler angles.
2. Generalized coordinates Hitherto we have considered only Cartesian coordinates, with an orthonormal base e ,e ,e ,.....,e . 1 2 3 n This means that the base fulfils the condition ei e j i j . The basic vectors are the same for every point in space. Because of the orthogonality the square of the infinitesimal distance element is the same in every point.
ds e1dx1 e2dx2 e3dx3 .... endxn
2 2 2 2 2 (2.1) ds ds ds dx1 dx2 dx3 ...dxn
In a generalized coordinate system x (x1, x2 , x3 ,..., xn ) , the base vectors are not necessarily an orthogonal system, and the base may (and usually does) wary from point to point. If x (x , x , x ,..., x ) represents a point in the generalized coordinates, we can define a base: 1 2 3 n (e1 ,e2 ,e3 ,...en ) along the axis of the coordinate system x . A differential displacement is given by:
dx (dx1,dx2 ,dx3 ,...,dxn )
And the displacement vector
(2.2) ds dx1e1 dx2e2 dx3e3 ... dxnen
The length of the distance element is:
2 2 ds ds ds ds (dx1e1 dx2e2 dx3e3 ... dxnen )(dx1e1 dx2e2 dx3e3 ... dxnen )
Covariant derivative and geodesics 6
2 (2.3) ds ei e j dxi dx j (summation over i and j is understood)
There is a tradition, which we shall substantiate presently, letting the index on the coordinate functions go upstairs. From now on we write (2.3) as:
2 i j (2.3) ds ei e j dx dx
Since the base vectors in the generalized coordinate system, are not (necessarily) orthogonal vectors, the scalar products e e does not vanish in general fori j . i j
We define the metric (fundamental) form gi j gi j (x) gi j (x1, x2 , x3 ,..., xn ) as
(2.4) gi j ei e j
And the distance element becomes
2 i j ds gi j dx dx
In the coordinate system x (x1, x2 , x3 ,..., xn ) we write of course:
2 i j (2.4) gi j ei e j and ds gi j dx dx
In a Cartesian coordinate system gi j ei e j i j
It is important to note that although the relation between generalized coordinate are (in general) not linear relations, the relations between the differentials dx (dx1,dx2 ,dx3 ,...,dxn ) are in fact linear. If
y f (x1, x2 , x3 ,..., xn ) Then f f f f (2.6) dy dx1 dx2 dx3 ... dxn x1 x2 x3 xn
Is in fact a linear form on the variables dxi
The transformation to the coordinates xi xi (x1, x2 , x3 ,..., xn ) goes as follows:
xk xk xk xk (2.7) dxk dx1 dx2 dx3 ... dxn x1 x2 x3 xn
Or in tensor notation with the indices raised:
x k (2.8) dx k dxi (summation over index i is implied) xi
Covariant derivative and geodesics 7
Similarly we may write the inverse transformation xi xi (x1, x2 , x3 ,..., xn )
x j (2.9) dx j dx k x k
The transformation between the two coordinate differentials can also be written in matrix form:
x x x x 1 1 1 . 1 x1 x2 x3 xn dx1 dx1 x2 x2 x2 x2 dx . dx 2 x x x x 2 1 2 3 n (2.10) dx x x x x dx 3 3 3 3 . 3 3 x1 x2 x3 xn . . . . . dxn dxn xn xn xn xn . x1 x2 x3 xn
The determinant of the transformation matrix T det(T) is called the Jacobian, and it is the scale factor between the two volume elements: dx1dx2dx3 dxn and dx1dx2dx3 dxn . This can be seen if we write the two vector columns as matrices.
x x x x 1 1 1 . 1 x1 x2 x3 xn dx1 0 0 0 0 dx1 0 0 0 0 x2 x2 x2 x2 0 dx 0 0 0 . 0 dx 0 0 0 2 x x x x 2 1 2 3 n (2.10) 0 0 dx 0 0 x x x x 0 0 dx 0 0 3 3 3 3 . 3 3 0 0 0 . 0 x1 x2 x3 xn 0 0 0 . 0 . . . . . 0 0 0 0 dxn 0 0 0 0 dxn xn xn xn xn . x1 x2 x3 xn
It is easy to convince one, that this represent the same relation between the differentials as before. From algebra we know that the determinant of a diagonal matrix is the product of the diagonal elements. Also we know that the determinant of the matrix products of two matrices is the product of the two determinants:
det(A B) = det(A)det(B)
The assertion then follows, if we take the determinant of both sides.
xi (2.11) dx1dx2dx3 dxn det dx1dx2dx3 dxn x j
If (x1, x2 , x3 ,..., xn ) are Cartesian coordinates we have according to (2.1)
Covariant derivative and geodesics 8
ds 2 ds ds dx1dx1 dx 2dx 2 dx3dx3 ...dx n dx n dxi dxi
x j Inserting (2.9) dx j dx k gives x k
n n i n n j 2 x k x l ds ( k dx )( l dx ) i1k 1 x j1l 1 x
n n n n i j 2 x x k l ds ( k l )dx dx k 1l 1i 1j 1 x x
From which it is seen that the metric form for the transformed coordinates is:
n n xi x j (2.10) gk l k l i1j 1 x x
The transformation formulas (2.8) and (2.9) can be written in matrix form:
x x x x 1 1 1 . 1 x1 x2 x3 xn dx1 dx1 x2 x2 x2 x2 dx . dx 2 x x x x 2 1 2 3 n (2.11) dx x x x x dx 3 3 3 3 . 3 3 . x1 x2 x3 xn . . . . . . dxn dxn xn xn xn xn . x1 x2 x3 xn
Or in tensor notation: x k (2.11) dx k dxi xi
The inverse transformation has the inverse matrix, which follows from the chain rule of differentiating: xi (2.12) dxi dx l x l and x k x k xi x k dx k dxi dx l dx l k l dx l dx k xi xi x l x l
Covariant derivative and geodesics 9
If we do not have an orthonormal base, which is often the case of generalized coordinates, one can always find a so called inverse base. In the inverse base and in any of the object expanded on the inverse base the indices go upstairs. Let the base be: (e1 ,e2 ,e3 ,.....,en ) These are linear independent unit vector, which are not (necessarily) mutually orthogonal. Our aim is now to establish another base: (e1 ,e 2 ,e 3 ,.....,e n ) where the following condition holds:
j j (2.13) ei e i
1 2 3 n Since (e1 ,e2 ,e3 ,.....,en ) is a base, any of the vectors (e ,e ,e ,.....,e ) can be expanded on that base, which we write it in the following way:
j e 1 je1 2 j e2 3 j e3 .... n j en or j e k j ek
using the summation convention. The condition (2.13) then gives
j j ei e 1 j ei e1 2 j ei e2 3 j ei e3 .... n j ei en i
We remember that gi j ei e j so the system of equations to determine the λi j is:
j (2.14) 1 j gi1 2 j gi 2 3 j gi3 .... n j gi n i j = 1..n
If λ is the matrix λi j , and g is the matrix gj k , (2.14) can be written:
(2.15) λ g = E λ = g-1
Not surprisingly, the inverse base has the inverse metric form.
i j i j From (2.4) gi j ei e j it then follows g e e
i j 1 Where g gi j is the inverse matrix to gi j Suppose we have two vectors:
a a1e1 a2e2 a3e3 .... anen and b b1e1 b2e2 b3e3 .... bnen
And we form the scalar product
(2.16) a b aib j ei e j gi j aib j
The same vectors expanded on the inverse base give correspondingly:
Covariant derivative and geodesics 10
a b aib je i e j g i j aib j
But if we expand the vector a on the base
a a1e1 a2e2 a3e3 .... anen
and expand the vector b on the inverse base
b b1e1 b2e 2 b3e 3 .... bne n we obtain the following expression for the scalar product
j j j j i i (2.17) a b aib ei e aib i aib so a b aib almost as in a Cartesian coordinate system
In the same manner the square of the length of a vector takes its usual form.
2 i (2.18) | a | a a ai a
3. Tensors It is required by the laws of physics, that they are covariant or form invariant i.e. “they look the same” under certain coordinate transformations. The laws of mechanics are invariant under translations, rotations and the Galileo transformation. Maxwell’s equations are also invariant under the Lorentz transformation of special relativity. In analytic mechanics one can show, that conservation of momentum is a consequence of invariance to translation, and conservation of angular momentum is a consequence of invariance under rotations. Thus the invariance of physical objects under various transformations plays an important role.
Mathematical objects which are invariant under certain coordinate transformations are called tensors. A scalar has only one component, and it is a tensor of rank 0. An object which has one index (i.e. a vector vi) is called a tensor of rank 1. An object which has two indices (i.e. a matrix gi j) is called a tensor of rank 2 Tensors can have arbitrary high rank.
When operating with tensors, one uses consequently the summation convention: that is, if the same index appears twice, summation is implied.
In section 2, we introduced the transformation of differentials from one coordinate system to another. Thus if xi xi (x1, x2 , x3 ,..., xn ) then
xk xk xk xk dxk dx1 dx2 dx3 ... dxn x1 x2 x3 xn
Covariant derivative and geodesics 11
Or in tensor notation, and with the index raised: x k (3.1) dx k dxi xi
Similarly we may write the inverse transformation of xi xi (x1, x2 , x3 ,..., xn )
x j (3.2) dx j dx k x k
A scalar has the same value in all coordinate systems:(x) (x)
Any object ai that transform as the coordinate differentials is called a contravariant tensor of rank 1.
x k (3.3) a k ai xi
On the other hand the gradient of a scalar, transforms as:
x j x k x j x k
Any object bi that transform as the gradient of a scalar is called a covariant tensor of rank 1.
x j (3.4) b b k j x k
A covariant tensor ci j has the transformation properties:
x k xl (3.5) c c i j k l x i x j
A mixed tensor is a tensor with having index both upstairs and downstairs.
Notice, with the transformation properties defined above, the scalar product is an invariant.
x j x i (3.6) a b a b i a bl j a bl a b j i j x i x l l j j
i j With the help of the metric tensor gi j or its inverse tensor g you may raise or lower one or several indices. This comes about, because the metric tensor is the transformation matrix between a base of unit vectors and the inverse base.
i j j i j ai g a c gi k g j l c j l
Covariant derivative and geodesics 12
For the same reason, you may also contract two tensors, by making equal a lower index in one with the upper index in the other.
j ci ai jb
That ci actually has the proper transformation properties of a covariant vector, can be shown by j writing the transformation properties for ai j and b , as we did for the scalar product.
4. Covariant derivative We have seen above, that for an object to be a tensor, it must have certain transformation properties. Since generalized coordinates require a base and its inverse base, which are orthogonal to each other, we have covariant and contravariant tensors, distinguished from each other by having lower and upper indices. x k A contravariant vector transforms in the same way, as the coordinate transformation: dx k dxi xi Any object ai that transform as the coordinate differentials is called a contravariant tensor of rank 1.
x k (4.1) a k ai xi
On the other hand the gradient of a scalar, transforms like
x j x k x j x k
Any object bi that transform as the gradient of a scalar is called a covariant tensor of rank 1.
x j (4.2) b b k j x k
One might expect that the derivatives of the components of a vector would transform like a tensor, but that is not the case. This is one of the reasons for introducing the concept of covariant derivative, a derivative which has proper tensor properties. The transformation of the i'th component of a contravariant vector vi goes as follows. According to (4.2) xi v i v k x k Consistently v i x i 2 x i x i v k xl ( v k ) v k x j x j x k x j x k x k xl x j
v i x i 2 x i x i xl vk ( vk ) vk ( ) x j x j xk x jxk x k x j xl
Covariant derivative and geodesics 13
If only the last term was present, the derivative of vi would transform as a proper mixed tensor, but the first term prevents this.
The reason for this, is that the components of a vector are the projection on a base vectors, and they are position dependent, e i e i (x) and vi v e i . Differentiating vi v e i , we get:
v j v e j (4.3) (v e j ) e j v xi x j xi xi
It is the second term that causes the trouble, and has to be dealt with.
The idea is now to try to construct a differential operator Di that has tensor properties, when applied on components of a vector v j . That is:
x k x j (4.4) D v j D vl i x i xl k
j Any vector v can be expanded either on the base ei or on the inverse base e . The problem arises, because the unit vectors themselves depend on position, which is not the case in a Cartesian coordinate system. i j j j j If we expand v on the base ei : v v ei then the projections are v v e since ei e i . The point is that both v and e j are good tensors, so when we differentiate v j written as v e j it might just be the differential operator we are looking for, since
v x j v x j and e j e k x i x i x j xk
By taking the scalar product of the two vectors we then obtain
v x j x j v e j ( ) e k x i x i xk x j
Which shows that it transforms like a proper mixed tensor, and it might be the covariant derivative D that we are looking for. i v (4.5) D v j e j i xi v This relation implies that D v j can be viewed as projections of the vectors on the vectors e j , i xi v we may therefore also write expanded on the basis vectors e xi k
v (4.6) D vk e xi i k
Covariant derivative and geodesics 14
Taking the scalar product with e j we get:
v (4.6) e j D vk e e j D vk j D v j xi i k i k i which is (4.5).
4.1 Christoffel symbols as expansion coefficients However we do not have a similar simple transformation property for the base vectors ek , since they by definition change under coordinate transformations. v e j As a result the corresponding expansion for (4.6) D vk e can not be copied for , xi i k xi but instead the derivative is written symbolically as an expansion on the vectors e j
e k (4.7) k e j xi i j
Or when forming the scalar product with v
e k v k v j xi i j
e k v v j v e j Putting v k v j and D v j e j into e j v we get: xi i j i xi xi xi xi
v j v e j (4.8) e j v D v j k v j xi xi xi i i j
j And after isolating Div , we finally arrive at. v j (4.9) D v j j vk i xi i k
Thus in order to produce a derivative of a vector component that conforms to the transformation properties of a tensor, we must add an extra term. Neither of the two terms on the right side of (4.9) are tensors, but the sum of them is. As we shall see later the concept of covariant derivative has affinities to finding the shortest path on a between to points on a manifold, and to the concept of parallel displacement.
One can show that de covariant derivative of a covariant vector is given by:
v j (4.10) D v k vk i j xi i j
Covariant derivative and geodesics 15
4.2 Expressing the Christoffel symbol by the metric tensor i You may find the covariant derivative for at mixed Tj tensor as the covariant derivative of the direct i product v u j . It is differentiated covariantly, treating each index separately. For a mixed tensor the result is.
T k (4.11) DT k j l T k k T m i j xi i j l i m j
A specific example is the differentiating of the metric tensor gi j
g (4.12) D g i j l g l g k i j xk k i l j k j il
k The Christoffel symbol i j is symmetric when permuting the two lower indices. This follows from the expression (4.17) below, where the Christoffel symbol is expressed by means of the derivatives of the metric tensor.
We shall now prove that although the metric tensor is position dependent, it is constant with respect to covariant differentiation. Dk gi j 0 (Identically zero).
One may prove this, by using the definition of the metric tensor in terms of the base vectors:
gi j ei e j
g e e i j (e e ) i e e j xk xk i j xk j i xk l l (k iel ) e j (k jel ) ei g i j l g l g xk k i l j k j l i g (4.13) i j l g m g 0 D g 0 xk k i l j k j mi k i j
We are now ready to express the Christoffel symbols, by the derivatives of the metric tensor. The starting point is (4.13)
g i j l g l g 0 xk k i l j k j il
Rewriting (4.13) cyclically permuting the indices i, j, k, and using the symmetry properties of the k k Christoffel symbol i j ji and the same symmetry of the metric form gi j g ji we get.
Covariant derivative and geodesics 16
g (4.14) k i l g l g 0 x j j k l i ji k l
g (4.15) j k l g l g 0 xi i j k l i k jl
Adding the two first equations, and subtracting the last (after having permuted some indices), we arrive at:
g g g i j k i j k 2l g 0 xk x j xi k j il
g g g (4.16) l g 1 ( i j k i j k ) k j il 2 xk x j xi
Finally by multiplying both sides by the inverse matrix of the metric form g lm , we get
g g g l g g l m 1 g l m ( i j k i j k ) k j il 2 xk x j xi
g g g l m 1 g l m ( i j k i j k ) k j i 2 xk x j xi
g g g m 1 g l m ( i j k i j k ) k j 2 xk x j xi
Or by renaming some of the indices
g g g (4.17) k 1 g k l ( il jl i j ) i j 2 x j xi xl
4.3 Geodesics and covariant derivatives In its origin, a geodesic is the name of the shortest path between two positions on earth. As we know that for a sphere it is a great circle connecting the two positions. In Differential geometry a geodesic is the shortest path connecting two points on a manifold.
In Differential Geometry 1, we restricted ourselves to surfaces i.e. two dimensional manifolds. But here we shall show that the results also apply to higher dimensions.
We shall do so by finding differential equations for the geodesic parameter curve.
We assume that x (x1(t), x2 (t), x3 (t),..., xn (t)) is a parameter curve with the parameter t. The distance element is:
Covariant derivative and geodesics 17
i j (4.18) ds gi jdx dx and therefore ds (4.19) s g xi x j dt i j
Where, as usual a bullet above a variable means differentiation with respect to t.
To derive the differential equations for xi (t) , which give the minimum value for s we use the classical Lagrangian approach.
Ldt 0 where L L(x, y, y,t) and the solution is the Euler-Lagrange equation.
L d L (4.19) 0 y dx y
If there are several variables x (x1(t), x2 (t), x3 (t),..., xn (t)) the equations become:
L d L (4.20) 0 with L g (xi , x j ) xi (t)x j (t) x k dt x k i j
i j Note that gi j (x , x ) does not depend explicitly on t.
If we furthermore choose t as a natural parameter, we have
ds ds g (xi , x j ) x i (t)x j (t) 1 dt ds i j
We may then drop the square root in the calculations, i.e.
g (xi , x j ) xi (t)x j (t) xk i j 1 g (xi , x j ) xi (t)x j (t) 1 (g (xi , x j ) xi (t)x j (t)) i j i j xk i j 2 xk i j 2 gi j (x , x ) x (t)x (t)
And by the same token, when differentiating with respect to t. Thus we find:
L g g (xi , x j ) x i (t)x j (t) 1 i j xi x j x k xk i j 2 xk and L g (xi , x j ) xi (t)x j (t) 1 (g x j g x i ) g xi (Since g = g ) x k x k i j 2 k j i k i k i k k i
Covariant derivative and geodesics 18
Multiplying by 2, the Euler-Lagrange equation becomes
g d i j xi x j 2 g xi 0 xk dt i k
d g g xi i k x j xi g xi dt i k x j i k
g g g And, because of the symmetry of the metric form: i k 1 ( i k j k ) x j 2 x j xi Putting it all together we get
L d L 0 xi dt xi
g g g i j xi x j ( i k j k )xi x j 2g xi 0 xk x j xi i k
Multiplying by ½ , changing sign and rearranging the terms we finally get:
g g g g xi 1 ( i k j k i j )xi x j 0 i k 2 x j xi xk
k l Finally multiplying by the inverse matrix to gi k , g we arrive at the equation.
g g g g k l g xi 1 g k l ( i k j k i j )xi x j 0 i k 2 x j xi xk g g g xi 1 g k l ( i k j k i j )xi x j 0 i l 2 x j xi xk
l l i j (4.21) x i j x x 0
Or, if renaming l to k
k k i j (4.21) x i j x x 0
This we shall compare to the covariant derivative
v j D v j j vk i xi i k put to 0.
Covariant derivative and geodesics 19
dx j We choose the vector vj, to be the tangent vector of a curve ds x x(s) (x1(s), x2 (s), x3 (s),..., xn (s))
We then change the formula for the covariant derivative, by differentiating down to s, remembering, dx j that when we differentiate with respect to xj, should be followed by multiplication with . The ds covariant derivative now writes, since it is the covariant derivative with respect to xj.
dv j dxi D v j j vk i ds i k ds dx j Insertion v j one finds: ds
dx j d 2 x j dxk dxi D j i 2 i k ds ds ds ds
j 2 j k i dx d x j dx dx (4.22) Di 0 i k 0 ds ds2 ds ds
Comparing with the equation for the geodesic
k k i j x i j x x 0
We see that it is in fact the same equation, apart from the name of the indices.. If you move in 3-dim space along a straight line, i.e. in the direction of a constant vector, the derivative of that vector is zero, at any point.
This suggests that, when you move in a manifold along a curve with zero covariant derivative, you move along a “straight line” meaning the shortest path between any two points on the curve. The geodesics in space-time are the paths followed by a light ray, since it is the geodesics for a body falling freely in the influence of gravity.
Parallel transport and the Riemann curvature tensor 20
4.4 Directional change under parallel transport of a vector Parallel transport is a fundamental notion in differential geometry. It comes about because the change of a vector has two components. v j One is the dependence of the vector on coordinates: xi The second is the coordinate themselves, which change with position, represented by the basic vectors e k e k (x) where x (x1, x2 , x3,.., xn ) .
We have already seen this, because this is precisely, what is reflected in the difference between the normal derivative, which only includes the first part, and the covariant derive which includes both i parts. As previously covariant differentiation with respect to x is denoted by a capital Di.
v j According to (4.9) D v j j vk i xi i k
Or when expressing it with total differentials:
v j Dv j (D v j )dxi And dv j dxi i xi So v j dv j dxi (D v j )dxi j vk dxi or xi i i k j j j j j j k i dv d(e v) e dv v de Dv i kv dx
The notion of parallel transport comes about, when moving a vector without changing the vector itself, that is, e j dv Dv j 0 . In a Euclidian space the notion of parallel transport is the same as parallel displacement: that is, the vector is unchanged.
Setting the “true change”, to zero
j j j k i (4.23) Dv 0 dv i k v dx 0 (True change)
We recall, that Di gi j 0 , which reflects the fact that the change of the metric tensor is due to coordinate changes only. This is however the same, as the covariant derivative is identically zero along a specific curve. The process of parallel transporting a vector vj along a parameter curve xi (t), can according to (4.12) be accomplished as:
Dv j dv j dxi (4.24) j vk 0 Dt dt i k dt
Parallel transport and the Riemann curvature tensor 21
In this manner, we may define the straightest possible curve by the condition of it being the “line” constructed by parallel transport of its tangent vector. dx j In this way the condition can also be formulated setting v j , so according to (4.24) dt D dx j d 2 x j dxi dx j (4.25) ( ) j vk 0 Dt dt dt 2 i k dt dt
An equation we (again) recognize as the geodesic equation. Covariant derivative, parallel transport, geodesics are basically the same geometrical concept. We may then conclude that when a vector vi is parallel transported along a geodesic, one can show that the angle subtended by the vector and the geodesic: that is the, tangent of the geodesic, is unchanged.
4.5 Angular excess and its correspondence to curvature Curvature measures the amount of deviation the manifold differs from Euclidian space. In a two dimensional space one can argue that for an infinitesimal polygon (a triangle) there is a remarkably simple relation between the angular excess E (which is the difference between the sum of the angles in the polygon and the sum of angles in the corresponding Euclidian polygon) and the area dA of the polygon. The relation is proportionality, and the scaling factor is the Gaussian curvature K. (4.27) E = K dA
How the angular excess can be determined is not quite obvious, but we shall use the concept of parallel transport to illustrate the method. This has the advantage that it will allow us to generalize the relation (4.27) from a two dimensional surface into n-dimensional manifolds.
It can be shown that the angular excess of a polygon E is equal to the directional change of a vector when parallel transported around the polygon. We shall first illustrate this by a simple example, where it is immediately clear. See the figure below. The figure shows a sphere, and a spherical triangle with three 900 angles. A vector initially placed at (1) is parallel transported to (2). Since the vector is perpendicular to the great circle arc 1 - 2, it does not change direction. But when it reaches (2) it is tangent to the arc 2 -3. When parallel transported: that is along the arc, the directional change is 900. When transported from (3) to (4), again there is no directional change, as the vector is perpendicular to the arc (3) – (4). We see that the directional change of the vector, when parallel transported from (1) to (4) is 900, which is equal to the angular excess of the triangle. For the directional change of a parallel transportation of a vector along an arbitrary triangle one may argue as follows: See the figure below.
Parallel transport and the Riemann curvature tensor 22
In the case for a constant curvature i.e. a sphere, the area does not need to be infinitesimal, and we shall now prove (4.27) for a spherical triangle. The figure shows a spherical triangle ABC, with the three angles α, β, γ. We notice, that it is only the part of a vector along a curve, which has a directional change. We start out at A, and move the tangent vector by parallel transport to B. The directional change is d1. Then we turn the vector an angle π - β, so the vector now is along a. We move the tangent vector by parallel transport to C, and the vector gets a directional change d2. Then we turn the vector an angle π - γ, so it becomes tangent to the arc AC and move the tangent vector by parallel transport to A.The directional change is d3. Finallly we turn the vector an angle π – α, so it coincides with the initial vector. The net result is that the vector has been turned an angle π – β + π – γ + π – α = π – (α +β + γ). At the same time, it has undergone a directional change d1 + d2+ d3. By construction the sum of the angular change, and the directional change must be zero.
(4.28) d1 + d2+ d3 + π – (α +β + γ) = 0 d1 + d2+ d3 = α +β + γ - π
Which illustrates the assertion that when a vector is transported parallel around a polygon, the directional change is equal to the angular excess.
If you look at the figure, one can see, that there are six lunes. ABA’C, AC’A’B’, BCB’A, BC’B’A’, CAC’B, CA’C’B’. Each point on the sphere belongs to exactly one of the six lunes, except for the two triangles ABC and A’B’C’, which belongs to three lunes. The area of a lune, with an opening angle θ is the fraction of θ to the perimeter 2πR times the area of the sphere 4πR2 Area of lune = 4R2 2R2 . 2 The area of the six lunes is therefore: 2( )R2 . This is the equal to the area of the sphere, minus two times the area of the triangles ABC an A’B’C’. For spherical triangles, however congruity of angles means congruity of the triangles themselves. Let Aabc be the common area of the two triangles, we can there write an equation for the areas.
2 2 4( )R 4R 4AABC By divding by 4R2, we obtain: A (4.29) ( ) ABC R2 Comparing this to (4.27) Angular excess = Curvature ∙ Area:
(4.29) E = K∙A,
Parallel transport and the Riemann curvature tensor 23
we see, that it gives the Gaussian curvature R-2, which is in fact the case for a sphere.
For a sphere, the result (4.29) can easily be extended to an arbitrary spherical polygon, since such a polygon can always be divided in triangles, where the sum of angular excesses equals the angular excess of the polygon, and the sum of the areas of the triangles equals the area of the polygon.
Also we have seen in (4.28) that the angular excess E of a polygon is equal to the directional change ∆v of a vector, when it is parallel transported along the edges of the polygon. dv The angular change of a vector is therefore (4.29) can also be written: v dv (4.30) KdA dv KvdA v
The above theorem is only valid, when the curvature is constant, and therefore it applies in generel only to infinitesimal polygons.
However when we no longer consider only 2-dimensional surfaces, and want to generalize to higher dimensions the concept of area, becomes a bit more circomstantial.
In 3-dimensional Euclidian space, the area spanned by two vectors a and b , can be calculated as the length of the cross product a b . Or using the Levi-Civitas antisymmetric tensor i j k .
i j (4.30) k i j k a b
Where has the magnitude | a || b | sin . (4.30) cannot immediately be generalized to higher dimensions in this form. For higher dimensions we shall need an antisymmetric tensor with a different number of indices.
In four dimensions the Levi Civitas symbol becomes i j k l and so on. In four dimensions we shall use a two index object i j to represent the area.
i j i j k i j k m n 1 i j j i (4.31) k mnk a b 2 (a b a b ) ,
i j k 1 i j i j Where we have made use of the identity: mnk 2 ( m n n m ) For this relation the index 1..3 is actually irrelevant, and we can readily generalize (4.31) to n dimensions: i j , where i, j = 1,2,3…,n. So in any case, the area element is represented by:
i j 1 i j j i (4.32) 2 (a b a b )
Parallel transport and the Riemann curvature tensor 24
4.7 Generalization of the Riemann tensor to n-dimensional space The equation (4.30) and (4.32): that is, the tensor expression for the area, suggest that one may express the change of a vector due to a parallel transport around a parallelogram spanned by the vectors ak and bl where the curvature K is replaced by the Riemann tensor.
i i j k l (4.33) dv R j k lv d
i Since all other parts of the equation are good tensors, it follows from the quotient theorem that R j k l is also a tensor.
By the rather intriguing calculation below (I suggest that you skip it), one can show that:
i i (4.34) Ri jl j k i m i m j k l xk xl mk jl ml j k
Since the Christoffel symbol being first derivative of the metric tensor, the Riemann curvature 2 g g tensor is a nonlinear function of and ( )2 (xk )2 xk Derivation of (4.34) The figure shows a parallelogram PP’Q’Q spanned by the infinitesimal vectors al and bm . The opposite vectors (a da)l and (b db)m , are obtained by parallel transport of al and bm respectively. From the figure we see that dbl al and dam bm Recall that parallel transport of a vector vj implies that the covariant j derivative is zero: Div = 0, meaning that the total change of the vector is due to coordinate changes only. i i j k (4.35) dv j k v dx
Our aim is to derive the expression above (4.34) from parallel transporting a vector around a closed path, given by the parallelogram PP’Q’Q. From (4.35) we then get:
l l l i j m m m i j (4.36) (a da) a i ja b and (b db) a i ja b
We then calculate from (4.35) the change of a vector vi due to parallel transport from P ->Q -> Q’.
i i i dv PQQ' dv PQ dv QQ' (4.36) i j l i j m (j lv )P a (j mv )Q (b db) Where P and Q denote the respective positions, where these functions are to be evaluated. Since the aim is to evaluate and compare all quantities at the same position P, we shall make a Taylor expansion around P for the quantities evaluated at Q.
i (4.37) (i ) (i ) ( j m ) al j m Q j m P xl P
Parallel transport and the Riemann curvature tensor 25
And according to (4.35) v j (vi ) (vi ) ( ) al (vi ) (i vk ) al Q P xl P P k l P
Substituting (4.36) and (4.37) into (4.36), and dropping the index P, since all quantities, are now evaluated at P. i i i j l i j m l j j k l m m p q dv PQQ' v a ( a )(v v a )(b a b ) j l j m xl k l p q Multiplying out the brackets, keeping only terms up to a pbq we find: i i i j l i j m i m j p q j m j l m i j k l m (4.38) dv PQQ' v a v b v a b v a b v a b j l j m j m p q xl j m k l The directional change by the path PP’Q’ can simply be obtained from the above expression by interchanging a with b. i i i j l i j m i m j p q j m j l m i j j l m (4.39) dv PQP' v b v a v b a v b a v a b jl j m j m pq xl j m k l To find the collected directional change from a tour around the parallelogram PQP’Q’ we subtract (4.39) from (4.38). i i i i i k m k l i l i j k l m (4.40) dv dv PP'Q' dv PQQ' ( )v a b xl xm j l k m j m k l Because the combination in the bracket is antisymmetric with respect to the indices l an m, only the antisymmetric combination ½(albm – ambl) contributes, but this is just the area tensor, we introduced in (4.31) Comparering with (4.33) with a minor change in the name of the indices, we realize, that (4.40) is identical with (4.31). And we have found an expression for the Riemann tensor.
i i j k l dv R j k lv d
4.8 Symmetries and contractions of the Riemann curvature tensor The Riemann curvature tensor has certain symmetry properties reducing the number of independent components in 4-dimensional space from 64 to 20. On a surface in a 2-dimensional space, the Riemann tensor has only one component which is the Gaussian curvature K. One can lower the upper index of the Riemann tensor by applying the metric tensor gi j.
m (4.41) Ri j k l gi m R j k l
The Riemann tensor is anti symmetric with respect to interchanging the first and the second index, and that of the third and the fourth index, respectively.
(4.42) Ri j k l Rj i k l and Ri j k l Ri j l k
It is symmetric with respect to interchange of the pair made up of the first and the second indices, and the pair made up of the third and the fourth indices.
Parallel transport and the Riemann curvature tensor 26
(4.43) Ri j k l Rk l j i
It also has a cyclic symmetry
(4.44) Ri j k l Ri l j k Ri k j l 0
Because of the symmetry properties stated above the contractions of the curvature tensor are essentially unique. After demonstrating the various contractions we show how we arrive at Einstein’s tensor, which is the left part of field equation in General Relativity.
The Ricci tensor is the Riemann curvature tensor, with the first and the third indices contracted-
l m m (4.45) Ri j g Rl i m j Ri m j
Which is a symmetric tensor:
Ri j R j i
The Ricci scalar is the Riemann curvature tensor contracted twice.
i j (4.46) R g Ri j
The Bianchi identities are a set of constraints on the curvature tensor
(4.47) Dk Ri j l m Dj Rk i l m Di R j k l m 0
We shall not prove these assertions, but rather perform some contractions with the metric tensor. l m Since the metric tensor is covariantly constant. Dk g 0 , the contraction can be “pushed through” the covariant differentiation.
j m i l (4.48) Dk R D j g Rk m Di g Rk l 0
At the last two terms, the contraction just raises the index.
j i j Dk R D j Rk Di Rk Dk R 2D j Rk 0
Pushing yet another gi k to raise the index in the last term, and putting the covariant derivative outside, we find:
i k ik (4.49) Dk (Rg 2R ) 0
We see, that the combination
i j i j 1 i j (4.50) G R 2 Rg
Parallel transport and the Riemann curvature tensor 27
Is covariantly constant, i.e. its covariant derivative is identically zero.
i j DiG 0
The reason for developing the last section is that Gi j is the tensor that appears on left side of Einstein’s famous field equation in The General Theory of Relativity, and for that reason, it is called the Einstein tensor.
To elaborate on this, is beyond a text on differential geometry and tensor analysis, here we shall just state Einstein’s geometrical interpretation of the dynamics of moving in a gravitational field.
(4.51) Gi j Ti j
The geodesic of a particle falling freely in a gravitational field is proportional to the energy momentum tensor.
If one takes the Newtonian limit of (4.51) the proportional factor κ can be shown to be:
8G (4.52) c4
Where G is the same constant, as in Newton’s law of gravitation, where F is the attracting force between two masses m1 and m2 , when separated by a distance r.
m m F G 1 2 r 2
References: Tai-Pei Cheng: Relativity, Gravitation and Cosmology. Jon Mathews and Robert L. Walker: Mathematical methods of physics. Albert Einstein: The meaning of relativity