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LECTURE NOTE Covariant Derivative

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Lecture Note: Covariant Derivative

KalyanmoyChatterjee
March 29, 2023

1 Introduction
Taking Derivative of a vector or a vector filed is prettymuch straight forward
in cartesian co-ordinate, where a set of basis vectors is constant through
out the space (e, g, î, ĵ, k̂ is same in magnitude and direction). However,
in case of curved coordinate system, such as spherical-polar or cylindracal
coordinate system, it is not so simple, because the basis vectors point to
different directions in different points in that space.
Consider a vector expressed in terms of general co-ordinate x1 , x2 , x3 and
basis e~1 , e~2 , e~3 :
~ = A1 e~1 + A2 e~2 + A3 e~3
A
= Ai ei

~ with respect to any coordinate xj is:


Derivative of A

~
∂A ∂Ai e~i
=
∂xj ∂xj
∂Ai i ∂e
~i
= e
~i + A (1)
∂xj ∂xj
The second term in this equation goes to zero in cartesian coordinate system,
but makes the life complecated in curvilinear coordinate system. Similar
problem occure for higher rank tensor aswell. Covariant derivative, takes
care for such change in the basis vectors and ensures that the derivative of a
tensor is always another tensor.

1
2 Christofell symbol
To understand the covariant derivative, first we have to understand the
Christofell symbol. Keeping the fact in mind that the derivative of a ba-
sis vectror will also be a vector, and could be written in terms of weighted
sum of the basis vecrotrs, we can write:
∂ e~i
= Γkij .e~k (2)
∂xj
where the weight factor Γkij is the Christofell symbol. Index i specifies the
basis vector for which the derivative is being taken, the index j denotes the
coordinate being varied to induce this change in the ith basis vector, and
the index k identifies the direction in which this component of the derivative
points.

Example: if e~r and e~θ are two basises in 2D, then:


∂ e~r
= Γkrθ .e~k
∂ e~θ
= Γrrθ .~
er + Γθrθ .e~θ

2.1 Relation between metric tensor and Christofell sym-


bol
It takes a bit algebra to find the realtion between metric tensor and the
Christofell symbol, which is given by :
 
l 1 kl ∂gik ∂gjk ∂gij
Γij = g + − k (3)
2 ∂xj ∂xi ∂x
We can aslo write:
 
1 ∂gik ∂gjk ∂gij
Γijk = + − k (4)
2 ∂xj ∂xi ∂x
Γlij = g kl Γijk (5)
Γlij is called the Christofell symbol of second kind where as Γijk is known as
the Christofell symbol of first kind.

Example In cylindrical coordinate system, differential lenth is given by


ds2 = dr2 + r2 dφ2 + dz 2 . Find out the the covariant metric tensor
and the contravariant metric tensor. From there, find out Γ111 and Γ122 .

2
3 Covariant Derivative
In Euclidean space, two vectors at different locations may be compared and
combined by dragging one of the vectors to the location of the other without
changing its magnitude or its direction. If the vector is expanded using
Cartesian coordinates, such “parallel transport” is accomplished simply by
keeping each of its components the same (because the Cartesian basis vectors
have the same magnitude and direction everywhere). But if the vector is
expressed in non-Cartesian coordinates, the length and direction of the basis
vectors may be different at the two locations. In such cases, the covariant
derivative provides a means of parallel-transporting one of the vectors to the
location of the other. [?]
You can understand the role of the covariant derivative by considering
a two-dimensional spherical surface embedded in a three- dimensional Eu-
clidean space. Imagine a series of tangent planes just touching the sphere at
each location, and picture a vector lying in one of those tangent planes. If
that vector is moved to a different location on the sphere while holding its
direction constant (as viewed in the larger three-dimensional space), it will
not lie in the tangent plane at the new location (you can think of the vector
as “sticking out” of the two-dimensional space of the sphere). In such cases,
the covariant derivative serves to project the derivative of the vector into the
tangent space of the sphere. [?].
Equation 1 and 2 could be writen as:
∂A~ ∂Ai ~i
= e + Ai Γkij e~k
∂xj ∂xj
Interchanging the dummy variable i and k:
~
∂A ∂Ai ~i
= e + Ak Γikj e~i
∂xj ∂xj
∂Ai

k i
= + A Γkj e~i (6)
∂xj
The covariant derivative is defined as the combination of the terms inside the
parentheses and denoed by asemicolon (;) in front of the index with respect
to which the covariant derivative is being taken.
∂Ai
Ai;j = + Ak Γikj (7)
∂xj
From a similar analysis, co-variant derivative of covariant components
will be:
∂Ai
Ai;j = j
− Ak Γkij (8)
∂x
3
Example In a cylindracal corodinate system a vecrotrs A~ is expressed by
the coordinates x = r, x = φ, x = z. find (a) A;φ and (b) Aφ;φ
1 2 3 r

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