Page 183 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes Now suppose we want to bind our tensor to some point in space, then another tensor to another
point and so on. Doing so we can fill our space with tensors, one per each point. In this case, we
say that we have a tensor field. In order to mark a point P to which our particular tensor is bound
we shall write P as an argument:
X = X(P) ...(1)
Usually the valencies of all tensors composing the tensor field are the same. Let them all be of
type (r, s). Then if we choose some basis e , e , e , we can represent any tensor of our tensor field
2
1
3
1 i ...i
as an array X 1. j ..j r s with r + s indices:
1 i ...i
X j1 1 i ...i s r  X 1 j ...j r s (P). ...(2)
. ..j
Thus, the tensor field (1) is a tensor-valued function with argument P being a point of three-
dimensional Euclidean space E, and (2) is the basis representation for (1). For each fixed set of
numeric values of indices i , ... , i , j , ... , j in (2), we have a numeric function with a point-valued
s
1
r
1
argument. Dealing with point-valued arguments is not so convenient, for example, if we want
to calculate derivatives. Therefore, we need to replace P by something numeric. Remember that
we have already chosen a basis. If, in addition, we fix some point O as an origin, then we get

Cartesian coordinate system in space and hence can represent P by its radius-vector r  OP and
p
by its coordinates x , x , x :
3
1
2
X 1 i ...i r  X 1 i ...i r (x ,x ,x ). ...(3)
3
2
1
1 j ...j s 1 j ...j s
Conclusion. In contrast to free tensors, tensor fields are related not to bases, but to whole
coordinate systems (including the origin). In each coordinate system they are represented by
functional arrays, i.e. by arrays of functions (see (3)).
A functional array (3) is a coordinate representation of a tensor field (1). What happens when we
change the coordinate system ? Dealing with (2), we need only to recalculate the components of
the array X 1 i ...i r s :
1 j ...j
3 3
...
r i
k
1 i
X 1 i ...i r (P)    T ...T S ...S X h 1 ...h r (P). ...(4)
s k

1
1 j ...j s h 1 r h 1 j s j k 1 ...k s
h 1 , r h
In the case of (3), we need to recalculate the components of the array X 1 i ...i r in the new basis
1 j ...j s
3 3
X 1 i ...i r (x ,x ,x )    T ...T S ...S X h k 1 ...k r (x ,x ,x ), ...(5)
...
1 ...h
1
1
2
r i
k
s k
1 i

3
2
3

 
1
1 j
h
r h
s j
1 j ...j
s
h 1 , r h 1 s
We also need to express the old coordinates x , x , x of the point P in right hand side of (5)
2
1
3
through new coordinates of the same point:
1
2
1
3
1
 x = x (x , x , x ),



 2 2 1 2 3



 x = x (x , x , x ), ...(6)
 x = x (x , x , x ).
1
2
3
3
3
   
Formula (5) can be inverted :
3 3
X 1 i ...i r (x ,x ,x )  ... S ...T S ...T X h k 1 ...k r (x ,x ,x ). ...(7)
3
1 i
s k 
r i
1 ...h
1
2
k
3  
 

1
h
r h
1 j
s j
1
2
1 j ...j
s
h 1 , r h 1 s
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