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Let the Cartesian coordinates , , be written as the , where runs from 1 to 3. In other words,
, , and . Incidentally, in the following, any lowercase roman subscript (e.g., , , ) is
assumed to run from 1 to 3. We can also write the Cartesian components of a general vector
as the . In other words, , , and . By contrast, a scalar is represented as
a variable without a subscript: e.g., , . Thus, a scalarwhich is a tensor of order zerois
represented as a variable with zero subscripts, and a vectorwhich is a tensor of order oneis
represented as a variable with one subscript. It stands to reason, therefore, that a tensor of order two is
represented as a variable with two subscripts: e.g., , . Moreover, an thorder tensor is represented as a variable with subscripts: e.g., is a thirdorder tensor, and a fourthorder tensor.
Note that a general thorder tensor has independent components.
Now, the
components of a secondorder tensor are conveniently visualized as a twodimensional matrix, just as
the components of a vector are sometimes visualized as a onedimensional matrix. However, it
is important to recognize that an thorder tensor is not simply another name for an dimensional matrix. A matrix is
just an ordered set of numbers. A tensor, on the other hand, is an ordered set of components
that have specific transformation properties under rotation of the coordinate axes. (See Section B.3.)
Consider two vectors and that are represented as and , respectively, in
tensor notation. According to Section A.6, the scalar product of these two vectors
takes the form

(1485) 
The above expression can be written more compactly as

(1486) 
Here, we have made use of the Einstein summation convention, according to which, in an expression
containing lower case roman subscripts, any subscript that appears twice (and only twice) in any
term of the expression is assumed to be summed from 1 to 3 (unless stated otherwise).
Thus,
, and
.
Note that when an index is summed it becomes a dummy index, and can be written as any
(unique) symbol: i.e., and are equivalent.
Moreover, only nonsummed, or free, indices count toward the order of a tensor expression. Thus,
is a zerothorder tensor (because there are no free indices), and is a firstorder tensor (because there
is only one free index). The process of reducing the order of a tensor expression by summing indices is known
as contraction. For example, is a zerothorder contraction of the secondorder tensor .
Incidentally, when two tensors are multiplied together without contraction the
resulting tensor is called an outer product: e.g., the secondorder tensor is the
outer product of the two firstorder tensors and . Likewise, when two tensors are multiplied
together in a manner that involves contraction then the resulting tensor is called an inner product:
e.g., the firstorder tensor is an inner product of the secondorder tensor and
the firstorder tensor . Note, from Equation (1486), that the scalar product of two
vectors is equivalent to the inner product of the corresponding firstorder tensors.
According to Section A.8, the vector product of two vectors and
takes the form
in tensor notation.
The above expression can be written more compactly as

(1490) 
Here,

(1491) 
is known as the thirdorder permutation tensor (or, sometimes, the thirdorder LeviCivita tensor). Note, in particular, that
is zero if one of its indices is
repeated: e.g.,
.
Furthermore, it follows from (1491) that

(1492) 
It is helpful to define the secondorder identity tensor (also known as the Kroenecker delta tensor),

(1493) 
It is easily seen that



(1494) 



(1495) 



(1496) 



(1497) 



(1498) 



(1499) 
etc.
The following is a particularly important tensor identity:

(1500) 
In order to establish the validity of the above expression, let us consider the various cases that arise.
As is easily seen, the righthand side of (1500) takes the values
Moreover, in each product on the lefthand side, has the same value in both factors. Thus,
for a nonzero contribution,
none of , , , and can have the same value as (because each factor is zero if any of its
indices are repeated). Since a given subscript
can only take one of three values (, , or ), the only possibilities that
generate nonzero contributions are and , or and , excluding (since
each factor would then have repeated indices, and so be zero). Thus, the lefthand side reproduces (1503), as well as the conditions on the indices in
(1501) and (1502). The lefthand side also reproduces the values in (1501) and (1502) since if and then
and the product
(no summation) is equal to , whereas
if and then
and the
product
(no summation) is equal to . Here, use has been made of Equation (1492).
Hence, the validity of the identity (1500) has been established.
In order to illustrate the use of (1500), consider the vector triple product
identity (see Section A.11)

(1504) 
In tensor notation, the lefthand side of this identity is written

(1505) 
where use has been made of Equation (1490). Employing Equations (1492) and (1500), this
becomes

(1506) 
which, with the aid of Equations (1486) and (1497), reduces to

(1507) 
Thus, we have established the validity of the vector identity (1504).
Moreover, our proof is much more rigorous than that given earlier (in Section A.11).
Next: Tensor Transformation
Up: Cartesian Tensors
Previous: Introduction
Richard Fitzpatrick
20120427