1) The document discusses tensors with multiple indices and the cross product of two vectors A and B. The components of the cross product vector C are given by Ai Bj - Aj Bi. 2) It describes how tensor components transform between coordinate systems using transformations of partial derivatives. The transformation property for cross products is derived. 3) Tensors are defined by their rank, with the number of covariant and contravariant indices specifying a tensor's rank. Vectors have a rank of 1. Examples calculate tensor components in different coordinate systems.

1. NPTEL – Physics – Mathematical Physics - 1
Lecture 27
Tensors with multiple indices – cross product
For two vectors and
𝐴⃑ 𝐵⃗⃑, the components of the vector 𝐶⃑(= 𝐴⃑ × 𝐵
⃗⃑) is
𝐴𝑖 𝐵𝑗 − 𝐴𝑗 𝐵𝑖. In the primed coordinate system, the components are 𝐴𝑖 𝐵𝑗 − 𝐴𝑗 𝐵𝑖.
Using the transformation discussed above,
𝐴̅𝑖𝐵̅𝑗 = 𝐴𝑘 𝐵ℎ
𝜕𝑥𝑘 𝜕𝑥ℎ
𝜕𝑥̅𝑖 𝜕𝑥̅𝑗
=
𝜕𝑥𝑘 𝜕𝑥ℎ
𝜕𝑥̅𝑖 𝜕𝑥̅𝑗 𝐴𝑘𝐵ℎ
Similarly 𝐴̅𝑗𝐵̅𝑖 =
𝜕𝑥𝑘 𝜕𝑥ℎ
𝜕𝑥̅𝑗 𝜕𝑥̅𝑖 𝐴𝑘𝐵ℎ
Changing the dummy indices k and h,
𝐴̅𝑗𝐵̅𝑖 =
𝜕𝑥ℎ 𝜕𝑥𝑘
𝜕𝑥̅𝑗 𝜕𝑥̅𝑖 𝐴ℎ𝐵𝑘
Thus 𝐴̅𝑖𝐵̅𝑗 − 𝐴̅𝑗𝐵̅𝑖 =
𝜕𝑥𝑘 𝜕𝑥ℎ
𝜕𝑥̅𝑖 𝜕𝑥̅𝑗 (𝐴𝑘𝐵ℎ − 𝐴ℎ𝐵𝑘)
Define 𝐶𝑘ℎ = 𝐴𝑘𝐵ℎ − 𝐴ℎ𝐵𝑘
Thus the transformation properly for a cross product is,
𝐶𝑖𝑗
=
𝜕𝑥𝑘 𝜕𝑥ℎ
𝜕𝑥̅𝑖 𝜕𝑥̅𝑗 𝐶𝑘ℎ
Rank of a tensor
A tensor T is said to have a rank (r + s) when 𝑛𝑟+𝑠 quantities 𝑇
𝑗 …………..
𝑗
1 𝑠
𝑖1…………𝑖
𝑟
define the components of a tensor which transform according to,
𝑇
𝑗 1…………..𝑗
𝑠
𝑖1…………𝑖𝑟
=
𝜕𝑥̅ 𝑖1
𝜕𝑥𝑛1
𝜕𝑥̅
… … … …
𝑖
𝑟
𝜕𝑥ℎ𝑟 𝜕𝑥̅ 𝑗1
… … … … …
𝜕𝑥̅ 𝑗𝑠
𝜕𝑥 𝜕𝑥
𝑘
𝑖
𝑘𝑠
𝑇ℎ1……………ℎ𝑟
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𝑘1……………..𝑘𝑠
{𝑖1 … … … … . 𝑖𝑟} and {𝑗1 … … … … 𝑗𝑠} are called the contravariant and covariant
indices respectively. Thus a vector is a tensor of rank 1.
Example 1.
Suppose that the components of a contravariant tensor, T of rank 2 in a
coordinate system (𝑥𝑖) of 𝑉2 are 𝑇11 = 1, 𝑇12 = 1, 𝑇21 = −1 and 𝑇22 = 2.

2. NPTEL – Physics – Mathematical Physics - 1
Find the components 𝑇̅ 𝑖𝑗 of T in the (𝑥̅𝑖 )- system, which is related to the (𝑥𝑖)
system by,
𝑥̅1 = (𝑥1)2 ≠ 0
𝑥̅2 = 𝑥1𝑥2
Further compute the value of 𝑇̅ 𝑖𝑗 at 𝑥1 = 1, 𝑥2 = −2
Solution
𝑇̅ 𝑖𝑗 = 𝑇𝑟𝑠 𝜕𝑥̅ 𝑖
𝑗
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𝜕𝑥 𝜕𝑥
𝑟 𝑠
𝜕𝑥̅
= 𝐽𝑟
𝑗
𝑖 𝑇𝑟𝑠𝐽
𝑠
where
𝐽 =
𝑟
𝑖 𝜕𝑥̅
𝜕𝑥𝑟 ; 𝐽 =
𝑗
𝑠 𝜕𝑥̅𝑗
𝜕𝑥𝑠
Thus, 𝑇̅ = 𝐽𝑇𝐽𝑇
= ( 2𝑥1 0
𝑥2 𝑥1) (
1 1
−1 2 0
) (
2𝑥1 𝑥2
𝑥1
)
4(𝑥1)2
= (
2𝑥1𝑥2 − 2(
𝑥1)2
2𝑥1𝑥2 + 2(𝑥1)2
2(𝑥1)2 + (𝑥2)2)
At (1, -2)
𝑇̅11 = 4, 𝑇̅12 = −2, 𝑇̅ 21 = −6, 𝑇̅ 22 = 6
Example: 2
Compute the components of a symmetric tensor 𝑇̅ 𝑖 in polar coordinates
𝑗
in terms of 𝑇𝑗 in rectangular coordinates.
Solution: The components of the tensor 𝑇̅ holds the following relation
with those of T,
𝑖
𝑇 = 𝑇
̅ 𝑗 𝑠
𝑖 𝑟 𝜕𝑥̅ 𝜕𝑥
𝑖 𝑠
𝜕𝑥 𝜕𝑥̅ 𝜕𝑥 𝑠
𝜕𝑥̅ 𝑗
𝑟 𝑗 𝑟
= 𝑇
𝜕𝑥̅𝑖 𝑠
𝑟 𝜕𝑥
The symmetric property implies
𝑇𝑗 = 𝑇
𝑖
𝑖 𝑗
Using the transformation,

3. NPTEL – Physics – Mathematical Physics - 1
𝑇̅ = 𝐽𝑇𝐽−1 = 𝐽−̅ 1𝑇𝐽̅
Here the transformation matrix J is written as
𝐽̅ = ( 𝑐𝑜𝑠 𝜃 − 𝑟 𝑠𝑖𝑛 𝜃
𝑟 𝑐𝑜𝑠 𝜃
𝑠𝑖𝑛 𝜃
)
Which transforms from the polar to the Cartesian coordinate system as,
𝑐𝑜𝑠 𝜃
𝑇̅ =
(−𝑠𝑖𝑛
𝜃
𝑠𝑖𝑛 𝜃
𝑐𝑜𝑠 𝜃
𝑟 𝑟
Usual matrix multiplication yields,
) (
𝑇1 𝑇1
1 2
2
𝑇1 𝑇
2
2
) (
𝑐𝑜𝑠 𝜃 − 𝑟 𝑠𝑖𝑛 𝜃
𝑠𝑖𝑛 𝜃
𝑟𝑐𝑜𝑠 𝜃
)
𝑇1 𝑐𝑜𝑠 𝜃 + 𝑇2 𝑠𝑖𝑛2𝜃 + 𝑇2 𝑠𝑖𝑛 𝜃
𝑇̅ = (
1 2 1 2 2
− 𝑇1 𝑠𝑖𝑛2𝜃 + 𝑟𝑇2 𝑐𝑜𝑠2𝜃 + 𝑇2 𝑠𝑖𝑛2𝜃
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𝑟
2 2
1 1 2
𝑟
−𝑇1
1 𝑠𝑖𝑛2𝜃
2𝑟
+ 𝑇1
2 𝑐𝑜𝑠2𝜃 2 𝑠𝑖𝑛2𝜃
𝑟 2𝑟
+ 𝑇2
𝑇1𝑠𝑖𝑛2𝜃 − 𝑇1𝑠𝑖𝑛2𝜃 + 𝑇2𝑐𝑜𝑠2𝜃 1
2 2
)