mathmatical physics

1. NPTEL – Physics – Mathematical Physics - 1
Lecture 26
Covariant and Contravariant Vectors
Let us remind ourselves of the scalar function. A scalar function is
physical quantity, such as height of a hill, temperature of a system etc.
which are function of coordinates in a particular coordinate system. Thus let us
define a scalar function 𝛷(𝑥1, 𝑥2, 𝑥3) in a 𝑉3 . In another (barred) coordinate
system, this is defined as 𝛷(𝑥̅1, 𝑥̅2, 𝑥̅3). But the value of the scalar function
should be independent of the coordinate system. Hence
𝛷(𝑥̅1, 𝑥̅2, 𝑥̅3) = 𝛷(𝑥1, 𝑥2, 𝑥3)
If we want to find the gradient,
𝜕𝛷̅ 𝜕𝛷 𝜕𝑥1 𝜕𝛷 𝜕𝑥2 𝜕𝛷 𝜕𝑥3 𝜕𝛷 𝜕𝑥𝑗
𝜕𝑥̅𝑖 =
𝜕𝑥1 𝜕𝑥̅𝑖 +
𝜕𝑥2 𝜕𝑥̅𝑖 +
𝜕𝑥3 𝜕𝑥̅𝑖 =
𝜕𝑥𝑗 𝜕𝑥̅𝑖
=
𝜕𝑥𝑗 𝜕
𝛷
𝜕𝑥̅𝑖 𝜕𝑥𝑗
(1)
Compare this with the following case, consider a curve in space parameterized
in a Cartesian Coordinate system, 𝑥𝑖 = 𝑓𝑖 (𝑡), i = 1, 2, 3
𝑓1(𝑡), 𝑓2(𝑡) and 𝑓3(𝑡) are smooth functions of the parameter t. The tangent to
𝑖
this curve which is a vector has components 𝑥𝑖 = 𝑑𝑥
= 𝑓 ′
(𝑡).
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𝑑𝑡 𝑖
Again consider a new coordinate system,
𝑥̅𝑖 = 𝑔𝑖(𝑥1, 𝑥2, 𝑥3)
The curve can be represented in terms of new coordinates,
𝑥̅𝑖 = 𝑔𝑖(𝑓1(𝑡), 𝑓2(𝑡), 𝑓3(𝑡))
= ℎ𝑖(𝑡)
The components to the tangent to the curve in the primed coordinate system is
given by,

2. NPTEL – Physics – Mathematical Physics - 1
𝑥̅𝑖 = hi (𝑡) =
𝜕𝑔𝑖 𝑑𝑓1 𝜕𝑔𝑖 𝑑𝑓2 𝜕𝑔𝑖 𝑑𝑓3
𝜕𝑥1 𝑑𝑡 𝜕𝑥2 𝑑𝑡 𝜕𝑥3 𝑑𝑡
+ +
= 𝜕𝑥̅
𝑖
𝜕𝑥
𝑗
𝑥
𝑗
(2)
Transformation according to Eq. (1) is definitely distinct than that of Eq. (2).
Based on such transformations, we can define two kinds of vectors, such as,
(a)One whose components transform according to Eq. (1)
(b)Other whose components transform according to Eq. (2)
To emphasize on the ongoing discussion, let us look at the dot product of two
vectors. Let 𝐴⃑ and 𝐵⃗⃑ be vectors that transform according to Eq. (2),
⃗𝐴
⃗⃗⃑𝑖
=
𝜕𝑥̅𝑖
𝜕𝑥𝑗 𝐴𝑗
;
⃗𝐵
⃗⃗⃑𝑖
=
𝜕𝑥̅𝑖
𝜕𝑥𝑘 𝐵𝑘
Then the dot product is 𝐴⃗⃗⃗⃑𝑖
𝐵⃗⃑𝑖
(with sum over repeated indices
understood). In terms of the unprimed coordinates,
⃗𝐴⃗⃗⃑𝑖 ⃗
𝐵⃗⃗⃑𝑖 =
𝜕𝑥̅𝑖 𝜕𝑥̅𝑖
𝜕𝑥𝑗 𝜕𝑥
𝑘
𝐴𝑗 𝐵𝑘
=
𝜕𝑥̅𝑖 𝜕𝑥̅𝑖
𝜕𝑥𝑗 𝜕𝑥
𝑘
𝐴𝑗 𝐵
𝑘
(3)
The RHS of Eq. (3) does not reduce to a dot product.
Next consider two vectors (𝑉⃗⃑ and 𝑈⃗⃑) where 𝑉⃗⃑ transforms according to
Eq. (1) and 𝑈⃗⃑ transforms according to Eq. (2). So,
𝑉̅𝑖 = 𝑉𝑘 ;
𝜕𝑥
𝑘
𝜕𝑥̅𝑖
⃗𝑈
⃗⃗⃑𝑖
=
𝜕𝑥̅𝑖
𝜕𝑥𝑗 𝑈
𝑗
Now take the dot product,
𝑈̅𝑖𝑉̅𝑖 = 𝑈𝑗 𝑉𝑘 =
𝜕𝑥̅𝑖
𝜕𝑥𝑗 𝜕𝑥̅𝑖
𝜕𝑥
𝑘
𝜕𝑥𝑘 𝜕𝑥̅𝑖
𝜕𝑥̅𝑖 𝜕𝑥𝑗
𝑈𝑗 𝑉
𝑘
Thus 𝑈̅𝑖𝑉̅𝑖 = 𝑈𝑗 𝑉𝑘 = 𝛿𝑗
𝑘
𝑈𝑗 𝑉
𝑘
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𝜕𝑥
𝑘
𝜕𝑥
𝑗
= 𝑈𝑗 𝑉
𝑗

3. NPTEL – Physics – Mathematical Physics - 1
Now the RHS is indeed a dot product. Thus we shall get a dot product only if
one of the vectors transform according to Eq. (1) and the other according to Eq.
(2).
It may be noted that the analysis can trivially be extended to an n- dimensional
space, namely, 𝑉𝑁. To be specific general theory of relatively demands a four
dimensional space time.
With the above notation in mind, we define two kinds of vectors namely,
𝐴⃑ = {𝐴1, 𝐴2 … … … . 𝐴𝑛} and 𝐵⃗⃑ = {𝐵1, 𝐵2 … … … 𝐵𝑛}
Which are defined as the contrvariant and covariant vectors respectively. They
transform according to,
𝐴̅𝑖 = 𝜕𝑥̅
𝐴𝑗 and
𝑖
𝜕𝑥𝑗 𝜕𝑥̅𝑖
𝐵 = 𝜕𝑥
𝐵
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̅𝑖
𝑗
𝑗
They are thus distinguished from placement of indices. Only when
a contravariant vector (with upper index) appears with a covariant vector (with
a lower index) in a sum, the result is independent of the coordinate system.