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Units, Constants and Useful Formulas
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Last modified: January 26, 2018
Contravariant and Covariant Components of a Vector
--------------------------------------------------
Consider cartesian coordinates.
^ X2
|
| -
x2|......c -
| / |c| = L
| /.
| L/ .
| / .
| / .
|/ .
--------------> X1
x1
L2 = (x1)2 + (x2)2
Now how do we get L when we have non-cartesian coordinates?
We need to introduce the concept of contravariant and
covariant components. In cartesian coordinates the
covariant and contravariant components are the same.
Contravariant components:
u1 = x1 - x2/tanα
u2 = x2/sinα
Covariant components:
u1 = x1
u2 = Lcos(α - β)
= Lcos(α)cos(β) + Lsin(α)sin(β)
Now Lcosβ = x1 and Lsinβ = x2
u2 = x1cosα + x2sinα
To get L2 try:
u1u1 = x1(x1 - x2/tanα) = (x1)2 - (x1x2/tanα)
u2u2 = (x1cosα + x2sinα)x2/sinα = (x1x2/tanα) + (x2)2
Therefore,
u1u1 + u2u2 = (x1)2 + (x1)2
Tranformation Matrices
----------------------
We can write:
- - - -
| u1 | = H | x1 |
| u2 | | x2 |
- - - -
where
- - - -
H = | ∂u1/∂x1 ∂u1/∂x2 | = | 1 -1/tanα |
| ∂u2/∂x1 ∂u2/∂x2 | | 0 1/sinα |
- - - -
and
- - - -
| u1 | = M | x1 |
| u2 | | x2 |
- - - -
where
- - - -
M = | ∂u1/∂x1 ∂u1/∂x2 | = | 1 0 |
| ∂u2/∂x1 ∂u2/∂x2 | | cosα sinα |
- - - -
H and M are transformation matrices (JACOBEAN MATRICES)
How do we convert between the covariant and contravariant
components and vice versa?
- - - -
| x1 | = H-1 | u1 |
| x2 | | u2 |
- - - -
and
- - - -
| x1 | = M-1 | u1 |
| x2 | | u1 |
- - - -
where H-1 and M-1 are the inverse of the JACOBEAN MATRICES
H and M. So,
- - - -
H-1 | u1 | = M-1 | u1 |
| u2 | | u2 |
- - - -
Therefore,
- - - -
| u1 | = HM-1 | u1 |
| u2 | | u2 |
- - - -
So,
- - - -
| u1 | = gij | u1 |
| u2 | | u2 |
- - - -
where gij is the INVERSE METRIC tensor.
In summary, we can construct a map as follows:
u1
u2
/ ^ ^
/ / | |
M-1/ /M | |
/ / | |
v / | |
x1 gij| |gij
x2 | |
^ \ | |
\ \ | |
H-1\ \H | |
\ \ | |
\ v v
u1
u2
From the map:
- -
gij = HM-1 where M-1 = (1/sinα)| sinα 0 |
| -cosα 1 |
- -
Therefore,
- -
gij = (1/sin2α)| 1 -cosα |
| -cosα 1 |
- -
Now gij = (gij)-1 (= H-1M from the map). Where gij is the
METRIC TENSOR. Therefore,
- -
gij = (1/det[gij])| 1/sin2α -cosα/sin2α |
| -cosα/sin2α 1/sin2α |
- -
det[gij] = 1/sin4α - cos2α/sin4α = sin2α/sin4α = 1/sin2α
So,
- -
gij = sin2α| 1/sin2α cosα/sin2α |
| cosα/sin2α 1/sin2α |
- -
- -
= | 1 cosα |
| cosα 1 |
- -
- - - -
Now gijgij = (1/sin2α)| 1 -cosα || 1 cosα |
| -cosα 1 || cosα 1 |
- - - -
- -
= (1/sin2α)| 1 - cos2α cosα - cosα |
|-cosα + cosα -cos2α + 1 |
- -
- -
= | 1 0 |
| 0 1 |
- -
Now gij = H-1M from the map so M = gijH
- - - -
= | 1 cosα || 1 -1/tanα |
| cosα 1 || 0 1/sinα |
- - - -
- -
= | 1 (-1/tanα + cosα/sinα) |
| cosα (-cosα/tanα + 1/sinα) |
- -
- -
= | 1 0 | which agrees with before.
| cosα sinα |
- -
Similarly, gij = HM-1 from the map so H = gijM
H = gijM
- - - -
= 1/sin2α | 1 -cosα || 1 0 |
|-cosα 1 ||cosα sinα |
- - - -
- -
= | 1 -1/tanα | which agrees with before.
| 0 1/sinα |
- -
In summary, the metric tensor enables us to convert
the contravariant components of a vector to their
covariant form, and vice versa, in the following
manner:
- - - - 2
gij | u1 | = | u1 | <--> Σgijuj = ui
| u2 | | u2 | j=1
- - - -
and
- - - - 2
gij | u1 | = | u1 | <--> Σgijuj = ui
| u2 | | u2 | j=1
- - - -
This process of going between contravariant and
covariant components, and vice versa, is referred
as the "raising and lowering of indeces".