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Relativity: Contravariant and covariant transforms



1. Lorentz Transformations Matrices

From the reference frame at rest (RF):
The velocity of the reference frame (RF) is 0
The velocity of the moving reference frame (MF) is v
The velocity of the moving object is u.

From the moving frame (MF):
The velocity of the moving reference frame (MF) is 0
The velocity of the reference frame (RF) is - v
The velocity of the moving object is u'


In the (MF):, Lorentz Transformations give:



ct' = γ(ct - βx)
x' = γ(x - β ct)
y' = y
z' = z

The contravariant 4-Vector V'(ct',x',y',z') is the transformed of he contravariant 4-Vector V(ct,x,y,z) by:

V' = Λ(- β) V

In the (RF), Lorentz Transformations give:


ct = γ(ct' + βx')
x = γ(x' + β ct')
y = y'
z = z'

The covariant 4-Vector V(ct,x,y,z) is the transformed of he covariant 4-Vector V(ct',x',y',z') by:

V = Λ(+β) V'

where
γ = 1/[1 - β2]1/2, and β = v/c



2. Contravariant Lorentz transformation

We have:
ct' = γ(ct - βx)
x' = γ(x - β ct)
y' = y
z' = z

Taking partial derivatives gives:

∂(ct')/∂(ct) = γ
∂(ct')/∂(x) = - γ β
∂(ct')/∂(y) = 0
∂(ct')/∂(z) = 0
∂(x')/∂(ct) = - γ β
∂(x')/∂(x) = γ
∂(x')/∂(y) = 0
∂(x')/∂(z) = 0
∂(y')/∂(ct) = 0
∂(y')/∂(x) = 0
∂(y')/∂(y) = 1
∂(y')/∂(z) = 0
∂(z')/∂(ct) = 0
∂(z')/∂(x) = 0
∂(z')/∂(y) = 0
∂(z')/∂(z) = 1


Therefore, the components of the matrix Λ(-β) are:

Λ(-β)00 = γ = ∂(ct')/∂(ct)
Λ(-β)01 = - γ β = ∂(ct')/∂(x)


Λ(-β)10 = - γ β = ∂(x')/∂(ct)
Λ(-β)11 = γ = ∂(x')/∂(x)

Λ(-β)22 = γ = ∂(y')/∂(y)
Λ(-β)33 = γ = ∂(z')/∂(z)
...
The others are zero.

If we denote by xi the contravariant components of V(ct,x,y,z) and by yi the contravariant component of V'(ct',x',y',z'), we can write the components of the matrix Λ(-β) as the following:

Λ(-β)ij = ∂(yi)/∂(xj)

i = 0,1,2,3 stands for line and j = 0,1,2,3 stands for column.

Therefore:
y0 = ΣΛ(-β)0j x0
y1 = ΣΛ(-β)1j x1
y2 = ΣΛ(-β)2j x2
y3 = ΣΛ(-β)3j x3


or:
yi = ΣΛ(-β)ij xi = Σ ∂(yi)/∂(xj) xi
or
V'i = Σ ∂(yi)/∂(xj) Vi

or
Vi(in the frame y) = Σ ∂(yi)/∂(xj) Vi(in the frame x)

Vi(y) = Σ ∂(yi)/∂(xj) Vi(x)

Using the Einstein summation convention (not writing the symbol Σ), we can write

Vμ(y) = (∂yμ/∂xν) Vν(x)



3. Covariant Lorentz transformation


WE have:
ct = γ(ct' + βx')
x = γ(x' + β ct')
y = y'
z = z'

Taking partial derivatives gives:



∂(ct)/∂(ct') = γ
∂(ct)/∂(x') = + γ β
∂(ct)/∂(y') = 0
∂(ct)/∂(z') = 0
∂(x)/∂(ct') = + γ β
∂(x)/∂(x') = γ
∂(x)/∂(y') = 0
∂(x)/∂(z') = 0
∂(y)/∂(ct') = 0
∂(y)/∂(x') = 0
∂(y)/∂(y') = 1
∂(y)/∂(z') = 0
∂(z)/∂(ct') = 0
∂(z)/∂(x') = 0
∂(z)/∂(y') = 0
∂(z)/∂(z') = 1


Therefore, the components of the matrix Λ(+β) are:

Λ(+β)00 = γ = ∂(ct)/∂(ct')
Λ(+β)01 = γ β = ∂(ct)/∂(x')


Λ(+β)10 = γ β = ∂(x)/∂(ct') Λ(+β)11 = γ = ∂(x)/∂(x')

Λ(+β)22 = γ = ∂(y)/∂(y')
Λ(+β)33 = γ = ∂(z)/∂(z')
...
The others are zero.

If we denote by xi the covariant components of V(ct,x,y,z) and by yi the contravariant component of V'(ct',x',y',z'), we can write the components of the matrix Λ(+β) as the following:

Λ(+β)ij = ∂(xi)/∂(yj)

i = 0,1,2,3 stands for line and j = 0,1,2,3 stands for column.

Therefore:
x0 = ΣΛ(+β)0j y0
x1 = ΣΛ(+β)1j y1
x2 = ΣΛ(+β)2j y2
x3 = ΣΛ(+β)3j y3


or:
xi = ΣΛ(+β)ij xi = Σ ∂(xi)/∂(yj) xi
or
V'i = Σ ∂(xi)/∂(yj) Vi

or
Vi(in the frame x) = Σ ∂(xi)/∂(yj) Vi(in the frame y)

Vi(x) = Σ ∂(xi)/∂(yj) Vi(y)

Using the Einstein summation convention (not writing the symbol Σ), we can write:


Vμ(x) = ∂(xμ/∂yν) Vν(y)



4. Contravariant and covariant transform: General case



If a vector A(x) has the contravariant components xμ in the curvilinear reference frame "x", its transformed contravariant vector A(y) in the curvilinear reference frame "y" has the components yν.

If A(x) is a scalar such as temperature, its related magnitude remains the same A(x) = A(y)

If A(x) is a vector, a little displacement dA(x) of A(x) in the coordinate reference frame "x" involves its related transformed displacement dA(y) of the transformed A(y) in the coordinate reference frame "y".

That is the displacements of the components dxμof dA(x) in the coordinate reference frame "x" involve dyν in the coordinate reference frame "y".

We can write the following total differential:
dA(x) = Σ (ν) (∂ A(x)/∂xμ) dxν =
(∂ A(x)/∂xμ) dxν

and

dA(y) = Σ (ν) (∂ A(y)/∂yμ) dxν =
(∂ A(y)/∂yμ) dyν

The transformed contravariant components of dA(y) are:
dyμ = Σ(ν)(∂ yμ)/∂xν) dxν =
(∂ yμ)/∂xν) dxν

The transformed contravariant components of A(y) are:
yμ = Σ(ν)(∂ yμ)/∂xν) xν =
(∂ yμ)/∂xν) xν

Therefore:
The transformed contravariant vector Aμ(y) is written as:
Aμ(y) = (∂ yμ)/∂xν) Aν(x)

Similarly;
The transformed covariant vector Aν(x) is written as:
Aν(x) = (∂ xν)/∂yμ) Aμ(y)

Aμ(y) = (∂yμ/∂xν) Aν(x)

Aν(x) = (∂xν/∂yμ) Aμ(y)







  


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