Transformation rule of contravariant vector

  • #1
403
14

Main Question or Discussion Point

In a spherical polar coordinate system if the components of a vector given be (r,θ,φ)=1,2,3 respectively. Then the component of the vector along the x-direction of a cartesian coordinate system is $$rsinθcosφ$$.
But from the transformation of contravariant vector $$A^{-i}=\frac{∂x^{-i}}{∂x^j}A^j$$ where $$A^1=1 ,A^2=2 ,A^3=3$$ from this transformation if $$x^{-1}=x$$ of the cartesian coordinate system then $$A^{-1}=sin\theta cos\phi+2rcos\theta cos\phi-3rsin\theta sin\phi$$(component along x of cartesian coordinate system)
Am I missing out something??
 

Answers and Replies

  • #2
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,451
6,255
I am sorry, but it is completely impossible to parse what you are trying to say here. The x-component of the position vector expressed in polar coordinates is ##x = r\sin(\theta)\cos(\varphi)##. The vector components of A that you are writing down are not the components of the position vector. It is also impossible to know what you are referring to with ##x^{-1}##. Are you intending to write ##\bar x^1##?
 
  • #3
403
14
Are you intending to write ¯x1x¯1\bar x^1?
Yes sir that is.....
 
  • #4
403
14
The vector components of A that you are writing down are not the components of the position vector.
But sir position is a contravariant vector
 
  • #5
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,451
6,255
But sir position is a contravariant vector
So what? You have chosen the conponents to be particular values that do not agree with the components of the position vector apart from in a single point.
 
  • #6
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
8,401
2,576
But sir position is a contravariant vector
Position is actually NOT a vector at all. Vectors transform linearly under coordinate transformations, while a position "vector" can transform nonlinearly.
 
  • #7
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,451
6,255
Position is actually NOT a vector at all. Vectors transform linearly under coordinate transformations, while a position "vector" can transform nonlinearly.
Well, the OP is dealing with spherical coordinates on ##\mathbb R^3##, which is an affine space and therefore has a position vector field (i.e., the vector field that at each point takes the value of the translation vector from the origin). What is important to note is that this position vector not necessarily has the coordinates ##x^i## as its components.
 
  • #8
haushofer
Science Advisor
Insights Author
2,255
590
But sir position is a contravariant vector
Can you show this explicitly by the transformation rules? Do the components of a position 'vector' transform covariantly?

Hint: they do not. Hence my '...'.
 
  • #9
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,451
6,255
Can you show this explicitly by the transformation rules? Do the components of a position 'vector' transform covariantly?

Hint: they do not. Hence my '...'.
In order for a position vector to exist, the space must be affine. Such a position vector does have components that transform covariantly. Those components are not the coordinates themselves, see #7.

Example: The ##\mathbb R^3## position vector is given by ##r\vec e_r## in spherical coordinates. This is a perfectly fine vector field.
 
  • #10
403
14
In a spherical polar coordinate system if the components of a vector given be (r,θ,φ)=1,2,3 respectively. Then the component of the vector along the x-direction of a cartesian coordinate system is $$rsinθcosφ$$.
But from the transformation of contravariant vector $$A^{-i}=\frac{∂x^{-i}}{∂x^j}A^j$$ where $$A^1=1 ,A^2=2 ,A^3=3$$ from this transformation if $$x^{-1}=x$$ of the cartesian coordinate system then $$A^{-1}=sin\theta cos\phi+2rcos\theta cos\phi-3rsin\theta sin\phi$$(component along x of cartesian coordinate system)
Am I missing out something??
Is it therefore the tangent vector to a curve which is contravariant transform this way....??
 
  • #11
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,451
6,255
I am sorry, it is still completely impossible to parse what you are asking because what you are asking.
 
  • #12
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
8,401
2,576
In order for a position vector to exist, the space must be affine. Such a position vector does have components that transform covariantly. Those components are not the coordinates themselves, see #7.

Example: The ##\mathbb R^3## position vector is given by ##r\vec e_r## in spherical coordinates. This is a perfectly fine vector field.
Okay. My comment was really about an ambiguity about what "position vector" means. You're right, that in a flat spacetime (or more generally, an affine space, I guess---does that just mean that there is a path-independent notion of parallel-transport?) you can associate a position with the vector "pointing" from the origin to the position. This is a true vector. But sometimes people sloppily talk about the n-tuple of coordinates ##(x^1, x^2, ..., x^n)## as a "position vector". The former is a true vector (assuming parallel transport is path-independent), while the latter isn't. In the special case of Cartesian coordinates in flat spacetime, the two types of n-tuples (components of the position vector and n-tuple of the coordinates) coincide.
 
  • #13
haushofer
Science Advisor
Insights Author
2,255
590
That was also my intent, Stevendaryl ;)
 
  • #14
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,451
6,255
Of course I agree that ##x^i \vec E_i## does not generally describe a coordinate independent vector field. It is of course possible to have a vector field that takes this form in a particular set of coordinates (e.g., the position vector in Cartesian coordinates), but the field will not be expressed like this in an arbitrary coordinate system.
 

Related Threads for: Transformation rule of contravariant vector

Replies
16
Views
8K
Replies
4
Views
2K
  • Last Post
Replies
19
Views
6K
Replies
8
Views
690
Replies
4
Views
3K
Replies
4
Views
2K
  • Last Post
Replies
7
Views
4K
Replies
10
Views
1K
Top