Contravariant Vector Transformation in Spherical Polar Coordinates

In summary, the conversation focuses on the concept of position vector and its components in different coordinate systems. It is clarified that position is not always considered as a vector, but in affine spaces, it can be represented as a vector field. The transformation of contravariant vectors and their components is also discussed, along with the difference between position vectors and coordinate tuples. There is a mention of parallel-transport and the importance of considering coordinate-independent vector fields.
  • #1
Apashanka
429
15
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??
 
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  • #2
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
Orodruin said:
Are you intending to write ¯x1x¯1\bar x^1?
Yes sir that is...
 
  • #4
Orodruin said:
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
Apashanka said:
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
Apashanka said:
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
stevendaryl said:
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
Apashanka said:
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
haushofer said:
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
Apashanka said:
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
I am sorry, it is still completely impossible to parse what you are asking because what you are asking.
 
  • #12
Orodruin said:
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.
 
  • #14
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.
 

What is a contravariant vector transformation?

A contravariant vector transformation is a mathematical operation that allows for the conversion of vectors between different coordinate systems. It is used to describe how a vector's components change when the coordinate system is changed.

What are spherical polar coordinates?

Spherical polar coordinates are a type of coordinate system used to describe points in three-dimensional space. They consist of a radial distance, an azimuthal angle, and a polar angle, and are often used in physics and engineering to describe spherical objects or phenomena.

Why is contravariant vector transformation important in spherical polar coordinates?

Contravariant vector transformation is important in spherical polar coordinates because it allows for the representation of vectors in this coordinate system. This is particularly useful in physics and engineering, where spherical objects or phenomena are common.

What is the formula for contravariant vector transformation in spherical polar coordinates?

The formula for contravariant vector transformation in spherical polar coordinates is:

Vr = Vxsinθcosφ + Vysinθsinφ + Vzcosθ
Vθ = Vxcosθcosφ + Vycosθsinφ - Vzsinθ
Vφ = -Vxsinφ + Vycosφ

What are some practical applications of contravariant vector transformation in spherical polar coordinates?

Some practical applications of contravariant vector transformation in spherical polar coordinates include describing the motion of objects in space, analyzing the forces acting on a spherical object, and calculating the electric or magnetic fields around a spherical charge or magnet. It is also used in fluid mechanics to describe the flow of fluids around a spherical object.

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