Contravariant Vector Transformation in Spherical Polar Coordinates

[Apashanka]

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??

 

[Orodruin]

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 $\bar x^1$?

 

[Apashanka]

Are you intending to write ¯x1x¯1\bar x^1?

Yes sir that is...

 

[Apashanka]

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

 

[Orodruin]

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.

 

[stevendaryl]

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.

 

[Orodruin]

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.

 

[haushofer]

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 '...'.

 

[Orodruin]

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.

 

[Apashanka]

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...??

 

[Orodruin]

I am sorry, it is still completely impossible to parse what you are asking because what you are asking.

 

[stevendaryl]

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.

 

[haushofer]

That was also my intent, Stevendaryl ;)

 

[Orodruin]

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.