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When we introduce coordinate systems, like spherical coordinates, one usually does it with respect to cartesian coordinates.

What would be the right way to derive the (for example) spherical coordinate basis of the tangent space at a point of a manifold(without using cartesian coordinates at all)? I mean, I have seen the definition of the tangent space and the coordinate basis, but how does one compute it in practice?

And deduce the metric tensor from the coordinate basis?

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This is not a proper answer to this post, but it might help the OP find out the answers they seek anyway. So, if we're doing differential geometry and we're considering the tangent space $T_P\mathcal{V}$ to some differentiable manifold $\mathcal{V}$, then any coordinate system $(q^1,\dots,q^n)$ on $\mathcal{V}$ the following vector basis on $T_P\mathcal{V}$, which we call the coordinate basis:

$$\left\{\left(\frac{\partial}{\partial q^1}\right)_P,\dots,\left(\frac{\partial}{\partial q^n}\right)_P\right\}.$$

When working in a manifold like the affine space $\mathbb{R}^n$, we can define a canonical coordinate system by taking $(0,\dots,0)$ as our origin and the canonical basis $\{\vec{u}_1,\dots,\vec{u}_n\}$ on the vector space $\mathbb{R}^n$ to associate the point $O+x^1\vec{u}_1+\cdots+x^n\vec{u}_n$ with the coordinates $(x^1,\dots,x^n)$.

We define the position vector on $\mathbb{R}^n$ as $\vec{x}=x^1\vec{u}_1+\cdots+x^n\vec{u}_n$, and we consider all the vectors in $T_P\mathbb{R}^n$ as al possible derivatives of this position vector. Now, every coordinate system $(q^1,\dots,q^n)$ which we would have defined from scratch in the former case can now be defined in terms of this canonical coordinate system, this is, as a system of equations of the form

$$x^i =\Phi^i(q^1,\dots,q^n), \space\space\space i=1,\dots,n.$$

I should note that the position vector only has components equal to the coordinates of the point it... points to, when we're considering cartesian coordinates. If you are familiar with the position vector in curvilinear coordinates, you OP must already know this.

Now, the coordinate basis on $T_P\mathbb{R}^n$ can be interpreted as the partial derivatives of $\vec{x} = \vec{\Phi}(q^1,...,q^n)$, which makes it

$$\left\{\left(\frac{\partial\vec{x}}{\partial q^1}\right)_P,\dots,\left(\frac{\partial\vec{x}}{\partial q^n}\right)_P\right\}.$$

This is how the coordinate basis is usually defined in physics. The fact that the position vector doesn't live in a tangent space whereas all the other important vectors (velocity, acceleration, momentum...) do is usually neglected, though. This two notations are not incompatible, in fact, they are the same if we write $\vec{u}_i$ (the canonical basis) as

$$\vec{u}_i = \left(\frac{\partial}{\partial x^i}\right)_P \implies \vec{e}_{\alpha} = \left(\frac{\partial\vec{x}}{\partial q^{\alpha}}\right)_P = \left(\frac{\partial x^i}{\partial q^{\alpha}}\right)_P\left(\frac{\partial}{\partial x^i}\right)_P=\left(\frac{\partial}{\partial q^{\alpha}}\right)_P$$

by virtue of the chain rule.

So, in principle, you can define spherical coordinates without going through canonical cartesian coordinates. Of course, you'll have to specify something about that coordinate system to know which is it. If we were to rely on canonical cartesian coordinates, we would define spherical coordinates with the coordinate transformation $\Phi\colon\space[0,\infty)\times[0,\pi]\times[0,2\pi) \longrightarrow \mathbb{R}^3$ such, that

$$\begin{cases} x^1=r\sin{\theta}\cos{\phi} \\ x^2=r\sin{\theta}\sin{\phi} \\ x^3=r\cos{\theta} \end{cases}$$

and from here we would obtain the coordinate basis doing

$$\vec{e}_i = \frac{\partial x^j}{\partial q^i}\vec{u}_j.$$

Is it even possible to define spherical coordinates without their change of basis to canonical cartesian coordinates? In this particular case, as $\theta$ and $\phi$ are defined as angles with respect to the canonical cartesian coordinate axis, it might be difficult, but we saw at the beginning of this answer that there exist manifolds without any canonical coordinate system, and they obviuosly can have coordinate systems and, thus, coordinate basis. So, in theory, this should be possible.

As for the metric tensor, you can find its components with respect to the coordinate basis (whenever you find it) by doing

$$g_{\alpha\beta} = g(\vec{e}_\alpha,\vec{e}_\beta)$$

thanks to bilinearity.

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