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I'm studying 'A Most Incomprehensible Thing - Notes towards a very gentle introduction to the mathematics of relativity' by Collier, specifically the section 'More detail - contravariant vectors'.

To give some background, I'm aware that basis vectors in tangent space are given by $\big\{\frac{\partial}{\partial x^i}\big\}$. I'm also aware that if we act these operators on the coordinate functions $x^i$, then we get a specific basis $\{\vec e_i\}$ whose elements are tangent to the coordinate curves that that point. This specific basis is commonly used as far as I understand.

Then in the 'More detail - contravariant vectors' section:

We can now state that a contravariant vector is a tangent vector to a parameterised curve. Let's see how this works. If the parameter of the curve is $\lambda$, and using a coordinate system $x^i$ the components of the tangent vector $\vec V$ are given by $$V^i=\frac{dx^i}{d\lambda}$$

I'm a little confused by this paragraph. Down the line there's another related paragraph:

In more advanced texts you may see the vector $\vec V$ written as $$\vec V=V^i\frac{\partial}{\partial x^i}$$ where $V^i$ are the vector's components and the partial derivative operators $\frac{\partial}{\partial x^i}$ are the coordinate basis vectors. In order to make sense of this formulation, consider an infinitesimal displacement $df$ at a point $p$ on the manifold, where $f$ is a function of some coordinate system $x^i$. If the displacement is along a curve parameterised by $\lambda$ we can drop $f$ into the above to get $$\vec V=V^i\frac{\partial f}{\partial x^i}=\frac{dx^i}{d\lambda}\frac{\partial f}{\partial x^i}=\frac{df}{d\lambda}$$ where $\frac{\partial f}{\partial x^i}$ are the coordinate basis vectors at $p$.

Again, I'm not able to make much sense of the whole parameterisation w.r.t. $\lambda$ thing.

First question: Why does he say that a contravariant vector is tangent to a parameterised curve? I can understand that a contravariant vector, being a member of the tangent space at some point $p$, must be tangent to some curve passing through $p$. But what makes us say that it should be parameterised?

Second question: Why does he claim that $\partial f/\partial x^i$ are basis vectors? I don't see a reason why, for an arbitrary function $f$, $\partial f/\partial x^i$ should form basis vectors. And why does he claim that $V^i=dx^i/d\lambda$ are the components w.r.t. this specific basis?

Would appreciate some guidance!

  • I'm not sure about Collier in particular, but typically $\partial_i$, $\vec{e}_i$ and $\frac{\partial}{\partial x^i}$ are just different notation for the same objects. – Kajelad Jun 19 '20 at 12:24
  • I'm reading two books - one Collier and another one on Relativity (Luscombe). In both of them $\vec e_i$ are explicitly defined as the operators acting on coordinate functions, i.e. $\vec e_i=\frac{\partial\mathbf{r}}{\partial x^i}$. So they're different objects from the operators $\frac{\partial}{\partial x^i}$ and $\partial_i$, at least in the context of those two books – Shirish Kulhari Jun 19 '20 at 12:29
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    I've been in the same situation as you right now, and if i could, i will advice myself in the past to read a proper material on smooth manifolds rather than physics version of this (which i found it confusing and unsatisfying). But some people fine with this so pick your path. – Kelvin Lois Jun 19 '20 at 12:46
  • @Eumenes: I mean what I'm doing currently is, if there's any claim that's taken for granted in Physics book (e.g. vector/covector transformation relations), then I try to study slightly more advanced stuff to understand the justification/derivation of such a claim (e.g. in case of transformation relations, I studied the basics of diff geom - tangent spaces, definition of vector/covector bases as operators/differentials, which gives an idea on how to derive transformation relations). Having said that, I don't want to get way too deep into math right at the outset, if that makes sense – Shirish Kulhari Jun 19 '20 at 12:51
  • @Eumenes: But you're right, some times the Physics version is unsatisfying/confusing - especially annoying when they throw a claim at you out of nowhere. – Shirish Kulhari Jun 19 '20 at 12:53

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