Solving for integral curves- how to account for changing charts?

[Shirish]

[Ref. 'Core Principles of Special and General Relativity by Luscombe]

Let $\gamma:\mathbb{R}\supset I\to M$ be a curve that we'll parameterize using $t$, i.e. $\gamma(t)\in M$. It's stated that:

If $\gamma(t)$ has coordinates $x^i(t)$ and [a vector field] $X$ has components $X^i$, finding the integral curve associated with $X$ reduces to solving a set of coupled first-order differential equations, $$\frac{d}{dt}x^i=X^i(x^1(t),\ldots,x^n(t))$$

Immediately after there's an example: if $X=x\partial_x+y\partial_y$, then $dx/dt=x$ and $dy/dt=y$, which gives the integral curve passing through $(a,b)$ at $t=0$ as $\gamma(t)=(ae^t,be^t)$.

1. Now from the context, provided we're talking about only one curve $\gamma$, shouldn't $X$ actually be the restriction of the vector field to the curve $\gamma$, rather than the vector field itself?
2. Referring to the phrase "If $\gamma(t)$ has coordinates $x^i(t)$...", I'm guessing it's unlikely that all the points on the curve belong to a single chart. So how can we claim only one coordinate system $x^i$ to represent the coordinates of all the points on the curve? Won't we have to adjust the coordinates according to the chart?
   
   e.g. if some $p,p'\in\gamma(I)$ are covered by different charts, and if the coordinates of $p$ are $x^i$, won't the coordinates of $p'$ have to be characterized by an entirely different coordinate system (e.g. some $y^i$)?

 

[Infrared]

1. No. You're starting with the vector field $X$ and solving for $\gamma.$ You don't have a curve to begin with.

2. Given $X$, you're looking for a curve $\gamma$ with $X(\gamma(t))=\gamma'(t)$ for all $t$. A solution to this equation will satisfy your quoted equation in every coordinate system. You're right that in general the image of the curve won't lie inside of a single coordinate chart.

 

[Shirish]

1. No. You're starting with the vector field $X$ and solving for $\gamma.$ You don't have a curve to begin with.

2. Given $X$, you're looking for a curve $\gamma$ with $X(\gamma(t))=\gamma'(t)$ for all $t$. A solution to this equation will satisfy your quoted equation in every coordinate system. You're right that in general the image of the curve won't lie inside of a single coordinate chart.

What do you mean by $X(\gamma(t))$? Is it that the coordinates are given by $X((x^i\circ\gamma)(t))$? That's the only way I can think of that'll make it chart independent.

 

[Infrared]

A vector field assigns to each $x\in M$ a tangent vector in $T_xM$. By $X(\gamma(t))$, I mean the tangent vector that $X$ assigns to the point $\gamma(t)$.

 

[Shirish]

A vector field assigns to each $x\in M$ a tangent vector in $T_xM$. By $X(\gamma(t))$, I mean the tangent vector that $X$ assigns to the point $\gamma(t)$.

Yes, that's clear to me so far.

In regards to the example in the OP, consider any point $p_0\in M$. So can I say that whatever coordinate system $\{x, y\}$ is used at $p_0$ (in accordance with whatever chart covers it), the coordinates of $X_{p_0}$ in the corresponding coordinate basis (corresponding to the coordinate system) will be $x,y$?

Essentially this means that if I use some other coordinate system $\{u,v\}$ at some other point $p_1$, then the coordinates of $X_{p_1}$ will now be $u,v$.

And what this implies for the integral curve that we calculate, i.e. $(ae^t,be^t)$, is that if $\gamma(t_0)=p_0$ and $\gamma(t_1)=p_1$, then $(ae^{t_0},be^{t_0})$ are the coordinates of the integral curve at $p_0$ in the coordinate basis $(x,y)$, while $(ae^{t_1},be^{t_1})$ are the coordinates of the integral curve at $p_1$ in the coordinate basis $(u,v)$. In essence, we're coming up with local solutions to the curve and "stitching them together".

Does that sound correct so far?