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

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  • Thread starter Shirish
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  • #1
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[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##)?
 

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  • #2
Infrared
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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.
 
  • #3
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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.
 
  • #4
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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)##.
 
  • #5
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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?
 

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