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Given a vector field $\mathcal{V}$ on a manifold $\mathcal{M}$ that assigns the vector $V_p$ to the point $p \in \mathcal{M}$, an integral curve is a curve

$$ \gamma : \mathbb{R} \to \mathcal{M} \\ t \mapsto \gamma(t) $$

such that $\forall p\in\mathrm{im}\left(\gamma\right),\ \gamma \in V_p$ (using the equivalence class definition of a vector).

Is this a valid definition of an integral curve? And if so, how does one recover the standard set of coordinate differential equations:

$$ \frac{\mathrm{d}x^{\mu}\left[\gamma(t)\right]}{\mathrm{d}t} = V^{\mu}(x) $$

from this definition?

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Okay I got it. A curve is a map

$$\begin{align} \xi:\mathbb{R} &\to \mathcal{M} \\ t &\mapsto \xi(t) \end{align}$$

and given a chart $x:\mathcal{M}\to\mathbb{R}^{n}$ on a coordinate patch $U_x \subset \mathcal{M}$, the coordinate representation of $\xi$ is

$$ \xi_x(t) \equiv x\left[\xi(t)\right] $$

and the derivative is

$$ \xi'_x(t) = \frac{\mathrm{d}\xi_x}{\mathrm{d}t}. $$

A vector, $V_p \equiv [\xi]$, at $p$ is the equivalence class of all curves tangent to one another at $p$:

$$ \xi_1, \xi_2 \in V_p \iff \left.\frac{\mathrm{d}\xi_{1x}}{\mathrm{d}t}\right|_{p} = \left.\frac{\mathrm{d}\xi_{2x}}{\mathrm{d}t}\right|_{p} $$

So then if there's some curve $\gamma$ such that $\forall p \in\mathrm{im}(\gamma),\ \gamma\in V_p \equiv [\xi_p]$ (where the $V_p$ are determined by a vector field $\mathcal{V}$ and $\xi_p$ is the representative curve of the class $V_p$):

$$ \left.\frac{\mathrm{d}\gamma_{x}}{\mathrm{d}t}\right|_p = \left.\frac{\mathrm{d}\xi_{px}}{\mathrm{d}t}\right|_p $$

Additionally, a vector can be defined as the directional derivative operator along a curve $\zeta$:

$$ \left.\frac{\mathrm{d}\zeta_{x}}{\mathrm{d}t}\frac{\partial}{\partial x}\right|_p $$

and this is invariant for two curves tangent at $p$. Hence, we can write:

$$ V_p = \left.\frac{\mathrm{d}\xi_{px}}{\mathrm{d}t}\frac{\partial}{\partial x}\right|_p $$

and identify $\frac{\mathrm{d}\xi_{px}}{\mathrm{d}t}$ as $V^{\mu}_p$, the components of $V_p$. Then the canonical form follows:

$$ \left.\frac{\mathrm{d}\gamma_{x}}{\mathrm{d}t}\right|_p = V^{\mu}_p\ \Box $$

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I really don't understand what you are doing. You say $\gamma(t) = p, \forall p \in \gamma(t)$, but this makes no sense! Has this been proposed as a new definition? Maybe you can correct it, but I'll have to know what was your intuition behind this definition? The regular defintion is clear enough though, $\gamma$ is a an integral curve of the vector field $X$ about $p \iff X(\gamma(t)) = \gamma'(t)$ for $|t|< \epsilon$.

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  • $\begingroup$ Not $\forall p \in \gamma(t)$, $\forall p \in \{\gamma(t)\}$. As in, the image of the curve $\gamma$... I just thought it was a definition that works better with the equivalence class definition of vectors and wanted to double check it was valid. $\endgroup$
    – gautampk
    Apr 29 '17 at 0:44
  • $\begingroup$ With that being said, how can $\gamma(t) = p$ for all $p \in \{\gamma(t)\}$? Here $t$ is a variable, and so are you varying $p$ as well. Does it make sense what I am saying? $\endgroup$ Apr 29 '17 at 0:48
  • $\begingroup$ Yeah, $p$ is just an element of the image of $\gamma$, so for some $p\ \exists t : \gamma(t) = p$. $\endgroup$
    – gautampk
    Apr 29 '17 at 0:51
  • $\begingroup$ Well then that's just definition of an injection. An integral curve is a curve who's trajectory follows the flow of an imposed vector field. Here you have said nothing about $\gamma'$, so you can't expect this to be an equivalent definition. $\endgroup$ Apr 29 '17 at 1:00
  • $\begingroup$ You know the definition of a vector is the equivalence class of curves tangent to one another, right? So $\gamma \in V_p$ is saying something about $\gamma'$, namely that $\forall \zeta \in V_p, \gamma'|_p = \zeta'|_p$. $\endgroup$
    – gautampk
    Apr 29 '17 at 1:08

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