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?


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:

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


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 $|t|< \epsilon$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.