Straight lines and flat surfaces

[birulami]

Suppose I have a parameterized line $\phi:\mathbb{R}\to\mathbb{R}^n$ given by $\phi(t) = (x^\mu(t))|_{\mu=1}^n$. How can I tell that the line is straight.

My best answer so far is that at every time $t$ the acceleration (2nd derivative) is parallel to the velocity (1st derivative), i.e. $\ddot{\phi}(t) = g(t)\cdot\dot{\phi}(t)$, for some function $g(t)$ (which likely should better not be zero anywhere).

Is this a valid description (necessary and sufficient) of a straight line? Are there different ones?

And a very similar question for a 2-dimensional surface, i.e. now we have $\phi:\mathbb{R}^2\to\mathbb{R}^n$. Assuming the above is true for the straight line description. Is there a similar condition for the surface?

 

[CompuChip]

Since you have posted this in the Differential Geometry forum, this may be a good time to read (up) on geodesics which is the generalization of the idea of "straight line".

 

[HallsofIvy]

If you really mean "straight" rather than "geodesic", a line is straight, at a given point, if and only if its second derivative, with respect to arc length, is 0.

 

[lavinia]

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My best answer so far is that at every time $t$ the acceleration (2nd derivative) is parallel to the velocity (1st derivative), i.e. $\ddot{\phi}(t) = g(t)\cdot\dot{\phi}(t)$, for some function $g(t)$ (which likely should better not be zero anywhere).

And a very similar question for a 2-dimensional surface, i.e. now we have $\phi:\mathbb{R}^2\to\mathbb{R}^n$. Assuming the above is true for the straight line description. Is there a similar condition for the surface?

This is correct. A straight line can be parameterized to have the equation c(t) = b + at for vectors a and b. Any path that follows this line will be a reparameterization of it i.e. t = f(s). You can prove your conclusion using the Chain Rule.

The same idea applies for the map of a plane into Euclidean space or for a map of R^m into R^n. The only difference is the number of parameters.