# What's the other way without the Euclidean approximations to prove that a geodesic lies on the great circle?

The geodesics are the shortest curves that can be drawn between two points in a space. If the surface is a spherical one on which we are trying to get the geodesic between two points then it is said to lie on the largest circle passing through the two points that can be drawn on the spherical surface and it's dubbed as great circle. To prove the thing the way I thought was driven by the approximation that tiny pieces or parts on the surface of sphere can be treated as being Euclidean (I may be wrong using the word, the thing I mean to say is that I can use the Pythagoras theorem up there). So if I define a coordinate system much like that on earth the latitudinal-longitudinal $(y-x)$ [such that the two points let's call them $P1$ and $P2$ lie on the curve $y=0$ ] so in tiny spaces I can treat it like being Euclidean (I can use Pythagoras theorem) so the distance measured moving along the curve $'S'$ is the summation of small changes in $S$ with respect to small changes in the $x$ and $y$ . So my approximation allows $(dS)^2= (dx)^2 +(dy)^2$ It will be how the function goes And $dS$ will be greater than or equal to $dx$ and thus the $dS=dx$ or $dy=0$ for having $S$ as minimum. And thus $S=ndx=x$ and it lies on the great circle as we have designed the system of coordinates. So my question is what is the other way the above thing can be proved or more generalized way (better if without the above approximation) and which can help to get a general property of a geodesic on any surface ?

(I checked a question first a bit similar one but I didn't understand the notations used there a comment on the question or any link will too be helpful)

One statement which is true of geodesics on a well-behaved surface in $\mathbb R^n$ is the following:

The second derivative of the geodesic is normal to the surface.

Which is true of great circles, where the second derivative always points towards the center, along a radius of the sphere. This basically means that geodesics "accelerate", but only as little as possible to stay on the surface. This fact is sufficient to determine the geodesic for any surface embedded in $\mathbb R^n$ for which the term "normal vector" makes sense. I will give a somewhat easy derivation of this below. One could fill in all the details to make this a proof, if they wanted to.

In particular, let us look at sufficiently implicitly defined surfaces, since they offer normal vectors most readily. That is, let $f:\mathbb R^n\rightarrow\mathbb R$ be a function such that your surface is the set of vectors $v$ such that $f(v)=0$ and such that $\nabla f$ is non-zero at all such $v$. We could, for instance, take $f(x,y,z)=x^2+y^2+z^2-1$ to represent the unit sphere thusly. Then, our question is, where $L(g)=\int_{0}^1 \|g'(x)\|\,dx$ is the length of a curve:

What functions $g:[0,1]\rightarrow\mathbb R^n$ are local minimums of $L$, subject to the constraint that $f(g(t))=0$ and with fixed endpoints?

To answer this question quickly, we can think about pushing $g$ around - i.e. is the some family of function $g_t$ such that $g_0=0$ but the derivative of the map $t\rightarrow L(g_t)$ is negative (i.e. the length is shrinking)? Informally speaking we might think of a function $h=\lim_{t\rightarrow 0}\frac{g_t-g}t$ as the "movement" represented here - a sort of small perturbation in $g$. We may then find that the derivative of $t\rightarrow L(g_t)$ is: $$\int_{0}^1\frac{g'(x)\cdot h(x)}{\|g'(x)\|}\,dx$$ which will be $0$ for any $h$ if $g$ is a local minimum. We may only find two constraints on $h$. Firstly, $h(0)=h(1)=0$, as the endpoints are fixed. Secondly, $h(x)\cdot \nabla f(g(x))=0$ as $h$ represents a movement along the surface and thus must be tangent to the surface. This immediately yields that if $g$ is everywhere parallel to a normal (i.e. parallel to $\nabla f$), then the above integral is $0$. I will leave it as an exercise to show that if $g$ deviates from this, there is some suitable $h$ for which the above integral is not zero (Hint: consider a sort of "tent map" pushing $g$ towards $\nabla f$ in the vicinity where it deviates, projecting vectors $h$ onto the appropriate tangent plane). This suffices to show the statement originally claimed - and by adding more constraints like $f$, we can do the same thing for any implicitly defined manifold.

However, one should note that approximating a space by a Euclidean one is more or less critical to the idea of a geodesic in general and even the above proof subtly uses it; geodesics naturally exist on a structure called a Riemannian manifold, which has a rather technical definition, but basically requires that the manifold more or less appears to be a plane when you look at only local properties.