Find the curve with the shortest path on a surface (geodesic)

[Westlife]

Homework Statement

Let $U$ be a plane given by $\frac{x^2}{2}-z=0$
Find the curve with the shortest path on $U$ between the points $A(-1,0,\frac{1}{2})$ and $B(1,1,\frac{1}{2})$
I have a question regarding the answer we got in class.

Homework Equations

Euler-Lagrange
$L(y)=\int L(x,y,y')dx$ has extremes when $L_y-\frac{d}{dx}L_{y'}=0$

The Attempt at a Solution

So how what we did in class was.
Let $\gamma (x)=(x,y(x),\frac{x^2}{2})$ then the shortest path is going to be the minimum of the functional
$$I(\gamma)=\displaystyle\int_{-1}^{1}\sqrt{dx^2+dy^2+dz^2}=\displaystyle\int_{-1}^{1}\sqrt{1+(y')^2+x^2}dx$$
Now using the Euler-Lagrange equation for the extremes of a functional we get that:
$L_y=0$
$L_{y'}=\frac{y'}{\sqrt{1+(y')^2+x^2}}$

Therefore we are going to have extremes when

$\frac{d}{dx}\frac{y'}{\sqrt{1+(y')^2+x^2}}=0$ which means that $\frac{y'}{\sqrt{1+(y')^2+x^2}}=C$

Now solving this DE we get
$y'=D\sqrt{x^2+1}\implies y=\frac{D}{2}( (\sqrt{x^2 + 1} x + \sinh^{-1}(x)) +E$

Which would mean that the shortes path on the curve would be

$\gamma(x)=(x,y(x),\frac{x^2}{2})$ where we could get $E,D$ from the initial conditions $y(-1)=0,y(1)=1$

All seems good. However last week in another class we said the a curve $\gamma\,\text{is a geodesic}\iff \gamma ''|| N$. Where $N$ is the normal of the surface $U$

I decided to check if this holds for the curve we got and got
$N=\nabla( \frac{x^2}{2}-z)=(x,0,-1)$
$\gamma ''(x)=(0, \frac{(D x)}{sqrt(x^2 + 1)}, 1)$

And here we see that $N\nparallel \gamma ''$. Can anyone explain me why this is the case? does that mean that the shortest path is not a geodesic?

 

[Hiero]

However last week in another class we said the a curve $\gamma\,\text{is a geodesic}\iff \gamma ''|| N$. Where $N$ is the normal of the surface $U$

I think you’re leaving out a key detail... this applies only to parametrizations with constant speed. (A straight line parametrized with variable speed accelerates along itself, but it’s still a straight line!)

I’m by no stretch an expert in this material, but everything else you did makes sense to me.

 

[Hiero]

Re-parameterizing γ to have a constant speed seems like a daunting task... I think we can take an alternative approach.

The idea of having constant speed is that there is no component of acceleration along the geodesic path. So we ought to be able to say (I’m making this up so someone please confirm) that [γ is a geodesic] ⇔ [γ’’ is in the span of N (surface normal) and γ’] where we no longer require |γ’| to be constant.
(@Orodruin do you agree?)

So instead of re-parametrizing γ we can just ask: are there coefficients a and b such that $γ’’ = aγ’ + bN$ ?
If I’m not mistaken, this question is equivalent to “is the curve a geodesic?” but without the constraint of constant speed.
(Again, someone please confirm or deny me.)

Now, from your work:

$γ’’= (0, \frac{Dx}{\sqrt{1+x^2}}, 1)$
$γ’= (1, D\sqrt{1+x^2}, x)$
$N = (x, 0, -1)$

A bit of inspection/algebra reveals that we can indeed solve the equation $γ’’ = aγ’ + bN$ if we choose $a = \frac{x}{1+x^2}$ and $b = \frac{-1}{1+x^2}$

So I believe all is well and γ is indeed a geodesic!

(Thanks for your question, I’ve learned from it.)

 

[Orodruin]

The idea of having constant speed is that there is no component of acceleration along the geodesic path. So we ought to be able to say (I’m making this up so someone please confirm) that [γ is a geodesic] ⇔ [γ’’ is in the span of N (surface normal) and γ’] where we no longer require |γ’| to be constant.
(@Orodruin do you agree?)

Yes. A geodesic with constant speed is called an affinely parametrised geodesic. You can find the differential equations describing it by using that curve parameter as the parameter instead of $x$. It is relatively easy to show that the geodesic equations with the additional condition of being affinely parametrised are equivalent to the EL equations you would get from extremising the same (reparametrised) functional as here but without the square root.

In this particular problem, it should not be too difficult to find an affinely parametrised geodesic as z only depends on x. You can therefore introduce a new coordinate $u$ such that $ds^2 = du^2 + dy^2$. This is a regular 2D Euclidean space and the geodesic is a straight line, which is easy to give an affine parametrisation.