The Geodesic Problem on a Plane | Calculus of Variations

In this short (hehe) video, I set up and solve the Geodesic Problem on a Plane. A geodesic is a special curve that represents the shortest distance between two points on a particular surface. Here, because the geodesic problem is being solved on a plane, I show that the geodesic on a plane is a straight line. To do this, I set up a length functional, and then using the Euler-Lagrange equation, I solve for the equation of the geodesic path. The computation here isn't that difficult, but I said I'd cover geodesics when I started this series so I figured I'd do this for the sake of completeness.

Now, I didn't mention this in the video because I thought it was a bit tangential to the topic, but geodesics become quite important in General Relativity. General Relativity essentially has two important (sets of) equations. The first are the Einstein Field Equations, which describe how spacetime (which can be thought of as a 4-D surface) curves under the influence of mass and energy. The second set are the geodesic equations, which describe how light and matter travel in this 'curved' spacetime. In other words, light travels along geodesics in spacetime (that's why light 'bends' in gravity).

The geodesic equation in General Relativity can, in fact, be derived using Euler-Lagrange. In this case, the dS isn't just sqrt(dx^2 + dy^2) but is a lot more complicated, and involves the metric tensor. I'll cover this in more depth once I start General Relativity, but I figured I'd briefly discuss an application for the interested folks out there.

ERRATA: At 3:29, I meant to say y' instead of y.

Special thanks to my Patrons:
- Tom
- Jennifer Helfman
- Justin Hill
- Jacob Soares
- Yenyo Pal
- Chi
- Lisa Bouchard

[NovaWarrior]

Thank you so much for this whole series! It is very hard to find content like this.

[_DD_]

Amazingly explained. I am glad you make these videos, THANK YOU!

[navilistener]

Great video and series! Thank you for taking the time to do this!

I would highliglht a small detail in 5:06: we arrive at y' = c/√(1-c²) = c₁. This means that c can not be ±1, so that the denominator √(1-c²) is not 0. Also, for c₁ to be a real number, we must have |c|<1, otherwise √(1-c²) would be the square root of a negative number and thus c₁ would become a complex number.

At first, I thought that maybe c₁ would not be able to assume all real values given these conditions and thus that this method hadn't given us the general straight line solution, as some values for the slope c₁ would not be possible.

However, when plotting the graph of the function c ↦ c/√(1-c²), we find that sweeping c in ]-1, 1[ produces all real values in ]-∞, +∞[ for c₁. Therefore, all slopes for the straight line solution are indeed possible.

[srujaniam]

Sir, I like the explanation... but there is a doubt at 3:18 ... why would the term dF/dy be 0? Yes... there is no y term but y prime is a derivative of y right? so can't we just derivate that?

[weirdonumber]

excellent！and your writing is so organized and beautiful

[lujason]

I came from Taiwan, and liked the way you explained how thing worked.

[AlirezaRezaee]

What a beautiful explanation, thank you

[TheLevano]

The formula of a geodesic on a plane, as shown at 6:43, is based on the second coordinates y2 and x2. Is there any preference for the usage of the later coordinate or could have we written the formula as y = y1 + k(x - x1), where k = C1 = (y2 - y1)/(x2 - x1), just as easily? Although I see no difference, there might be some practical reason, that is why I ask the question.

[abusufian]

At 4:45 how did make sure real square root exist? The constant [C^2/(1-c^2) ] can be negative and we are working on real space.

[Hornswoggled_Engineer]

This is a really excellent and helpful video, but I think I have found an error in you algebra around 5:45:

By solving 2E2U starting off by solving for C2 using the y2 equation, you should get the boxed equation except with (x2-x) instead of (x-x2).

By solving for C2 using the y1 equation, you get the boxed equation except with y2 and x2 directly replaced by y1 and x1.

I've checked this several times and believe I am correct. I could even share my algebra with you via email if you want to discuss it further.

[kennywong]

Thank you very much for the explanation! I am interested in the 2nd order equation for determining the max and min. Is it the Geodesic Differential equation?

[TheViolaBuddy]

By assuming our answer is a function y(x), are we implicitly ignoring the possibility that the shortest-path function we're looking for is vertical or loops back on itself or anything else that's not a function? Obviously it's not in this case, as we can tell intuitively, but if we were to set up a different situation, we could imagine the path that minimizes the functional is a path that goes in a wide circle left, up, right, and then down to the ending point off to the right, which would not be a function. How would we handle that situation?

[John-qtem]

@2:00 you said taking the dx^2 out of the sqrt, which I assumed you meant to say divided through dx^2 to give Is that correct?

[JuvenileLeech]

Gr8 video sir khan. Wat software do u use to make? Paint? Thx.

[highlightedreply]

Great video! Would it make sense to use Beltrami Identity insteat of the Euler-Lagrande-Eq since dF/dx = 0?
Thanxs a lot

[p.z.]

In the last video you mentioned that the Euler-Lagrange is just a necessary condition for y being a extremal. So even if y fullfils the E.L. equation, it isnt proven that y is a extremal ? Is that valid ?

[PaoloSita]

I've already got my master degree. Well, better later than never ❤️ thank you

[blzKrg]

So, the proof that the geodesic on a plane is a straight line is the same as proving the shortest distance between 2 points is a straight line by Euler-lagrange equation?

[hajramughal]

sir kindly share the problem of simple pendulum using lagrange equation of motion

[apurbamandal]

Sir in this video you have shown how to find the extrema but how I shall understand that the extrema is either a maxima or a minima??

Please give the full procedure.