The Geodesic Problem on a Plane | Calculus of Variations

[@NovaWarrior77]

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

[@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|

[@_DD_15]

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

[@lujason677]

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

[@srujaniam9762]

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?

[@scientificresearch1400]

d/dy (dy/dx) d(1)/dx =0

[@cihangwang8044]

excellent！and your writing is so organized and beautiful

[@hajramughal11]

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

[@FacultyofKhan]

I will do that soon, stay tuned!

[@AlirezaRezaee]

What a beautiful explanation, 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?

[@JannikEggert-fj8vv]

yes, a geodesic is by definition a line between to points with minimal length ( over a given metric and boundary conditions ).So, to calculate that line one could minimize the length of an arbitrary path between these two points ( over a given metric and boundary conditions ). Thats what he did with lagrange

[@blzKrg]

@@JannikEggert-fj8vv but lagrange just gives the stationary point of the function so how do we know that stationary point is a maximum or minimum in spheres or other odd shapes?

[@JannikEggert-fj8vv]

MR. NOOB Since E-L is just a first derivative test on functionals, i guess you could just go for the second derivative test like in basic analysis. But idk how that would look like on functionals. As a physicist I‘d just say by intuition

[@blzKrg]

@@JannikEggert-fj8vv yeah second derivatives is what i thought about but i didn't know how to differentiate the functional. And intuition works well for 2d surfaces and some general 3d surfaces but what about surfaces like distorted spheres?

[@dr.swatirai4572]

@@blzKrg you should look it up in mathematical methods for physicists by Arfken,Weber and Harris.

[@apurbamandal6535]

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.

[@TonyKaku-g8n]

thanks to give explanation to who do not have phys background

[@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?

[@FacultyofKhan]

As you said, the solution you mentioned isn't a function, so we probably would not be able to get it from solving the E-L (Euler-Lagrange) equation (just like how I didn't find any other weird solutions in this video). However, what you said raises an interesting point when it comes to the geodesic problem on a sphere: if I have two points A and B on a sphere (e.g. New York and London), the geodesic between those points is an arc along the great circle passing through those two points. But when I solve the geodesic problem on the sphere, I'll basically get an equation which I could interpret as going in one of two directions. The first direction is the *actual* geodesic from New York to London (directly through the Atlantic Ocean - the shorter distance), and the second direction is a decoy and obviously way too long to be the geodesic (i.e. going West from New York and curling all the way around the globe to finally reach London). How can I distinguish which one is the actual geodesic and which one is fake (I believe that's your question)? Well, I'll have to either use common sense/intuition (like you said), or I can use the second variation (analogous to the second derivative) to find which path is minimum. It's usually more common to use intuition, since the second variation is a bit cumbersome and hard to evaluate. Hopefully that answers your question, and if you'd like a reference on second variations as it applies to the planar geodesic problem, this should help (see page 12): www.math.uconn.edu/~gordina/NelsonAaronHonorsThesis2012.pdf

[@TheViolaBuddy]

Oh, interesting. Thanks!

[@vanadium1021]

They way my professor found the geodesic without this being a problem was by defining a parameter λ. Then he expressed ds= dλsqrt[ (dx/dλ)^2 + (dy/dλ)^2) by using chain rule and then taking 2 euler-langrange equations, one for x and one for y. After a few calculations it ends up at dy/dx = c. You probably don't have this question anymore after 3 years, but I am writing this here in case anyone is watching in the future and has the same problem.

[@kennywong7850]

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?

[@Psicose7]

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.

[@FacultyofKhan]

I don't know, I think the equation in the video seems right. It also makes sense if you apply the boundary conditions. Here's all the algebra for the solution, perhaps you might find a mistake in there or perhaps you might find something you did wrong in your own work: imgur.com/a/yWPFduK In any case, let me know, and thank you for the feedback!

[@Psicose7]

So you're right, and it's because (20 years of school on and in to a PhD) I had a misconception. (y1-y2)/(x2-x1) = - (y2-y1)/(x2-x1), not - (y2-y1)/(x1-x2) Wrote a quick matlab script to confirm. You learn something new every day!

[@John-qt5em]

@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 sqrt(1+dy^2/dx^2)=sqrt(1+[dy/dx]^2)=sqrt(1+[y']^2). Is that correct?

[@FacultyofKhan]

Yes, exactly! Remember that when you take the dx^2 out of the sqrt, you get dx on the outside as well!

[@John-qt5em]

@@FacultyofKhan Awesome. All thanks, mate.

[@TheLevano22]

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.

[@p.z.8355]

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 ?

[@FacultyofKhan]

Yup, it's not necessary that y be a 'local maximum' function for the functional or a 'local minimum' function if it satisfies the EL equation. The EL equation is analogous to setting the first derivative of f(x) = 0: you might get a maximum; you might get a minimum; you might get a saddle point. There's no guarantee from just the first derivative.

[@PaoloSita]

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

[@JuvenileLeech]

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

[@FacultyofKhan]

SmoothDraw, though Paint could probably work just as well.

[@necatigoren9187]

Great derivation clear and precise thank you

[@abusufian5391]

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.

[@FacultyofKhan]

C is automatically less than 1, because it's specified as (y')^2/(1+(y')^2) = C^2 (see 4:20). Because C < 1, we can take the square root of [C^2/(1-C^2)] without worrying about imaginary terms.

[@lagrangian143]

That was super cool

[@JanKowalski-zz8ef]

Very good video

[@marrytesfu3163]

Is x in this problem the epsilon (we integrated with respect to x) if so how to identify it in the case of multivariable functional

[@highlightedreply6440]

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

[@wbcboxing3538]

of all things I'm getting stuck on the Algebra. I get y=(y2-y1/x2-x1)x+(y2x1-y1x2/(x1-x2)). Have I gone wrong or do i need to simplify further? awesome video by the way.

[@FacultyofKhan]

Nope, you're correct! If you take my boxed equation at 6:02, expand out the second term and combine the fractions, you should get the same answer. And thank you!

[@wbcboxing3538]

Faculty of Khan ah got it! Thanks again!!

[@sesangse]

Great. So much helpful.

[@structuralanalysis6885]

very fine video, but i disagree that dF/dy=0, maybe we can use chain rule:dy'/(dx*dy/dx)=y''/y'. Although. the final result would still be a straight line as y''=0.

[@sarwank6308]

Nice I understand thanks

[@ArthurMorganFTW_RDR]

You are a legend thank you so much!!!!!

[@FacultyofKhan]

Glad it helped!

[@thevegg3275]

One minute and 47 seconds where you give the definition of the line between a and B why is that not automatically the geodesic?

[@valor36az]

Awesome!

[@FacultyofKhan]

Thank you! Cheers!

[@iaggocapitanio7909]

thanks

[@nagamamaniyar6999]

Hello sir... Plz.. make video for a spherical surface the geodesic are the great circle

[@FacultyofKhan]

Thank you for the suggestion!

[@dreamcatgamedev1891]

Technically, geodesics are not the shortest paths between two points. For example, on a sphere, there can be either infinitely many geodesics between two points (if they are antipodal) or two geodesics otherwise. The two geodesics between two non-antipodal points would be the shortest one and the longest one (the path going the other direction through the great circle). The real definition of a geodesic is rather technical. math.stackexchange.com/questions/1432985/why-may-geodesic-not-be-the-shortest-path-on-a-surface

[@yarooborkowski5999]

You made it very clear to undersand, thank You. By the way.... Could You show us in similar way "Geodesic problem on a sphere", please?

[@FacultyofKhan]

Here you go: kzbin.info/www/bejne/Z5S9dGafgtV3rrM

[@yarooborkowski5999]

Yes, I was waiting for it. Thanks to You so much.

[@j.k.sharma3669]

Pls explain why dF/ dy= 0 in detail

[@averagegamer9513]

If you still didn’t figure it out, it’s because here F is sqrt(1+y’). As you can see, it only explicitly depends on y’, and does not contain the variable y. Therefore, dF/dy=0.

[@pennpaper9310]

Geodesic on a plane is straight line..bt circle is also lie in plane so it's geodesic is also a straight line???pls tell

[@FacultyofKhan]

I don't understand your question. A circle is not considered a straight line on a plane. If you're referring to the great circle, that's the geodesic on a sphere (not a plane). I'd check this video out for more details on that: kzbin.info/www/bejne/Z5S9dGafgtV3rrM

[@kevinchou4147]

god like man ! THX

[@memesdk6121]

Sir, geodesic In 2 plane In R3 are

[@memesdk6121]

??

[@philipphoehn3883]

Geodeeeesic

[@srishtikdutta8946]

Nice 😁

[@ttttt_]

I think your definition is a little bit inaccurate, as a geodesic is a path which locally minimizes the distance between two points.

[@FacultyofKhan]

Yeah, perhaps my definition was a bit too global here, but for these cases, it's all the same. Thanks for pointing that out! Incidentally, wikipedia basically gives my definition, while wolfram mathworld calls it a 'locally length-minimizing curve'.

[@Madhugowda.S]

Geodesic right circular cylinder

[@jameslongstaff2762]

This makes sense. hahaha. I did study math in college, but you explained this very clearly.

[@Peter_1986]

My lecture notes on this subject are really annoying - they are those kinds of "serious" notes that go on forever about formal theorems all over the place and never seem to get to the freaking point, and the handful of actual examples that they do give are always as formal and complicated as possible.