Sure :), I'll plan out a video for going over that and place it in my schedule. Thanks for your support, and feel free to send me any questions you may have. Also, please subscribe to our channel to ensure that you don't miss the video.
Thanks a million! However, can you please explain a bit about why is z a function of theta. Cause what I'm thinking is sort of x and y lying in our conventional 2D plane, and theta being the angle between (I hope you get what I want to ask, I'm kinda bad at describing my problem). Thanks again!
Well, in this instance, it's convenient to express Z as a function of theta in order to derive our distance functional. However, you can technically define the curve we're trying to find in terms of whatever variable you want; polar coordinates just make our lives easier. If you think of the conventional x and y terms for a function of a circle (the one you're referring to in your post), the theta described in that representation is no different than the theta described here. By saying that Z is a function of theta, we're asserting that the vertical distance from some 2D plane (perhaps the same plane containing that circle defined by x and y)varies as theta change. If you picture a helix (think of a spring), that's exactly what's going on; as you revolve around the helix's axis, the height of the function (or z term) also changes (check out en.wikipedia.org/wiki/Helix#/media/File:Rising_circular.gif ). In fact, to define a helix, the x and y functions are, indeed, those used for a circle; we're just adding elevation in the z-direction. Hope that helps :)!
Aren't you mixing up the directions of x? in one picture it's some direction perpendicular to the surface of the cylinder, but in the small triangle on the surface of the cylinder, it's the 'horizontal'. How do I understand this?
I think I understand now. You just did a weird thing by drawing the x and y axes like that in the 3-d space which the cylinder lives in. We don't even really talk about x and y in that context! We exclusively talk about x and y (what you call x and z) in the 2-dimensional surface of the cylinder.
I was thinking a little more about this. In particular, the direction of travel. If (theta2 - theta1) > pi, does it automatically fall out of the solution that you go backwards? Imagine if the points were pi apart. There are two equal minimal paths, one going clockwise, the other counter clockwise. Is the direction implied by the limits of integration? Is there a "plus-or-minus" left out of all the working with square roots that I missed? The same would be true for a sphere. In the final step where you take the square root to determine z', a slope of -m would also be valid, right? There are two solutions to the equations z'^2 = C. Also near the beginning where you are determining F(theta,z,z') where you factor dtheta^2 outside the radical, there could also be a minus sign there too. I suppose this is where you implicitly decide that dtheta is positive and you're always going counter-clockwise.
Marco, that's an excellent question; I'm glad you're thinking really deeply about this :)! When the points are at an angle pi apart, then you're correct; there are two equal minimal paths. In fact, that reality is captured in the general form that I derived in the video (ie z = m theta + b). What's missing is the final step of defining the actual unknowns in this equation. In other words, what we have is a GENERAL form of the final equation, and we need initial/boundary conditions to fully constrain it. It tells us that stationary points are a "straight line" if the curved surface was unwound into a plane. Now, if we consider the case in which the two points are at an angle pi apart, you will find that we yield both results. For example, try to define it for (theta1 = 0, z1 = 0; theta2 = pi, z2 = 5) and (theta1 = 0, z1 = 0; theta2 = -pi, z2 = 5). You'll find that the solutions z = 5/pi * theta and z = - 5/pi * theta give us what you described above; there are technically two solutions. What the Euler-Lagrange equation yields here is the general equation for the shortest path between the points, not necessarily the final result. That's partially why I stopped the discussion there and didn't fully define the equation. In regards to your square root question, defining the slope as -m is a little superfluous. After all, multiplying a constant by -1 is still a constant. However, it would lead to the same result we obtained, and you can certainly do that if it makes you feel more comfortable. What's interesting is when the angle between the points is < pi and > pi. In these cases, you can end up with the "wrong" curve depending on how the boundary conditions are chosen. What's important to realize is that the Euler-Lagrange equation yields STATIONARY points for the functional, not necessarily minima. So, you still have to use your intuition and think about the solution once you obtain it from the Euler-Lagrange equation; never blindly trust what it yields (the result could be a maximum!). Hope that helps you out a bit more!
The Kaizen Effect. Would be brilliant to draw a straight line on a flat piece of paper at an angle to one edge and then roll it into a cylinder to show the spiral created. [ for demonstration purposes]. Great videos as usual, starting to get to grips with the generality of the Euler Lagrange , it's truly amazing.
That's actually a really good idea! Even better, one could put together a graphical representation of that idea (sort of like the video thumbnail but in motion); it would just require much more time investment lol :).
@@TheKaizenEffect . Could you just clear up something for me. The final expression derived is the equation of a spiral, but what is the length of this spiral between any two given points. We appear to have found some function of z, i.e, height of cylinder, z[theta]=mtheta +b that minimises the length of the line on a cylinder. What about the integration? Don't quite follow. Any advice would be gratefully received , and thanks again for great videos.
Hey Barry, I'm not sure that I fully understand your question, but you can determine the length between the two points pretty easily by "unwrapping" it to see that it's a line on a 2D plane . Once you do that, you can determine the length through simple trigonometry (it's essentially a triangle!). Here, delta z is the dy distance, and R * dtheta is the dx distance. Hope that helps!!
In this case, our functional f(theta,z(theta),z'(theta)) = sqrt(R^2+(z'(theta))^2 does not have any z within its expression. Rather, it has the DERIVATIVE of z (ie z' here) - which it sounds like you're getting confused about. When you compute the partial derivatives found in the Euler-Lagrange equation, you hold all other variables constant - aside from the variable that we're differentiating in terms of. For example, if we want to find the partial df/dz, then any z' terms are like constants here! That's why df/dz = 0 in this case; the entire expression is essentially treated as a constant (there are no z terms!). However, if we want to compute df/dz', we differentiate in terms of the varaible z', and any z variables are treated as constants. That's why the df/dz' partial yielded an expression further along in the video. Hopefully that helps clear that up, but, if you still have questions, we can further address that in our session tomorrow.
I didn't see the geometric intuition for the length formula, but why, WHY go throught taylor series instead of using the definition of the radian. Also, great video, I could keep up much more than on the first one
Thanks again for the geometric intuition. : D Also, you seem to have made a subtle connection to Linear Algebra when you said that we're trying to restrict the space of all possible solutions to the subspace that is constrained by the euler langrange equation, is there something else that I'm missing? So in any geometry, the shortest path is the one that conforms to the geometry of the object, cool.
Awesome, I'm glad that you picked up the linear algebra reference :); you're spot-on with that interpretation (good job!!!). We're simply trying to find the subspace of solutions which satisfy the Euler-Lagrange Equation. The question you should ask yourself for each problem then is this: how many solutions are actually in that subspace?
If you enjoyed the video, please consider supporting the channel through one of the options below. Thanks for your support!!!: ►►►Check out our most popular, BEST-SELLING Udemy courses and get a special LIMITED DISCOUNT just for my followers: udemy.thekaizeneffect.com/ relativity.thekaizeneffect.com/ explore.thekaizeneffect.com/ ►►►Join our Facebook community to sharpen your mind and interact with others: facebook.com/groups/SharpenYourMind/ ►►►Book one of our legendary programs and get extra help: www.thekaizeneffect.com/Services.html ►►►Subscribe to the channel (and hit that notification bell!!) to get notifications of new, free content: ruclips.net/channel/UC3wV-FhrsdZkstGj3wcnRVA ►►►Donate to support our underlying mission and goals: www.paypal.com/cgi-bin/webscr?cmd=_s-xclick&hosted_button_id=P97KV2MXAWJLG
Great question! Actually, you most definitely can; that's another way of solving it. In fact, I resolve it again using that (aka polar coordinates) as a starting point (you may want to check out later in the video to see that; I also explain where R dtheta comes from too). However, I used the first method initially to provide a solution methodology that will ALWAYS work no matter what - regardless of the geometry and appropriate coordinate systems. If you jump to R dtheta, you're technically skipping steps, because you've done/have been shown this stuff before :), which is fine and great for efficiency. Just make sure you understand where it's coming from! Hope the video helped you out, and thanks for the question.
The Kaizen Effect I've never done this stuff before but I have some experience in line integrals in other coordinate systems. mostly from electromagnetism. anyway just started learning this stuff for the heck of it. thanks for the hard work and the passing on of knowledge. I'm pushing 60 and this is a good way to stay sharp.
Awesome, I definitely respect and support that a lot; learning tends to keep us young and sharp. If you need any help/assistance, please feel free to reach out. I'm currently wrapping up a Udemy course (on this same subject) and will get back to posting videos once that's finished. Cheers!
I have struggled over the past week to find geodesics of an ellipsoid (x/a)^2 + (y/b)^2 + (z/c)^2 =1 using this method. The obvious change of variables (u=x/a, v=y/b, w=z/c) does not help completely. The computations get enormous.
If you use this method, you will be able to obtain the geodesics for an ellipsoid; keep trying! Also, from a quick google search, it looks like this has been done by many others before; check some of that information out if you need more help. You could try setting up this method in a program like MATLAB too to streamline the results, but make sure you understand what you're trying to do beforehand (and always have a way to check what the program is providing). Good luck!
First I thought that geodesic curve may be a part of the biggest ellipse that passes through these 2 points (sort of "the Earth equator") . But from "geometrical" definition of a geodesic curve (curve with the normal that is common to the surface) I understood that ellipses can not be geodesic. Sorry for them :)
No problem, Stav :)! I'm glad that you're thinking independently about the material and questioning everything you consume; that's the best kind of mindset you can have in physics! Hope the video helped to clear up and confusion, and hope to see you around in the future.
No problem, Amin; glad to hear that you enjoyed the video :)!!! Please subscribe and hit the notification bell for more video lectures, if interested!!!
You might want to look over your "thetas". The central line should mostly be horizontal, as in " or maybe " ϑ". Yours at times look more like small phis or maybe danish Great job otherwise!
@@ShaunJW1 Wow, thank you very much for that; I really do appreciate it and am honored. If there's anything I can help out with regarding your teaching, please feel free to reach out. What are you teaching right now?
Your conversational skills are far beyond the old fellas out there, love you man.. Physics is life..😄
Physics is the philosophy of life translated into the language of mathematics. --- The Kaizen Effect guy, 2016
Haha, yeah, thanks for pointing that out!! That's actually one of my favorite phrases from this video/series :).
how come Z is a function of theta?Or did you simply assume it to be so?
Your explanation is really good.
Thanks; I'm glad to hear that the video seems to have helped you :)!
Yes it did help me understand it better thx to ur explanation. 👍
Thank you for explaining this clearly!
No problem, I'm glad that it was useful for you. Please feel free to subscribe to the channel for more videos in the future :)!!!
Could you please work through the geodesic of a sphere! Thank you :)
Sure :), I'll plan out a video for going over that and place it in my schedule. Thanks for your support, and feel free to send me any questions you may have. Also, please subscribe to our channel to ensure that you don't miss the video.
31:58 square root, which you fix @ 34:23
Could you also do dy instead of dtheta ?
Thanks a million! However, can you please explain a bit about why is z a function of theta. Cause what I'm thinking is sort of x and y lying in our conventional 2D plane, and theta being the angle between (I hope you get what I want to ask, I'm kinda bad at describing my problem). Thanks again!
Well, in this instance, it's convenient to express Z as a function of theta in order to derive our distance functional. However, you can technically define the curve we're trying to find in terms of whatever variable you want; polar coordinates just make our lives easier. If you think of the conventional x and y terms for a function of a circle (the one you're referring to in your post), the theta described in that representation is no different than the theta described here. By saying that Z is a function of theta, we're asserting that the vertical distance from some 2D plane (perhaps the same plane containing that circle defined by x and y)varies as theta change. If you picture a helix (think of a spring), that's exactly what's going on; as you revolve around the helix's axis, the height of the function (or z term) also changes (check out en.wikipedia.org/wiki/Helix#/media/File:Rising_circular.gif ). In fact, to define a helix, the x and y functions are, indeed, those used for a circle; we're just adding elevation in the z-direction. Hope that helps :)!
Aren't you mixing up the directions of x? in one picture it's some direction perpendicular to the surface of the cylinder, but in the small triangle on the surface of the cylinder, it's the 'horizontal'.
How do I understand this?
I think I understand now. You just did a weird thing by drawing the x and y axes like that in the 3-d space which the cylinder lives in. We don't even really talk about x and y in that context! We exclusively talk about x and y (what you call x and z) in the 2-dimensional surface of the cylinder.
How would you handle a curve on a cone, where R (distance from z axis to the surface) may vary?
How about the geodesics of an ellipsoid?
Thanks, You really healp me in this.
I'm so grateful 😊🙆
Excellent video.
I was thinking a little more about this. In particular, the direction of travel. If (theta2 - theta1) > pi, does it automatically fall out of the solution that you go backwards? Imagine if the points were pi apart. There are two equal minimal paths, one going clockwise, the other counter clockwise. Is the direction implied by the limits of integration? Is there a "plus-or-minus" left out of all the working with square roots that I missed? The same would be true for a sphere.
In the final step where you take the square root to determine z', a slope of -m would also be valid, right? There are two solutions to the equations z'^2 = C.
Also near the beginning where you are determining F(theta,z,z') where you factor dtheta^2 outside the radical, there could also be a minus sign there too. I suppose this is where you implicitly decide that dtheta is positive and you're always going counter-clockwise.
Marco, that's an excellent question; I'm glad you're thinking really deeply about this :)! When the points are at an angle pi apart, then you're correct; there are two equal minimal paths. In fact, that reality is captured in the general form that I derived in the video (ie z = m theta + b). What's missing is the final step of defining the actual unknowns in this equation. In other words, what we have is a GENERAL form of the final equation, and we need initial/boundary conditions to fully constrain it. It tells us that stationary points are a "straight line" if the curved surface was unwound into a plane. Now, if we consider the case in which the two points are at an angle pi apart, you will find that we yield both results. For example, try to define it for (theta1 = 0, z1 = 0; theta2 = pi, z2 = 5) and (theta1 = 0, z1 = 0; theta2 = -pi, z2 = 5). You'll find that the solutions z = 5/pi * theta and z = - 5/pi * theta give us what you described above; there are technically two solutions. What the Euler-Lagrange equation yields here is the general equation for the shortest path between the points, not necessarily the final result. That's partially why I stopped the discussion there and didn't fully define the equation.
In regards to your square root question, defining the slope as -m is a little superfluous. After all, multiplying a constant by -1 is still a constant. However, it would lead to the same result we obtained, and you can certainly do that if it makes you feel more comfortable. What's interesting is when the angle between the points is < pi and > pi. In these cases, you can end up with the "wrong" curve depending on how the boundary conditions are chosen. What's important to realize is that the Euler-Lagrange equation yields STATIONARY points for the functional, not necessarily minima. So, you still have to use your intuition and think about the solution once you obtain it from the Euler-Lagrange equation; never blindly trust what it yields (the result could be a maximum!).
Hope that helps you out a bit more!
Thank you. That makes sense. I should have thought harder.
As you mentioned that the shortest path on a cylinder would be a helix, Does it have a direction like clockwise or counter-clockwise helix?
The Kaizen Effect. Would be brilliant to draw a straight line on a flat piece of paper at an angle to one edge and then roll it into a cylinder to show the spiral created. [ for demonstration purposes]. Great videos as usual, starting to get to grips with the generality of the Euler Lagrange , it's truly amazing.
That's actually a really good idea! Even better, one could put together a graphical representation of that idea (sort of like the video thumbnail but in motion); it would just require much more time investment lol :).
@@TheKaizenEffect . Could you just clear up something for me. The final expression derived is the equation of a spiral, but what is the length of this spiral between any two given points. We appear to have found some function of z, i.e, height of cylinder, z[theta]=mtheta +b that minimises the length of the line on a cylinder. What about the integration? Don't quite follow. Any advice would be gratefully received , and thanks again for great videos.
Hey Barry, I'm not sure that I fully understand your question, but you can determine the length between the two points pretty easily by "unwrapping" it to see that it's a line on a 2D plane . Once you do that, you can determine the length through simple trigonometry (it's essentially a triangle!). Here, delta z is the dy distance, and R * dtheta is the dx distance. Hope that helps!!
I came for the hand slapping
And hand slapping ye shall get XD
At min 30:37, why did you say df/dz = 0? S= integral(sqrt(R^2+(dz/d*theta)^2) dtheta. So we have dz here.
In this case, our functional f(theta,z(theta),z'(theta)) = sqrt(R^2+(z'(theta))^2 does not have any z within its expression. Rather, it has the DERIVATIVE of z (ie z' here) - which it sounds like you're getting confused about. When you compute the partial derivatives found in the Euler-Lagrange equation, you hold all other variables constant - aside from the variable that we're differentiating in terms of. For example, if we want to find the partial df/dz, then any z' terms are like constants here! That's why df/dz = 0 in this case; the entire expression is essentially treated as a constant (there are no z terms!). However, if we want to compute df/dz', we differentiate in terms of the varaible z', and any z variables are treated as constants. That's why the df/dz' partial yielded an expression further along in the video. Hopefully that helps clear that up, but, if you still have questions, we can further address that in our session tomorrow.
Thanks for making a video on this. Great explanation.
No problem, Sagar :)!! Thanks for your comment and support; hope you learned a lot from the video!!!
I didn't see the geometric intuition for the length formula, but why, WHY go throught taylor series instead of using the definition of the radian. Also, great video, I could keep up much more than on the first one
Thanks again for the geometric intuition. : D Also, you seem to have made a subtle connection to Linear Algebra when you said that we're trying to restrict the space of all possible solutions to the subspace that is constrained by the euler langrange equation, is there something else that I'm missing?
So in any geometry, the shortest path is the one that conforms to the geometry of the object, cool.
Awesome, I'm glad that you picked up the linear algebra reference :); you're spot-on with that interpretation (good job!!!). We're simply trying to find the subspace of solutions which satisfy the Euler-Lagrange Equation. The question you should ask yourself for each problem then is this: how many solutions are actually in that subspace?
If you enjoyed the video, please consider supporting the channel through one of the options below. Thanks for your support!!!:
►►►Check out our most popular, BEST-SELLING Udemy courses and get a special LIMITED DISCOUNT just for my followers:
udemy.thekaizeneffect.com/
relativity.thekaizeneffect.com/
explore.thekaizeneffect.com/
►►►Join our Facebook community to sharpen your mind and interact with others:
facebook.com/groups/SharpenYourMind/
►►►Book one of our legendary programs and get extra help:
www.thekaizeneffect.com/Services.html
►►►Subscribe to the channel (and hit that notification bell!!) to get notifications of new, free content:
ruclips.net/channel/UC3wV-FhrsdZkstGj3wcnRVA
►►►Donate to support our underlying mission and goals:
www.paypal.com/cgi-bin/webscr?cmd=_s-xclick&hosted_button_id=P97KV2MXAWJLG
Why can't you just say that the distance around the circle is the arc-length, namely Rdθ?
Great question! Actually, you most definitely can; that's another way of solving it. In fact, I resolve it again using that (aka polar coordinates) as a starting point (you may want to check out later in the video to see that; I also explain where R dtheta comes from too). However, I used the first method initially to provide a solution methodology that will ALWAYS work no matter what - regardless of the geometry and appropriate coordinate systems. If you jump to R dtheta, you're technically skipping steps, because you've done/have been shown this stuff before :), which is fine and great for efficiency. Just make sure you understand where it's coming from! Hope the video helped you out, and thanks for the question.
The Kaizen Effect I've never done this stuff before but I have some experience in line integrals in other coordinate systems. mostly from electromagnetism. anyway just started learning this stuff for the heck of it. thanks for the hard work and the passing on of knowledge. I'm pushing 60 and this is a good way to stay sharp.
Awesome, I definitely respect and support that a lot; learning tends to keep us young and sharp. If you need any help/assistance, please feel free to reach out. I'm currently wrapping up a Udemy course (on this same subject) and will get back to posting videos once that's finished. Cheers!
I have struggled over the past week to find geodesics of an ellipsoid (x/a)^2 + (y/b)^2 + (z/c)^2 =1 using this method.
The obvious change of variables (u=x/a, v=y/b, w=z/c) does not help completely. The computations get enormous.
If you use this method, you will be able to obtain the geodesics for an ellipsoid; keep trying! Also, from a quick google search, it looks like this has been done by many others before; check some of that information out if you need more help. You could try setting up this method in a program like MATLAB too to streamline the results, but make sure you understand what you're trying to do beforehand (and always have a way to check what the program is providing). Good luck!
yeah please do a video on a geodesic of a sphere , please please!!
please make a video for geodesic of sphere
Ajudou muitooo
Thank you.
First I thought that geodesic curve may be a part of the biggest ellipse that passes through these 2 points (sort of "the Earth equator") . But from "geometrical" definition of a geodesic curve (curve with the normal that is common to the surface) I understood that ellipses can not be geodesic. Sorry for them :)
No problem, Stav :)! I'm glad that you're thinking independently about the material and questioning everything you consume; that's the best kind of mindset you can have in physics! Hope the video helped to clear up and confusion, and hope to see you around in the future.
Thank you
No problem, Amin; glad to hear that you enjoyed the video :)!!! Please subscribe and hit the notification bell for more video lectures, if interested!!!
You might want to look over your "thetas". The central line should mostly be horizontal, as in " or maybe " ϑ". Yours at times look more like small phis or maybe danish
Great job otherwise!
Theta is also important... A teacher of mine insists that the letter "eta" is called "neta". Am I sick from losing concentration?
I learned a lot thank u
Very helpful vedio thanku
You made it understand too easy for us
Thankuu
Please make Vedio on geodesic of cone and some main figures please
Thanks for the kind words, chief; I'm glad that the video ended up helping you :)!!
Good, very good
Thanks for the kind words, and I'm glad that you enjoyed the video!!! Hope to see you around the comments section again in the future :)
@@TheKaizenEffect Thank you, will support and thumbs up your other vids, and let my other students know about your vids
@@ShaunJW1 Wow, thank you very much for that; I really do appreciate it and am honored. If there's anything I can help out with regarding your teaching, please feel free to reach out. What are you teaching right now?
@@TheKaizenEffect studying applied maths physics stream final year. We're 10 students, so passing word around ✌️
I've only seen two different derivations of the Euler-Lagrange equation, and neither one is 100% clear to me.