Vector calculus in curvilinear coordinates! (a helpful intro)

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  • Опубликовано: 25 ноя 2024

Комментарии • 40

  • @itskarudo
    @itskarudo 2 года назад +5

    I love your energy! best video I've seen on the subject by far, thank you 😄

    • @JenFoxBot
      @JenFoxBot  2 года назад

      Aww yay that makes me so happy to hear, thank you for sharing!!

  • @priyanshusorout234
    @priyanshusorout234 3 года назад +11

    Benefits of having less subscribers - mostly everyone will get answer of their questions 😁😁

  • @isnafrehman7132
    @isnafrehman7132 Месяц назад

    Good teaching method

  • @selamselam9006
    @selamselam9006 2 года назад

    I would join every class with your lecture

    • @JenFoxBot
      @JenFoxBot  2 года назад

      awww thank you!! that makes me very happy to hear.

  • @onyiboemmanuel6060
    @onyiboemmanuel6060 Год назад

    Thanks a lot, this is so helpful.

  • @blakely1317
    @blakely1317 Год назад

    thank you, this was helpful indeed!

  • @EricPham-gr8pg
    @EricPham-gr8pg 8 месяцев назад

    Maybe newton method on each dimension separately or differential equations systems of equations may easier

  • @danielzlotin6607
    @danielzlotin6607 2 года назад

    Hey Jen thanks for the videos, they are great and easy to understand. would be nice if you will arrange them in the right order.

    • @joeboxter3635
      @joeboxter3635 2 года назад

      No - not translate. Transport. Beautiful people transport. Translate is for the so-so people. Lol. Ooopsie?! You sound like one of my friends.
      Help me start a campaign to have all text books swap phi and theta in spherical coordinates making the definition of theta consistent in polar, cylindrical, and spherical coordinates. Thus z in cylindrical coordinates or phi in spherical coordinates. And z = R*cos (phi). In fact, let's change phi to be the angle from the x-y plane to R. Not the angle between R and z. This makes z = R sin (phi). This is better, because y = r sin theta. And x is r cos theta. So the odd sin is now with new y and new z. This leave x and cos happy being associated with each other.
      Now what's happening in the x-y plane is the same in all coordinate systems and consistent with polar coordinates.
      See then there are far fewer formulas and the formulas are consistent and the definitions are consistent.
      Then this will solve so many problems in physics including dark matter, why QM and GR don't play well with each other, and why the speed of light is a constant.
      Well maybe not. But it'd make my life easier. Lol.
      PS
      And in honor of beautiful, we will now call these the beautiful spherical coordinates leaving the so-so people behind with their old spherical coordinates. And we enjoy the beauty of consistent mathematical forms.

  • @teenpatti8294
    @teenpatti8294 Год назад

    merry christmas mam❤

  • @adamholmer5693
    @adamholmer5693 Год назад

    Great video!

  • @nachoswithnyesha201
    @nachoswithnyesha201 10 месяцев назад

    thanks a lot

  • @binita4672
    @binita4672 2 года назад

    Thank you so much. This was very helpful 💓

    • @JenFoxBot
      @JenFoxBot  2 года назад

      I'm so glad to hear!!

  • @thevegg3275
    @thevegg3275 3 месяца назад

    Hi, you make math fun! So let's see if you can make answering my question fun. :)
    I'm an extremely visual learner. That being said I'm trying to visualize Christoffel symbol 'values'.
    So if I have a very messy curvilinear csys, how can I visually see or plot the coordinates of a derivative vector (one that represents the change in a given vector before and after being parallel transported an infinitetesimal distance dx.
    Should I...
    a) plot the coordinates of the derivative vector on the messy curvilinear csys (no idea how to do that)
    or
    b) plot the coordinates of the derivative vector on cartesian coordinate csys, aka known as the tangent plane...(I know how to do that).
    The issue is that a cartesian csys has a nice grid to count lengths of coordinates...but messy curvilinear cys is usually not represented as such (just four curvy lines.
    Where are my cury line grids!!! How do I make them. And, on the off chance that both of them are required, how do we size them such that they match? ie. If my cartesian csys has small grid units of one inch, but my messy curvy csys had them as one mile...you see what I mean. I feel like the metric tensor has something to do with matching them up.
    Thank you if you try to answer this.
    Rocky (big fan)

  • @mnada72
    @mnada72 3 года назад +2

    It's helpful. What's your book? Are you going to make a separate video for each coordinate system ?

    • @JenFoxBot
      @JenFoxBot  3 года назад +1

      Just the one! I have a short book on microcontrollers but thats all.. for now 🙃

    • @mnada72
      @mnada72 3 года назад

      @@JenFoxBot 6:40 But choosing t as the scalar function confused me a little because later on I think it's time and ended asking my self how she is differentiating time with respect to position 😆

  • @EricPham-gr8pg
    @EricPham-gr8pg 8 месяцев назад

    Instrumentation may simulate better because it interactively and limited of integration unless...

  • @chandan_Lal_69
    @chandan_Lal_69 8 месяцев назад

    Thanks mam🎉🎉🎉

  • @AbDullAHMoHAAmeD
    @AbDullAHMoHAAmeD 8 месяцев назад

    This is a lil advanced class for me I'll come back later but thanks

  • @smftrsddvjiou6443
    @smftrsddvjiou6443 7 месяцев назад

    silly question. is it not dl suqared ? Or is dl not the length ?

  • @mdmouzakkirhossain3263
    @mdmouzakkirhossain3263 3 года назад +2

    If F(x_,y_,t_) is a function in Cartesian coordinate system, then can you tell me please, what will be the form of del F/del x_ and del F/del y_ in curvilinear coordinate system.
    The curve is (x_, cos(x_)).
    x_,t_ in Cartesian coordinate,
    "
    x,t in curvilinear coordinate.
    Please help me to rectify this problem.

    • @JenFoxBot
      @JenFoxBot  2 года назад

      You'll need to convert F(x,y,t) from Cartesian into spherical (or cylindrical, whichever one you need), then use the spherical (or cylindrical) form to take the gradient (or divergence/curl). Note that you can't convert time, bc its not a spatial coordinate, so you are only converting x and y.
      For spherical, the gradient of a function, F, is: (dF/dr)rhat + (1/r)(dF/dθ)θhat + 1/(r*sinθ)(dF/dφ)φhat
      Where all of those are *partial* derivatives and the "hat" term indicates the unit vector.

  • @chamelious
    @chamelious 3 года назад +2

    Your vids are really good, but the camera constantly autofocusing is distracting, maybe it has a manual mode that will stop it doing that?

    • @JenFoxBot
      @JenFoxBot  3 года назад +1

      Ugh yea I have tried everything, its currently on manual mode and still does the weird attempt at auto focus. I gave up trying to figure it out 🤷‍♀️

  • @priyanshusorout234
    @priyanshusorout234 3 года назад +2

    Can anyone do phd in particle physics after getting master's in chemistry??

    • @JenFoxBot
      @JenFoxBot  3 года назад +2

      Unsure, but worth a try if you're passionate!

    • @priyanshusorout234
      @priyanshusorout234 3 года назад +1

      @@JenFoxBot okay thanks

  • @lucasf.v.n.4197
    @lucasf.v.n.4197 Год назад

    it's much better (and easier) to learn this in general curvilinear coordinates rather than orthogonal ones; also, u need to stress the importance of using unit basis vectors rather than unscaled basis vectors; and if I may compliment u, u r hot;

  • @Sumit-zb5uj
    @Sumit-zb5uj Год назад

    We can confirm that she is not confirm when when she was laughin but we weren't.

  • @deepanshuiitbhu1928
    @deepanshuiitbhu1928 2 года назад

    IITBHU

  • @griffithfimeto3387
    @griffithfimeto3387 3 года назад +1

    Are you doctor ?

    • @JenFoxBot
      @JenFoxBot  3 года назад +3

      no, just a master :)

  • @lucientdaemon2773
    @lucientdaemon2773 3 года назад

    I don't remember subscribing here