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.
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.
@@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 😆
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)
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.
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;
![Div, Grad, and Curl: Vector Calculus Building Blocks for PDEs [Divergence, Gradient, and Curl]](http://i.ytimg.com/vi/lKXW7DRyyro/mqdefault.jpg)
I love your energy! best video I've seen on the subject by far, thank you 😄
Aww yay that makes me so happy to hear, thank you for sharing!!
Good teaching method
I would join every class with your lecture
awww thank you!! that makes me very happy to hear.
Thanks a lot, this is so helpful.
thank you, this was helpful indeed!
Glad it was helpful!
Maybe newton method on each dimension separately or differential equations systems of equations may easier
Great video!
Thank you!
Benefits of having less subscribers - mostly everyone will get answer of their questions 😁😁
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.
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.
silly question. is it not dl suqared ? Or is dl not the length ?
It's helpful. What's your book? Are you going to make a separate video for each coordinate system ?
Just the one! I have a short book on microcontrollers but thats all.. for now 🙃
@@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 😆
Thank you so much. This was very helpful 💓
I'm so glad to hear!!
Thanks mam🎉🎉🎉
Instrumentation may simulate better because it interactively and limited of integration unless...
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)
merry christmas mam❤
thanks a lot
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.
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.
Can anyone do phd in particle physics after getting master's in chemistry??
Unsure, but worth a try if you're passionate!
@@JenFoxBot okay thanks
This is a lil advanced class for me I'll come back later but thanks
Your vids are really good, but the camera constantly autofocusing is distracting, maybe it has a manual mode that will stop it doing that?
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 🤷♀️
We can confirm that she is not confirm when when she was laughin but we weren't.
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;
Are you doctor ?
no, just a master :)
IITBHU
I don't remember subscribing here