Quick clarification: when I talk about the spherical coordinates starting at 10:00, the x^2 and x^3 coordinates should be switched on the diagram: that is, the angle relative to the +x-axis should be x^3 and the angle relative to the +z-axis should be x^2. The equation for the arc length element will be correct this way.
10:55 I'm pretty sure that the x2 and x3 coordinates are the wrong way round either in the diagram of the spherical coordinate system or in the following equations, because the sin term should be the sin of the polar angle and it should go with the derivative of the azimuthal angle.
I was under the impression that in any smooth manifold (M,g), the metric g should already be given to you in some coordinate system. If you don't like it, you can change it. But in the form it is given, you figure out the components g_ij in a particular tangent space T_pM by evaluating g(x_i,x_j) for any pair of basis vectors x_i and x_j. This should be the same when restricting g to any submanifolds (such as curves in R^3). I'm still not quite familiar with how to do this in practice, though.
The typical ones (spherical, Cartesian, cylindrical, and even the Minkowski & Schwarzschild metrics) are usually represented by only diagonal elements (i.e. no cross-terms), but there's more complicated situations (e.g. en.wikipedia.org/wiki/Alcubierre_drive#Mathematics) where you will encounter cross terms.
What is the relationship between the tensor and the jacobian of the curve w.r.t. to x^1, x^2,..,x^n? Is the tensor showing the structure of the jacobian?
Not quite...the Jacobian matrix helps you transform between different coordinate systems while the metric tensor contains information about a particular coordinate system/space that essentially relates to the underlying curvature of that coordinate system. They are related in the sense that when written as a matrix, the determinant of the metric tensor is the square of the determinant of the Jacobian matrix.
Now the problem is how do we take the derivative of that when the Lagrangian is L=sqrt[g(x)(dx^i/dt)(dx^j/dt)] using the Euler Lagrange equation? I have no idea how the derivative is taken because those index stuff is very confusing.
Don’t get too ahead of yourself! We’ll need to learn the Christoffel symbols before we get there so we can apply Euler Lagrange and eventually get to the geodesic equation!
0:09 Where are the arrows of the axes x and y? Did the American teaching system lost the arrows? In French schools, there are always arrows on the axes. I thick it's better with arrows. It's also better for health, specially for Americans (writing more needs more energy, so this helps for loosing weight). Pro tips for Americans: don't write 1 as I, but with the little straight line at the end, this also helps for loosing weight. Appart for this, thanks for the video.
Quick clarification: when I talk about the spherical coordinates starting at 10:00, the x^2 and x^3 coordinates should be switched on the diagram: that is, the angle relative to the +x-axis should be x^3 and the angle relative to the +z-axis should be x^2. The equation for the arc length element will be correct this way.
For those looking for part 2, here it is! th-cam.com/video/2lqE9my0ch8/w-d-xo.html
OH MY GOD THE GOAT IS BACK
Fantastic discussion
thank you so much, I've been struggling with this for so long and now understand, thank you
10:55 I'm pretty sure that the x2 and x3 coordinates are the wrong way round either in the diagram of the spherical coordinate system or in the following equations, because the sin term should be the sin of the polar angle and it should go with the derivative of the azimuthal angle.
Good catch! I'll put in a pinned comment with that clarification! Spherical coordinates can be annoying with the notation sometimes...
Thank you very much.
So this is just the Kronecker delta in Cartesian?
Yes, exactly!
I was under the impression that in any smooth manifold (M,g), the metric g should already be given to you in some coordinate system. If you don't like it, you can change it. But in the form it is given, you figure out the components g_ij in a particular tangent space T_pM by evaluating g(x_i,x_j) for any pair of basis vectors x_i and x_j. This should be the same when restricting g to any submanifolds (such as curves in R^3). I'm still not quite familiar with how to do this in practice, though.
What would be an example of a coordinate system with cross-terms in the metric tensor?
The typical ones (spherical, Cartesian, cylindrical, and even the Minkowski & Schwarzschild metrics) are usually represented by only diagonal elements (i.e. no cross-terms), but there's more complicated situations (e.g. en.wikipedia.org/wiki/Alcubierre_drive#Mathematics) where you will encounter cross terms.
What is the relationship between the tensor and the jacobian of the curve w.r.t. to x^1, x^2,..,x^n? Is the tensor showing the structure of the jacobian?
Not quite...the Jacobian matrix helps you transform between different coordinate systems while the metric tensor contains information about a particular coordinate system/space that essentially relates to the underlying curvature of that coordinate system. They are related in the sense that when written as a matrix, the determinant of the metric tensor is the square of the determinant of the Jacobian matrix.
I am not getting the use of parameter 't' here, you can show the same just by using only dx and dy, ins't it?
The parameter t is used to parameterize the curve, making it easier to generalize to higher dimensions.
Now the problem is how do we take the derivative of that when the Lagrangian is L=sqrt[g(x)(dx^i/dt)(dx^j/dt)] using the Euler Lagrange equation? I have no idea how the derivative is taken because those index stuff is very confusing.
Don’t get too ahead of yourself! We’ll need to learn the Christoffel symbols before we get there so we can apply Euler Lagrange and eventually get to the geodesic equation!
Waiting for the follow-up video(s)
They'll be here next week!
0:09 Where are the arrows of the axes x and y? Did the American teaching system lost the arrows? In French schools, there are always arrows on the axes. I thick it's better with arrows. It's also better for health, specially for Americans (writing more needs more energy, so this helps for loosing weight). Pro tips for Americans: don't write 1 as I, but with the little straight line at the end, this also helps for loosing weight. Appart for this, thanks for the video.
Least insufferable frenchie
Sir, this is a Wendy's...
No, there are arrows in the American system. Khan is just lazy (probably drawing with a mouse)