What Are the Challenges in Using Tensor Notation for Cartesian Coordinates?

[Shackleford]

Maybe this is better here. We're in the first chapter - Matrices, Vectors, and Vector Calculus. I'm not sure why I'm having trouble. I got an A in Vector Analysis in the spring. I think some of the notation is throwing me off for some reason.

I know a bold letter represents a vector, but what about the italicized letter? Is it the norm of the vector or simply a scalar? Please, check my notation carefully and my work. I apologize for the blurry picture. I'm having to use my camera phone right now as I haven't moved my multi-function printer to the new house.

26. Haven't got a damn clue.

http://i111.photobucket.com/albums/n149/camarolt4z28/q26.jpg?t=1282967742

http://i111.photobucket.com/albums/n149/camarolt4z28/26bc.jpg?t=1282967403

http://i111.photobucket.com/albums/n149/camarolt4z28/26.jpg?t=1282967514

28. I think I'm close to the right solution, but I still need clarification on some things. Also, I realize it should be d/dxi not d/dri.

31. I started (a) and haven't done (b). I just worked out (c), so I hope it will elucidate the other ones.

http://i111.photobucket.com/albums/n149/camarolt4z28/31.jpg?t=1282967605

http://i111.photobucket.com/albums/n149/camarolt4z28/q28.jpg?t=1282967285

http://i111.photobucket.com/albums/n149/camarolt4z28/2831a.jpg?t=1282966751

http://i111.photobucket.com/albums/n149/camarolt4z28/31c.jpg?t=1282966772

 

[lanedance]

i think the italicized letter means magnitude of the vector (a positive scalar) in this context

for the 2nd part have you seen the form of the gradient in spherical coordinates, using that representation will greatly simplify the problems

 

[lanedance]

see
http://en.wikipedia.org/wiki/Gradient#Non-Cartesian_Coordinate_Systems

 

[lanedance]

or just click on the word gradient

 

[Shackleford]

i think the italicized letter means magnitude of the vector (a positive scalar) in this context

So, would I represent the italicized letter as [(x-sub-i)^2]^1/2 ?

for the 2nd part have you seen the form of the gradient in spherical coordinates, using that representation will greatly simplify the problems

Sorry. For which problem? 26?

We derived the grad, div, and curl in cylindrical and polar coordinates last semester in my vector analysis course.

 

[Shackleford]

I'm getting -1/r^2 for 31.c.

 

[Shackleford]

Here's my work for 31.a. I'm not close.

http://i111.photobucket.com/albums/n149/camarolt4z28/31a.jpg?t=1283118115

Here's my work for 38. I had to bust out my calculus textbook to look up the procedure for Stokes's Theorem for surface integral evaluation.

38. Find the value of the integral - surface integral curl A dot da, where A = yi + zj + xk and S is the surface defined by the paraboloid z = 1 - x^2 - y^2.

http://i111.photobucket.com/albums/n149/camarolt4z28/38a.jpg?t=1283118127

http://i111.photobucket.com/albums/n149/camarolt4z28/38b.jpg?t=1283118138

And as a side note, we're going to start this material tomorrow. I'm hoping it'll clear mostly everything up for me.

 

[Shackleford]

Any bright ideas?

 

[lanedance]

31(a)
easiest - why not try the spherical form of the gradient operator

harder - are you trying to do it in cartesian coords & index notation?

if so i would probably write its as
r = \sqrt{x_i x_i}
r^n = (x_i x_i)^{n/2}

then
(\nabla r^n)_j <br /> = \frac{\partial }{\partial x_j}(x_i x_i)^{n/2} <br /> = \frac{n}{2} (x_i x_i)^{(\frac{n}{2}-1)} \frac{\partial }{\partial x_j}(x_k x_k)

<br /> = \frac{n}{2} (x_i x_i)^{(\frac{n-2}{2})} \frac{\partial }{\partial x_j}(x_k x_k) <br /> = \frac{n}{2} r^{n-1} \frac{\partial }{\partial x_j}(x_k x_k)

see if you can carry on from there

 

[lanedance]

updated for 31 a) to r^n

 

[Shackleford]

updated for 31 a) to r^n

Ah. I wasn't sure how to write that in tensor notation. I tried r-sub i-to the n, then squared that which gave me r-sub i-to the 2n, which is incorrect. I didn't realize it was the norm of r to the n power.

Yes, I was doing it in Cartesian with tensor notation. I'm still trying to get used to it.

Why did you change dummy indices?