• Support PF! Buy your school textbooks, materials and every day products Here!

Cartesian space

  • Thread starter zheng89120
  • Start date
  • #1
149
0

Homework Statement



Let a surface be define by z = x^(a=3) = f[x^(a=1,2)]

Show that the Christoffell sybols of the 2nd kind are:

[Christoffell symbol]^abc = { fafbc }/ { f^[tex]\alpha[/tex] f_sub_[tex]\alpha[/tex] }

where indices on f indicates partial derivatives

Homework Equations



(d^2 x/dt^2)^[tex]\alpha[/tex] + [Christoffell symbol]^[tex]\alpha[/tex]BC (dx/dt)^B (dx/dt)^C = 0

compare with:

Euler-Lagrangian Equation

The Attempt at a Solution



E-L equatiion: x** - m dz*/dx* = -g dz/dx

compare with the first relevant equation...

how? what is x*^B and x*^C in the first equation
 
Last edited:

Answers and Replies

  • #2
213
8
have you got an online version of the question or picture because it is hard to make sense of this thing
 
  • #3
149
0
http://www.flickr.com/photos/59383047@N05/5436668502/ [Broken]
 
Last edited by a moderator:
  • #4
213
8
you have a 2d metric that only depends on the derivatives of f . You can derive it to be

[tex] g = \begin{pmatrix}

1+(\partial_1 f)^2 & \partial_1 f \partial_2 f \\
\partial_1 f \partial_2 f & 1+(\partial_2 f)^2 \end{pmatrix} [/tex]

or in component form and using your notation

[tex] g_{ab} = \delta_{ab} +f_a f_b [/tex]

then you can use the standard formula for the Christoffel symbols

[tex] \Gamma^{a}_{bc} = \frac{1}{2} g^{ad} (\partial_c g_{bd} +\partial_b g_{cd} - \partial_d g_{bc} ) [/tex]

the inverse metric is a bit harder to write down in component form but it is possible
 

Related Threads on Cartesian space

  • Last Post
Replies
2
Views
4K
Replies
0
Views
1K
  • Last Post
Replies
1
Views
1K
Replies
1
Views
743
  • Last Post
Replies
3
Views
2K
Replies
0
Views
804
Replies
7
Views
644
Replies
5
Views
493
Replies
5
Views
1K
Replies
7
Views
7K
Top