Calculating Christoffell Symbols of Second Kind for Cartesian Space

[zheng89120]

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]^a_bc = { f_af_bc }/ { f^\alpha f_sub_\alpha }

where indices on f indicates partial derivatives

Homework Equations

(d^2 x/dt^2)^\alpha + [Christoffell symbol]^\alpha_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

 

[sgd37]

have you got an online version of the question or picture because it is hard to make sense of this thing

 

[zheng89120]

http://www.flickr.com/photos/59383047@N05/5436668502/

 

[sgd37]

you have a 2d metric that only depends on the derivatives of f . You can derive it to be

g = \begin{pmatrix}<br /> <br /> 1+(\partial_1 f)^2 &amp; \partial_1 f \partial_2 f \\<br /> \partial_1 f \partial_2 f &amp; 1+(\partial_2 f)^2 \end{pmatrix}

or in component form and using your notation

g_{ab} = \delta_{ab} +f_a f_b

then you can use the standard formula for the Christoffel symbols

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

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