# 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]^abc = { fafbc }/ { 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

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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/ [Broken]

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sgd37
you have a 2d metric that only depends on the derivatives of f . You can derive it to be

$$g = \begin{pmatrix} 1+(\partial_1 f)^2 & \partial_1 f \partial_2 f \\ \partial_1 f \partial_2 f & 1+(\partial_2 f)^2 \end{pmatrix}$$

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