General Relativity Geodesic Problem

AI Thread Summary
The discussion focuses on proving that the curve x1 = a sec x2 is a geodesic in a 2D Euclidean metric expressed in polar coordinates. The Euclidean metric is defined as ds^2 = dr^2 + r^2 dθ^2, and the geodesic equations involve calculating derivatives and Christoffel symbols. The calculations show that the second derivatives lead to equations that can be solved for the derivatives of r and θ, ultimately demonstrating that r = a and θ = ln(λ) + b satisfy the original equation. Thus, the curve is confirmed to be a geodesic. The conclusion emphasizes the simplicity of the result once the correct approach is taken.
regretfuljones
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Show that x1=asecx2 is a geodesic for the Euclidean metric in polar coordinates.

So I tried taking all the derivatives and plugging into polar geodesic equations. Obviously, bad idea.

Now I'm thinking I need to use Dgab/du=gab;cx'c and prove that the lengths of some vectors and their dot product are invariants under parallel transport, but I don't know how to go about doing that. Any advice on how to relate these concepts would be appreciated.
 
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this turned out to be very simple.
since it is in 2D space, it is a straight line therefore a geodesic.
 


First, let's define the Euclidean metric in polar coordinates as:

ds^2 = dr^2 + r^2dθ^2

Now, we want to show that x1 = asecx2 is a geodesic for this metric. To do this, we need to use the geodesic equations:

d^2x^a/dλ^2 + Γ^a_bc(dx^b/dλ)(dx^c/dλ) = 0

where λ is an affine parameter, Γ^a_bc are the Christoffel symbols, and x^a are the coordinates. In our case, we have x^1 = r and x^2 = θ.

We can start by calculating the first derivatives of x^1 and x^2 with respect to λ:

dx^1/dλ = dr/dλ = r'
dx^2/dλ = dθ/dλ = θ'

where ' denotes differentiation with respect to λ.

Now, let's calculate the second derivatives:

d^2x^1/dλ^2 = d(r')/dλ = r''
d^2x^2/dλ^2 = d(θ')/dλ = θ''

Next, we need to calculate the Christoffel symbols for our metric. For the Euclidean metric in polar coordinates, we have:

Γ^1_11 = 0
Γ^1_22 = -r
Γ^2_12 = Γ^2_21 = 1/r

Now, let's plug everything into the geodesic equations:

r'' + (-r)(r')^2 = 0
θ'' + (1/r)(r')^2 = 0

Since we have two equations and two unknowns (r and θ), we can solve for r' and θ':

r' = 0
θ' = 1/r

Integrating with respect to λ, we get:

r = a
θ = ln(λ) + b

where a and b are constants of integration.

Now, let's substitute these values back into our original equation x^1 = asecx^2:

r = a
θ = ln(λ) + b
x^1 = a
x^2 = ln(λ) + b

We can see that this satisfies our original equation, so x^1 = a
 
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