General Relativity Geodesic Problem

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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.
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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
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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