We will show thereafter that for a surface of a sphere like the Earth:

- - the part of a meridian joining the equator to the north pole is a geodesic
- - a great circle as the equator is also a geodesic.
- - a line of latitude on the top part of the sphere is not a geodesic

We know from our previous article Geodesic exercise part I: calculation for 2-dimensional Euclidean space that the geodesic equation can be written as

In spherical polar coordinates for a two dimensional surface where r = cste, the index i equals 1 or 2 and **x ^{1 }= θ**,

**x**.

^{2}= φWe then need the connection coefficients for a surface of a sphere as calculated previously in Christoffel symbol exercise: calculation in polar coordinates part II

In the geodesic equation, if we replace x^{i} by θ, we have to sum over four elements in the form of Γ^{θ}_{jk}.

But **the only connection coefficient not null in the form of Γ ^{θ}_{jk} is Γ^{θ}_{φφ} = -sinθ cosθ** so that the geodesic equation becomes (i=θ, j=φ, k=φ)

**Equation 1**

Now we have to replace x^{i} by φ In the geodesic equation, and therefore we have to sum over four elements in the form of Γ^{φ}_{jk}.

**As the only two connection coefficients in the form of Γ ^{φ}_{jk} which are not null are Γ^{φ}_{φθ} = Γ^{φ}_{θφ} = cosθ/sinθ** then the geodesic equation becomes (i= φ , j=θ, k=φ and i= φ , j=φ, k=θ ) :

**Equation 2**

##### Checking that the meridian is a geodeisc

If we** parameterise our meridian** by saying that **θ = λ where 0 <= λ <= π/2**, and** φ = 0**

then dθ/dλ = 1 => d^{2}θ/dλ^{2} = 0 also as φ = 0, then dφ/dλ = d^{2}φ/dλ^{2} = 0

So equation 1 becomes 0 - sinθcosθ x 0 = 0 so 0 = 0 which holds true

In the same way, equation 2 becomes 0 + 2(cosθ/sinθ) x 1 x 0 = 0 <=> 0 = 0 which is verified

Both equations 1 and 2 are true (left-hand side equals the right hand side), therefore they satisfy the geodesic equations, meaning we have shown that **the meridian is a geodesic**.

##### Checking that the equator is a geodeisc

If we **parameterise our equator** by saying that **θ = π/2** and **φ = λ where 0 <= λ <= 2π**

then dφ/dλ = 1 => d^{2}φ/dλ^{2} = 0 also as θ = π/2, then dθ/dλ = d^{2}θ/dλ^{2} = 0 and cosθ = cos (π/2) = 0 sinθ = sin(π/2)=1

So equation 1 becomes 0 - 1 x 0 x 1^{2} = 0 so 0 = 0 which holds true

In the same way, equation 2 becomes 0 + 2 x 0 x 0 x 1 = 0 <=> 0 = 0 which is verified

Because equations 1 and 2 are true (left-hand side equals the right hand side), therefore they satisfy the geodesic equations, meaning we have shown that **the equator is a geodesic**.

##### Checking that a top latitude line is not a geodeisc

Let's take now the example of a circle on the top part of the sphere defined by **θ = π/4** and **φ = λ where 0 <= λ <= 2π**

Then dφ/dλ = 1 => d^{2}φ/dλ^{2} = 0 also as θ = π/4, then dθ/dλ = d^{2}θ/dλ^{2} = 0 and cosθ = cos (π/4) = sinθ = sin(π/4)= √ 2

So equation 1 becomes 0 - √ 2 x √ 2 x 1^{2} = -2 ! = 0 (not true!)

Equation 2 becomes 0 + 2 x 1 x 0 = 0 <=> 0 = 0 which is verified

As equation 1 is not verified even if equation 2 Is verified, it means this top circle of latitude** is a not geodesic**.

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