# Einstein Relatively Easy

### What's Up If you like this content, you can help maintaining this website with a small tip on my tipeee page 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 x1 = θ, x2 = φ. We 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 xi by θ, we have to sum over four elements in the form of Γθjk.

But the only connection coefficient not null in the form of Γθjk is Γθφφ = -sinθ cosθ so that the geodesic equation becomes (i=θ, j=φ, k=φ)

Equation 1 Now we have to replace xi by φ In the geodesic equation, and therefore we have to sum over four elements in the form of Γφjk.

As the only two connection coefficients in the form of Γφjk which are not null are Γφφθ = Γφθφ = 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 => d2θ/dλ2 = 0   also as φ = 0, then dφ/dλ = d2φ/dλ2 = 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 => d2φ/dλ2 = 0   also as θ = π/2, then dθ/dλ = d2θ/dλ2 = 0  and cosθ = cos (π/2) = 0 sinθ = sin(π/2)=1

So equation 1 becomes 0 - 1 x 0 x 12 = 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 => d2φ/dλ2 = 0   also as θ = π/4, then dθ/dλ = d2θ/dλ2 = 0  and cosθ = cos (π/4) =  sinθ = sin(π/4)= √ 2

So equation 1 becomes 0 - √ 2 x √ 2 x 12 = -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.

### Quotes

"The essence of my theory is precisely that no independent properties are attributed to space on its own. It can be put jokingly this way. If I allow all things to vanish from the world, then following Newton, the Galilean inertial space remains; following my interpretation, however, nothing remains.."
Letter from A.Einstein to Karl Schwarzschild - Berlin, 9 January 1916

"Quantum mechanics is certainly imposing. But an inner voice tells me that it is not yet the real thing. The theory says a lot, but does not really bring us any closer to the secret of the 'old one'. I, at any rate, am convinced that He is not playing at dice."
Einstein to Max Born, letter 52, 4th december 1926