Assignment 7:

Following the lectures, we will practice the mathematical concepts developed in class in this and in

later homeworks. Our favorite example is the manifold known as S 2 , which in math lingo is the two- sphere, i.e. the surface of your ordinary sphere (the 2 sets the dimensionality of the “surface”). Notice that the language is based on a larger definition of the word “sphere” such that S n is an n-dimensional spherical surface, defined as the set of all points that are at a common distance with a point that is not

on S n . Another example is S 1 which is just the ordinary circle.

The usual starting point of any geometric calculation is choosing a metric on the manifold in question. What this really means is that generally there is more than one unique metric for any manifold. We choose the metric:

22222µν ds==+ gµν dx dx R() dθθsin dϕ

where the indices run over µ,1,2ν = such that xx12==θ, ϕ . Although we are working with spatial coordinates, i.e. there is no time, we will still use Greek indices, just to get used to them. Now, the constant R is obviously useless, so we simply set it to be R = 1, in other words we are studying the unit-sphere (a sphere with radius=1). The metric is now: ds2222=+ dθθsin dϕ (1) One last thing to note, just for future reference, is that the metric we have chosen does not cover the entire manifold. There are two points on S 2 that are ill-defined by this metric (it is said that they have a “pathological behavior”, and yes, this is a technical term). Can you tell which ones?

1) Convince yourself that this metric satisfies the demand of Riemannian geometry that manifolds behave locally as if they were flat. In other words

show that on a very small part of the manifold, the metric (1) looks like the flat Euclidean metric in Cartesian coordinates.

µν 2) Write down the components of the metric tensor g µν and its inverse g .

µνν Check that ggµρ= δ ρ .

ρ 3) Calculate all the non-vanishing Christofell connections Γµν . You may use

Γ2 Γϕ either the notation 12 or θϕ both of which mean the same thing. 4) Write down the geodesic equations (there are two of them) for the sphere by

2 µαβ d xµ dx dx substituting in +Γαβ =0. Choose the arbitrary parameter λ to dddλλλ2 be λ = ϕ and show that the geodesic equations are satisfied only for a point on the north pole (θ = 0 ) and the equator (θπ= 2 ) which is a great circle. Since we can arbitrarily change our coordinates such that any great circle on the sphere can be the equator, the conclusion is that all great circles are geodesics! Notice that the equations are NOT satisfied by any latitude lines θ = constant . For the choice λθ= , one can show that the equations are satisfied by “vertical” great circles, i.e. line of longitude ϕ = constant , but you don’t have to show this (unless you really want to – it is only a couple of lines). 5) In problem four we proved that lines of constant latitude (θ = constant ) are not generally geodesics. Let us use one of those to illustrate parallel transport. Imagine that you are planning to walk around the world, starting from South Hadley, making a full circle and back, provided that you follow a constant latitude line (for South Hadley θ =42.3º North, measured from the equator, which is θ =90-42.3=47.7º in our choice of coordinates). Let us also assume that you are holding in your hands a flag pole. Because of the curvature of the earth, if you maintain the orientation of the flag pole with respect to the latitude line at all times as you walk around the world, you will return with the flag pole pointing in a

µθϕ different direction!! Choose a vector Ag==+AAAˆˆˆµ gθϕ g along the flag pole, parallel transport this vector A around the curve using the parallel

µν transport equation VA∇=µ 0 and find the angle it makes with its original self after you return to South Hadley. We do this over three steps: (a) First show that the parallel transport equation gives

θ ∂A ϕ = sinθθ cos A ∂ϕ

∂ ϕθ AA=− ∂ϕθtan

(b) These are coupled partial differential equations, to uncouple them, differentiate them with respect to ϕ and use the results to reduce these two first order coupled equations to two second order uncoupled equations, i.e. each in a single function A θ and A ϕ . Show that the resulting equations are similar to the form of the very familiar simple df2 harmonic motion equation +=κ 2fq() 0 . Solve (or look up the dq 2 solution in any mechanics textbook). You may choose initial conditions for the vector A in order to set the constants of integration that arise. (i.e., choose what direction the flag pole started out pointing towards, and how long it is) For example, you could choose: At ϕ = 0 :

A θ = 0 and A ϕ ===AAi A 1. In other words, the vector A starts

off pointing exactly tangent to the direction of the latitude line and the magnitude is equal to 1 unit of length! (c) Finally, look at the solutions. We chose the vector to point along the line of constant latitude at ϕ = 0 . Making a full circle, i.e. at ϕ = 360 , what does the vector look like now? Can you figure out the smaller angle ω between it and its original form? Also note that if one chooses the line of constant latitude to be the equator (θ = 90 ) then parallel transport brings a tangent vector back to itself, because the equator is the only line θ = constant that is also a geodesic.