What do the three indices in \Gamma ^a_{bc} represent in the unit sphere?

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You are saying that [tex]v^{\phi} = A e^{i \cos \theta_0 \phi} + B e^{-i \cos \theta_0 \phi}[/tex] should be [tex]v^{\phi} = A e^{i \cos ^2\theta_0 \phi} + B e^{-i \cos ^2\theta_0 \phi}[/tex] ?

I disagree because it is a solution to

[tex]\frac{d^2v^{\phi}}{d\phi^2} = - \cos^2 \theta v^{\phi}[/tex]
Also, in post 21 (could you tell me what post I differentiated theta in?) the first equation is just the parallel transport equation for phi and I just plugged in the expression for v^phi.

So, if we collect everything up to know we get that A = B =2, do you agree?
 
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Ok, right. I agree with you. My mistake. And I think I was mistaking a typo for a differentiation. Don't worry about it. I get that A=B=v^phi(0)/2.
 
Dick said:
Yes you can. It has unit length and it's pointed in the phi direction. And if you are having fun with exponentials, then those equations just say A=B with an initial condition of v^theta(0)=0. And stop differentiating theta. It's a constant.

How do you know that v^theta(0)=0?
 
It really doesn't matter what initial conditions you take. But if you want to follow the instructions of the problem to the letter, then (I've just reread it) it says to take the initial unit vector parallel to the phi=0 circle. I guess this would mean we actually should have had the initial conditions to be v^phi(0)=0 and v^theta(0)=1.
 
Dick said:
It really doesn't matter what initial conditions you take. But if you want to follow the instructions of the problem to the letter, then (I've just reread it) it says to take the initial unit vector parallel to the phi=0 circle. I guess this would mean we actually should have had the initial conditions to be v^phi(0)=0 and v^theta(0)=1.

I would think that it does matter what the initial conditions are but we'll see. So, with v^phi(0)=0 and v^theta(0)=1, I get:

A = -B = i/2 csc theta_0
 
Plugging in 2 pi for theta gives v^phi (phi = 2pi) = 0, right? So, now I need to solve the DE for v^theta and hopefully get v^theta (phi = 2pi) = 1, right?
 
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Nooo. I get v^phi=constant*sin(cos(theta)*phi). Putting in phi=2pi doesn't give you zero for a general value of theta. And you should expect this - the problem asks you to show that after transport the direction v is not generally the same, but the length is the same.
 
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Sure, but it's e^(i*cos(theta)*2pi), not e^(i*2pi).
 
Sorry, I deleted a post in between there.

Anyway, e^(i*cos theta 2pi) = (e^(i*2pi))^cos(theta) = 1^{cos theta} = 1, right?
 
Wrong. That rule only works for positive real numbers. i=exp(i*2pi*(1/4))=exp(i*2pi)^(1/4)=1^(1/4)=1.

What went wrong there? It's true that i and 1 are both fourth roots of 1 but they aren't equal.
 
Dick said:
Wrong. That rule only works for positive real numbers. i=exp(i*2pi*(1/4))=exp(i*2pi)^(1/4)=1^(1/4)=1.

What went wrong there? It's true that i and 1 are both fourth roots of 1 but they aren't equal.

Interesting!

I would need to see a counterexample for negative real numbers, though.

Yes, so then I get:

[tex]v^{\phi}(\phi = 2 \pi) = \sin ( \cos (\theta) 2 \pi) \csc \theta[/tex] .

I am not really sure how to simplify that.

So, now I need to solve the DE for v^theta and hopefully get v^theta (phi = 2pi) = sqrt( 1- sin^2 ( cos (theta) 2 pi) csc^2 theta ), right?
 
Something like that. Just find v^phi and v^theta and put them into the metric. BTW ((-1)^2)^(1/2)=1 which is NOT EQUAL to (-1)^(2*(1/2))=(-1)^1=-1. In the real case you can allow the exponents to be negative, but not the base (or whatever you call the -1 part).
 
OK. I plugged the expression for v^phi into the DE for v^theta and integrated and plugged in 2 pi for the phi parameter and got:

[tex]v^{\theta}( \phi = 2 \pi) = \cos^2 \theta \sin (\cos (\theta) 2 \pi)[/tex]

When you say plug it into the metric, do you mean find g_ab v^a v_b?
 
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I don't get the cos(theta)^2 and do I get that its a cos function of the argument. Check your derivation.
 
Do you get:

[tex]v^{\theta}( \phi = 2 \pi) = -\cos (\cos (\theta) 2 \pi)[/tex]

No way, that actually does give 1 when we plug into the metric. I think that completes the problem.
 
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If you plug phi=0 in you get v^phi(0)=0, v^theta(0)=(-1) (we wanted +1). I think you have another sign error lurking around. But, yeah, that's basically it, once you correct your derivation of v^theta.