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Connection of the unit sphere 
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#37
Oct207, 04:10 PM

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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?



#38
Oct207, 04:17 PM

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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.



#39
Oct207, 04:27 PM

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Sure, but it's e^(i*cos(theta)*2pi), not e^(i*2pi).



#40
Oct207, 04:32 PM

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Sorry, I deleted a post in between there.
Anyway, e^(i*cos theta 2pi) = (e^(i*2pi))^cos(theta) = 1^{cos theta} = 1, right? 


#41
Oct207, 04:39 PM

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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. 


#42
Oct207, 04:53 PM

P: 1,996

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? 


#43
Oct207, 05:02 PM

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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).



#44
Oct307, 05:36 PM

P: 1,996

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? 


#45
Oct307, 05:37 PM

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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.



#46
Oct307, 05:47 PM

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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. 


#47
Oct307, 05:56 PM

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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.



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