Register to reply

Connection of the unit sphere

by ehrenfest
Tags: connection, sphere, unit
Share this thread:
ehrenfest
#37
Oct2-07, 04:10 PM
P: 1,996
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?
Dick
#38
Oct2-07, 04:17 PM
Sci Advisor
HW Helper
Thanks
P: 25,251
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.
Dick
#39
Oct2-07, 04:27 PM
Sci Advisor
HW Helper
Thanks
P: 25,251
Sure, but it's e^(i*cos(theta)*2pi), not e^(i*2pi).
ehrenfest
#40
Oct2-07, 04:32 PM
P: 1,996
Sorry, I deleted a post in between there.

Anyway, e^(i*cos theta 2pi) = (e^(i*2pi))^cos(theta) = 1^{cos theta} = 1, right?
Dick
#41
Oct2-07, 04:39 PM
Sci Advisor
HW Helper
Thanks
P: 25,251
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.
ehrenfest
#42
Oct2-07, 04:53 PM
P: 1,996
Quote Quote by Dick View Post
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?
Dick
#43
Oct2-07, 05:02 PM
Sci Advisor
HW Helper
Thanks
P: 25,251
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).
ehrenfest
#44
Oct3-07, 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?
Dick
#45
Oct3-07, 05:37 PM
Sci Advisor
HW Helper
Thanks
P: 25,251
I don't get the cos(theta)^2 and do I get that its a cos function of the argument. Check your derivation.
ehrenfest
#46
Oct3-07, 05:47 PM
P: 1,996
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.
Dick
#47
Oct3-07, 05:56 PM
Sci Advisor
HW Helper
Thanks
P: 25,251
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.


Register to reply

Related Discussions
Multiple integrals - volume of part of a unit sphere Calculus & Beyond Homework 13
Intersection of concentric, unit area sphere and square General Math 10
Unit sphere arcwise connected? Calculus & Beyond Homework 4
In which frame does the unit 2-sphere look locally flat? Differential Geometry 3
Metric tensor on unit sphere Differential Geometry 3