Re: geodesics in hyperbolic plane

In article <1133406829.956824.170340@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
<berthuffman@xxxxxxxxx> wrote:
>berthuffman@xxxxxxxxx wrote:
>> Its kind of hard to show my computations, but here is most of it:
>>
>> First, metric and its inverse on H
>>
>> g_11 = y^{-2} g_12 = g_21 = 0 g_22 = y^{-2}
>> g^11 = y^2 g^12 = g^21 = 0 g^22 = y^2
>>
>> chirstoffel symbols
>>
>> G_12^1 = G_22^2 = -y^{-1}
>> G_11^2 = y^{-1}
>> G_11^1 = G_12^2 = G_22^1 = 0
>>
>> (not listing the symmetric symbols)
>>
>> if V=d(gamma)/dt=(v_1(t),v_2(t)) then the covariant derivative is
>>
>> DV/dt = \sum_k (dv_k/dt + \sum_{i,j} G_ij^k v_i v_j) X_k
>>
>> we see that (v_1(t),v_2(t)) = (0,1)
>>
>> so dv_k/dt = 0. also since v_1=0 we see that the G_ij^k v_i v_j will be
>> zero unless i=j=2. so it seems we only get one contribution from the
>> sum:
>>
>> DV/dt = G_22^2 v_2 v_2 X_2 = -y^{-1} X_2
>>
>> where X_2 is the standard basis vector in the "y" direction i.e. @/@y.
>>
>> I know my notation is really sloppy but I think the basic idea is
>> there.
>>
>> does this have anything to do with gamma not being parameterized
>> correctly, or do I simply have a mistake in my computation?
>
>
>Can anyone help with this computation? I know my work is really sloppy
>so if someone could just point me the right direction (first of all, is
>it true that the covariant derivative of the curve is zero?)

I think the problem is that your curve gamma(t) = (0, t) is not a
geodesic. In fact you can't have t in (-infty, infty) here because
y <= 0 is not part of the "plane" (really a half-plane). What I
get for the geodesic equations is

x'' - 2 x' y'/y = 0, y'' + ((x')^2 - (y')^2)/y = 0

and if x = constant, a solution would be y = a exp(b t) for arbitrary
constants a and b.

Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.

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