Geodesic in a curved space

1. Nov 1, 2007

ledol83

Hi there:

i have a question on geodesics in a Eculidean space equipped with a metric tensor \lambda(x)*I, where I is the identity matrix. Is any general statement that can be made towards the geodesic between two points in this modified space?

My feel is that this space is quite special and should have some good properties but don't know how to address it.

Thanks for any suggestion!!

Last edited: Nov 1, 2007
2. Nov 1, 2007

Aren't the geodesics just lines in that case? Incidentally, I don't think that space is curved.

3. Nov 1, 2007

ledol83

hi, why do you think geodesics are lines?

According to Theodore Shifrin's book on line pp.86, for hyperbolic plane (u,v) equipped with metric tensor 1/v^2*I, the geodesics are not lines.

I was just curious if any analytic relation between the geodesics and the lambda(x) function can be made (except for the PDE thing).

I agree, it should not be called 'curved'.

Last edited: Nov 1, 2007
4. Nov 1, 2007

I thought $\lambda$ was constant here

5. Nov 3, 2007

Chris Hillman

Providing a buzzword

of form $g_{ab} = \lambda \, \eta_{ab}$, where $\lambda$ is a scalar functon and where $\eta_{ab}$ is the metric tensor of Euclidean three-space, which in a Cartesian chart will appear as

Such a metric tensor is said to define a Riemannian geometry which is conformally related to Euclidean geometry.

It is, but I know more about the Lorentzian case than the Riemannian case so I'll limit myself to mentioning the buzzword "conformally flat".

Last edited: Nov 3, 2007
6. Nov 6, 2007

ledol83

7. Nov 16, 2007

WWGD

in SW and pasted them here, would the program read it effectively?.Again, sorry
for not knowing to use Tex yet.

I don't know what assumption we may be making here, but one way of seeing it
(tho, I admit, I think I am missing an assumption here to have a full response)
is that a geodesic in a manifold is a curve with acceleration 0, i.e, a curve whose
covariant derivative is 0, and, using the Euclidean connection , lines, when
parametrized by arc-length, have second derivative zero, i.e, acceleration
zero (d^2/ds^2=d/ds(d/ds)=0 , means derivative of velocity with resp. to
arc-length is zero, i.e, acc. is zero):

Write your line in terms of arc-length (any curve, with few conditions
can be parametrized by arc-length)

L(s)=x_1=a_1+b_1s....x_n=a_n+b_1s , Sum(b_i)^2=1 (bcse. of arc-length par.)

It follows that d^L(s)/ds^2==0

The doubt I have is where we are making use of the Euclidean connection
in here. (which I think is the only connection that is compatible with the
metric dx^2+dy^2 and has zero torsion, i.e, the only Levy-Cine-Cita
connection. I will try to prove this and get back )