# Killing Vectors (question on Problem)

1. Feb 23, 2013

### Morgoth

Well my main questions are colored in red....

Set of problem
Let's suppose we have a metric gab(x)
and the Killing vectors
Ia
( Ia;k=0, gμαIμIα= 0)

question
Show that those killing vectors are the gradient of a Scalar field, and that it satisfies the equation IαRαβγρ=0
Show that the new metric bellow describes the same space.
Gab= gab+IaIb

Attempt to solve
1. Since Ia;k=0, Ik;a=0
we get the equations
Ia,k - Γρak Iρ= 0 (1)
Ik,a - Γρka Iρ= 0 (2)

from (1)-(2) we get
Ia,k-Ik,a=0
can I say now that the Iα(x) is the gradient of a scalar?
I think I can.

2.
IαRαβγρ=0
by using the commutation of covariant derivative:
Ia;k;c-Ia;c;k=IdRdakc
but we know that Ia;b=0 so
0;c - 0;k =IdRdakc
IdRdakc=0
I think that this was an easy question.

3.
I have that
Gab= gab+IaIb
how can I show that it describes the same space?
one way I thought of is to calculate the R' riemann tensor of metric G and show that it is equal to R tensor of the metric g. But things get lost in calculations...

2. Feb 23, 2013

### WannabeNewton

Were these two conditions given to you as part of the problem? Or are you claiming this is the definition of a killing vector?
One way (the brute force but straightforward way) is you can start with the arbitrary metric and the coordinate transformation law for the metric tensor. Then try to play around with it until you show that the new components of the metric tensor are in fact what the problem wants.

3. Feb 23, 2013

### Morgoth

1. yes they were given to me as part of the problem. they are what I know... One question was to show that I is indeed a killing vector, which I proved.
2. OK I will try this now...

4. Feb 23, 2013

### WannabeNewton

Ok. Note that your original plan would not necessarily have to work because in general the components of the Riemann tensor change under a change of coordinates from one form of the metric tensor to another.

By the way, in terms of definitions, two metrics describe the same space if they are equivalent up to an isometry.

Last edited: Feb 23, 2013
5. Feb 23, 2013

### Morgoth

Yes but from the Rieman Tensor I can evaluate the Ricci tensor and the scalar curvature. The same S.curvature would describe the same space.
But the whole procedure is tough, tiring and of course might contain mistakes (when I tried it, just for the Riemann tensor I filled 3 pages and I was not sure if I calculated it correctly). So I wanted to know if there is a Killing vector identity that I was missing (since the new metric is your previous one + the term of two multiplied killing vectors).

6. Feb 23, 2013

### Ben Niehoff

Just because two spaces have the same scalar curvature does NOT mean they are the same space.

You need to do what WBN suggested: Show that the new metric is a coordinate transformation of the old one.

7. Feb 24, 2013

### Morgoth

One more question, in case anyone knows...
If you have an I:
ds2= a (dθ2+sin2θ dφ2)
you can show that a is not absorbable parameter if the below equation has no solution:
Lξgαβ=∂gαβ/∂a
where L is the operator of Lie Derivative...

Is there an alternative way to show it?

Last edited: Feb 24, 2013