Killing Vectors (question on Problem)

[Morgoth]

Well my main questions are colored in red...

Set of problem
Let's suppose we have a metric g_ab(x)
and the Killing vectors
I_a
( I_a;k=0, g^μαI_μI_α= 0)

question
Show that those killing vectors are the gradient of a Scalar field, and that it satisfies the equation I_αR^α_βγρ=0
Show that the new metric bellow describes the same space.
G_ab= g_ab+I_aI_b

Attempt to solve
1. Since I_a;k=0, I_k;a=0
we get the equations
I_a,k - Γ^ρ_ak I_ρ= 0 (1)
I_k,a - Γ^ρ_ka I_ρ= 0 (2)

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

2.
I_αR^α_βγρ=0
by using the commutation of covariant derivative:
I_a;k;c-I_a;c;k=I_dR^d_akc
but we know that I_a;b=0 so
0_;c - 0_;k =I_dR^d_akc
I_dR^d_akc=0
I think that this was an easy question.

3.
I have that
G_ab= g_ab+I_aI_b
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...
what is your opinion?

 

[WannabeNewton]

I_a
( I_a;k=0, g^μαI_μI_α= 0)

Were these two conditions given to you as part of the problem? Or are you claiming this is the definition of a killing vector?

3.
I have that
G_ab= g_ab+I_aI_b
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...
what is your opinion?

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.

 

[Morgoth]

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.

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...

 

[WannabeNewton]

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.

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.

 

[Morgoth]

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.

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).

 

[Ben Niehoff]

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.

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.

 

[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:
L_ξg_αβ=∂g_αβ/∂a
where L is the operator of Lie Derivative...

Is there an alternative way to show it?