Finding killing vector fields of specific spacetime

In summary: K^v = 0 needs to hold. But it doesn't. In fact, it seems like \partial_x K^x = 0 and \partial_y K^y = 0 are actually opposite. If that is the case, then my solution is wrong.
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Aemmel
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Homework Statement
Find explicit expressions for a complete set of Killing vector fields for the space [itex]ds^2 = -(dudv+dvdu)+a^2(u)dx^2+b^2(u)dy^2[/itex] (with coordinates [itex]\{u,v,x,y\}[/itex]) where [itex]a[/itex] and [itex]b[/itex] are unspecified functions of [itex]u[/itex]. (Carroll Exercise 3.13)
Relevant Equations
Hints from the exercise: there are five Killing vectors in all and all of them have a vanishing [itex]u[/itex] component [itex]K^u[/itex]
I have been at this exercise for the past two days now, and I finally decided to get some help. I am learning General Relativity using Carrolls Spacetime and Geometry on my own, so I can't really ask a tutor or something. I think I have a solution, but I am really unsure about it and I found 6 Killing Vectors instead of 5. Maybe someone could take a look at it and tell me where I went wrong or if my answer is actually correct and I am not seeing it.

Straight up we can see 3 Killing vectors: [itex]
K^{(1)\mu} = \left(0, 1, 0, 0\right) ,\
K^{(2)\mu} = \left(0, 0, 1, 0\right) ,\
K^{(3)\mu} = \left(0, 0, 0, 1\right)
[/itex] which we will also find again later on.
From the Killing equation [itex]\nabla_{(\mu} K_{\nu)} = g_{\nu\sigma} \nabla_\mu K^\sigma + g_{\mu\sigma} \nabla_\nu K^\sigma = 0[/itex] (I tried doing it with lower indices on [itex]K[/itex] but got completely stuck. Raising the index seems to make it a bit easier) we can find the Killing vectors. Maybe there's a clever trick I am not seeing, but I can't find it. So I tried to just write out all the equations and go from there.
Writing them all out and using [itex]K^u = 0[/itex] we get
[tex]
(I) \ \ \ \ \ \partial_u K^v = 0 \\
(II) \ \ \ \ \ \partial_x K^x = 0 \\
(III) \ \ \ \ \ \partial_y K^y = 0 \\
(IV) \ \ \ \ \ \partial_v K^v = 0 \\
(V) \ \ \ \ \ a^2 \partial_u K^x - \partial_x K^v = 0 \\
(VI) \ \ \ \ \ b^2 \partial_u K^y - \partial_y K^v = 0 \\
(VII) \ \ \ \ \ \partial_v K^x = 0 \\
(VIII) \ \ \ \ \ \partial_v K^y = 0 \\
(IX) \ \ \ \ \ b^2 \partial_x K^y + a^2 \partial_y K^x = 0
[/tex]

From these it's directly clear that [itex]K^v=K^v(x,y)[/itex] (from I, IV), [itex]K^x=K^x(u,y)[/itex] (from II, VII) and [itex]K^y=K^y(u,x)[/itex] (from III, VIII). Using this and differentiating V, VI, IX we find [itex]{\partial_x}^2 K^v = 0[/itex], [itex]{\partial_y}^2 K^v = 0[/itex], [itex]{\partial_x}^2 K^y = 0[/itex] and [itex]{\partial_y}^2 K^x = 0[/itex]. Integrating these we find
[tex]
K^v(x,y) = V_1 + V_2 x + V_3 y + V_4 xy \\
K^x(u,y) = X_1(u) y + X_2(u) \\
K^y(u,x) = Y_1(u) x + Y_2(u)
[/tex]
where the [itex]V_i[/itex] are constant with regard to any coordinate.
Plugging these into V and VI, rearranging with regard to [itex]x[/itex] or [itex]y[/itex] and comparing coefficients we can find that
[tex]
\partial_u X_1 \equiv X_1' = \frac{V_4}{a^2} \\
X_2' = \frac{V_2}{a^2} \\
Y_1' = \frac{V_4}{b^2} \\
Y_2' = \frac{V_3}{a^2}
[/tex]
Now, using [itex] X_2'=\frac{V_2}{V_4}X_1'[/itex] we can write [itex] X_2=\frac{V_2}{V_4}X_1 + C_x [/itex] (where [itex]C_x[/itex] is a constant regarding any coordinate) and a smiliar expression for [itex] Y_2 [/itex]. So that we get
[tex]
K^x = X_1 y + \frac{V_2}{V_4} X_1 + C_x \\
K^y = Y_1 y + \frac{V_3}{V_4} Y_1 + C_y
[/tex]
Putting this into V again and using the known derivative for [itex] X_1 [/itex] we find
[tex]
X_1(u) = V_4 \frac{b}{a(ab'-a'b)}
[/tex]
which is where I actually find the first problem. This would imply that the relation [itex] a = b a'' / b'' [/itex] has to hold (because we know the derivative of [itex] X_1 [/itex]). But in the problem it was mentioned they are arbitrary. So I thought my problem had to be there, but I just can't see why my derivation up to this point is wrong (or the spacetime in general implies that the relation has to hold, otherwise those Killing vectors wouldn't exist. That would also be possible). Let's continue with this result for now.
Plugging our [itex] K^x [/itex] and [itex] K^y [/itex] into IX we get [itex] Y_1 = -\frac{a^2}{b^2}X_1 [/itex] Thus arriving at
[tex]
K^u = 0 \\
K^v = V_1 + V_2 x + V_3 y + V_4 xy \\
K^x = V_4 \frac{b}{a(ab'-a'b)} y + V_2 \frac{b}{a(ab'-a'b)} + C_x \\
K^y = - V_4 \frac{a}{b(ab'-a'b)} x - V_3 \frac{a}{b(ab'-a'b)} + C_y
[/tex]
which finally gives us 6 Killing vectors, by setting one of the 6 constants to 1 and the rest to 0
[tex]
K^{(1)\mu} = \left(0, 1, 0, 0\right) \\
K^{(2)\mu} = \left(0, 0, 1, 0\right) \\
K^{(3)\mu} = \left(0, 0, 0, 1\right) \\
K^{(4)\mu} = \left(0, x, \frac{b}{a(ab'-ba')}, 0\right) \\
K^{(5)\mu} = \left(0, y, 0, -\frac{a}{b(ab'-a'b)}\right) \\
K^{(6)\mu} = \left(0, xy, \frac{b}{a(ab'-ba')} y, -\frac{a}{b(ab'-a'b)} x\right)
[/tex]

Now there are two problems with my solution.
First: there are 6 vector fields instead of 5. [itex] K^{(6)} [/itex] could be crafted with [itex] K^{(6)\mu} = yK^{(4)\mu} + xK^{(5)\mu} - xyK^{(1)\mu} [/itex] but that would go against the knowledge I have of "linear independence". I had the idea that it might be different for vector fields (meaning non constant coefficients are fine) but that would mean there are only 3 independent Killing vectors in [itex] \mathbb{R}^3 [/itex] which is not the case (as shown in Carrol 3.8).
Second: for [itex] K^{(4)} [/itex] to [itex] K^{(6)} [/itex] to be Killing vector fields, the equation [itex] a=ba''/b'' [/itex] has to hold, which, again, I don't see why that would be the case for arbitrary [itex] a, b [/itex].
 
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  • #2
I did not check your work carefully but I'd like to share what's the recipe I use when I want to get the Killing vectors out of a given metric describing a spacetime.

Let's work with the 2D anti-de Sitter metric as an example

$$(ds)^2 = -cosh^2 (\eta) d \tau^2 + d \eta^2$$

You should be able to follow the same logic to get your problem done.

Here we get one Killing vector by just having a look at the metric; as its components are independent of ##\tau## we get ##K_0=\partial_{\tau}##

1- Get all non-zero Christoffel symbols out of the given metric

For the ##AdS_2## we find there are only two (remember that the Christoffel symbols are symmetric under interchange of lower indeces):

$$\Gamma^{\tau}_{\tau \eta} =\tanh(\eta), \qquad \Gamma^{\eta}_{\tau \tau} = \cosh (\eta) \sinh (\eta)$$

2 - Have clear what's the most general equation for the Killing vectors you're looking for

In this case, the spacetime coordinates look like ##x^{\mu} = (\eta, \tau)##, so we expect a Killing vector equation of the form

$$K = f(\eta, \tau) \partial_{\mu} + g(\eta, \tau) \partial_{\tau} \ \ \ \ (1)$$

Based on it, we expect the components of the killing vector equation to be

$$K_{\eta} = f(\eta, \tau), \qquad K_{\tau} = -\cosh^2 (\eta) g(\eta, \tau)$$

3- Determine how many independent components your Killing vector has and get their corresponding equations

##\nabla_{(\mu}K_{\nu)}=\partial_{(\mu}K_{\nu)}-\Gamma_{\mu \nu}^{\rho}K_{\rho}## has three independent components in this case: ##(2) (\eta, \eta), (3) (\tau, \tau), (4) (\eta, \tau)##. Explicitly we get these three equations:

$$\partial_{\eta} K_{\eta} = 0 \ \ \ \ (2)$$

$$\partial_{\tau} K_{\tau} - \Gamma_{\tau \tau}^{\eta} K_{\eta}= 0 \ \ \ \ (3)$$

$$\partial_{\tau} K_{\eta} + \partial_{\eta} K_{\tau} - 2\Gamma_{\tau \eta}^{\tau} K_{\tau}= 0 \ \ \ \ (4)$$

Plugging the components of the Killing vector and and the Christoffel symbols into ##(2), (3), (4)## we get

$$\partial_{\eta} f(\eta, \tau) = 0 \ \ \ \ (5)$$

$$\partial_{\tau} g(\eta, \tau) + \tanh(\eta) f(\eta, \tau) = 0 \ \ \ \ (6)$$

$$\partial_{\tau} f(\eta, \tau) - \cosh^2 (\eta) \partial_{\eta} g(\eta, \tau) = 0 \ \ \ \ (7)$$

4- Solve such equations

To solve ##(5)## we set ##f(\eta, \tau)=f(\tau)##

Then plug ##f(\tau)## into ##(6)## and get

$$\partial_{\tau} g(\eta, \tau)=-\tanh\eta f(\tau)$$

Which can be solved by setting

$$g(\eta, \tau)=\tanh(\eta)h(\tau), \qquad h'(\tau)=-f(\tau) \ \ \ \ (*)$$

We now can go to equation ##(7)## and get

$$f'(\tau) - \cosh^2 (\eta) \partial_{\eta} (\tanh (\eta) h(\tau)) = f'(\tau)-h(\tau)$$

As ##h'(\tau) = -f(\tau)## then ##h''(\tau) = -f'(\tau)##

Thus equation ##(7)## ends up having the form

$$h''(\tau)=-h(\tau)$$

Which is a linear homogeneous second order differential equation whose solutions are

$$h_1(\tau)=e^{i\tau}, \qquad h_2(\tau)=e^{-i\tau}$$

Plug these solutions into ##(*)## and get

$$f_1(\tau) = -ie^{i\tau}, \qquad g_1(\eta, \tau)=\tanh(\eta)e^{i\tau}$$

$$f_2(\tau) = ie^{-i\tau}, \qquad g_2(\eta, \tau)=\tanh(\eta)e^{-i\tau}$$

5- Plug the functions you've got into (1)

In this case we get two Killing vectors

$$K_1 = e^{i\tau}(-i\partial_{\eta} + \tanh (\eta) \partial_{\tau})$$

$$K_2 = e^{-i\tau}(i\partial_{\eta} + \tanh (\eta) \partial_{\tau})$$

6- Conclude

Thus we get 3 Killing vectors out of the ##AdS_2## metric

$$K_0=\partial_{\tau}$$

$$K_1 = e^{i\tau}(-i\partial_{\eta} + \tanh (\eta) \partial_{\tau})$$

$$K_2 = e^{-i\tau}(i\partial_{\eta} + \tanh (\eta) \partial_{\tau})$$
 
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1. What is a killing vector field?

A killing vector field is a vector field on a manifold that preserves the metric tensor. In other words, it is a vector field that represents a symmetry of the manifold, meaning that the manifold looks the same at any point along the vector's path.

2. How do you find killing vector fields?

Finding killing vector fields involves solving a set of differential equations known as the Killing equations. These equations relate the metric tensor to the vector field and can be solved using various mathematical techniques, such as Lie algebra or group theory.

3. What is the significance of finding killing vector fields in specific spacetimes?

Finding killing vector fields in specific spacetimes can provide insight into the symmetries of the spacetime and the underlying physical laws that govern it. It can also be useful in solving problems in general relativity and other areas of physics.

4. Can killing vector fields be used to simplify calculations in physics?

Yes, killing vector fields can be used to simplify calculations in physics by reducing the number of parameters needed to describe a system. This is because the existence of a killing vector field implies a symmetry, which can be used to simplify equations and make calculations more manageable.

5. Are killing vector fields unique in a given spacetime?

No, killing vector fields are not unique in a given spacetime. In fact, there can be an infinite number of killing vector fields in a single spacetime. However, some spacetimes may have a special or unique killing vector field that represents a particularly important symmetry.

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