- #1
Aemmel
- 1
- 1
- 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].
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].