What has gone wrong with this argument?!

The original question

A space-time such that $$ds^2=-dt^2+t^2dx^2$$ has Killing vectors $(0,1),(-\exp(x),\frac{\exp(x)}{t}), (\exp(-x),\frac{\exp(-x)}{t})$.

Given that $$\dot x^b\frac{\partial}{\partial x^b}(\dot x^a\xi_a)=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(*)$$ where $x^a=(t,x)$, $\dot x^a$ is the tangent vector to a geodesic and $\xi^a$ is a Killing vector, then $$a\exp(x)+b\exp(-x)+c\frac{1}{t}=0$$ where $a,b,c$ are constants.

So I tried directly using $(*)$. What I got are $$t\dot t\dot x=0\\ \exp(x)(2\dot t\dot x+t\dot x^2)=0\\ \exp(-x)(2\dot t\dot x-t\dot x^2)=0$$ for each of the Killing vectors given.

(Right so far?)

Then multiplying the last 2 equations by $t$ and using the first equation, I get $$\exp(x)t^2\dot x^2=0\\ -\exp(-x)t^2\dot x^2=0$$ So any linear combination of the 2 LHS's must vanish, giving $$\alpha\exp(x)t^2\dot x^2+\beta\exp(-x)t^2\dot x^2=0$$ for arbitrary constants $\alpha,\beta$.

Either I have gone wrong somewhere, or somehow we must have $$t^2\dot x^2=1+\frac{\gamma}{t}$$

Unfortunately, I can't see why it is so. Any insight? Thanks.


Let us denote \begin{align} \xi_1 = (0,1), \qquad \xi_2 = (-e^x, e^x/t), \qquad \xi_3 = (e^{-x}, e^{-x}/t) \end{align} Each of these killing vectors leads to a conserved quantity \begin{align} c_1 &= \dot x_\mu\cdot (\xi_1)^\mu = \dot x t^2 \\ c_2 &= \dot x_\mu\cdot (\xi_2)^\mu = \dot t e^x +\dot x te^x \\ c_3 &= \dot x_\mu\cdot (\xi_3)^\mu = -\dot t e^{-x} + \dot x t e^{-x} \end{align} From the first conservation equation we obtain $$ \dot x t = \frac{c_1}{t} $$ Plugging this into the second two gives \begin{align} c_2e^{-x} &= \dot t+\frac{c_1}{t} \\ c_3e^{x} &= -\dot t+\frac{c_1}{t} \\ \end{align} and adding them together then gives the desired result.

  • $\begingroup$ Thank you, josh. I have double-checked the question and I'm sure I have transcribed it correctly... $\endgroup$ – Jill May 14 '13 at 23:19
  • $\begingroup$ @Jill Yes! I found my mistake as well; I had, in fact, made an algebraic error. I edited the response; glad to be of service. $\endgroup$ – joshphysics May 15 '13 at 0:52

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