Solution Generating Transformations

Action (7.20) is invariant under the transformation

$$M \to \Omega M \Omega^T \quad\mbox{and}\quad {\cal A}\mu{}^i \to \Omega_{ij} {\cal A}_\mu{}^j,$$ (7.32)

where $\Omega \in O(2,2)$ satisfying

$$\Omega^T L \Omega = L$$

In order to get a convenient parametrisation of $\Omega$ it is easier to work with the diagonal form of $L$. In appendix B we defined the orthogonal matrix $U$ that diagonalises $L$, so in this case $U$ becomes

$$U = \frac{1}{\sqrt{2}} \left( \begin{array}{cccc} 1&0&1&0\\ 0&1&0&1\\ 1&0&-1&0\\ 0&1&0&-1 \end{array} \right),$$ (7.33)

so

$$ULU^T = L_d = \left( \begin{array}{cccc} 1&&&\\ &1&&\\ &&-1&\\ &&&-1 \end{array} \right).$$ (7.34)

Then $U\Omega U^T$ preserves $L_d$ if $\Omega$ preserves $L$.

The main idea of this chapter is to apply these transformations on a known classical solution and generate new classical solutions of the equations of motion. We shall restrict ourselves to solutions with a fixed asymptotic configuration of the fields, respresenting asymptotically flat space. Explicitly this means we assume the following asymptotic behaviour for the scalars [27]

$$\Phi_\infty=0, \quad \sigma_\infty = \tau_infty = 1, \quad (\varphi_A)_\infty = (\varphi_B)_\infty = 0,$$ (7.35)

and for the vectors

$${\cal A}_\mu{}^i \to 0\quad\mbox{if}\quad r\to\infty$$

This means

$$M_\infty = \left( \begin{array}{cccc} 1&&&\\ &-1&&\\ &&1&\\ &&&-1 \end{array}\right).$$ (7.36)

We can choose this asymptotic configuration because given a solution with an other arbitrary asymptotic configuration, we can always transform it in to the form (7.35) with a combination of $O(2,2)$ transformations. Thus we do not suffer from any loss of generality by restricting the asymptotic configuration of our solutions to (7.35). [27] [20] [30]

We start with the simplest static solution of our action, the Schwarzschild solution. Expressed in our three-dimensional fields Schwarzschild is

$$\tau = 1 - \frac{2m}{r}, \quad \sigma=1,$$ (7.37)

with all other fields equal to zero. The scalar matrix then becomes

$$M_S = \mbox{diag}\left( (\frac{r}{r-2m}),-1,(\frac{r-2m}{r}),-1 \right),$$ (7.38)

We can apply to this the $O(2,2)$ transformations, but we want to restrict ourselves to the aymptotic configuration (7.35). We therefore not only have to demand that the transformation matrix $\Omega$ keeps $L$ invariant, but $M_\infty$ as well.

$$\Omega^T L \Omega = L \quad\mbox{and}\quad \Omega M_\infty \Omega^T = M_\infty$$

The subgroup of $O(2,2)$ that satisfies both these demands is

$$O(1,1) \times O(1,1) \subset O(2,2)$$ (7.39)

The continuous part of each of these $O(1,1)$'s is a one parameter boost. We already encountered this group in section 3.4. The solution generating transformation can now most conveniently parametrised as

$$\Omega_{SGT} = \Omega_1 \Omega_2,$$

with the two boosts given in the diagional basis, $L_d$

$$U \Omega_1 U^T = \left( \begin{array}{cccc} \cosh \alpha &0&0&\sinh \alpha\\ 0&1&0&0\\ 0&0&1&0\\ \sinh\alpha &0&0&\cosh \alpha \end{array} \right),$$ (7.40)

$$U \Omega_2 U^T = \left( \begin{array}{cccc} 1&0&0&0\\ 0&\cosh \beta &\sinh \beta&0\\ 0&\sinh\beta &\cosh \beta&0\\ 0&0&0&1 \end{array} \right).$$

Apply these transformations on the Schwarzschild solution

$$M' = \Omega_{SGT} M_S \Omega_{SGT}^T$$

and read off from $M'$ the new solution, expressed in the four dimensional fields:

$$A_t = \frac{-m\sinh(\alpha+\beta)}{r-m+m\cosh(\alpha+\beta)},$$

$$B_t = \frac{m\sinh(\alpha-\beta)}{r-m+m\cosh(\alpha-\beta)},$$

$$\sigma = \frac{r-m+m\cosh(\alpha+\beta)}{r-m+m\cosh(\alpha-\beta)},$$

$$e^{-\phi} = \frac{1}{r}\left[\left( r-m+m\cosh(\alpha+\beta) \right) \left( r-m+m\cosh(\alpha-\beta) \right) \right]^{\frac{1}{2}}$$

$$= \frac{\Delta}{r},$$

$$\lambda (r) = \frac{r-2m}{\Delta} ,$$

$$R^2 (r) = r \Delta.$$

This is precisely the same solution we found in section 5.3. It is a electrically charged Dilaton/Modulus black hole solution with electric charges

$$Q_A = -m \sinh(\alpha+\beta) \quad\mbox{and}\quad Q_B= m\sinh(\alpha-\beta).$$ (7.41)

If we had used only one of the two boosts, then the charges would have been equal and the modulus field would have vanished.

We want to construct dyonic solutions, with independent electric and magnetic charges. We can do this by making successive $SL(2,\hbox{\mybb R})$ and $O(2,2)$ transformations. These two transformations do not commute so we expect this strategy will produce four independent charges. But there is a more systematic way of producing these solutions. The two symmetry groups can be united in one single group containing all the features. [27] [26]