Two Vector Fields

Let's now turn to the case of two vector fields. We now consider the Modulus field again and thus (when $a=1$) we are left with two constraints on four possible charges. It is not very interesting to look at singly charged solutions because they will be the same as the previous one vector field solution. So we'll focus on dyonic solutions and first keep the extra parameter $a$. Let's see what happens when we use the form of the metric from eq.5.16 and the electric vector fields (eq.5.17) and put this in the equation for $R_{tt}$ (first equation of eq.5.4)

$$(\lambda ' R^2)' = {\textstyle{1\over 2}}\left\{ (c_A^2 + c_B^2) e^{-a\phi} + (G_A^2 + G_B^2) e^{a\phi} \right\} R^{-2}$$

In the previous case we used this equation to express the index $b$ in the 'known' index $a$. But now we find much more restricting conditions:

$$2 a \sqrt{b(1-b)} = b \land 2 a \sqrt{b(1-b)} = 1 - b$$

$$\mbox{so} \quad b = {\textstyle{1\over 2}} \mbox{and} a=\pm 1 \enspace !$$

This means we can only find a doubly charged solution for $a=\pm 1$.

This is the proof we promised for the fact that dyonic solutions only exist for $a=1$ ($a=-1$ is the same as $\phi \rightarrow -\phi$, so it corresponds to the purely magnetic case). This proof is given in the case of two vector fields but is equally valid for the case of one vector field.

So, from now one we take $a$ to be one and thus consider the string effective action again. Under the same assumptions as in the one vector field case we can solve the equations and we find the dyonic two vector field solution

$$R^2 (r) = r^2 - \Sigma^2 \, ,$$ (5.21)

$$\lambda (r) = \frac{(r-r_+)(r-r_-)}{r^2 - \Sigma^2} \, ,$$ (5.22)

$$e^\phi = e^{\phi_0} \frac{r+\Sigma}{r-\Sigma} \qquad \Delta_{\phi}=2\Sigma \,$$ (5.23)

where $\Delta_{\phi}$ is the Dilaton scalar charge defined through it's one over $r$ behaviour at infinity. The metric parameters must be expressed in terms of the mass an charges as follows

$$\Sigma ( r_+ + r_- ) = {\textstyle{1\over 4}}\left[ (c_A^2+c_B^2) e^{-\phi_0} - (G_A^2+G_B^2) e^{\phi_0} \right] \, ,$$ (5.24)

$$\Sigma^2 + r_+r_- = {\textstyle{1\over 4}}\left[ (c_A^2+c_B^2) e^{-\phi_0} + (G_A^2+G_B^2) e^{\phi_0} \right] \, ,$$ (5.25)

and

$$Q_{A(B)} = Q_{A(B)} e^{-\phi_0} \quad \mbox{and} \quad 2M = r_+ + r_- \, .$$

This $M$ can again be interpreted as the mass.

When we consider purely electric or magnetic solutions the solution reduces to the one vector field case (just work out the parameter relations and shift the solution in $r$). So the one vector field solution can be obtained from this more general dyonic solution.

Until now we explicitly wrote the Dilaton solutions multiplied with a constant, the Dilaton value at infinity. From now on we will take this value to be one. Afterwards it is always possible to consider other Dilaton values at infinity, just by identifying the squared electromagnetic charges as the previous ones multiplied with the Dilaton value at infinity ( $\phi \rightarrow \phi +\phi_0$, $F^2 \rightarrow e^{\phi_0} F^2$, this is a symmetry of the action and this transformation is a subgroup of the $SL(2,R)$ symmetry group).