Equations of Motion

Before we are forced to use solution generating techniques we are going to try solving the equations of motion. We consider the four dimensional action obtained after reduction of a five dimensional string effective action discussed in the previous chapter. Varying with respect to the metric field we get

$$R_{\mu\nu} = {\textstyle{1\over 4}}\sigma^{-2} \partial_\mu \sigma \partial_\n... ...u \phi + {\textstyle{1\over 2}}e^{-2 \phi} \partial_\mu \Psi \partial_\nu \Psi+$$

$$-{\textstyle{1\over 2}}\sigma e^{\phi} \left\{ F_{\mu\rho}(A) F_\nu{}^\rho(A)-{\textstyle{1\over 4}}g_{\mu\nu} F(A)^2 \right\}$$

$$-{\textstyle{1\over 2}}\sigma^{-1} e^{\phi} \left\{ F_{\mu\rho}(B) F_\nu{}^\rho(B)-{\textstyle{1\over 4}}g_{\mu\nu} F(B)^2 \right\}.$$ (5.1)

The vector fields give:

$$\nabla_\mu \left( \sigma e^{\phi} F^{\mu\nu}(A) + \Psi \tilde{F}^{\mu\nu}(B) \right) = 0$$

$$\nabla_\mu \left( \sigma^{-1} e^{\phi} F^{\mu\nu}(B) + \Psi \tilde F^{\mu\nu}(A) \right) = 0$$ (5.2)

and the scalar fields must satisfy:

$$\sigma^{-1}\Box \sigma -\sigma^{-2}(\partial \sigma)^2 + {\textstyle{1\over 2}} e^{\phi} \left( \sigma F(A)^2 - \sigma^{-1} F(B)^2 \right) = 0$$

$$\Box \phi + e^{-2\phi} (\partial \Psi)^2 + {\textstyle{1\over 4}}e^{\phi} \left( \sigma F(A)^2 + \sigma^{-1} F(B)^2 \right) = 0$$

$$e^{-2\phi}\Box \Psi - 2 e^{-2\phi} \partial \phi \partial \Psi + {\textstyle{1\over 2}}F(A) \tilde F(B) = 0$$ (5.3)

Where $\phi$ is the Dilaton scalar, $\sigma$ is the Modulus scalar and $\Psi$ is the Axion scalar.

Using the simplifications obtained by demanding spherical symmetry (eq.2.1 and eq.2.2) we get

$$( \lambda' R^2)' = {\textstyle{1\over 2}}e^{\phi} \sigma \left\{ F_{rt}^2 (A) R^2 + ... ...r 2}}e^{\phi} \sigma^{-1} \left\{ F_{rt}^2 (B) R^2 + \frac{G_B^2}{R^2} \right\}$$

$$- \frac{4R'' }{R} = {\textstyle{1\over 2}}\sigma^{-2}(\sigma')^2 + (\phi')^2 + e^{-2\phi}(\Psi')^2$$

$$( \lambda R R')' -1 = - {\textstyle{1\over 2}}( \lambda' R^2)'.$$ (5.4)

The last equation can be integrated:

$$( \lambda R^2 )'' = 2$$

$$\lambda R^2 = ( r - r_+ ) ( r - r_- )$$ (5.5)

,the vector equations give

$$\partial _\theta \left( \frac{1}{\sin \theta} F_{\theta\phi}(A) \right) = 0 \Rightarrow F_{\theta\phi}(A) = G_A \sin \theta$$

$$\partial _\theta \left( \frac{1}{\sin \theta} F_{\theta\phi}(B) \right) = 0 \Rightarrow F_{\theta\phi}(B) = G_B \sin \theta$$ (5.6)

,the two other vector equations can be integrated to give

$$e^{\phi} \sigma R^2 F_{rt}(A) + G_B \Psi = c_A$$

$$e^{\phi} \sigma^{-1} R^2 F_{rt}(B) + G_A \Psi = c_B$$ (5.7)

, where $c_{A/B}$ are integration constants. The three remaining scalar equations simplify to:

$$( \lambda R^2 \frac{\sigma'}{\sigma})' + e^{\phi} \left\{ \sigma ... ...) - \sigma^{-1} \left( F_{rt}^2 (B) R^2 - \frac{G_B^2}{R^2}\right) \right\} = 0$$

$$( \lambda R^2 \phi' )' + e^{-2\phi} (\Psi')^2+ {\textstyle{1\over... ...t) + \sigma^{-1} \left( F_{rt}^2 (B) R^2- \frac{G_B^2}{R^2}\right) \right\} = 0$$

$$e^{ -2\phi} ( \lambda R^2 \Psi' )' - 2 e^{-2\phi} \lambda R^2 (\phi')(\Psi') + G_B F_{rt}(A) + G_A F_{rt}(B) = 0$$

,where all the primes denote taking the derivative with respect to $r$.

We see that the magnetic fields are fixed, as is $\lambda R^2$, and so we are left to solve for two metric equations, two vector equations and three scalar equations. Generally, because the equations are non-linear, these equations are too difficult to solve. Only in the case we consider the Modulus and Axion scalar fields to be trivial (constant) the equations can be solved. Putting the Axion and Modulus field constant we get constraints on the electromagnetic charges from the scalar equations. The constraints are

$$\sigma'=0 \Rightarrow F(A)^2 = F(B)^2 \leftrightarrow Q_{-}^2 =G_{-}^2$$ (5.8)

$$\Psi'=0 \Rightarrow F(A){\tilde F}(B)=0 \leftrightarrow Q_A G_B +Q_B G_A =0$$ (5.9)

, where $Q_{A/B}=c_{A/B} e^{-\phi_0}$ are the electric charges ($e^{\phi_0}$ is the Dilaton value at infinity) and $Q_{\pm}^2=Q_A ^2 \pm Q_B ^2$ (same for $G_{\pm}^2$).

So, we now only consider the Dilaton scalar. At first sight we could now easily consider only one vector field (both vector fields couple in the same way to the dilaton and metric). But there is a subtlety here, considering only one vector field by making $F(A)=F(B)=\frac{1}{\sqrt{2}}F$ (putting one vector field zero means you lose the Axion) will mean that the charges ($Q,G$) are zero because of the constraints (one vector field means two charges, two constraints on two charges means no free charges). The way we'll circumvent this is by not considering the Modulus scalar at all in the one-vector field case. This means we'll put the Modulus field one in our string effective action and in this way lose the Modulus constraint. We are then only left with an Axion constraint which tells us that either $Q$ or $G$ is zero, so we only have to consider singly charged solutions in this case.

You could wonder, why bother solving the equations for the one vector field case. Two vector fields don't complicate the equations of motion because they couple in the same way to the dilaton and metric. Well, the first reason is that the action with only Dilaton and one vector field was historically the first one considered and we would like to reproduce these results. Secondly, in the one vector field case we only have to consider one charge (electric or magnetic, because of the Axion constraint) which simplifies the equations. Finally, the properties of this solution nicely demonstrate the most important features of more general solutions.

In the next section we are explicitly going to solve the equations of motion for the Dilaton black holes, as well for one vector field as for two vector fields.