# $M^{+}_4$ Randall-Sundrum Brane Calculation

The basic Randall-Sundrum model is given by the metric,

$$\mathrm{d}s^2 = e^{-2|\sigma|}\left[ \mathrm{d}t^2 -\mathrm{d}x^2-\mathrm{d}y^2 - \mathrm{d}z^2 \right]-\mathrm{d}\sigma^2$$

where $\sigma$ denotes the additional fifth dimension. Notice the brane is localized at $\sigma=0$; this 'slice' is precisely Minkowski spacetime. To compute the stress-energy tensor, I define a vielbien,

$$\omega^\mu = e^{-|\sigma|}\mathrm{d}x^\mu \qquad \omega^\sigma = \mathrm{d}\sigma$$

where $\mu=0,..,3.$ Taking exterior derivatives and expressing in the orthonormal basis yields,

$$\mathrm{d}\omega^\mu = \epsilon(\sigma) \, \omega^\mu \wedge \omega^\sigma$$

where we have defined,

$$\epsilon(\sigma)=\theta(\sigma)-\theta(-\sigma)$$

which arises because of the absolute value function in the exponent, and $\theta(\sigma)$ is the Heaviside step function. By Cartan's first equation, the non-vanishing spin connections $\gamma^a_b$ are,

$$\gamma^{\mu}_{\sigma}= \epsilon(\sigma)e^{-|\sigma|} \mathrm{d}\omega^\mu$$

Taking exterior derivatives once again, and expressing in terms of the basis yields,

$$\mathrm{d}\gamma^\mu_\sigma = \left[ \epsilon^2 (\sigma)-2\delta(\sigma)\right]\, \omega^\mu \wedge \omega^\sigma$$

which arises by applying the product rule, and noting that,

$$\frac{\mathrm{d}\epsilon(\sigma)}{\mathrm{d}\sigma} = 2 \delta(\sigma)$$

because the delta function is the first derivative of the step function. From Cartan's second equation,

$$R^a_b=\mathrm{d}\gamma^a_b + \gamma^a_c \wedge \gamma^c_b$$

the components of the Ricci tensor are,

$$R^\mu_\sigma = \left[ \epsilon^2 (\sigma)-2\delta(\sigma)\right]\, \omega^\mu \wedge \omega^\sigma$$ as the second term vanishes. By the relation,

$$R^a_b = \frac{1}{2}R^a_{bcd} \omega^c \wedge \omega^d$$

we may deduce the Riemann tensor components,

$$R^\mu_{\sigma \mu \sigma} = 2\epsilon^2 (\sigma)-4\delta(\sigma)$$

I believe, in this case, both tensors in the coordinate basis and orthonormal basis are identical. Therefore we obtain the rank $(0,2)$ Ricci tensor,

$$R_{\sigma \sigma}=8\epsilon^2 (\sigma)-16\delta(\sigma)$$

As the only diagonal component, the Ricci scalar is identical to the Ricci tensor at $(\sigma,\sigma)$. Using the Einstein field equations, the stress-energy tensor is given by,

$$T_{55} = \frac{1}{8\pi G_5}\left[ 4\epsilon^2 (\sigma)-8\delta(\sigma) + \Lambda\right]$$

where $\Lambda$ is the cosmological constant, and $G_5$ is the five-dimensional gravitational constant. The function $\epsilon^2(\sigma)$ is given by,

$$\epsilon^2(\sigma)=\theta^2(\sigma)+\theta^2(-\sigma)-2\theta(\sigma)\theta(-\sigma)$$

The last term appears to be the delta function, as it is zero everywhere, but singular at zero. The first terms are unity everywhere, but undefined at zero, therefore,

$$T_{55}= \frac{1}{8\pi G_5}\left[ 4\theta^2(\sigma)+4\theta^2(-\sigma) -16\delta(\sigma) + \Lambda\right]$$

However, this disagrees with Mannheim's Brane-Localized Gravity which states,

$$T_{ab}=-\lambda \delta^\mu_a \delta^\nu_b \eta_{\mu\nu}\delta(\sigma)$$

where $\lambda = 12/\kappa^2_5$. In his text, $T_{\mu\nu}\propto -\eta_{\mu\nu}$, but in my calculation the entire purely 4D stress-energy vanishes. I can only assume I've done something wrong.