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

+ 2 like - 0 dislike

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,


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,


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.

This post imported from StackExchange Physics at 2014-05-04 11:34 (UCT), posted by SE-user user2062542
asked Apr 27, 2014 in Theoretical Physics by user2062542 (90 points) [ no revision ]

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