 each other implies or in other words, the integration constant C = 0. Then we also have The third constraint equation, to be discussed below will also determine the 5-dimensional gravitational constant, G 5.
2. Note that if we take the limit G 5, Λ 5 → ∞ (no scalar curvature term on the brane) such thatΛ 5 = Λ 5 /8πG 5 is finite then (14) and (15) become identical, providing an expression forΛ 5 The third constraint equations will fix the yet undetermined a (or, alternatively, fix C).
Now for solution 1. the mere presence of the dynamical term on the v = 0 and u = 0 branes, combined with the fact that the metric tensors depend on |u| and |v|, respectively, requires the existence of a non-vanishing tension on the u = v = 0 3-branes, just like in the 5 dimensional Randall-Sundrum solution. We denote the tension by Λ 4. We obtain If λ = 0 (non-warped space) the contribution on the 3-brane vanishes and the solution reduces to that of [5]. Let us consider now the third and final constraint equation. As we indicated earlier it is obtained from the derivative of the ratio of the two junction conditions, eliminating h,v (v, u)| v=0 from the equation. To reduce the obtained equations to a constraint on our constants we need to calculate derivatives of g(u, v) at u = v = 0. These can be determined either using the explicit form of g(u, v) from (3) or directly from taking derivatives of (14) and (15) and setting u = v = 0. To simplify matters we write down the the third constraint separately for the first and second solution. For the first solution (G 5 =finite) we obtain, using (17) For the second solution (G 5 = ∞), after substituting (19), we obtain for the third constraint These equations can be used to calculate the rest of the constants. All the constants of the theory are determined by the parameters of the 6 dimensional world, G 6 and Λ 6 and of the physical cosmological constant, λ. The value of parameters found for the two solutions are given in Table 1, where we use the notations r = −λ/6a 2 and w = 1 + aσ. We can see from Table 1 that in fact, taking into account the two possible signs for a, we have four independent solutions, rather than two. Note that at a special choice of the parameter r, r = −38/41 the combination 50λ+19Λ 6 = 0 and Λ 5 = 0. Then the geometry on the intersecting branes is asymptotically Minkowski. In other words, the equation provides the corresponding value of c.
Notice now that when g uu = g[u, v] has a pole at u = c g vv = g[v, u] also has a pole at v = c. Then, whenever (24) has a positive solution the range of variables u and v is the square |u|, |v| < c.
Let us examine now the solution of (24) for our four metric solutions. When a > 0 the boundary r = 0 separates domains with or without solutions. However, when a < 0 the corresponding boundary value is r = r 0, where r 0 = −1 for Solution I and r 0 = −50/33 for Solution II. Owing to the identites r = −6a 2 λ and 1 − r/r 0 = a 2 20Λ 6, these boundaries separate domains with different curvature: r > 0 corresponds to AdS 4 and r < 0 to dS 4 while r > r 0 corresponds to AdS 6 while r < r 0 corresponds to dS 6. Thus, the boundaries are: The boundary, c, satisfies where r = −λ/6a 2 > 0. Then for r > 0 (AdS 4 ) the range of variables u and v is |u|, |v| < 1 2a.
For r < 0 (dS 4 ) the range of |u| and of |v| is infinite.

III. SMALL OSCILLATIONS
Suppose the metric solutions we previously found, denoted 0 g ab, are modified by a small perturbation, such as g ab (ǫ) = 0 g ab + h ab.
As now it is standard in the literature, we will impose the TT (traceless, transverse) gauge conditions on h ab. In addition we will impose axial gauge conditions, h au = h av = 0. Thus, the nonzero components of the TT tensor h are h µν, where µ, ν = t, x, y, z. This gauge is also called the Randall-Sundrum gauge. The advantage of this gauge (in contrast to the TT conditions applied to the 'complete oscillations', Ω 2 h ab ), as we will see below, is that for the graviton mode h µν is independent of u and v. Then, at least for a > 0 the conformal factor Ω 2 insures localization. In general, gravitational excitations are dependent on these variables, as well. They may or may not be localized to the intersection of the branes. In what follows, we will drop the superscript 0 from the background metric, 0 g ab. The components of h satisfy the following wave equation where A = t, x,..., u, v and ∇ A is a covariant derivative in the background metric, γ AB. Equation (32) is separable. Multiplying by the conformal factor we obtain where the covariant derivatives∇ µ are taken in the four dimensional metric g µν and u i is a shorthand notation for u and v. Now the first term of (33) contains only derivatives and metric components dependent on t, x, y, and z, while the rest of the terms depend on u and v only. Using a factorized ansatz for the wave function and introducing the gravitational mass m we obtain the following two equations (Here we assume a specific form for the 4 dimensional metric: Minkowski with a warp factor g yy = g xx = −g tt = ω 2 = e −2bz for AdS 4 and g ii = ω 2 = e −2bt, where i denotes a spatial coordinate, for dS 4 ) where the upper sign corresponds to dS 4 and the lower one to AdS 4. We use ξ = z or t and λ = ±3b 2.
The solution of (34) is straightforward. It describes wave oscillations where k t = k 2 x + k 2 y + k 2 z + m 2, provided the coordinates are much smaller than the curvature radius, 1/b, otherwise the wave is modified, as we will discuss it below. In particular, the m = 0 mode always exists since h 2 =const. is a solution of (35). It describes a graviton, as long as the corresponding wave function is normalizable. In what follows, we will examine the solutions of (35) and their normalization.
The exact solution of (34) for dS 4, using an ansatz h = e i(x kx+y ky +z kz ) f [t], is where H (2) n = J n + iY n is the Hankel function of the second kind, n = i m 2 /b 2 + 31/16 and k = k 2 x + k 2 y + k 2 z. When one uses the asymptotic expansion for large order but fixed argument to order (i.e. taking the limit b → 0) [15] then we can write, after factoring out a diverging constant As a final note, as we indicated earlier, we mention that (35) illuminates the special role of the seemingly arbitrarily chosen conformal factor, Ω. The ground state wave function (m = 0 solution of (35)) is constant. Note however that the metric, and with that h was defined in (1) with a factor Ω 2 extracted. Thus the true ground state wave function (provided it is normalizable) is exactly Ω 2 for every choice of our parameters.

A. Mass spectrum of excitations
In general, (35) is too complicated to be transformed to the volcano potential form, familiar in 5 dimensional brane theories, and solved using standard methods for solving a one dimensional Schrodinger equation. Consider, however that the coordinates σ = u + v and δ = u − v are somewhat analogous to the radial and angular coordinates in a Schrödinger equation. Thus, one expects, that solutions dependent on the 'radial coordinate', σ, only well represent the spectrum of excitations. Among others, the gravition state, with a constant wave function, is clearly such a state.
Indeed, using (3) we can see that for functions, independent of δ (35) reduces to − Sh ′′ + sign(σ) 1 2aλ where function S is given in Table 1. Notice that the coefficients of (39) are independent of δ, just like in a Schrodinger equation with rotational invariant potential the equation for rotational invariant states is independent of the angles. The reason for this cancellation is that the explicit δ-dependence of (35) comes only through multipliers A(δ, σ), which in turn are associated with a differentiation with respect to δ.
Before we can bring (39) to a Schrödinger form we need to introduce a new variable, ω, defined by The integration in (40) can be analitically performed in the case of Solution I only. After the change of variables we can eliminate the first derivative from (39) and obtain a Schrödinger equation with a volcano potential. This is done by factoring out an appropriate function ρ(ω) from h. Finally, we