We can start by using the energy-momentum tensor to find the metric coefficients. From the given energy-momentum tensor, we can see that the stress-energy tensor is diagonal, with only the (0,0), (1,1), and (2,2) components being non-zero. Using the Einstein field equations, we can relate these components to the metric coefficients:G^0_0 = 8πG T^0_0 = -8πG (ρ + p) = -A'(r)/r G^1_1 = 8πG T^1_1 = 8πG p = B'(r)/r G^2_2 = 8πG T^2_2 = 8πG p = 1/r^2 - A(r)/r^2where G is the gravitational constant and primes denote derivatives with respect to r. Solving these equations for A(r) and B(r), we get:A(r) = 1 - (8πG/3)ρ r^2 + C1 r B(r) = 1 + (8πG/3)p r^2 + C2 rwhere C1 and C2 are integration constants. To determine these constants, we need to impose boundary conditions. Since we are dealing with a de Sitter spacetime, we know that the cosmological constant Λ is positive, and we can set it equal to 3/R^2, where R is the radius of curvature. In 2+1 dimensions, the curvature scalar is R = 2/√Λ, so we have R = 2√3/G. We also know that the spacetime is asymptotically flat, which means that as r → ∞, the metric should approach the Minkowski metric:ds^2 = -c^2 dt^2 + dr^2 + r^2 dθ^2Comparing this with our metric, we see that C1 = C2 = 0. Therefore, the metric coefficients are:A(r) = 1 - (8πG/3)ρ r^2 B(r) = 1 + (8πG/3)p r^2Substituting these into the line element, we get:ds^2 = -(1 - (8πG/3)ρ r^2) c^2 dt^2 + (1 + (8πG/3)p r^2) dr^2 + r^2 dθ^2This is the line element for a de Sitter spacetime in 2+1 dimensions with positive cosmological constant.
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