Appearance
question:Consider a smooth manifold with a metric given by g=dx^2+dy^2+dz^2. Determine the Ricci curvature of the manifold at the point (1,2,3).
answer:To compute the Ricci curvature, we first need to compute the Christoffel symbols, which are given by: Gamma^k_{ij} = frac{1}{2} g^{kl} (partial_i g_{jl} + partial_j g_{il} - partial_l g_{ij}) Since the metric is given by g=dx^2+dy^2+dz^2, we have a diagonal metric tensor: g_{ij} = begin{pmatrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 end{pmatrix} The inverse metric tensor g^{ij} is the same as g_{ij} in this case. Now, we compute the partial derivatives of the metric tensor components: partial_i g_{jl} = partial_j g_{il} = partial_l g_{ij} = 0 This is because the metric tensor components are constants. Therefore, all the Christoffel symbols are zero: Gamma^k_{ij} = 0 Now, we compute the Riemann curvature tensor components, which are given by: R^l_{ijk} = partial_i Gamma^l_{jk} - partial_j Gamma^l_{ik} + Gamma^l_{im} Gamma^m_{jk} - Gamma^l_{jm} Gamma^m_{ik} Since all the Christoffel symbols are zero, the Riemann curvature tensor components are also zero: R^l_{ijk} = 0 Finally, we compute the Ricci curvature tensor components, which are given by: R_{ij} = R^k_{ikj} Again, since the Riemann curvature tensor components are zero, the Ricci curvature tensor components are also zero: R_{ij} = 0 Thus, the Ricci curvature of the manifold at the point (1,2,3) is zero. This is expected, as the given metric corresponds to a flat Euclidean space.
question:Consider a Riemannian manifold M with metric tensor given by g_{ij}=begin{bmatrix}3&22&2end{bmatrix} for a local coordinate system x,y. Find the Ricci curvature scalar for the manifold M.
answer:To find the Ricci curvature scalar, we first need to compute the Christoffel symbols, the Riemann curvature tensor, and the Ricci tensor. Let's start with the Christoffel symbols. The Christoffel symbols are given by: Gamma^k_{ij} = frac{1}{2} g^{kl} (frac{partial g_{il}}{partial x^j} + frac{partial g_{jl}}{partial x^i} - frac{partial g_{ij}}{partial x^l}) First, we need to find the inverse of the metric tensor g^{ij}: g^{ij} = begin{bmatrix}3&22&2end{bmatrix}^{-1} = frac{1}{(3)(2) - (2)(2)} begin{bmatrix}2&-2-2&3end{bmatrix} = begin{bmatrix}1&-1-1&frac{3}{2}end{bmatrix} Now, we can compute the Christoffel symbols: Gamma^1_{11} = frac{1}{2} g^{11} (frac{partial g_{11}}{partial x^1} + frac{partial g_{11}}{partial x^1} - frac{partial g_{11}}{partial x^1}) = 0 Gamma^1_{12} = frac{1}{2} g^{11} (frac{partial g_{11}}{partial x^2} + frac{partial g_{12}}{partial x^1} - frac{partial g_{12}}{partial x^1}) = 0 Gamma^1_{22} = frac{1}{2} g^{11} (frac{partial g_{12}}{partial x^2} + frac{partial g_{12}}{partial x^2} - frac{partial g_{22}}{partial x^1}) = 0 Gamma^2_{11} = frac{1}{2} g^{22} (frac{partial g_{21}}{partial x^1} + frac{partial g_{21}}{partial x^1} - frac{partial g_{11}}{partial x^2}) = 0 Gamma^2_{12} = frac{1}{2} g^{22} (frac{partial g_{21}}{partial x^2} + frac{partial g_{22}}{partial x^1} - frac{partial g_{12}}{partial x^2}) = 0 Gamma^2_{22} = frac{1}{2} g^{22} (frac{partial g_{22}}{partial x^2} + frac{partial g_{22}}{partial x^2} - frac{partial g_{22}}{partial x^2}) = 0 Since all the Christoffel symbols are zero, the Riemann curvature tensor will also be zero: R^i_{jkl} = frac{partial Gamma^i_{jl}}{partial x^k} - frac{partial Gamma^i_{jk}}{partial x^l} + Gamma^i_{mk} Gamma^m_{jl} - Gamma^i_{ml} Gamma^m_{jk} = 0 Now, we can compute the Ricci tensor: R_{ij} = R^k_{ikj} = 0 Finally, the Ricci curvature scalar is given by: R = g^{ij} R_{ij} = 0 So, the Ricci curvature scalar for the manifold M is 0.
question:Consider a two-dimensional manifold specified by the following metric in polar coordinates: ds^2 = dr^2 + f(r)^2 dtheta^2 where f(r)=frac{1}{sqrt{1-r^2}}. Determine the Ricci curvature tensor R_{ij} and the Ricci scalar R of this manifold.
answer:To compute the Ricci curvature tensor and the Ricci scalar, we first need to find the Christoffel symbols of the second kind, which are given by: Gamma^k_{ij} = frac{1}{2} g^{kl} (partial_i g_{lj} + partial_j g_{il} - partial_l g_{ij}) For our metric, we have: g_{ij} = begin{pmatrix} 1 & 0 0 & f(r)^2 end{pmatrix} and its inverse: g^{ij} = begin{pmatrix} 1 & 0 0 & frac{1}{f(r)^2} end{pmatrix} Now, we compute the non-zero Christoffel symbols: 1. Gamma^r_{rr} = frac{1}{2} g^{rr} (partial_r g_{rr} + partial_r g_{rr} - partial_r g_{rr}) = 0 2. Gamma^r_{rtheta} = frac{1}{2} g^{rr} (partial_r g_{rtheta} + partial_theta g_{rr} - partial_r g_{rtheta}) = 0 3. Gamma^r_{thetatheta} = frac{1}{2} g^{rr} (partial_r g_{thetatheta} + partial_theta g_{rtheta} - partial_theta g_{rtheta}) = -frac{1}{2} partial_r f(r)^2 = r(1-r^2) 4. Gamma^theta_{rr} = frac{1}{2} g^{thetatheta} (partial_r g_{rtheta} + partial_r g_{rtheta} - partial_theta g_{rr}) = 0 5. Gamma^theta_{rtheta} = frac{1}{2} g^{thetatheta} (partial_r g_{thetatheta} + partial_theta g_{rr} - partial_r g_{rtheta}) = frac{1}{2} frac{1}{f(r)^2} partial_r f(r)^2 = frac{r}{1-r^2} 6. Gamma^theta_{thetatheta} = frac{1}{2} g^{thetatheta} (partial_theta g_{thetatheta} + partial_theta g_{thetatheta} - partial_theta g_{thetatheta}) = 0 Next, we compute the components of the Riemann curvature tensor: R^k_{ ijl} = partial_j Gamma^k_{il} - partial_l Gamma^k_{ij} + Gamma^k_{mj} Gamma^m_{il} - Gamma^k_{ml} Gamma^m_{ij} We only need to compute the non-zero components: 1. R^r_{ theta rtheta} = partial_theta Gamma^r_{rtheta} - partial_r Gamma^r_{thetatheta} + Gamma^r_{mtheta} Gamma^m_{rtheta} - Gamma^r_{mr} Gamma^m_{thetatheta} = -frac{1}{(1-r^2)^2} 2. R^theta_{ rtheta r} = partial_theta Gamma^theta_{rr} - partial_r Gamma^theta_{theta r} + Gamma^theta_{mtheta} Gamma^m_{rr} - Gamma^theta_{mr} Gamma^m_{theta r} = frac{1}{(1-r^2)^2} Now, we compute the Ricci curvature tensor: R_{ij} = R^k_{ ikj} We have: 1. R_{rr} = R^r_{ rrr} + R^theta_{ rtheta r} = 0 + frac{1}{(1-r^2)^2} = frac{1}{(1-r^2)^2} 2. R_{rtheta} = R^r_{ rrtheta} + R^theta_{ rthetatheta} = 0 + 0 = 0 3. R_{theta r} = R^r_{ theta r r} + R^theta_{ thetatheta r} = 0 + 0 = 0 4. R_{thetatheta} = R^r_{ theta rtheta} + R^theta_{ thetathetatheta} = -frac{1}{(1-r^2)^2} + 0 = -frac{1}{(1-r^2)^2} So, the Ricci curvature tensor is: R_{ij} = begin{pmatrix} frac{1}{(1-r^2)^2} & 0 0 & -frac{1}{(1-r^2)^2} end{pmatrix} Finally, we compute the Ricci scalar: R = g^{ij} R_{ij} = g^{rr} R_{rr} + g^{thetatheta} R_{thetatheta} = frac{1}{(1-r^2)^2} - frac{1}{f(r)^2} frac{1}{(1-r^2)^2} = 0 Thus, the Ricci scalar is R = 0.
question:Consider a Riemannian manifold whose metric tensor is given by: g = begin{pmatrix} 1 & 0 & 0 0 & 2 & -1 0 & -1 & 3 end{pmatrix} Determine the Ricci curvature of this manifold at the point (0,1,1).
answer:To compute the Ricci curvature, we first need to compute the Christoffel symbols, then the Riemann curvature tensor, and finally contract the indices to obtain the Ricci tensor. Let's start with the Christoffel symbols. The Christoffel symbols are given by: Gamma^k_{ij} = frac{1}{2} g^{kl} (partial_i g_{lj} + partial_j g_{li} - partial_l g_{ij}) where g^{kl} is the inverse of the metric tensor g_{ij}. First, we need to find the inverse of the given metric tensor. We have: g^{-1} = begin{pmatrix} 1 & 0 & 0 0 & 2 & 1 0 & 1 & 1 end{pmatrix} Now, we can compute the Christoffel symbols. Since the metric tensor is constant, its partial derivatives are all zero. Therefore, all the Christoffel symbols are also zero: Gamma^k_{ij} = 0 Next, we compute the Riemann curvature tensor, which is given by: R^l_{ijk} = partial_j Gamma^l_{ik} - partial_k Gamma^l_{ij} + Gamma^l_{jm} Gamma^m_{ik} - Gamma^l_{km} Gamma^m_{ij} Since all the Christoffel symbols are zero, the Riemann curvature tensor is also zero: R^l_{ijk} = 0 Finally, we compute the Ricci curvature tensor by contracting the indices of the Riemann curvature tensor: R_{ij} = R^k_{ikj} = 0 Thus, the Ricci curvature tensor is zero at the point (0,1,1).