Page 2 of 2
Chapter 8: Ricci Tensor and Scalar
Applications and Computations
This page derives the contracted Bianchi identity in detail, explains why it guarantees energy-momentum conservation, presents the trace-reversed Einstein equations, and works through the Ricci tensor computations for both the Schwarzschild and FLRW spacetimes.
Contracted Bianchi Identity: Full Derivation
Starting from the differential Bianchi identity:
$$\nabla_\lambda R_{\rho\sigma\mu\nu} + \nabla_\rho R_{\sigma\lambda\mu\nu} + \nabla_\sigma R_{\lambda\rho\mu\nu} = 0$$
Step 1: Contract with \( g^{\rho\mu} \) (raise \( \rho \)and contract with \( \mu \)):
$$\nabla_\lambda R_{\sigma\nu} - \nabla_\rho R^\rho_{\ \sigma\lambda\nu} + \nabla_\sigma R_{\lambda\nu} = 0$$
Wait - let us be more careful. Using \( g^{\rho\mu} \) on each term:
First term: \( g^{\rho\mu}\nabla_\lambda R_{\rho\sigma\mu\nu} = \nabla_\lambda R^\mu_{\ \sigma\mu\nu} = \nabla_\lambda R_{\sigma\nu} \)
Second term: \( g^{\rho\mu}\nabla_\rho R_{\sigma\lambda\mu\nu} = \nabla^\mu R_{\sigma\lambda\mu\nu} \)
Third term: \( g^{\rho\mu}\nabla_\sigma R_{\lambda\rho\mu\nu} = \nabla_\sigma R_{\lambda\ \nu}^{\ \mu\ } = -\nabla_\sigma R_{\lambda\nu} \)
Using \( R_{\sigma\lambda\mu\nu} = -R_{\sigma\lambda\nu\mu} \), the second term becomes \( -\nabla^\mu R_{\sigma\lambda\nu\mu} = \nabla^\mu R_{\sigma\lambda\mu\nu} \). After simplification:
$$\nabla_\lambda R_{\sigma\nu} + \nabla^\mu R_{\sigma\lambda\mu\nu} - \nabla_\sigma R_{\lambda\nu} = 0$$
Step 2: Contract again with \( g^{\sigma\nu} \):
First term: \( g^{\sigma\nu}\nabla_\lambda R_{\sigma\nu} = \nabla_\lambda R \)
Second term: \( g^{\sigma\nu}\nabla^\mu R_{\sigma\lambda\mu\nu} = \nabla^\mu R^\nu_{\ \lambda\mu\nu} = -\nabla^\mu R_{\lambda\mu} \)
Third term: \( -g^{\sigma\nu}\nabla_\sigma R_{\lambda\nu} = -\nabla^\nu R_{\lambda\nu} \)
Combining (and noting \( \nabla^\mu R_{\lambda\mu} = \nabla^\nu R_{\lambda\nu} \)):
$$\nabla_\lambda R - 2\nabla^\mu R_{\lambda\mu} = 0$$
Rearranging:
$$\nabla^\mu R_{\mu\nu} = \frac{1}{2}\nabla_\nu R$$
This can be rewritten as:
$$\boxed{\nabla^\mu\left(R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R\right) = \nabla^\mu G_{\mu\nu} = 0}$$
The contracted Bianchi identity
Why This Guarantees Conservation
The Einstein field equations state:
$$G_{\mu\nu} = \frac{8\pi G}{c^4}\, T_{\mu\nu}$$
Taking the covariant divergence of both sides:
$$\nabla^\mu G_{\mu\nu} = \frac{8\pi G}{c^4}\, \nabla^\mu T_{\mu\nu}$$
Since the left side vanishes identically (Bianchi identity), we get:
$$\nabla^\mu T_{\mu\nu} = 0$$
Covariant conservation of energy-momentum
This is remarkable: energy-momentum conservation is not imposed as an additional law but follows automatically from the geometric structure of the Einstein equations. In a sense, the conservation law is encoded in the Bianchi symmetries of the Riemann tensor.
Note: The 4 equations \( \nabla^\mu G_{\mu\nu} = 0 \) also reduce the number of truly independent Einstein equations from 10 to 6. This is related to the diffeomorphism invariance of the theory: 4 components of the metric can always be chosen freely (gauge freedom), so only 6 equations are needed to determine the remaining 6 physical degrees of freedom.
Trace-Reversed Einstein Equations
Taking the trace of \( G_{\mu\nu} = 8\pi G\, T_{\mu\nu} \) (setting \( c = 1 \)):
$$g^{\mu\nu}G_{\mu\nu} = R - 2R = -R = 8\pi G\, T$$
where \( T = g^{\mu\nu}T_{\mu\nu} \) is the trace of the stress-energy tensor
Therefore \( R = -8\pi G\, T \). Substituting back into \( G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R \):
$$R_{\mu\nu} = 8\pi G\left(T_{\mu\nu} - \frac{1}{2}g_{\mu\nu}T\right)$$
Trace-reversed form of the Einstein equations
This form is sometimes more convenient because the Ricci tensor is simpler than the Einstein tensor. In vacuum (\( T_{\mu\nu} = 0 \)), it immediately gives \( R_{\mu\nu} = 0 \), which is the vacuum Einstein equation.
Worked Example: Ricci for Schwarzschild
For the Schwarzschild metric with \( f = 1 - 2M/r \), we compute the Ricci tensor by contracting the Riemann tensor components from the previous chapter.
The diagonal Ricci components are (using \( R_{\mu\nu} = R^\rho_{\ \mu\rho\nu} \)):
$$R_{tt} = R^t_{\ ttt} + R^r_{\ trt} + R^\theta_{\ t\theta t} + R^\phi_{\ t\phi t}$$
$$= 0 + \frac{2M(r-2M)}{r^4} - \frac{M(r-2M)}{r^4} - \frac{M(r-2M)}{r^4} = 0$$
Similarly, computing each component:
$$R_{tt} = 0, \quad R_{rr} = 0, \quad R_{\theta\theta} = 0, \quad R_{\phi\phi} = 0$$
$$R_{\mu\nu} = 0 \quad \text{for all } \mu, \nu$$
Verification
The vanishing of \( R_{\mu\nu} \) confirms that Schwarzschild is a vacuum solution. The Ricci scalar also vanishes: \( R = 0 \). The Einstein tensor vanishes:\( G_{\mu\nu} = 0 \). Yet the spacetime is curved because the Weyl tensor (= Riemann tensor in vacuum) is non-zero, with the Kretschner scalar \( K = 48M^2/r^6 \neq 0 \).
Worked Example: Ricci for FLRW (Friedmann Equations)
For the flat FLRW metric \( ds^2 = -dt^2 + a^2(t)(dr^2 + r^2 d\Omega^2) \), the Ricci tensor components are:
$$R_{00} = -3\frac{\ddot{a}}{a}$$
$$R_{ij} = \left(a\ddot{a} + 2\dot{a}^2\right)g_{ij}/a^2 = \left(\frac{\ddot{a}}{a} + 2\frac{\dot{a}^2}{a^2}\right)g_{ij}$$
The Ricci scalar is:
$$R = 6\left(\frac{\ddot{a}}{a} + \frac{\dot{a}^2}{a^2}\right) = 6\left(\dot{H} + 2H^2\right)$$
The Einstein tensor components are:
$$G_{00} = 3\frac{\dot{a}^2}{a^2} = 3H^2$$
$$G_{ij} = -\left(2\frac{\ddot{a}}{a} + \frac{\dot{a}^2}{a^2}\right)g_{ij} = -(2\dot{H} + 3H^2)g_{ij}$$
Setting \( G_{\mu\nu} = 8\pi G\, T_{\mu\nu} \) with a perfect fluid \( T^{\mu\nu} = \text{diag}(-\rho, p, p, p) \)yields the Friedmann equations:
$$H^2 = \frac{8\pi G}{3}\rho \qquad \text{(First Friedmann Equation)}$$
$$\frac{\ddot{a}}{a} = -\frac{4\pi G}{3}(\rho + 3p) \qquad \text{(Second Friedmann Equation)}$$
These two equations, derived from the Ricci tensor of the FLRW metric, govern the expansion history of our universe. The first relates the expansion rate to the energy density; the second describes the acceleration/deceleration of the expansion. The conservation equation \( \dot{\rho} + 3H(\rho + p) = 0 \) follows from \( \nabla^\mu T_{\mu\nu} = 0 \).
Fortran: Numerical Ricci Calculation
Numerically compute the Ricci tensor, Ricci scalar, and Einstein tensor for the Schwarzschild metric at \( r = 6M \). Verify that all components vanish (within numerical precision).
Ricci Tensor Verification
PythonVacuum verification: Ricci = 0 across all radii
Click Run to execute the Python code
Code will be executed with Python 3 on the server
Key Concepts Summary
Ricci tensor: \( R_{\mu\nu} = R^\rho_{\ \mu\rho\nu} \) - the trace of the Riemann tensor, encoding how volumes change under geodesic flow.
Ricci scalar: \( R = g^{\mu\nu}R_{\mu\nu} \) - measures the deviation of geodesic ball volumes from flat space.
Einstein tensor: \( G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R \) - the unique divergence-free tensor linear in second derivatives of the metric.
Contracted Bianchi identity: \( \nabla^\mu G_{\mu\nu} = 0 \) - guarantees \( \nabla^\mu T_{\mu\nu} = 0 \) (energy-momentum conservation).
Trace-reversed equations: \( R_{\mu\nu} = 8\pi G(T_{\mu\nu} - \frac{1}{2}g_{\mu\nu}T) \) - gives \( R_{\mu\nu} = 0 \) in vacuum.
Schwarzschild: \( R_{\mu\nu} = 0 \) (vacuum). Curved but Ricci-flat; curvature lives in the Weyl tensor.
FLRW: Non-zero Ricci tensor leads to the Friedmann equations\( H^2 = 8\pi G\rho/3 \) governing cosmic expansion.