Question

If we multiply both sides of equation 21.10 by Bav, we get 8pv (21.33) Notice that since the metric with upstairs indices is the matrix inverse of the ver- sion with downstairs indices, gavg㎛-δμμ 4. Exercise 21.3.1. Why is this equal to 4? You can use this and the definitions of R and T to show that-R + 4A = KT. Exercise 21.3.2. Verify this.

Definitions:

Answer 1

1) In the equation

$\small g_{\mu\nu}g^{\mu\nu}=\delta_\mu^{\mu}=4$

4 is the dimension of the space-time. This means the metric tensor $\small g_{\mu\nu}$ is a 4-dimensional square matrix.

2) In general relativity, a scalar can be defined using metric tensor and a rank 2 tensor as follows

$\small g_{\mu\nu} T^{\mu \nu}=T$

T is called trace of the energy-momentum tensor $\small T^{\mu \nu}$

similarly

$\small g_{\mu\nu} R^{\mu \nu}=R$

R is the Ricci scalar which is obtained taking the trace of the Ricci tensor

Hence with these definitions, the equation 21.33 becomes

$\small R-\frac{1}{2}\times 4 \times R+\Lambda \times 4=\kappa T$

This can be written as

$\small R(1-2)+4\Lambda =\kappa T$

which gives

$\small -R+4\Lambda =\kappa T$