# Einstein Tensor

1. Dec 12, 2008

### Philosophaie

G_ab = R_ab - 1/2 * R * g_ab

and

R = g^(ab) * R_ab

then

G_ab = R_ab - 1/2 * R_ab * (g^(ab) * g_ab)
=1/2 * R_ab

for this abomination someone please correct me.

2. Dec 12, 2008

### George Jones

Staff Emeritus
No. a and b are free indices on the right, so a and b have to be free indices on right, i.e., they cannot be repeated and summed over.

Use $R = g^{cd} R_{cd}$ instead of $R = g^{ab} R_{ab}$.

Also, what does $g^{ab} g_{ab}$ equal?

3. Dec 12, 2008

### Philosophaie

g^(ab) * g_ab = I, the Identity Matrix only if a square matrix and when dotted with R_ab gives you back R_ab.

4. Dec 12, 2008

### George Jones

Staff Emeritus
No.

It is true that

$$g^{ab} g_{bc} = \delta^a_c.$$

5. Dec 12, 2008

### GDogg

That term doesn't appear in the equation. What you have is:

$$g^{cd} g_{ab}$$

Where the left metric tensor is contracted with the Ricci tensor. Remember not to repeat indices more than twice. It messes up the summation convention. Use different letters like George Jones said.

Last edited: Dec 12, 2008
6. Dec 17, 2008

### Philosophaie

If you use the equation:

G_ab = R_ab - 1/2 * R _cd

You will get all the elements of the Ricci Tensor in each cell. Is really what the equation is?