Reverse-tracing Einstein's field equation

[TheMan112]

How do I rewrite Einsteins famous field equation

R_{ab} - \frac{1}{2} g_{ab} R = \frac{8 \pi G}{c^4} T_{ab}

into:

R_{ab} = \frac{8 \pi G}{c^4} (T_{ab} - \frac{1}{2} g_{ab} T)

I've tried experimenting with reverse-tracing the Ricci-scalar, but I just don't get the right equation. The trace from these equations would yield R=-T. Is this really correct?

 

[Mentz114]

Multiply both sides by g^{ab} to get

\frac{1}{2}R = \kappa T

then substitute back.

 

[TheMan112]

Multiply both sides by g^{ab} to get

\frac{1}{2}R = \kappa T

then substitute back.

Ok, I insert the expression and get:

R_{ab} - \frac{8 \pi G}{c^4} g_{ab} T = \frac{8 \pi G}{c^4} T_{ab}

R_{ab} = \frac{8 \pi G}{c^4} \left(T_{ab} + g_{ab} T \right)

Which as you can see is not entirely right. Have I missed something very obvious?

 

[smallphi]

Take trace of both sides of Einstein eq. Trace of the metric is 4. Trace of Ricci tensor is R. You get

R = - k T

then substitute that back for R and take the metric term on the other side.

 

[Mentz114]

I made a mistake, g^{ab}g_{ab} = -1.

 

[country boy]

...
R_{ab} - \frac{1}{2} g_{ab} R = \frac{8 \pi G}{c^4} T_{ab}

Check for an errant sign in this equation.

 

[country boy]

I made a mistake, g^{ab}g_{ab} = -1.

Okay, that fixes it.

 

[smallphi]

g^{ab}g_{ab} = \delta^a_a = 4

 

[Mentz114]

g^{ab}g_{ab} = \delta^a_a = 4

Yes, you're correct. It's |g| = -1.

 

[TheMan112]

Yes, you're correct. It's |g| = -1.

How can |g| = abs(g) = -1 ? An absolute value can't be negative. For example a radius cannot be negative.

 

[smallphi]

|g| = 1 only in Special Relativity. In GR, the value of |g| changes with the coordinate system used but it's signature (the signs of the eigenvalues when you diagonalize it) doesn't change, depending on the sign convention it's either (-, +, +, +) or (+, -, -, -).

 

[TheMan112]

Got it right now, thanks everybody.