Solutions to Field Equations with Einstein Tensor = 1

[edgepflow]

Suppose there is a solution to the field equations with the Einstein Tensor = 1:

Gtt = 1

and/or,

Gxx = Gyy = Gzz = 1

This would leave for the stress energy tensor:

T = 1 / 8 pi G

Now for stress, it seems to get physical units of pressure, you would apply:

Txx = Tyy = Tzz = c^2 / 8 pi G tp^2

where tp is the planck time.

And for the time component:

Ttt = 1 / 8 pi G tp^2

Please let me know if I have these conversions straight.

[jfy4]

Suppose there is a solution to the field equations with the Einstein Tensor = 1:

Gtt = 1

and/or,

Gxx = Gyy = Gzz = 1

This would leave for the stress energy tensor:

T = 1 / 8 pi G

Now for stress, it seems to get physical units of pressure, you would apply:

Txx = Tyy = Tzz = c^2 / 8 pi G tp^2

where tp is the planck time.

And for the time component:

Ttt = 1 / 8 pi G tp^2

Please let me know if I have these conversions straight.

The Gravity field equations are

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

I'm not sure why the plank time worked it's way in there. The units are fine the way they are. Maybe I'm not getting something though.

[Bill_K]

- The metric tensor gμν is dimensionless.
- The curvature tensors Rμν, Gμν and Rμνστ are second derivatives of the metric and have dimension L-2. (This is assuming your coordinates have dimension L. If you use polar coordinates or something like that, the dimension of those components will be different.)
- The stress-energy tensor Tμν has dimensions of energy density, which is M(L/T)2/L3 = ML-1T-2.
(Pressure has the same dimensions as energy density: force/area = M(LT-2)/L2 = ML-1T-2.)
Like jfy4 says, Einstein's Equations are Gμν = (8πG/c4) Tμν. What are the dimensions of that constant G/c4?
Well the Schwarzschild radius is 2 Gm/c2 ~ L, so G/c4 ~ M-1L-1T2, and with that I'll leave you to verify that the dimensions on both sides agree.

[edgepflow]

Thank you for the replies. I think the "natural units" were messing me up.