5-dimensional space with a 4-dimensional FRW spacetime as subspace

[1]We will wish to find five-d spaces with 4-d subspaces which are familiar to us. For example the metric for a five-dimensional space with a four-dimensional subspace which is homogeneous and isotropic:

ds² = - dt² + [Rt²/(1+kr²)²](dr² + r²dW²) + 5t² 5r² dX₅²

The Einstein Tensor for Five-Dimensional Spaces

The Einstein tensor components for spherically symmetric 4 and 5 d spaces have a simple relationship to each other (see appendix for derivation).

For the G^r_rtensor component: the relationship between the r parts of the 4 and 5-d components:

⁵_rG^r_r= ⁴_rG^r_r + 2(1/Rt²)(1/Rr²)(5'_r/5_r)(q_r'/q_r) ,

which for this metric with Rr= (1+kr²)and q_r = r/(1+kr²) is:

⁵_rG^r_r= ⁴_rG^r_r + (2/Rt²)(1+kr²)²(5'_r/5_r)(1-kr²)/[r(1+kr²)]

= ⁴_rG^r_r + (2/Rt²)(5'_r/5_r)(1-k²r⁴)/r

Homogeneous Isotropic Space

For the 5-d tensor component to be homogeneous, it must be independent of r, and since the 4-part is homogeneous and r-independent it must be that:

(5'_r/5_r)(1-k²r⁴)/r

is r-independent. That is:

(5'_r/5_r)(1-k²r⁴)/r = c ,

i.e:

5'_r/5_r= cr/(1-k²r⁴).

5'_r/5_r = c [r/(1-k²r⁴)]

ln5_r= c [r/(1-k²r⁴)] + K

The integral can be easily performed using the substitution R = r² , dR=2rdr.

The solution is:

5_r= K [(1-kr²)/(1+kr²)]^-c/4k

^{so
that the 5-d metric for a homogeneous isotropic space is (squaring}5_r^{makes the exponent -c/2k)}

ds²= -dt²+Rt²/(1+kr²)²(dr²+r²dW²) + 5t²K [(1-kr²)/(1+kr²)]^-c/2kdX₅²

^{It
is amazing that this solution can be obtained so simply - for an appreciation
of how simple our method is, see
Chatterjee paper: he bravely solved this using the usual long and tedious
calculations.}

^{The other results derived there are
also easily found in this way.}