Why is there an 8 instead of a 4 in Einstein's gravity formulation?

[snoopies622]

I have read that Einstein's formulation of gravity G_{ab}=\frac{8 \pi G}{c^4}T_{ab} is analogous to the differential form of Newton's version \nabla ^2 \phi = 4 \pi G \rho with the metric tensor and energy-momentum tensor in the modern form playing the same roles as gravitational potential and density in the classical one, respectively.

My question: why did the 4 become an 8?

 

[HallsofIvy]

In simple, though rough terms, Newton's version is in a 3 dimensional space and measures the field at the surface of a sphere and that area is given by 4\pi r^2.

Einstein's equation is in a 4 dimensional space and measures the field at the surface of a 4-sphere and that area is given by 8\pi r^3.

 

[snoopies622]

I may be miscalculating but I'm getting the 'surface volume' of a 3-sphere to be 2 \pi ^2 R^3 (instead of 8 \pi R^3).

 

[tiny-tim]

Hi snoopies622!

(have a pi: π and a rho: ρ )

I may be miscalculating but I'm getting the 'surface volume' of a 3-sphere to be 2 \pi ^2 R^3 (instead of 8 \pi R^3).

Yes, according to http://en.wikipedia.org/wiki/4-sphere#Volume_of_the_n-ball, the 4-ball has volume π²r⁴/2, and surface area 2π²r³

(the surface area is always n/r times the volume of the n-ball)

My question: why did the 4 become an 8?

The mathematical reason:

we require R₀₀ = 4πGρ.

But R₀₀ = constant(T₀₀ - 1/2 T g₀₀),

and T₀₀ - 1/2 T g₀₀ = ρc⁴ - 1/2 ρc⁴,

so the constant must be 8πG …

(i got this from http://en.wikipedia.org/wiki/Einstein_Field_Equations_(EFE)#The_correspondence_principle )

… but I'd still like someone to explain the physical significance of this!

 

[snoopies622]

Thanks, tiny-tim. I'll spend some time looking through that Wikipedia derivation.