1. Nov 22, 2008

### 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?

2. Nov 22, 2008

### 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.

3. Nov 22, 2008

### 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 $$8 \pi R^3$$).

Last edited: Nov 22, 2008
4. Nov 22, 2008

### tiny-tim

Hi snoopies622!

(have a pi: π and a rho: ρ )
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)
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

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

5. Nov 22, 2008

### snoopies622

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