This binding energy may be viewed as energy in the gravitational field, and is negative because gravity is attractive. To obtain a general expression for the energy density we assume it is proportional to the square of the gravitational field ⃗ g = −∇ φ . That is we set This is in direct analogy with the energy density of the electric field — except of course for the opposite sign! The proportionality constant λ can be determined by integrating ρ g in Eq.(30) over the volume between R and infinity in Fig.6 to get the total field energy. Equating this energy expression with the total binding energy given by Eq.(29) we obtain Thus the energy density of the Newtonian gravitational field is written as In general relativity the problem of gravitational field energy is notoriously more subtle and complex. 37 This is due to the nonlinearity of the field equations, which in turn is related to the fact that gravity carries energy and is thus a source of more gravity. In this sense gravity differs fundamentally from the electric field, which does not carry charge and thus is not the source of more electric field. For our present purpose we will content ourselves with the rough estimate given by Eq.(32). We now consider a space region of size l that is nominally empty and free of gravity, except for fluctuations allowed by the energy-time uncertainty relation Eq.(7). As in section VI we attempt to measure the gravitational energy in the region in time l/c , with accuracy limited to Δ E ≈ $_{ℏ}$c/l . We thus cannot verify that the region is truly free of gravity, but only that the gravitational energy in the region is no more than about Δ E ≈ $_{ℏ}$c/l . From Eq.(32), this limit implies the following limiting relation for the Newtonian potential field As a rough estimate ( ∇ φ ) 2 ≈ (Δ φ/l ) 2 , where Δ φ is the uncertainty or fluctuation in the nominally zero Newtonian potential field, so from Eq.(33) we obtain ρ g = λ ( ∇ φ ) 2 (30) λ (4 πG $^{2}$M$^{2}$/R ) = − GM $^{2}$/ 2 R, λ = − 1 / 8 πG (31) ρ g = − ( ∇ φ ) $^{2}$/ 8 πG (32) l $^{3}$( ∇ φ ) $^{2}$/ 8 πG ≈ $_{ℏ}$c/l (33) Δ φ ≈ √ $_{ℏ}$cG/l (34) 14