In our discussion of rigidity, we agreed to set external differential forces equal to zero, or to correct our instruments for the effects of such forces. Let us suppose that we were in the presence of an electrostatically charged gravitational body. That is, the gravitational body (e.g. the sun) would have some electrostatic charge (as is probably the case for the sun). If we corrected our rods and clocks for the effect if any of this differential force (which would certainly be negligible in any case, so far as spatial geometry is concerned), we might then expect that the geometry of the field would be the same as when the gravitational body is uncharged. But is this the case? Not at all! And the explanation is not far to seek: the charge contributes energy to the field; but energy, by relativity theory, is equivalent to mass, and mass always modifies the gravitational potential. Hence, the geometry of the charged gravitational field (which depends on the potential) must be different from the uncharged.