I am trying to understand how the Schwartzschild metric works and coming to the conclusion that if I drop an object that it will not fall to the earth but just stay there when I open my hand. Therefore, I'm confident I have made a mistake, but I don't see where it is. Here is the metric directly from Schwartzschild's paper. This agrees with various other sources I've seen although the notation is different. ds2 = (1 - α/R)dt2 - dR2/(1 -α/R) - R2(dθ2 + sin2θ d∅) where α is the Schwartzschild radius, t is the time coordinate, and R is the radial coordinate. Please note that Schwarzschild uses units where c = 1. I'm just interested in dropping a clock, so I set dθ and d∅ to zero, reducing the equation to ds2 = A dt2 - dR2/A where I have defined A = (1 - α/R) for brevity. The traveler along this line element will see himself "at rest" and so ds2 will equal dτ2, where τ is the proper time. dτ will equal f dt for some function f of R that reflects slower ticking of the clock as it falls. Thus, we have f2 dt2 = A dt2 - dR2/A Rearranging a bit, I get dR2/A = A dt2 - f2 dt2 Solving for dR/dt yields dR/dt = (A2 - Af2).5 I then take the derivative to get the acceleration. d2R/dt2 = .5 (A2 - Af2)-.5 (2A dA/dt - A 2f df/dt - f2 dA/dt)) But, dA/dt = -α/R2 dR/dt, and by hypothesis dR/dt = 0 (we are just dropping an object, whose velocity is initially 0). Therefore, we get d2R/dt2 = .5 (A2 - Af2)-.5 (- A 2f df/dt)) Since f is a function of R, df/dt = df/dR dR/dt = 0 also initially. But that means d2R/dt2 = 0 initially as well. So the object does not move! What's wrong?