Schwartzschild metric difficulty

[Thinkor]

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?

 

[Mentz114]

You're using trying to use dynamics, but GR is a kinematic theory. This means that the curvature predefines trajectories that will be followed by test particles with given initial conditions and/or proper accelerations. So it is possible to find the 4-velocity of a dropped object but not the way you're attempting.

You should look this up in a textbook, or maybe start with this (very good) article in Wiki.

http://en.wikipedia.org/wiki/Schwarzschild_geodesics

 

[yuiop]

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

It might be worth noting that:

f² = A - dR²/(dt²A)

so f is not just a function of R, but a function of R and dt.

These links show how the Schwarzschild acceleration is obtained:

http://www.mathpages.com/rr/s6-07/6-07.htm
http://www.mathpages.com/rr/s6-04/6-04.htm

 

[Thinkor]

These links show how the Schwarzschild acceleration is obtained:

http://www.mathpages.com/rr/s6-07/6-07.htm
http://www.mathpages.com/rr/s6-04/6-04.htm

Thanks very much. I'll look carefully at these pages.
My Newtonian intuition is evidently not working well here.