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.(adsbygoogle = window.adsbygoogle || []).push({});

Here is the metric directly from Schwartzschild's paper. This agrees with various other sources I've seen although the notation is different.

ds^{2}= (1 - α/R)dt^{2}- dR^{2}/(1 -α/R) - R^{2}(dθ^{2}+ sin^{2}θ 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

ds^{2}= A dt^{2}- dR^{2}/A

where I have defined A = (1 - α/R) for brevity.

The traveler along this line element will see himself "at rest" and so ds^{2}will equal dτ^{2}, 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

f^{2}dt^{2}= A dt^{2}- dR^{2}/A

Rearranging a bit, I get

dR^{2}/A = A dt^{2}- f^{2}dt^{2}

Solving for dR/dt yields

dR/dt = (A^{2}- Af^{2})^{.5}

I then take the derivative to get the acceleration.

d^{2}R/dt^{2}= .5 (A^{2}- Af^{2})^{-.5}(2A dA/dt - A 2f df/dt - f^{2}dA/dt))

But, dA/dt = -α/R^{2}dR/dt, and by hypothesis dR/dt = 0 (we are just dropping an object, whose velocity is initially 0). Therefore, we get

d^{2}R/dt^{2}= .5 (A^{2}- Af^{2})^{-.5}(- A 2f df/dt))

Since f is a function of R, df/dt = df/dR dR/dt = 0 also initially.

But that means d^{2}R/dt^{2}= 0 initially as well. So the object does not move!

What's wrong?

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# Schwartzschild metric difficulty

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