Tension on a cable in a gravitational field

Question

Consider a mass 'm' suspended in the gravitational field of a massive star. Assuming the Schwarzschild metric it is easy to calculate the gravitational acceleration at the location of the mass and thus the tension in the cable. The question is: how does this tension propagate up the cable?

I've tried to apply the stress-energy tensor, but I'm not convinced I know the correct principles to apply. In thinking about this I've come up with a thought experiment that gives a surprising result and would like some comments on it, and also ideas about the "correct" way to do this via the stress-energy tensor. My thought experiment:

-Consider a long loop of cable with pulleys at each end, one directly above the other. A generator is attached to the upper pulley, and the generator is operated to provide constant tension on the cable on one side of the pulley. An operator at the bottom pulley turns a crank and makes the pulley turn one full turn and stops.

-The work done by the lower operator is $2 \pi RT$. If we assume the tension is constant up the cable (i.e. the cable is massless and there is no transform of the tension by the metric) then the work received at the generator is the same.

-If we now convert this work into a photon and send it back down to the lower pulley, when it arrives there it will be blue shifted and have additional energy, proportional to the square root of the ratio of the $g_{00}$ components of the metric. This would violate conservation of energy and allow a perpetual motion machine, so we assume that this can't happen.

-My conclusion is that the tension on the cable must vary with the square root of the time metric component. I have not seen this described anywhere, however. Does someone know the correct answer, or see the fallacy in this thought experiment?

Answer 1

Your naïve interpretation of the work equation doesn't quite make sense in this context. Consider that the standard formula for work gives that $W=\int {\vec F}\cdot d{\vec x}$. At minimum, the presence of the dot product in the above equation should not be ignored, and we should interpret, for a radial force, the work to be equal to $\int \frac{F\,dr}{\sqrt{1-\frac{2M}{r}}}^{1}$. We should then note that this factor will exactly cancel the effect you cite.

${}^{1}$Note that, properly, we are defining the line integral in terms of three unit vectors normal to the line, so, the integral would have the form $\int \sqrt{|g|}\epsilon_{abcd}{\hat t^{a}}{\hat \theta^{b}}{\hat \phi^{c}}F^{d}$. When this is completely simplified, you'll find that the measure of the integral $\sqrt{|g|}$ has a factor of $\sqrt{1-\frac{2M}{r}}$ that cancels against the factor of $\frac{1}{1-\frac{2M}{r}}$ in $g_{rr}$, producing the term seen above.

Answer 2

Your question relates somewhat to space elevator science and technology. http://en.wikipedia.org/wiki/Space_elevator

For rods and ropes in a gravitational field you may also find useful to read ch. 5 and problems 5.5, 5.6, 5.7 of Problem book in relativity and gravitation by Lightman et al. http://www.amazon.com/Problem-Book-Relativity-Gravitation-Lightman/dp/069108162X

Answer 3

"a photon and send it back down to the lower pulley, when it arrives there it will be blue shifted and have additional energy, proportional to the square root of the ratio of the g00g00 components of the metric. This would violate conservation of energy and allow a perpetual motion machine, so we assume that this can't happen."

An f-ing brilliant observation.

May I offer a slightly different model for interpretation? A similar thought experiment. Lowering a mirror toward a Schwarzschild event horizon from a stationary platform.

Ideal components of infinite strength & negligible mass. On the platform are a laser and receiver (the blueshift of the descent will be reversed by the redshift of the return with the remaining redshift due to dopplering from mirror recession) that counts received light waves. The amount of rope payed out would be equal to (1/2)(laser wavelength)(difference btwn number of light waves leaving the laser - number received back)... cummulative difference due to Doppler frequency reduction, which is constant.

Answer 4

It's very simple: When the crank operator is lowered down, his muscle strength decreases. When the cable is lowered down, its tensile strength decreases.

here is a derivation of that result