B Calculating Mass for Gravitational Time Dilation

1. May 31, 2016

marksrobby

Hello all. I'm trying to determine the mass of an object required to make it so that a traveler on a massive planet experiences 1 day but on earth, or some infinitely distance away from the planet, 1000 years passes by. I'm using the following equation:

t0 = tf*root(1-(2GM/rc^2))
Found here: https://en.wikipedia.org/wiki/Gravitational_time_dilation#Outside_a_non-rotating_sphere

I'm trying to solve for M. I end up with the following equation:

M = rc^2 * (1 - (t0/tf)^2)/2G

I arbitrarily picked r = 1 * 10^6 meters. When I run through this equation I get a negative mass. What am I doing wrong?

P.S. This is not a homework assignment, I'm just a curious guy trying this equation out.

2. May 31, 2016

Yukterez

The terms "1-" and ".../2G" are wrong and need to be replaced with ".../2/G".

$r=\frac{2\cdot T^-2}{T^-2-1}\cdot \frac{G\cdot M}{c^2}$

with T=dt/dτ (Note that it is T-2, not T-2). So if you want the time dilation factor to be T = 10000, you place the observer who's clock you want to tick slower in a distance of 200000000/99999999 GM/c² from the center of mass. At r = 2GM/c² (the Schwarzschildradius) the time dilation factor gets infinite; below that radius it will be imaginary.

If you have a fixed radius and want to solve for the mass use

$M = \frac{c^2\cdot r-c^2\cdot r \cdot T^-2}{2\cdot G}$

So if you want again a time dilation factor of T = 10000 you choose the mass to be 99999999/200000000 c²/G r.

$r > \frac{2\cdot G\cdot M}{c^2}$

or alternatively

$M < \frac{c^2 \cdot r}{2\cdot G}$

And make sure that T ≥ 1 but never T < 1 since if the mass is positive the clocks near the mass can only tick slower, but never faster than the clock at infinity.

If you don't want to compare the clock near the mass to a clock at infinity but to a clock on earth you have to use U = T·1.00000000007 instead of T alone (which would be the dilation factor with respect to infinity in the equations) since the clocks on earth are already running slower by a factor of 1.00000000007 with respect to infinity. Multiplied with that factor you can use T as the dilation factor with respect to earth and plug U into the equation instead of T to get your mass.

Last edited: May 31, 2016
3. Jun 1, 2016

4. Jun 1, 2016

Ibix

$r=\frac{2\cdot T^{-2}}{T^{-2}-1}\cdot \frac{G\cdot M}{c^2}$ is how you do that, fyi. Or multiply top and bottom by $T^2$ to duck the issue.

5. Jun 1, 2016

Yukterez

Ah, I remember, one has to write x^{-y} instead of x^-y for the Latex Interpreter to display that readable, or better yet, x^{^{-y}} to raise the index even higher. Unfortunately I can't edit the first posting anymore, so here again both equations in a better notation:

$r=\frac{2\cdot G\cdot M}{(1-1/T^2)\cdot c^2}$

$M = \frac{c^2\cdot r \cdot (1-1/T^2)}{2\cdot G}$

Last edited: Jun 1, 2016