Proper time in Schwarzschild Space

1. Apr 17, 2012

Nilupa

Please explain me how to find the proper time taken by one twin to travel around a massive body while his other twin stays on the earth in Schwarzschild space?

2. Apr 17, 2012

A.T.

3. Apr 17, 2012

dipole

You need to calculate the time dilation using the g00 component of the metric tensor, integrated along the path. You can assume a radial path from r = r1 to r = r2 (and because the metric is diagonal in the schwarzschild space, the total time dilation is twice this integral).

The SR time dilation is symmetric so I think you can ignore it.

4. Apr 17, 2012

Staff: Mentor

First of all, I'm not sure the OP's question was intended to imply a purely radial path. "Travel around a massive body" implies a path that goes to the other side of the body, then back again. However, that is a more complicated problem; perhaps the OP would be willing to revise his question to fit with purely radial motion. Something like "what is the proper time taken by one twin to drop down close to a black hole, then come back up again, compared to the other twin who stays at the same place, far away from the hole, the whole time." That would be simpler to solve.

However, even in the purely radial case, you can't just use the g_00 component of the metric, because dt is not the only coordinate differential in the line element that is nonzero. You have to also know g_rr, because dr is also nonzero (the radial coordinate is changing with time). So to do the integral, you have to make some assumption about dr/dt--how the radius changes with time, i.e., how the traveling twin actually moves. Different assumptions will lead to different answers.

For the "stay at home" twin, yes, you can find his total elapsed proper time just from g_00, since his radius does not change. If you assume he is far enough away from the black hole that his g_00 is approximately 1, then his elapsed proper time is approximately the same as elapsed coordinate time.

No, it isn't, for the same reason it isn't in the standard SR twin paradox. However, doing the integral I described above, with any reasonable assumption for dr/dt, automatically takes into account "time dilation" due to relative motion as well as that due to the change in the strength of gravity with radius.

5. Apr 17, 2012

dipole

I assumed radial motion because it sounded like the OP was referring to the twin paradox where one twin zooms off and then returns, and the difference in their ages is a result of the gravitational time dilation, but on a second look it seems he's referring to a different problem.

6. Apr 18, 2012

yuiop

The Earth is also a massive body and as you may have read in other threads, calculating exact solutions for two massive bodies is extremely difficult. However, if the body the travelling twin goes around is a black hole and he is sufficiently close to the black hole, the effect of the mass and motion of the Earth can be considered negligible. In this case we can treat the Earth twin as being equivalent to the Schwarzschild observer at infinity. In the most simplified case, if the orbiting twin measures his local velocity to be v and he is orbiting at a radius of r, then the proper time t measured by the orbiting twin compared to the time T measured by the Earth twin is given by:

$$t = T \sqrt{1-\frac{2GM}{rc^2}} \sqrt{1-\frac{v^2}{c^2}}$$

In Schwarzschild coordinates, the orbital velocity for an object in a circular orbit is (surprisingly) exactly the same as the that obtained from the ancient Kepler laws, (i.e. $v=\sqrt{GM/r}$). The above equation for proper time requires the local velocity so the Kepler laws have to be transformed by the gravitational gamma factor to give the local orbital velocity as

$$v = \sqrt{\frac{GM}{r}}* \sqrt{\frac{1}{1-2GM/(rc^2)}}$$

We can now equate the proper time of the free falling twin orbiting at radius r, relative to the time of the Earth twin, by substituting for v in the first equation, to obtain the following relation:

$$t = T \sqrt{1-\frac{2GM}{rc^2}} \sqrt{1-\frac{GM}{rc^2-2GM}}$$

when the radius = $3GM/c^2$ the term on the right goes to zero so the proper time goes to zero, but this is of course the radius of the photon sphere where only photons can orbit at the speed of light.

Now this is a twin experiment and we like them to meet up again and compare ages. We let the travelling twin descend to r and then he starts orbing at radius r. After 90 years measured by the Earth twin, he also descends to r and is reunited with his twin who might still be a young boy if r is sufficiently close to the photon sphere radius.

Last edited: Apr 18, 2012
7. Apr 18, 2012

PAllen

It would be interesting to know what the OP really intended. My guess would be they meant to consider only one gravitating body, and that by earth, they meant a static SC observer some distance from the 'sun', and for the other observer, various paths going around the sun. If anything similar to this is what was intended, it is not correct to treat the 'earth' observer as an observer at infinity in SC coordinates.

Taking the OP more literally, I would still see no intent to treat the earth's gravitation in the problem. Instead you would be comparing a the proper time along two events on an orbital geodesic path with proper time along a path between the same events which goes around the sun one more time than the geodesic path.

8. Apr 18, 2012

yuiop

In which case the proper time t of the twin circumnavigating the sun at velocity v relative to the time T measured by the twin on the Earth (with the Earth orbiting the Sun at velocity u), is simply:

$$t = T \frac{\sqrt{1-v^2/c^2}}{\sqrt{1-u^2/c^2}}$$

where I am assuming the circumnavigating twin is at the same radius as the Earth from the Sun and ignoring the Earth's gravitation. However the time dilation effects due to natural Solar system orbital speeds are small and the effects of the Earth's gravity as the circumnavigating twin varies his distance from the Earth would not be negligible in this case. v and u are both measured by a hovering observer that is stationary in the Sun's gravitational field.

For a further simplification, still ignoring the Earth's gravitational effects on time dilation, the relation is:

$$t = T \sqrt{1-w^2/c^2}$$

where w is the speed of the circumnavigating twin relative to the Earth.

However, the OP did mention a massive body, so I assume he was interested in gravitational effects on time dilation.

Last edited: Apr 18, 2012
9. Apr 18, 2012

Nilupa

Thanks everyone.
Actually this is not a radial motion. So, angle ∅ should be integrated from 0 to 2∏.

10. Apr 18, 2012

PAllen

As always, it is so useful for the OP to be clear on intent. I assumed, e.g. the massive body was the sun, and the 'earth' was meant to be effectively the name of one of the world lines.

11. Apr 18, 2012

yuiop

That is fine, but for simplicity we can talk in terms of instantaneous tangential velocity at a point instead of angular velocity. You have still not made it clear how complex the scenario you have in mind is. Do you intend to find out the proper time of an observer with both radial and angular velocity, spiralling in towards a massive body and back out again? Did you intend for the gravitational mass of the Earth to significant to the calculations or not?