# Gravitational time dilation question

1. Oct 25, 2009

### ezezez

i'm 21, my bf is 20, but my parents won't allow me to marry a younger man...so is there anyway, i can use the idea of general relativity to slow down my aging relative to my bf, so that we both reach 25 at the same time?

here's what i know so far:
i understand that if i'm one floor above my friend on earth surface, i would observe my time runs faster than my friend: Tfriend=Tme[1-(gh/c^2)], with g being 9.8 and h being the distance between us. this is all i understand so far but i can't relate it to the original question. perhaps i need to move closer to the center of the earth to slow down my aging??

any help would be greatly appreciated.

2. Oct 25, 2009

### mathman

As a practical matter, the apporach you are suggesting would not result in any observable difference in aging between you and your bf. If you are 21 you don't need your parents permission to marry (in most parts of the world).

3. Oct 25, 2009

### ezezez

i'm sure there's a mathmatical solution to the problem by locating myself in a place with a different gravitational potential

4. Oct 25, 2009

### JesseM

If you could find yourself a large black hole and a ship that could hover close to the event horizon for a while, you could age slower than a person far from the black hole by a few years. But with the Earth the difference in aging at different distances from the center is never going to get above some tiny fraction of a second even if you hang out at different distances for years.

5. Oct 25, 2009

### ezezez

could you elaborate on the blackhole method? hand how this can quantatively answer the original question? is there a formula?

6. Oct 25, 2009

### matheinste

I think JesseM is being a bit extreme. Might I suggest the more realistic method of you taking a space trip and so, with the correct calculations of distance and speed, you can return at the required age.

Matheinste.

7. Oct 25, 2009

### JesseM

The time dilation formula for a person outside the event horizon of a nonrotating black hole (and this formula works for people outside the surface of a nonrotating sphere like a planet too...I don't think the rotation of the Earth is fast enough to change things much) is given on this wikipedia page, which compares the time elapsed for a person at radius r to the time for a person "at an arbitrarily large distance" from the black hole, but you can use it to figure out the ratio between times for two people at smaller distances r1 and r2, which would just be $$\frac{\sqrt{1 - \frac{2GM}{r_1 c^2}}}{\sqrt{1 - \frac{2GM}{r_2 c^2}}}$$. To be outside the event horizon, both people have to have a radius larger than the event horizon's radius which is 2GM/c^2. This means that if you want to write the radius of each person in units where the event horizon radius = 1, then the formula for the ratio of their clock ticks becomes a lot simpler, just $$\frac{\sqrt{1 - \frac{1}{r_1}}}{\sqrt{1 - \frac{1}{r_2}}}$$. For example, if one person is at 30 times the radius of the event horizon, and the other is at 1.5 times the radius of the event horizon, then the ratio should be$$\frac{\sqrt{1 - \frac{1}{30}}}{\sqrt{1 - \frac{1}{1.5}}}$$ and plugging sqrt(1 - 1/30) / sqrt(1 - 1/1.5) into http://www.math.sc.edu/cgi-bin/sumcgi/calculator.pl [Broken] gives about 1.7, so the the farther person has aged 1.7 years for every year that the closer person ages (the closer person is always the one aging slower). You can try playing around with the numbers to get different answers for the ratios of their aging, for example if one is only at 1.01 times the event horizon radius and the other is at 30, then the farther person will have aged 9.89 years for every year the closer person ages.

Of course if you want to get close to a black hole it needs to be a big one or you'll get spaghettified by tidal forces (gravity pulling more strongly on your feet than your head and ripping you apart)...This wikipedia page says the black hole thought to be at the center of the galaxy probably has a mass of about 8.2 * 1036 kilograms, and the gravitational constant G = 6.673 * 10-11 meters3 kilograms-1 seconds-2, and the speed of light c = 299792458 meters/second. So, the event horizon would have a radius of 2GM/c^2 = 2*(6.673 * 10-11)*(8.2 * 1036)/(299792458)^2 = about 6 billion meters.

There is also the method matheinste suggested of just traveling away from Earth at some high speed v, using velocity-based time dilation rather than gravitational time dilation. In this case you will have aged less than people on Earth by a factor of $$\sqrt{1 - v^2/c^2}$$. For example, if you travel away from Earth and back at 0.8c (80% the speed of light), then you will only have aged $$\sqrt{1 - 0.8^2}$$ = 0.6 years for every year aged by those on Earth. I didn't mention this one before because you only seemed to be asking about gravitational time dilation.

Last edited by a moderator: May 4, 2017
8. Oct 25, 2009

### Creator

Ha, ha...
Mathman, true; but the poster is testing your knowledge of how gravitational potential varies with distance AWAY FROM and INTO the earth's surface. (Probably a homework question).

Creator

9. Oct 25, 2009

### JesseM

For that the poster could take a look at the two posts of George Jones which he links to in post #2 of this thread.

10. Oct 25, 2009

### Creator

I think that would be a good link Jessie.

I think the basic idea that ezezer needs to get is:

1. Its not gravitational field that affects time dilation, but rather the gravitational potential, namely, -GM/R....( at distances > or = earth surface).

2. The potential is highest at greatest distance away from earth surface (zero at infinity), and decreases (becomes increasingly negative) toward the earth surface, and continues to become (lower) increasingly more negative toward the earth's center.

Creator

11. Oct 26, 2009

### ezezez

i think i have the answer now.

tie myself and my bf on the end of a string separated by 1m:

me---bf---------------------center

spin the string really really fast around the center until gravity is created towards the outside of the outside:

me---bf---------------------center
<----gravity----

then i use v=ah/c, with h=1m. use doppler's effcect formula 4=5sqrt[(1-v/c)/(1+v/c)], plug in ah/c for v, and solve for a. i got 1.973e16 m/s^2 for a, so if i spin the string fast enough until i create this gravity, then i can age 4 yrs while my bf ages 5 yrs

does this make sense? can anyone verify?

12. Oct 26, 2009

### JesseM

No, it doesn't work that way, "artificial gravity" created by the centrifugal force doesn't actually curve spacetime like real gravity created by mass, so the only time dilation in this situation would be the velocity-based time dilation seen in SR, with the person farther from the center aging slower because they have a constant higher speed in the inertial rest frame of the center.

Last edited: Oct 26, 2009
13. Oct 26, 2009

### DaveC426913

Read Haldeman's classic novel "The Forever War". The main character rides a relativistic shuttle until his mate catches up in age.