# Gravitational Time Dilation

1. Aug 24, 2004

### gonzo

I have a few questions about time dilation due to gravity. It seems to me there are some strange/interesting consequences of this that I've never seen discussed. I'm not sure what signifigance it has, but it seems like it should have some.

Basically, since gravity is lower on the surface of Mars than on the surface of Earth, then time should procede faster on Mars than on Earth (relative to each other). I know this difference is too small for a human lifetime to be significant, but it seems like it should add up to something serious on a geological time scale. Even more so for smaller planets (or non-planets) like Pluto. Doesn't this mean that they are some large number of years ahead of us in geological development in some sense? That the surface of those planets is in some sense a lot older than the surface of Earth (relative to the start of the solar system and each other)?

Like I said, I don't know if this has any meaning other than a possible science fiction story plot, but it seems interesting at least. It might have implications though if we find some old Martian rock on Earth (which has happened) and try and figure out "when" it left Mars and came to Earth. If that was long enough ago, that "when" could be radically different from an Earth and from a Mars perspective.

Also, this just let me to another thought. Time speeds up as you get closer to the center of the Earth for the same reason. Doesn't this have some sort of weird implications for geological development of the planet, where the center of the Earth is going to be quite a bit older geologically than the surface of the planet? Over several billion years this difference has to be pretty serious. Do geologists ever consider this when thinking about planetary geology theories?

2. Aug 24, 2004

### Ray Eston Smith Jr

There has been a science-fiction novel with a similar theme - Larry Niven's Neutron Star.

I don't know but my guess is that the time difference between Earth's surface and Mar's surface would only add up to a few minutes per billion years.

I don't think the sppeding of time inside the Earth as you move toward lower gravitation at the the center is geologically significant, but it might make a big difference inside a neutron star.

3. Aug 24, 2004

### Severian596

Why is this? I've never heard it, but thinking it through as you approach the center of the earth (from the surface), a couple things happen:

1. Gravity on a per-point basis decreases, because as you approach the center of the earth its mass distribution "above" you is roughly the same as its mass distribution "below" you.

2. Your speed relative to the speed of the surface decreases, because you're no longer on its outer rim and no longer moving at the speed of its surface.

I think these effects cancel out to some degree, making the effect (that's already very small, as pointed out by Ray Eston Smith Jr) and making it even smaller.

4. Aug 24, 2004

### Janus

Staff Emeritus
Over 4+ billion years of the Earth's age, the surface of Mars would age less than 3 years more than the surface of the Earth due to gravitational time dilation, not exactly a "large number of years".
As pointed out above, the age difference would be of no significance.
No, it wouldn't. Gravitational time dilation is not related to the local force of the gravitational field, but rather the gravitational potential For any given body this relates to how deep you are in the gravity field> As you move towards the center of the Earth, you are moving deeper into Earth's Gravity field, and thus time will move slower, not faster.
Again, Over the age of the Earth the age difference will be quite small. We are talking about the core being less than 1 year younger than the surface, insignificant on geological time scales.

5. Aug 24, 2004

some scientists think that this would be the best way for futureisitc time travel. To build a shperical hollow machine which has no gravity in the centre but has a hudge amount of gravity on the outside. Making the time on the ouside pass faster than the time on the inside.

6. Aug 25, 2004

### gonzo

I did some number checking finally, and it does seem that on the surface of the earth compared to no gravity it's 1 part in a billion, so that is only 1 year ever billion not very much as has been pointed out.

However, I'm confused as to why time would slow down towards the center of the Earth. Gravity starts getting less and less as soon as you go inward from the surface, where it is maximum, until it reaches zero at the center of mass, where there should be no time dilation from the Earth. What am I missing?

I plugged in some numbers for an arbitrary neutron star, and it seem you can easily get dilation ranging from 10 to 100 percent. So here's a question. Assuming you have a many billion year old neutron star, with a time dilation at the surface of 20 percent. So the surface is lagging some point in space by several hundred million years over the life time of the star.

What happens if you were to land a ship on the surface? Would you be "going back in time" several hundred million years, and thus get a radically different view of the sky when you land from when you were in orbit? Or would you only experience time change related to the amount of time it actually took you to land the ship (a few hours say)? I hope it's at least clear what I'm asking.

7. Aug 25, 2004

### pervect

Staff Emeritus
In the weak field case, you can think of gravitational time dilation as being caused by gravitational potential energy. You are apparently thinking of gravitational time dilation as being caused by the local acceleartion of gravity, 'g'. This is not true.

see for instance

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/gratim.html#c4

Also note, though, that g is closely related to the gravitational potential energy U=-GmM/R. Specifically, g=dU/dr.

8. Aug 26, 2004

### gonzo

I looked at that link and am still confused. According to that formula, it still seems that time would go faster towards the center of the Earth compared with the surface. Taking the first formula (not the one using "g" on the surface) the only thing you have changing as you go towards the center are M and R, mass and radius (unless these mean somthing else here?). Mass changes faster than radius, which means that bottom of the equation approaches 1 as you approach the center of the Earth, and the two times approach each other.

9. Aug 26, 2004

### pervect

Staff Emeritus
If you look at the text, T is the time at infinity and T0 is the time on the clock in the gravity well. They are related by the expression

$$T = \frac {T_0} {\sqrt{1-\frac {2 G M}{R c^2}}}$$

Thus T > T0 always, which means the clock at infinity runs faster, and the clock in the gravity well runs slower.

There is an unfortunate typo in their series expansions, which are convenient when the time dilation is very small (as is usually the case)

$$T \approx T_0 (1 + \frac{GM}{R c^2}) \hspace{.5 in} T_0 \approx T(1-\frac{GM}{Rc^2})$$

Note that U = GM/R is the Newtonian potential energy per unit mass. Thus U/c^2 has the dimensions of Energy / (mass * c^2), i.e. it's dimensionless.

If you want the reading on a clock in a gravity well, the second approximation will give it as a function of the time of a clock at infinity, and as I said before, the clock in the gravity well always indicates that less time has passed than the clock at infinity.

10. Aug 26, 2004

### pmb_phy

Note: The Newtonian potential energy per unit mass is also known as the Newtonian potential.

Pete

11. Aug 26, 2004

### Severian596

I think I understand why you're confused. As you head toward the center of the earth, you don't really think you're leaving its gravitational field, do you? I mean...you're sitting right in the middle of it!!

The net force on your body approaches zero (as you're pulled in all directions by mass that's approximately equal on all sides), but the force is there nonetheless. In order to leave the gravitational field you'd have to move far enough away from the mass.

12. Aug 26, 2004

### gonzo

Thank you Severian, that was exactly what I needed to here to understand why I was confused. I got it now.

13. Aug 26, 2004

### mee

Thank you Janus this is very interesting!

14. Aug 26, 2004

### Severian596

15. Aug 26, 2004

### DW

That formula is only valid for the region exterior to the earth. Inside the Earth the corresponding formula would be
$$t = \frac{\tau}{\sqrt{1 + \frac{2\Phi}{c^2}}}$$
$$\Phi = \frac{1}{2}\frac{GM_{tot}r^{2}}{R^{3}} - \frac{3}{2}\frac{GM_{tot}}{R}$$
where $$M_{tot}$$ is the mass of the planet, R is its radius, r is the distance of the clock inside the earth from the center with $$\tau$$ as its time and t is the time for a far remote clock. According to this gravitational time dilation is greatest at the center.

16. Aug 29, 2004

### Chronos

Gack! Remember once you reach the center of earth [or mars] the gravitational potential is exactly.. zero.

17. Aug 29, 2004

### Janus

Staff Emeritus
Zero GP is defined at an inifinite distance from the body for which you are measuring it and becomes increasingly negative as you approach the body in question. It is at its most negative at the center of the body.

18. Aug 29, 2004

### DW

No. The gravitational potential for the interior is given by
$$\Phi = \frac{1}{2}\frac{GM_{tot}r^{2}}{R^{3}} - \frac{3}{2}\frac{GM_{tot}}{R}$$
At the center this becomes
$$\Phi = - \frac{3}{2}\frac{GM_{tot}}{R}$$
and that is where the magnitude of the potential and gravitational time dilation is at their greatest.

19. Aug 29, 2004

### pmb_phy

Janus is correct. The potential decreases from zero outside a spherical body to -GM/R at the surface where R is the radius of the sphere. Inside the body the gravitational force is given by F = -GMr/R3 which means its still directed towards the center. For this to be true the gravitational potential must still decrease as you get to r = 0.

This can be derived using Gauss's law for gravitation, i.e.

$$\int \bold g_{inc} \bullet d\bold s = -4\pi G M_{inc}$$

where Minc is the mass inclosed by the surface of integration and gin is the gravitational acceleration at the point on the surface of the integration. This yields

$$\bold g_{>} = -\frac{GM}{r^2}\bold e_r$$

$$\bold g_{<} = -\frac{GM_{in}}{r^2} \bold e_r= -\frac{GMr}{R^3}\bold e_r$$

where g> is the gravitational acceleration for r > R, g< is the gravitational acceleration for r < R and M is the total mass of the sphere. er = r/r. The corresponding gravitational potential for r > R is

$$\Phi(r)_{>} = -\frac{GM}{r}$$

and for r < R is

$$\Phi(r){<} = -\frac{Mr^2}{2R^3} + C$$

The gravitational potential must be a continuous function of r which means for r > R

$$\Phi(R)_{>} = \Phi(R)_{<}$$

This means that

$$C = -\frac{3GM}{2R}$$

So

$$\Phi(r)_{<} = -\frac{Mr^2}{2R^3} -\frac{3GM}{2R}$$

When r = 0

$$\Phi(0)_{<} = -\frac{3GM}{2R}$$

Pete

Last edited: Aug 29, 2004
20. Aug 29, 2004

### pervect

Staff Emeritus
It's necessary to make some assumptions as to the variation of density with radius (depth) to compute the gravitational potential inside a body. Assuming that the density is constant as a function of r lead's too a "Hooke's law" force which is directly proportional to the distance, like a giant spring. This assumption leads to the results previously presented in detail by both pmb and DW.