# I Gravity Time Dilation

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1. Jun 13, 2017

If we assume I live in Jupiter and there is one year passed in my clock how many years passes in the earth?

And how can I use that Equation in the attached?

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2. Jun 13, 2017

### Staff: Mentor

How are you going to compare the readings on the clocks? You can't do it directly since Jupiter is very far away.

3. Jun 13, 2017

### Janus

Staff Emeritus
I should also point out that any such comparison would need to take into account that Earth and Jupiter are at different locations in the Sun's gravity well and moving at different orbital speeds.

4. Jun 13, 2017

I means if I spend a lot time there and go back to the earth and I had a twin there what the deference between my age and my brother age @Janus @PeterDonis

5. Jun 13, 2017

### Staff: Mentor

That is a different question. By having you and your twin separate and rejoin you take care of the problem that PeterDonis points out but you still have all the complications that Janus mentions, as well as some additional ones because it makes a difference which path you follow to Jupiter and back to Earth.

An easier question to answer would be how the rate of a clock on the surface of Jupiter compares with the rate of a clock in orbit safely above the Jovian atmosphere.

6. Jun 13, 2017

### Staff: Mentor

Ah, ok, this is a well-defined problem. The short answer is, not much difference, because all of the gravitational fields involved (to a first approximation, the Earth's, the Sun's, and Jupiter's) are really weak, in the sense that their time dilation effects are really small. There are also effects due to your motion relative to your twin, both because you have to get in a spacecraft and travel to Jupiter and back, and because as @Janus said, Jupiter is moving relative to the Earth; those are also small, because all of the relative speeds involved are small compared to the speed of light (at least if you use current space travel technology), but that doesn't mean they're small relative to the gravitational effects.

The equation you give is relevant, but incomplete, because it only takes into account gravitational effects, not the effects of relative motion, and only from one source, whereas, as I said, there are three. There are a number of ways to construct a formula that takes all that into account; I'll only describe one here, because it seems to me to be the easiest to grasp conceptually.

Imagine a hypothetical observer who is far outside the solar system, looking in with a very powerful telescope and recording things. This observer can adopt coordinates in which, to a good approximation, the Sun is at rest, and we can treat all other objects (the Earth, Jupiter, and your spaceship) as moving by using their speeds relative to the Sun, and assuming that all of these speeds are small compared to the speed of light. This observer also carries a clock with him whose time can be treated as the coordinate time in these coordinates, i.e., as a common reference.

The only other assumption we need is that gravity is very weak everywhere that you and your twin will be, so that the gravitational fields of the three massive objects in question (the Sun, Earth, and Jupiter) superpose linearly--i.e., their effects can just be added together. This also allows us to ignore terms of higher than first order in $GM / c^2 r$ and $v^2 / c^2$, and to expand out the square root that would normally be in the formulas using the binomial theorem (dropping quadratic and higher order terms).

With all these approximations as given, the general formula for the instantaneous time dilation of an observer who is at coordinates $\vec{r}$ and moving with speed $v$ is:

$$\frac{d\tau}{dt} = 1 - \frac{G M_S}{c^2 \vert \vec{r} \vert} - \frac{G M_E}{c^2 \vert \vec{r} - \vec{r}_E \vert} - \frac{G M_J}{c^2 \vert \vec{r} - \vec{r}_J \vert} - \frac{1}{2} \frac{v^2}{c^2}$$

where $\tau$ is the proper time of the observer, $t$ is coordinate time, $M_S$, $M_E$, and $M_J$ are the masses of the Sun, Earth, and Jupiter, and $\vec{r}_E$ and $\vec{r}_J$ are the position vectors of the Earth and Jupiter (so the differences are just the distances from the observer to Earth and Jupiter). Note that the speed $v$ of the observer is relative to the Sun, so, for example, if the observer is your twin, at rest on Earth, his speed is Earth's speed in its orbit.

Now, this is an instantaneous time dilation, which means that, in order to get the total elapsed time for the observer from one event to another (for example, from the event of you and your twin parting to the event of you coming back together again), you have to know, first, the coordinate times $t$ for the two events, and second, the observer's position $\vec{r}$ and speed $v$ as functions of $t$. Then you plug those functions into the above formula and integrate with respect to $t$.

All that is complicated, and it would be nice if we could make some more approximations to simplify it. Fortunately, we can. First, by running numbers, you can see that the term in $M_E$ is negligible unless the observer is close to Earth, and the term in $M_J$ is negligible unless the observer is close to Jupiter. (Note carefully that the term in $M_S$ is not negligible in either case.) Second, at any speed achievable by current space technology, it will take years for you to get from Earth to Jupiter (and to return, when you return), so during the transit we can ignore both the $M_E$ and the $M_J$ terms, because you won't be close to Earth or Jupiter for practically all of the transit time. That gives us two much simpler integrals to do.

First, for your twin, who stays on Earth the whole time, if we idealize the Earth's orbit as circular, and ignore the Earth's rotation (for extra credit, you should check for yourself that that is a good approximation here), and observe that the Earth's orbital velocity about the Sun is just $G M_S / \vert \vec{r}_E \vert$, we simply have

$$\tau_\text{twin} = \int_0^T \left[ 1 - \frac{G M_E}{c^2 R_E} - \frac{3 G M_S}{2 c^2 r_E} \right] dt$$

where $R_E$ is the radius of the Earth (since the twin is on the Earth's surface) and $r_E$ is the Earth's mean distance from the Sun (i.e., it's $\vert \vec{r}_E \vert$). The factor of $3/2$ in the last term in the integrand comes from the equation for the Earth's orbital velocity. All of the factors in the integrand are constant, so we just have

$$\tau_\text{twin} = \left[ 1 - \frac{G M_E}{c^2 R_E} - \frac{3 G M_S}{2 c^2 r_E} \right] T$$

where $T$ is the total coordinate time elapsed from when you leave your twin to when you return.

For you, we divide your trip into three legs: outbound to Jupiter, stay on Jupiter, return to Earth. For the second of these, the formula will look similar to what we just did for your twin, but with the figures for Jupiter instead of Earth. For outbound and returning, the only mass term will be for the Sun, and there will be a term for your speed, and the formula will be the same for both since the trips are mirror images of each other. We can idealize your motion as being purely radial and at constant speed $v$, which means your distance from the Sun will be $r_E + vt$, where $t$ is the coordinate time from the trip start. The total coordinate time for the trip will be $T_\text{trip} = (r_J - r_E) / v$. So we have $\tau_\text{you} = 2 * \tau_\text{trip} + \tau_\text{Jupiter}$, where

$$\tau_\text{Jupiter} = \left[ 1 - \frac{G M_J}{c^2 R_J} - \frac{3 G M_S}{2 c^2 r_J} \right] T_\text{Jupiter}$$
$$\tau_\text{trip} = \int_0^{T_\text{trip}} \left[ 1 - \frac{G M_S}{c^2 \left( r_E + v t \right)} - \frac{1}{2} \frac{v^2}{c^2} \right] dt$$

Only one term actually is a function of $t$, and looking in a table of integrals and substituting for $T_\text{trip}$ gives

$$\tau_\text{trip} = \frac{1}{v} \left[ \left( 1 - \frac{1}{2} \frac{v^2}{c^2} \right) \left( r_J - r_E \right) - \frac{G M_S}{c^2} \ln \frac{r_J}{r_E} \right]$$

Putting all of the above together, and using $T = 2 T_\text{trip} + T_\text{Jupiter}$ to relate the coordinate times for you and your twin, we end up with

$$\tau_\text{twin} = \left( 1 - \frac{G M_E}{c^2 R_E} - \frac{3 G M_S}{2 c^2 r_E} \right) \left( T_\text{Jupiter} + \frac{2 \left( r_J - r_E \right)}{v} \right)$$
$$\tau_\text{you} = \left( 1 - \frac{G M_J}{c^2 R_J} - \frac{3 G M_S}{2 c^2 r_J} \right) T_\text{Jupiter} + \frac{2}{v} \left[ \left( 1 - \frac{1}{2} \frac{v^2}{c^2} \right) \left( r_J - r_E \right) - \frac{G M_S}{c^2} \ln \frac{r_J}{r_E} \right]$$

I'll leave it to you to plug in numbers to these formulas and see what comes out for various possible values of $T_\text{Jupiter}$ and $v$.

Last edited: Jun 13, 2017
7. Jun 13, 2017

### PAllen

A simplification could be to imagine very long lived twins. Traveling twin goes to Jupiter, stays put for a million years his time, then returns to earth. Then the impact of the rocket trips can be taken to be negligible.