# Time Dilation-Frames-Details and hypotheticals

1. Sep 12, 2009

### JackAtl

I'm working on the basis of a scifi book and 2 Questions are nagging me about accelerated frames for time dilation;

1 Is the dilation determined by the net velocity vector of a frame, or some composition of all the many acceleration vectors exerted by every gravitational field in the solar system/galaxy? Namely would an object in a frame with the same net velocity characteristics
experience the same time as Earth frame, or do all the other solar objects which exert various magnitudes of forces play a part in Earth frame.

2 Since I can assume the lower acceleration/velocity frame = relative acceleration of time, would a hypothetical object (deep space station?) in a solar stationary orbit (moving with the solar-system but appears stationary far outside the plane) experience a relative acceleration of time compared to due to a lower net accel/vel? (or less sum forces creating accel/vel?)

2. Sep 13, 2009

### Staff: Mentor

JackAtl:

These questions are a little bit vague, so I'm not sure I'm answering exactly the questions you asked, but I'll give it a try. If the following doesn't help, it might be helpful for you, instead of asking general questions about time dilation, to post a description of a scenario you have in mind and ask more specific questions about what particular observers in that scenario would experience.

First of all, I suggest that you read the Usenet Physics FAQ entry on http://math.ucr.edu/home/baez/physics/Relativity/SR/clock.html" [Broken]. The key point there that I think will help with your questions is: "an accelerating clock will count out its time in such a way that at any one moment, its timing has slowed by a factor $\gamma$ that only depends on its current speed; its acceleration has no effect at all."

Now the above observation, technically, applies only in flat spacetime; whereas I think you're asking about the situation in curved spacetime, specifically the curved spacetime surrounding the Sun, which I'll call a "gravity well" because it sounds nice and sci-fi-ish. However, the clock postulate still has a role to play here too, because it is one of the postulates we need in order to show the following:

If you are at rest in a gravity well, your time dilation factor (the factor by which your clock runs slow, relative to observers who are far out in space away from the gravity well) depends *only* on your "height" in the gravity well: the deeper you are in the gravity well, the slower your clock runs relative to observers far away. Your time dilation factor does *not* depend on how you got to that height in the gravity well.

The reason the clock postulate is relevant to the above is that, in order to be at rest in a gravity well, you have to accelerate: either you have to be in a rocket ship firing its engines downward to avoid falling, or you have to be standing on something (like the surface of a planet) that pushes up on you and keeps you from falling.

It's important to realize that this effect, the slowing of clocks deep in a gravity well, is a *separate* effect than the time dilation in special relativity, where moving clocks appear to run slow relative to clocks at rest. The statement I gave just now applies to clocks which are *at rest* in the gravity well. To get fully correct predictions about how the rates of various clocks will compare when the clocks are moving, you need to take *both* effects into account. (This is what the GPS system needs to do, for example, to figure out how clock rates on the ground relate to the clock rates on its satellites, which are orbiting the Earth, which slows them relative to ground-based clocks, but are also higher in Earth's gravity well, which speeds them up relative to ground-based clocks.)

With all that as preamble, let me try to cobble together answers to your two questions:

(1) If I have two clocks that are both at the same height in a given gravity well, and are both moving at the same speed (i.e., they are at rest relative to one another), they will both run at the same rate.

(2) If I have two clocks, one at rest deep in a gravity well, the other at rest (relative to the first) far outside the gravity well, the clock far outside will run faster than the clock deep inside the well.

Last edited by a moderator: May 4, 2017
3. Sep 13, 2009

### JackAtl

Thanks for the reply! OK- I think I have a better way to explain those two questions. SO apparently I was sort of combining both forms, the one based purely on relative velocity, as well as the dilation based on curvature due to Gravity. I now understand that they are both separate and significant.

1 Is the gravity well at Earth frame composed of all the different interacting gravity field vectors from all the solar bodies? Or is the closest most relevant body, Earth, all that needs to be considered. i.e. Does Jupiter pulling you a bit this way, the sun a bit that, factor in to the net gravity well effect? If true, then doesn't the whole galaxy add to the effect slightly?

2 Is there a decently straightforward way (or Java simulation?) where I can roughly calculate that x height in the solar system gravity well would equate to y dilation factor (2:1, 4:1 etc) between two clocks at rest relative to each other.

Specifically, these apply to two questions/plot issues I'm wrangling with:

One, when traveling between solar systems, at low velocities (50 years a parsec), wouldn't a starship's clocks pass that 50 years much faster than Earth's clocks with the gravity effects- because for the majority of time the "r" from either the originating or destination gravity well is huge for the starship. It's average velocity only a small fraction of c, would the velocity effect slow the ship clock enough relative to Earth to counter it's presence far from a gravity source?

Two, if Earth placed a self sufficient deep space research colony out of the solar plane, at rest relative to the sun, how high in the well would it need to be to experience significant factors of dilation?

Last edited: Sep 13, 2009
4. Sep 13, 2009

### Staff: Mentor

JackAtl:

You appear to be thinking that Earth's gravity well, considered as a source of time dilation, is much stronger than it actually is. It's actually very weak. Even the Sun's gravity well isn't that strong, considered as a source of time dilation. So for an interstellar spaceship, there's almost no difference in clock rates due to being out of the gravity well of either the Earth or the Sun.

Just to give you the numbers: the time dilation factor at a given radius r from the center of a gravity well, relative to observers very far away, is given by

$$\gamma = \sqrt{1 - \frac{2 G M}{c^2 r}}$$

where M is the mass of the source, G is Newton's gravitational constant, and c is the speed of light. In SI units, G = 6.67 x 10^-11, and c^2 = 9 x 10^16. This is a general formula that will work for any gravity well, as long as the objects you are considering are at rest relative to one another.

For the surface of the Earth, M is 5 x 10^24 kg and r is 6.38 x 10^6 m, so

$$\gamma_{Earth} = \sqrt{ 1 - \left( 1.16 * 10^{-9} \right)} \approx 1 - \left( 5.8 * 10^{-10} \right)$$

For the Sun, at the distance of Earth's orbit, M is 2 x 10^30 kg and r is 1.5 x 10^11 m, so

$$\gamma_{Sun} = \sqrt{ 1 - \left( 1.98 * 10^{-8} \right)} \approx 1 - \left( 0.99 * 10^{-8} \right)$$

As you can see, the Sun's effect, at Earth's position, is about ten times the Earth's effect, so the Sun's effect will be the one to use when comparing clock rates on Earth to clock rates far out in interstellar space. But even the Sun's effect is very small, about 1 part in 100 million; that means that, relative to a clock far out in interstellar space, a clock anywhere in Earth's orbit about the Sun will run slow by about 1 second in 3.3 Earth years (since there are 30 million seconds in a year) due to the Sun's gravity well.

Regarding your other question, all gravitating bodies contribute to the field felt by an object at any given point. So, for example, the Earth's motion *is* affected by the other planets as well as the Sun. But for many practical problems, the field of one body is so much stronger than all the others that it's all that needs to be considered. For example, if you're in low Earth orbit, the Earth's field is all that is needed, in practical terms, to determine your motion; the effects of all other gravitating bodies (even the Sun and the Moon) are negligible.

Part of the reason for that is simply that the other bodies are either much smaller in mass, or much farther away, or both. However, there's another reason as well. Suppose you're in orbit about the Earth. Since you and the Earth are almost the same distance away from the Sun, you both will respond with almost identical accelerations to the Sun's field. So even though the *force* exerted by the Sun on you and the Earth might be rather large, the *difference* in the force between the two of you is almost zero. And it's the *difference* in force on you and the Earth that determines what effect the Sun will have on your orbit about the Earth.

Something similar applies to the Earth itself in orbit about the Sun. The other planets do exert a gravitational force on the Earth, but they also exert a force on the Sun; the Sun's trajectory "wiggles" depending on where the various planets are in their orbits. The Earth is much farther from the Sun than you are from the Earth if you're in low Earth orbit, so the differences in the force that Jupiter, say, exerts on the Sun and Earth can still be significant; but they're still very small compared to the Sun's force that keeps the Earth in orbit. Also, the planets' forces in general pull in different directions (they aren't all on the same side of the Earth relative to the Sun), so they cancel each other out to a large extent. (This happens to an even larger extent with the forces exerted by objects in the rest of the galaxy; since they are in all directions, the forces they exert on any object in the solar system pretty much sum to zero.)

5. Sep 13, 2009

### JackAtl

Thanks for the detailed reply. The only question that leaves me with is: What is "very far away"-Where are the assumed observers of that equation located? In a location with no gravity well (as close to none as possible)? Otherwise you have answered my question completely. The effect of time dilation due to these effects is not drastic enough to be used as a tool (or as a cool plot twist :P) Nor does it play a huge factor in inter-stellar travel. Makes me wonder what the implications might be in intergalactic travel. Is a frame in the dead space between galaxies significantly accelerated compared to one in our galaxy? Would you try and use the mass of the galaxy or just the galactic center to compute that... hrrm...

6. Sep 13, 2009

### Staff: Mentor

Technically, "very far away" means "at r = infinity". In practice, "very far away" means "far enough away that the effects of the gravity well are negligible". You can use the formula I gave to figure out, given some appropriate definition of "negligible", in terms of how small the time dilation effect needs to be, how far away "very far away" needs to be.

Of course, there is no place in the universe where there is "no gravity well"--there are gravitating bodies everywhere. But in most situations, that doesn't matter, for the reasons I gave in my last post. For example, if you're out in interstellar space, halfway between the Sun and Alpha Centauri, you're still well within the gravity well of the Milky Way galaxy; but you and the Sun and Alpha Centauri are all practically at the same distance from the galactic center, so its effects cancel out as I explained before, and your position can count as "very far away" compared to the gravity wells of the Sun and Alpha Centauri.

Speaking of the galaxy, we can use the formula I gave to compute the time dilation effect at the Sun's distance from the center, due to the galactic gravity well. You're right in thinking that the fact that the galaxy is spread out, with parts of it further from the center than the Sun, introduces a complication. It turns out that you can approximate by just using the mass of that portion of the galaxy that's closer to the center than the Sun, and pretending it's all at the center. (We don't know the total mass of the galaxy to more than one significant figure anyway, so the answer we get is only going to be within about an order of magnitude. As we'll see, that's enough for our purposes.)

Given that, the numbers for the galaxy are: M = 10^41 kg (about 50 billion times the mass of the Sun--again, this is just to within an order of magnitude or so), r = 3 x 10^20 m (about 30,000 light years). That gives:

$$\gamma_{galaxy} = \sqrt{ 1 - \left( 4.9 * 10^{-7} \right)} \approx 1 - \left( 2.45 * 10^{-7} \right)$$

So the galaxy's time dilation factor is about 25 times that of the Sun--but you have to get out into intergalactic space for it to make any difference. And even that is only a few parts in 10 million--a clock out in deep intergalactic space would gain only about 10 seconds per Earth year compared to a clock moving with the Sun in orbit about the galactic center.