1. Mar 3, 2006

### ubavontuba

As we know, the twin paradox states that an astronaut that accelerates away from the Earth, stops, turns around, and then accelerates back to the Earth ages less than his twin that remains behind (on Earth). How about in the case where the acceleration isn't linear?

Let's say our astronaut accelerates away from the Earth and immediately begins a long arc in a path that returns him to the earth in a circular trajectory. Does he age the same as if he went linearly to the farthest point of the arc?

How do you reconcile the fact that he is both, always accelerating away from the Earth and accelerating toward it at the same time? (he never has to stop and turn around)

P.S. I know that on the face of it this seems similar to the closed universe version, but consider that space-time is flat in this universe. Are the effects the same?

Last edited: Mar 3, 2006
2. Mar 3, 2006

### JesseM

Probably not, but it depends on the details of his speed and the total time of both paths. If you assume his speed is constant in the circular case, and that his speed is also constant on the inbound and outbound legs of the straight-line paths (and that the acceleration is negligibly brief in this case), then in both cases the total time elapsed on the travelling clock will be $$( t_1 - t_0) \sqrt{1 - s^2/c^2}$$ where s is the speed, t0 is the time they depart and t1 is the time they reunite as seen in the earth-frame (or whichever inertial frame you use to analyze the problem, you'll get the same answer no matter what).
In both relativity and newtonian physics, when you travel in a circle at constant speed your acceleration vector always points in the direction of the center of the circle. So when he started his acceleration vector would point away from earth, at the farthest point it would point back towards the earth, and between those times it would be rotating at a constant rate.

3. Mar 3, 2006

### ubavontuba

You're suggesting an axis that is being crossed by the astronaut and not by the Earth, right? To the astronaut's perception wouldn't the Earth be accelerating symmetrically to his acceleration and therefore wouldn't the Earth also pass behind this axis (relative to the astronaut)?

Last edited: Mar 3, 2006
4. Mar 3, 2006

### ubavontuba

JesseM,

I'm not sure you caught my edit in the first post. Here it is again, just in case:
Anyway, I had another thought. In your model, wouldn't the inner planets in our solar system be aging slower than the outer planets?

5. Mar 3, 2006

### yogi

Einstein considered this in his 1905 paper - as long as the tangential velocity is constant, it doesn't make any difference what polygonal path the traveler takes - he can fly directly to a distant object, quickly turn around and return, or he can fly a curved path that embraces a continuous centripetal acceleration. The time difference upon return will depend upon the length of the path and the velocity

Planets closer to the Sun are in a greater gravitational potential well - their clocks will run slower than the outer planets primarily due to the G potential - slowing of time due to the their velocity is less and the gravitational effect.

Last edited: Mar 3, 2006
6. Mar 3, 2006

### JesseM

The astronaut is not in an inertial reference frame, so you can't assume any symmetry here--in particular, if you come up with a coordinate system for the astronaut and figure out the earth's velocity v(t) in this coordinate system, you can't assume that the rate that the earth's clock is ticking in this coordinate system is $$\sqrt{1 - v(t)^2 /c^2}$$ the rate of the astronaut's clock at any given moment. Also, there is no "standard" way to define the coordinate system of a non-inertial observer, so there isn't any one standard answer to what things would look like "from his perspective", assuming the phrase "what things would look like" means what is happening in a coordinate system where he is at rest as opposed to what he sees using light-signals.

7. Mar 3, 2006

### JesseM

Spacetime is also flat in the standard version of the closed universe twin paradox--it is only its topology that makes it closed, not its curvature like in the usual big bang cosmological models.
As yogi points out, you also have to consider gravitational time dilation...but if you ignore this, then yes, the inner planets are aging slightly slower than the outer ones. The effect would be pretty tiny, though.

8. Mar 3, 2006

### ubavontuba

JesseM and Yogi,

Thanks. This has been very interesting.

9. Mar 4, 2006

### pervect

Staff Emeritus
On the question of the orbiting satellites: there is a very elegant way to find the time dilation vs height of the satellite by use of the virial theorem.

https://www.physicsforums.com/showpost.php?p=920092&postcount=117

one can express the approximate elapsed proper time for a clock following a geodesic trajectory governed by a potential function V(r) in geometric units as:

$$\frac{\tau}{\tau_s} = 1 + \bar{V} - \bar{T}$$

This is an approximation that's good to the first order only (but because the corrections are so small, a first order approximation is good enough for most purposes).

Here $\tau$ is the elapsed proper time of the moving clock

$\tau_s$ is the proper time of the stationary clock at a constant potential V=0

$\bar{V}$ is the average over time of the potential energy of the moving clock

$\bar{T}$ is the average over time of the kinetic energy of the moving clock.

Now the virial theorem relates average over time of the kinetic energy to average potential energy, allowing us to simplify this equation. The usual form of the virial theorem assumes that

V(r) = a r^n+1

where n = -2 for an inverse square law force like gravity.

In this application though, we wish to write

V(r) = -M/r + M/r_0

so that we are free to set V(r) = 0 at the surface of the Earth at $r=r_0$, as we wish to use the clock at the surface of the Earth for the reference. The modified form of the virial theorem then gives us

$$\bar{T} = \frac{\int (M/2r) dt} { \int dt}$$

Note that this is a function of r(t) but not of r_0, as we would expect.

Let us assume that r(t) is a constant, i.e. we have a clock in a circular orbit. Then the process of finding the time average becomes simple, and we simply write

$$\bar{T} = \frac{M}{2r}$$

Then total expression for time dilation is just

$$\frac{\tau}{\tau_s} = 1 + \bar{V} - \bar{T} = 1 - \frac{3M}{2r} + \frac{M}{r_0}$$

If one wishes to use non-geometric units, putting the factors of G and c back into place gives the following useful result:

$$\frac{\tau}{\tau_s} = 1 - \frac{3GM}{2rc^2} + \frac{GM}{r_0 c^2}$$

Here r is the height of the orbiting clock (assumed to be in a circular orbit) and $r_0$ is the height of the observer (the radius of the Earth).

Thus we can see that a clock orbiting directly at the surface of the Earth would experience only the slowing due to it's velocity.

At a height of 1.5 Earth radii, when $r = 3 r_0/2$, the two effects cancel, and the orbiting clock keeps the same time as the stationary clock.

For $r > 3 r_0 / 2$, the orbiting clock will tick faster.

At r=infinity, the time acceleration is just 1 + GM/c^2 r_0, as one would expect, i.e. the Earth clock ticks slow by 1 - GM/c^2 r_0. (Note that 1/(1-x) $\approx$ 1+ x when x is very small)

Last edited: Mar 4, 2006
10. Mar 6, 2006

### yogi

Most interesting pervect - another formalism that reveals a connection between time (clock rate) and the potential/kinetic energy state.