# Circling spaceship

1. May 14, 2009

### ibc

If there's a spaceship going in a circle (constant speed), and I want to know the time elapsed for me and for the spaceship (say, when it returns after a complete cycle).

So, the ship is always traveling at the same speed v, therefore each time I can go to a certain inertial reference frame and say dt' = dt*(gamma), and gamma always stays the same. so if I integrate it I get that the total time in one system is gamma times the time in the other one.
However, for the ship to travel at constant speed in a circle, it must have a constant acceleration, and I'd expect it have some influence on the calculation (for example, by the previous one I get that the time for the circling ship is the same as I'd get for a ship going in a constand velocity)

But I don't see what's wrong at the first calculation, or how to add the effect of acceleration if it's speed is not changing.

thanks
ibc

2. May 14, 2009

### JesseM

The amount of time dilation at each moment depends only on speed in whatever inertial frame you're calculating it in, not acceleration, so your first calculation was right.

3. May 14, 2009

### ibc

This doesn't make any sense, I know that if a ship "sits" above me in a gravitational field, I'll see its clocks different than mine, yet if I go back to the inertial frame calculation, each time the inertial frame is at the same speed, which is zero. therefore by that calculation there won't be any time differences between me and the ship above, but I know there will be, and I know that a constant acceleration acts the same.

4. May 14, 2009

### DrGreg

What Jesse said is true when you, the observer, are inertial and you are measuring another object, regardless of whether the object you are measuring is inertial or not.

It gets more complicated if you, the observer, are not inertial, i.e. accelerating (or, to be pedantic, properly-accelerating), whatever you are measuring. As well as the usual dilation due to relative motion, there is an additional "gravitational" dilation to consider --remembering that acceleration is equivalent to gravity.

5. May 14, 2009

### JesseM

What do you mean "the inertial frame is at the same speed"? I was talking about the speed of the circling ship in the inertial frame, not the speed of the inertial frame itself (what would you be taking the speed of the inertial frame relative to? Itself?)
You're talking about a ship at rest in some frame and experiencing gravitational time dilation? Gravity involves the curved spacetime of general relativity, whereas "inertial frames" are defined in the uncurved gravity-free spacetime of special relativity. In general relativity it's impossible for any coordinate system defined over a large region of curved spacetime to be "inertial", although the http://www.aei.mpg.de/einsteinOnline/en/spotlights/equivalence_principle/index.html [Broken] says that a freefalling observer can have a "locally inertial" coordinate system in an infinitesimally small region of spacetime around an event on their worldline, and that in this locally inertial coordinate system the laws of physics will look the same as in an inertial one in SR.

It is actually possible to have a "pseudo-gravitational field" in uncurved spacetime, as discussed in this section of the twin paradox page. This is what's seen in a non-inertial coordinate system where an accelerating observer is at rest (the laws of physics don't obey the same equations in non-inertial systems as they do in inertial ones; in particular, the velocity-based time dilation equation that is used in inertial frames cannot be assumed to work in non-inertial ones). Here, a variation on the equivalence principle says that the "gravitational" time dilation seen by clocks at different positions in the non-inertial Rindler coordinate system is equivalent to the time dilation between clocks undergoing Born rigid acceleration in inertial coordinates. But from the perspective of the inertial frame, the time dilation is explained entirely in terms of the velocities of the the different accelerating clocks--the key is that in order to satisfy the condition of "Born rigidity" the clocks must accelerate at different rates in the inertial frame. Born rigidity basically means that the distance between the clocks in either clock's instantaneous inertial rest frame remains constant from one moment to another (even though each clock will have a different instantaneous inertial rest frame at each moment)...the page on Rindler coordinates above explains it like this:
It also shows a diagram with the worldlines of different members of the flotilla drawn from the perspective of an inertial frame, you can see that their velocities are different at any given moment (shown by the slope of each worldline at a given value of t):

Last edited by a moderator: May 4, 2017
6. May 14, 2009

### Staff: Mentor

This is correct, in an inertial frame only the speed matters, not acceleration or higher derivatives. This is known as the http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html#Clock_Hypothesis" and has been experimentally validated for accelerations up to 10^18 g.

Last edited by a moderator: Apr 24, 2017
7. May 15, 2009

### ibc

I mean that in order to measure the time dilation, I go each time to an inertial frame going at the velocity the ship is going at that moment, and measure another dt, and because the ship is always at the same speed for me, the inertial reference frame which I'll measure through will always have the same speed, therefore the small time dilation will be the same.

And the problem comes because the ship is constantly accelerating therefore should experience acceleration/gravity phenomenon as mentioned.

So what you guys are saying is that there won't be any gravity/acceleration phenomenon because I am not accelerating/at a gravity field, it doesn't matter what the ship is doing, and so the total time dilation would simply be t'=t*gamma?

8. May 15, 2009

### JesseM

I don't understand what you mean. You don't calculate the time dilation of the ship in your rest frame by looking at the frame where the ship is at rest, you use the velocity of the ship in your own rest frame. For each dt in your frame, since the ship was moving at velocity v in your frame, the ship's clock only ticks forward by dt*sqrt(1 - v^2/c^2).
Right. More generally, if you have a clock which is moving with some known speed as a function of time v(t) relative to some specific inertial frame, and you want to know how much time will elapse on the clock between two events on the clocks worldline that occur at time coordinates t0 and t1 in this inertial frame, you'd look at the integral $$\int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt$$. In the case where v(t) is constant in the inertial frame you're using, this reduces to (t1 - t0)*sqrt(1 - v^2/c^2).