Calculating Time Elapsed in Circling Spaceship

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In summary, the time dilation at each moment depends only on speed in the inertial frame, not acceleration. However, if the observer is not inertial, there may be additional "gravitational" time dilation to consider. In a non-inertial frame, the time dilation between clocks can be explained by differences in velocity, but in order to satisfy the condition of "Born rigidity", the clocks must accelerate at different rates in the inertial frame.
  • #1

ibc

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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
 
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  • #2
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
JesseM said:
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.

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
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
ibc said:
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.
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?)
ibc said:
therefore by that calculation there won't be any time differences between me and the ship above
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:
We can imagine a flotilla of spaceships, each remaining at a fixed value of s by accelerating at 1/s. In principle, these ships could be physically connected together by ladders, allowing passengers to move between them. Although each ship would have a different proper acceleration, the spacing between them would remain constant as far as each of them was concerned.
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):

Coords.gif
 
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  • #6
ibc said:
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)
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.
 
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  • #7
JesseM said:
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?)
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
ibc said:
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.
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).
ibc said:
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?
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 [tex]\int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt[/tex]. 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).
 

1. How do you calculate time elapsed in a circling spaceship?

In order to calculate time elapsed in a circling spaceship, you will need to know the radius of the spaceship's orbit and the speed at which it is traveling. Once you have these values, you can use the formula t = 2πr/v, where t is the time elapsed, r is the radius, and v is the speed. This formula assumes that the spaceship is moving at a constant speed.

2. Does the size of the spaceship affect the time elapsed?

Yes, the size of the spaceship does affect the time elapsed. The larger the spaceship, the longer it will take to complete one orbit and therefore the longer the time elapsed will be. This is because the larger spaceship will have a larger circumference and therefore will have to travel a greater distance to complete one orbit.

3. How does the speed of the spaceship affect the time elapsed?

The speed of the spaceship directly affects the time elapsed. The faster the spaceship is traveling, the shorter the time elapsed will be. This is because the faster the spaceship is moving, the smaller its circumference will be and therefore it will take less time to complete one orbit.

4. Can the time elapsed be calculated if the spaceship is not moving at a constant speed?

No, the formula for calculating time elapsed in a circling spaceship only works if the spaceship is moving at a constant speed. If the speed is changing, the calculation becomes more complex and will require additional information such as the acceleration of the spaceship.

5. How does the gravitational force of a planet affect the time elapsed in a circling spaceship?

The gravitational force of a planet can affect the time elapsed in a circling spaceship by altering the speed at which the spaceship is traveling. The stronger the gravitational force, the faster the spaceship will need to travel to maintain its orbit, resulting in a shorter time elapsed. Additionally, the gravitational force can also affect the shape and size of the spaceship's orbit, which can also impact the time elapsed.

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