Time dilation formula for constant acceleration on a circular path

[Passionflower]

Time dilation formula for accelerated circular and spherical motion

A traveler goes in an accelerated circular motion of radius r from spacestation A to spacestation B with a constant proper acceleration alpha, e.g. there is also a tangential acceleration.

[PLAIN]http://img204.imageshack.us/img204/4314/presentation1o.gif

What is the formula to express the differential aging when he flies by space station B (the space stations clocks are Einstein synchronized).

And what if he goes around full circle?

We assume that at all times:

$$r < {c \over 2 \sqrt{\pi \alpha}}$$

Of course we are interested in generalizing this idea up to spherical accelerated motion where A goes a round a sphere with a constant tangential acceleration of alpha where the angular velocity is identical in both dimensions.

 

[tiny-tim]

Hi Passionflower!

Time dilation does not depend on acceleration.

SR https://www.physicsforums.com/library.php?do=view_item&itemid=166" depends only on the instantaneous speed …

(and GR time dilation depends only on position, but presumably is the same all the way round your circle)

simply integrate over the whole path.

 

[Passionflower]

To get the maximum dilation for a given alpha, r and time we need to approach r*alpha*time =1.

For r=t=1 (t is coordinate time) we get:

$$1/2\,\sqrt {1-{\alpha}^{2}}+1/2\,{\frac {\arcsin \left( \alpha \right) }{\alpha}}$$

For alpha is 1 we get:
$${1 \over 4}\pi$$

Approximately 0.7853981635 for the traveling clock.

 

[tiny-tim]

what's α?

What happened to v?

 

[Passionflower]

what's α?

What happened to v?

alpha is the proper acceleration tangential to the circular motion.

It appears you misread the problem, we are talking about accelerated circular (and spherical) motion.

 

[tiny-tim]

nooo … you misread the answer …

Time dilation does not depend on acceleration.

 

[John232]

Would an object in free fall experience any time dilation due to it falling?

 

[tiny-tim]

Hi John232!

Yes, SR https://www.physicsforums.com/library.php?do=view_item&itemid=166" of √(1 - v²/c²),

and GR time dilation of √(g₀₀) or approximately 1 - 2gh/c².

 

[pervect]

I think Passionflower worked out a different problem than the one you'd think he worked. Since he didn't include any details of his work, it's hard to be sure.

One thing that's missing from his problem statement, is in fact the initial velocity of the traveller at time zero.

A traveler goes in an accelerated circular motion of radius r from spacestation A to spacestation B with a constant proper acceleration alpha, e.g. there is also a tangential acceleration.

A simple clarification of this missing information in his original post (especially when asked "what happened to the velocity", which one would think would key one to notice that one had omitted to state explicitly what the initial velocity of the traveller was!) would have done a lot to fix the communication problem. Saying "read what I wrote does not clarify the situation, as what he wrote wasn't quite complete.

Also, he states that the proper acceleration is constant - which makes no sense if taken literally.

Both of these issues can be fixed with plausible assumptions. If one assumes that he meant that the MAGNITUDE of the proper acceleration was constant, and also assumes that the velocity of the traveller was initially zero at time zero (in the inertial frame), the problem starts to make sense, given his explicit statement that there is tangential acceleration as well as acceleration towards the center.

So a few minor omissions in his initial problem statement turn into a big argument.

He also seems to be interested in some other problem which he calls "spherical acceleration", I currently have no idea what the difference is, however.

 

[yuiop]

Assuming that Passion is talking about the angular acceleration (e.g initial instantaneous tangential velocity is zero at A and 0.8c at B) then one way to attack the problem is to linearise it and use the relativistic rocket equations http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html as if we were talking about motion in a straight line. The time dilation factors will be correct and it would be easy to adapt the answers back to circular motion. Any "centrifugal" radial acceleration can be ignored as far time dilation is concerned.

Let's try a numerical example. let us say the travel accelerates from zero to 0.8c (magnitude of instantaneous tangential velocity) in ten seconds according to an inertial clock at rest rest with the centre of the system. The equation (using the same symbols as Baez) for the acceleration is (if I have got it right):

$$a = \sqrt{\frac{t^2}{v^2} - \frac{t^2}{c^2}}$$

so using units of c=1 the magnitude of the constant proper tangential acceleration a is 0.133333 c/s.

The elapsed proper time of the traveller is 8.23959 seconds.

The distance traveled by the traveller (in a straight line) is 5 light seconds.

Now if the traveller completes a quarter turn in this time, the radius of the circle must be r = 5*2/pi = 3.183 light seconds.

Now if another traveller travels around a smaller circle (r = 0.79577 light seconds) with exactly the same constant magnitude of tangential acceleration and completes a full circle in 10 seconds according to the clock at the centre (initial and final instantaneous tangential velocity magnitudes of 0 and 0.8c respectively as before) then the elapsed proper time will be the same as in the first case and time dilation is independent of the radius or radial acceleration.

 

[Passionflower]

Let's try an example for circular motion (circular motion is motion around a circle, while spherical motion is obviously motion around a sphere):

Let's say we have a clock going around in a circle in pi seconds according to an inertial clock. The accelerating clock starts from a zero velocity and has a tangential constant proper acceleration of alpha, how many seconds does the moving clock measure for the roundtrip if we maximize the radius of the circle?

Since the roundtrip time in coordinate time is pi seconds we have to solve alpha:

$$2\,\sqrt {{\frac {\pi }{\alpha}}}=\pi$$

Which gives an alpha of:

$${4 \over \pi}$$

The radius of the circle must be smaller than 1/4 (otherwise the traveling clock reaches the speed of light)

Approaching r=1/4 we get an elapsed time of:

$${1 \over 4} \pi^2$$

Or a coordinate over proper time ratio of

$${\pi \over 4}$$

Am I correct?

 

[yuiop]

Since the roundtrip time in coordinate time is pi seconds we have to solve alpha:

$$2\,\sqrt {{\frac {\pi }{\alpha}}}=\pi$$

How did you obtain this equation?

P.S. For constant proper acceleration, you never get to light speed no matter how many revolutions of the circle you complete, no matter how long you maintain the proper acceleration and no matter what the radius of the circle is. You can accelerate with constant proper acceleration (circular or in a straight line) forever without reaching the speed of light.

 

[Passionflower]

How did you obtain this equation?

P.S. For constant proper acceleration, you never get to light speed no matter how many revolutions of the circle you complete, no matter how long you maintain the proper acceleration and no matter what the radius of the circle is. You can accelerate with constant proper acceleration (circular or in a straight line) forever without reaching the speed of light.

It is a good thing we are discussing this topic as it seems that many (including me of course) are not very familiar with this class of problems.

I disagree with you that we can continue forever to undergo constant tangential proper acceleration or that the radius of the circle does not matter as the total velocity depends on both the radial and tangential components.

 

[yuiop]

It is a good thing we are discussing this topic as it seems that many (including me of course) are not very familiar with this class of problems.

I disagree with you that we can continue forever to undergo constant tangential proper acceleration or that the radius of the circle does not matter as the total velocity depends on both the radial and tangential components.

Well just for me, how did you obtain this equation?

As far as I can tell it is essentially:

$$\alpha = \frac{4\pi}{t}$$

where t is the coordinate time to complete one rotation.

[EDIT] Ok I see you are using the equation (assume an initial velocity of zero):

$$\theta = \frac{1}{2} \alpha t^2$$

(See http://en.wikiversity.org/wiki/Angular_acceleration)

where $$\theta$$ is the angular displacement so that:

$$2 \pi = \frac{1}{2} \alpha t^2$$

$$\rightarrow t =2 \sqrt{\frac{\pi}{\alpha}}$$

This is a Newtonian equation and will not work. It would be like using the Newtonian linear equation d = 1/2 at^2 from which you could falsely conclude that the speed of light can be exceeded.

 

[Passionflower]

Well just for me, how did you obtain this equation?

As far as I can tell it is essentially:

$$\alpha = \frac{4\pi}{t}$$

where t is the coordinate time to complete one rotation.

Hmmm, looks like you are correct, it seems I am talking about constant tangential coordinate acceleration as w=at but only for constant tangential coordinate acceleration.

Back to square 1.

 

[yuiop]

Hmmm, looks like you are correct, it seems I am talking about constant tangential coordinate acceleration as w=at but only for constant tangential coordinate acceleration.

Back to square 1.

As I mentioned in post #10 you can work out anything you want about time dilation in the circular case by examining the straight line case for accelerated motion. The radial acceleration makes no contribution at all to the time dilation.

 

[John232]

Hi John232!

Yes, SR https://www.physicsforums.com/library.php?do=view_item&itemid=166" of √(1 - v²/c²),

and GR time dilation of √(g₀₀) or approximately 1 - 2gh/c².

That seems strange because an object in orbit is in constant free fall around the planet. Well, in another post someone mentioned that satillites in orbit maintain the same clock as each other even though they are traveling at a velocity around the planet. I thought if two satallites maintained the same clock relative to each other that maybe they wouldn't dialate due to free fall. Or, is it that they DO have their clocks all run slower if orbiting around a planet at the same speed?

I don't know anyone who works on this stuff but the educated guess here is that they should all observe each others clock as running slow and have to be sync with each other as well as the Earth clock... Or do we just not have to care about what the satallite itself measures other clocks and only what we measure them to be.

I don't see how they would ever get them all in sync if ever satallite measured the other to be something different along with them being measured differently on Earth.

 

[yuiop]

That seems strange because an object in orbit is in constant free fall around the planet. Well, in another post someone mentioned that satillites in orbit maintain the same clock as each other even though they are traveling at a velocity around the planet. I thought if two satallites maintained the same clock relative to each other that maybe they wouldn't dialate due to free fall. Or, is it that they DO have their clocks all run slower if orbiting around a planet at the same speed?

Yes, all the satellite clocks are running at the same rate as each other (if they are orbiting at the same height and speed and direction, i.e. on the same orbital path).

I don't know anyone who works on this stuff but the educated guess here is that they should all observe each others clock as running slow and have to be sync with each other as well as the Earth clock...

No, if the satellites are all on the same orbit as each other then they will appear stationary to each other and they will see neighbouring satellite clocks as running at the same rate as their clocks. However...

Or do we just not have to care about what the satallite itself measures other clocks and only what we measure them to be.

... there might be situations where orbital paths are criss-crossing each other at various angles and then they might see there relative clock rates as different. As you suggest this does not really matter. All we need to know is their altitude and velocity relative to a hypothetical non rotating Earth to work out their clock rates relative to surface based clocks.

 

[John232]

No, if the satellites are all on the same orbit as each other then they will appear stationary to each other and they will see neighbouring satellite clocks as running at the same rate as their clocks. However...

I guess your saying that they would if they orbited in opposite directions at the same orbit?

 

[yuiop]

I guess your saying that they would if they orbited in opposite directions at the same orbit?

Nah, I mean if they all going in the same direction at the same speed on the same orbital path, just staggered behind each other. Some communications satellites are arranged like this. The GPS satellites have (as far as I know) criss crossing paths to give greater global coverage, so this would not apply to them.

 

[tiny-tim]

Hi John232!

(just got up :zzz: …)

Just to concur with and summarise what yuiop has been saying …

Between two satellites on circular orbits of the same radius, only the relative speed matters.

If they're on the same orbit, in the same direction, then their relative speed will be constant, and there will be no https://www.physicsforums.com/library.php?do=view_item&itemid=166" between them (though both will be slower than a clock on Earth, with SR and GR time dilations "in opposite directions").

In any other case, there will be time dilation between them (though that dilation will cancel out once every orbit, as can be seen by comparison with any Earth clock).

 

[Passionflower]

As I mentioned in post #10 you can work out anything you want about time dilation in the circular case by examining the straight line case for accelerated motion. The radial acceleration makes no contribution at all to the time dilation.

So let's go to the 3d case, assume we have an object with a constant proper acceleration (starting at a zero velocity) of equal amounts in the x, y, and z direction, we can make this equivalent to the straight line? There are no Thomas precession like considerations necessary?

 

[yuiop]

So let's go to the 3d case, assume we have an object with a constant proper acceleration (starting at a zero velocity) of equal amounts in the x, y, and z direction, we can make this equivalent to the straight line? There are no Thomas precession like considerations necessary?

Imagine we have a rocket that is initially at rest with an inertial observer which then accelerates off in a random curving path. If we calculate the time dilation of each small interval simply as a function of its instantaneous tangential velocity, then the total elapsed proper time of the rocket is simply the sum of all those infinitesimal intervals, which is normally done using integration.

However, I am not sure how the observer on board the the rocket would determine what component of his proper acceleration is parallel to his instantaneous velocity and which component of his total proper acceleration is orthogonal to his tangential velocity and how Thomas precession would impinge on those calculations. That obviously takes a lot more working out.

P.S. If an observer has equal acceleration in the x, y and z directions, then according to an inertial observer, they would in fact be traveling in a straight line! :tongue2:

 

[tiny-tim]

just came across "Does a clock's acceleration affect its timing rate?" by john baez (answer: no), at http://math.ucr.edu/home/baez/physics/Relativity/SR/clock.html"