Acceleration and special relativity?

How do you work out the time dilation of a clock that is experiencing a constant acceleration?

For example:
If I constantly accelerate a clock at 10 m/s^2
for 25 million seconds from standstill
it will attain a velocity of 250,000 km/ps
and cover a distance of 3,125 billion km.

However, the accelerating clock will not show 25 million seconds have passed. How do I work out the dilation for that clock over that time and distance with that acceleration?

I know d = sqrt(1-v^2/c^2) works for uniform velocities but have no idea how to apply this to accelerating clocks. Or if it actually applies!!! (Hence my question mark in the title).

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Just to let you know, I'm not a qualified scientist. More an interested layman.

thanks in advance for any help, pointers or insights.

 

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The behavior of two coordinate systems in relative motion is the Lorentz transform, for which the time coordinate transforms as

(1) t = (t' - vx/c²) / sqrt(1 - v²/c²)

in which v is the relative velocity and x is the displacement.

If you compare the difference in time for two events by two different observers in relative motion, you will discover that the measured time difference is dilated for each observer by

(2) T = T₀ / sqrt(1 - v²/c²).

where T is the time measured by the stationary observer and T₀ is the time measured by the moving observer. The time T₀ is called the proper time, which is why I will denote it with a naught.

Since you have given in your problem that the body experiences a constant acceleration, I assume that you understand an increasing force will be required as the body continues to gain velocity.

We can then simply accept v(T₀) = aT₀, and substitute that into the special-relativistic time dilation equation. We'll have to do an integral, though, since the velocity is always changing.

The time T measured by the stationary observer can be represented as an integral over the infinitesimal dT. When the moving observer measures dT₀, the stationary observer measures a time

(3) dT = dT₀ / sqrt(1 - a²T₀²/c²)

Integrating both sides of this differential equation yields the equation

(4) T = (c / a) sin^-1 (a T₀ / c)

Plugging in T₀ = 25 x 10⁶ s and a = 10 m/s, we discover that the stationary observer will measure T = 29.5 x 10⁶ s, or a dilation of roughly 18%, over the entire T₀ = 25 million second interval. Near the end of the trip, more than 1.8 seconds elapse on the stationary observer's clock for every second that elapses on the traveller's clock.

edit: and, of course, if you want the stationary clock to read 25 million, and want to know what the moving clock will read, just swap the T and T₀. Doing it the other way around shows that the moving clock would show 22.2 million seconds when the stationary clock shows 25 million seconds.

- Warren

 

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Warren,
Wow! Thank you very much for typing that up for me. I only expected equation (4) not how you got there too.

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However, when I tried it out it didn't work.

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So I sat there and thought about it for a minute (more like ten, ok, I'm slow :bugeye: ) and realised that you had used inverse hyperbolic sin (arcsinh) rather than the inverse sin that you posted.

When I tried it with arcsinh it worked a treat.

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Yes, I am aware that as the clock gets faster it needs more energy to attain the same acceleration. Thanks for mentioning it though.

BTW, how did you do the "-1" when you typed in sin-1?

kind regards
Paul

 

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In what way didn't it work?

Er, no... I meant exactly what I typed. Not arcsinh. If the operation in (3) was addition, it would end up as an arcsinh. Since the operation in (3) was substraction, it ends up as an arcsin.

Put < sup > < /sup > around stuff you want to be shown as a superscript.

- Warren

 

Hi Warren,
Now you have me confused again. I'm no expert mathematician so please forgive me if I'm barking up the wrong tree here.

If I plot the velocity of the clock against my time I get a straight line, but if I plot the velocity of the clock against the clocks time then it would surely describe a curved line?

Again, let me stress I am not an expert mathematician however, I'm pretty sure that sin and sin^-1 are for straight lines, and that for curves you use sinh and arcsinh.

Well, using your equation (NB. I'm using an approximation of c = 3x10⁸) where a=10, T₀=25x10⁶

T = 1693280707.14 = (c / a) sin^-1 (a T₀ / c)

where you made T=29.5x10⁶

Aaaaaah. Here we have the problem, I believe, and why we have disagreement. You wrote sin^-1 and meant sinh.

sin^-1 is the inverse sin, and sinh is the hyperbolic sin (sin and arcsinh for their respective "opposites").

Thanks for the info. on superscript.

kind regards
Paul

 

Yes, it does.

You don't know what you're talking about. The integral in (3) is an inverse sine.

No. (3 x 10⁸ / 10) sin^-1 (10 * 25 x 10⁶ / 3 x 10⁸) = 29.5 x 10⁶, exactly as I wrote it. Perhaps you are doing the arithmetic wrong.

NO. I WROTE SIN<SUP>-1</SUP> AND MEANT SIN<SUP>-1</SUP>, THE INVERSE SINE, FOR THE LAST TIME.

- Warren

 

Then I suggest a new line of work, because §lîñk€¥™ is right.

The accelerating clock will show 22.75 million seconds elapsed, 91% of the time elapsed on the stationary clock. The Relativistic Rocket

 

Zanket, thank you for confirming my suspicion and for the excellent link. I made it 22.75 million seconds too.

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Warren. We all make small mistakes from time to time. No hard feelings.

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kind regards
Paul

 

Excuse me?

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I derived my results by considering the proper time to be that measured by the moving observer. The person in your link derived his results by assuming that the proper time is that measured by the stationary observer.

The two are related through a Lorentz transform. If you take my assertion in (2) and apply a Lorentz transform to it to switch coordinate systems, you will change the sign of the terms under the square root from a minus to a plus. You will then integrate (3) and arrive at a result with a sinh.

As given, with the assumptions I made, the result (4) is correct -- an inverse sine.

- Warren

 

You are SERIOUSLY out of line.

- Warren

 

§lîñk€¥™ - My pleasure and welcome to sciforums.

 

Warren,
what is the inverse of the hyperbolic sin?

No, I'm a curved line. We are related by sinh

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kind regards
Paul

 

Look it up. While you're at it, look at an integral table, find form (3), and note that its integral is (4)... for future reference.

- Warren

 

Warren,

Firstly, I did look it up - arcsinh is the inverse of sinh.

Secondly, I also looked at the link that Zanket gave. They use sinh and arcsinh too.

Thirdly, using sin^-1 gives the wrong answer.

Fourthly, that's the last I will say on the matter lest this turn into a needless and meaningless mudsling.

kind regards
Paul

 

:applause:

They derived their formulae under the coordinate system of the stationary observer. I believe I explained this. This doesn't mean that (4) is wrong.

You're asking the wrong question.

Did you look up the integral I asked you to look up? Mm?

- Warren

 

Maybe you should take a look at the page again, because they give the formula for the proper motion too. Guess what? They use sinh not sin^-1.

Secondly, you agree that they would use arcsinh for the stationary observer, and that the inverse of sinh is arcsinh.

Yet, you persist in saying you're right and it is sin^-1?

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Ok, so I guess it is really pointless talking to you any further.

This really is my final post on this.

kind regards.
Paul

 

all are welcome

 

Actually, I'm becoming a bit confused myself. I was 99% sure a Lorentz coordinate change would allow me to get (3) into a form with a + under the square root, and would be able to use hyperbolics. I tired it, but it didn't work out....

There's nothing wrong with my step (3) -> (4). The integral of (3) CERTAINLY is (4). I've used Maple and the integral table in Thomas & Finney to verify it. There is no question.

Now... there's nothing wrong with my step (2) -> (3) either.

This means something is wrong with my step (2). This is the usual form for the time dilation between two coordinate systems, but obviously something is squirrely.

I'm going to keep looking around. I know that my answer and the answer provided in the Usenet Physics FAQ are different; now I'm just trying to do damage control to figure out why. The difference in the normal trig and hyperbolic trig functions is just the sign of the operation under the square root....

- Warren

 

Warren,
Rock on. Happy to see that you are a reasonable guy.

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thank you
kind regards
Paul

 

I'm a resonable guy. The bottom line, though, is that if (2) is correct, then (4) is correct too... without question. It just really got my goat when you and zanket were trying to tell me the integral of (3) is not (4) -- when textbooks and CA systems agree with me.

So something's wrong with (2)..

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- Warren