How to Calculate Proper Time

1. Jan 18, 2010

jacksonwalter

Is this the correct way to calculate the proper time that has passed for an object under constant acceleration over the time interval [0,t1]? I just downloaded and started teaching myself how to use LaTex today so bear with me here. I can get the output PDF file but I don't know how to insert everything into the thread here, so I just attached it, and https://docs.google.com/fileview?id=0BxjDgugfV3GMYTdhNjI5OWYtNjUyZi00MzcwLTlkNjUtOWQ1NDU0NjNkYzFj&hl=en" [Broken] a link to it in Google Docs.

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Last edited by a moderator: May 4, 2017
2. Jan 18, 2010

George Jones

Staff Emeritus
Careful; it is not true that $v = at$, as this would give $v > c$ after sufficient time.

3. Jan 18, 2010

jacksonwalter

No, I don't think so because it's going to take more and more energy to maintain that constant acceleration, and it would take an infinite amount of energy to go past c. I'm assuming that the thing accelerating is increasing it's power in order to maintain constant acceleration.

4. Jan 18, 2010

atyy

I didn't check, but this site is usually very good:
http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]
"The proper time as measured by the crew of the rocket (i.e. how much they age) will be denoted by T, and the time as measured in the non-accelerating frame of reference in which they started (e.g. Earth) will be denoted by t."

Last edited by a moderator: May 4, 2017
5. Jan 18, 2010

Staff: Mentor

George is correct. In your formula for t>c/a you get FTL travel. Your expression is wrong at that point, but up until then it should be fine. It would be constant coordinate acceleration, not constant proper acceleration. Also, at t>c/a you get imaginary results for your proper time which would clue you in that there is a problem.

6. Jan 18, 2010

George Jones

Staff Emeritus
And consequently, the acceleration as measured by an accelerometer (e.g., bathroom scales) carried by the object would not be constant.

7. Jan 18, 2010

jacksonwalter

Yea, that website covers pretty much everything. I'm curious where the hyperbolic trig functions came from, though.

So, if someone could check my calculation and also tell me how to insert things from LaTex that would be awesome.

8. Jan 18, 2010

Staff: Mentor

They come from the assumption of constant proper acceleration instead of your assumption of constant coordinate acceleration.

9. Jan 19, 2010

MikeLizzi

I just did a quick scan of that reference and it looks pretty awkward. The deriviation starts with the rocket having a constant acceleration wrt the inertial observer. That would mean an ever increasing acceleration wrt the people in the rocket. Not realistic.

The derivation I am familiar with starts with a constant acceleration wrt the people in the rocket. That gives an ever diminishing acceleration wrt the inertial observer.

Last edited by a moderator: May 4, 2017
10. Jan 19, 2010

Staff: Mentor

Hi MikeLizzi, I think you misread the Baez page. The linked derivation uses constant proper acceleration. Jacksonwalter's derivation uses constant coordinate acceleration.

11. Jan 19, 2010

Mentz114

MikeLizzi :
That being the case can we deduce that the interial observer will always be able to see the accelerating rocket, forever approaching c but not getting there. While the rocket will eventually lose the inertial observer behind an horizon ?

12. Jan 19, 2010

George Jones

Staff Emeritus
Look at section 2.10, which starts on page 35, of Gron and Hervik,

If you want to insert Latex in a line of prose, surround your LateX code by the the tags itex and \itex, where the tags are inside square brackets. For example,

Code (Text):
The energy of an individual photon, itex E = \hbar \omega /itex is used to derive
produces:

The energy of an individual photon, $E = \hbar \omega$, is used to derive

If you want to insert Latex in a stand-alone line, surround your LateX code by the the tags tex and \tex, where the tags are inside square brackets. For example,

Code (Text):

The energy of an individual photon,

tex E = \hbar \omega, /tex

is used to derive

produces:

The energy of an individual photon,

$$E = \hbar \omega,$$

is used to derive

tex, \tex, itex, \itex all should be surrounded by square bracket, but if I did this in my examples, the actual Latex would not appear.

13. Jan 19, 2010

MikeLizzi

Don't know, never extrapolatedd the transformation out that far. Good question for somebody to answer.

Last edited: Jan 19, 2010
14. Jan 19, 2010

MikeLizzi

You are right. I scanned it to fast. So the derivation must be the same as the one I'm famliar with. It's just that I can't read the format of the equations. I guess the author didn't know Latex either.

added a few minutes later: Ok, I can now see that the equations are the same. My bad.

Last edited: Jan 19, 2010
15. Jan 19, 2010

George Jones

Staff Emeritus
This depends on how things are set up, but, in the usual set-up, this is close to what happens. In this set-up, the inertial observer does cross the horizon, but, similarly to what happens with a black hole event horizon, the the rocket observer never quite sees the inertial observer cross the horizon. The rocket observer see the inertial observer more and more redshifted. See

Originally, this was written for an internet discussion group in 1996, and LateX was not implemented by this discussion group.

16. Jan 19, 2010

jacksonwalter

Yea, it only works for t<c/a, but I only wanted it to work on the interval (0,t), as that's the interval where it wouldn't require an infinite amount of energy. I think the other guy's derivation is better though.

Thanks for the responses, and thanks for the LaTex help George.

Last question: So the difference between proper acceleration and proper time is similar to the difference between between proper time and 'coordinate' time? Could I say proper acceleration $$= \frac{d^{2}x}{dT}$$ if T = proper time? i.e proper acceleration is the acceleration in the ship's reference frame and and coordinate acceleration is the acceleration as seen from earth?

17. Jan 19, 2010

DrGreg

Actually it's a bit more complicated than that.

You might find my derivation within the thread Questions about acceleration in SR helpful, in posts #13, #14, #15 of that thread and a correction in post #28 to equation (5).

In fact, the proper acceleration is given by

$$\frac{d^2x}{dt^2} \left( \frac{dt}{d\tau} \right) ^3$$​

which follows from equations (2) and (12) in my derivation. $dt/d\tau$ is the Lorentz factor $\gamma = 1 / \sqrt{(1 - v^2 / c^2)}$.

18. Jan 19, 2010

DrGreg

By the way, I recently found out there's a little-publicised way of getting the characters "[noparse][tex][/noparse]" to appear in a post.

[noparse][noparse][tex][/noparse][/noparse]

19. Jan 19, 2010

jacksonwalter

Yea, it sucks having not quite taken a formal course in SR + GR yet, just a lot of reading and hand waving. Thanks though, lot of good insight.