0
$\begingroup$

A particle undergoing constant proper acceleration $\alpha$ will appear to have a hyperbolic worldline in the Minkowski spacetime $(ct,x)$ of an inertial frame given by: $$x=\frac{c}{\alpha}(\sqrt{c^2+\alpha^2t^2}-c)$$ Clearly, this shows that the particle appears to travel arbitrarily far to the inertial observer since $$\lim_{t\to+\infty}\frac{c}{\alpha}(\sqrt{c^2+\alpha^2t^2}-c)=+\infty$$ However, I have just seen a calculation that purports to demonstrate that the accelerating observer doesn't think they've actually gone that far. The calculation says that the distance $x'$ that the accelerated observer thinks they've travelled is just found through length contraction, so that: $$x'=\frac{x}{\gamma}$$ and using this, it can be shown that $x'\to c^2/\alpha$ is bounded. My question is what does this have anything to do with length contraction? I'm just not getting why we can casually use the above formula like that.

$\endgroup$
9
  • 1
    $\begingroup$ In my opinion, it's a good idea to provide a reference to things that one has seen or read somewhere. It provides a useful context. $\endgroup$
    – robphy
    Commented Jun 15, 2022 at 13:44
  • $\begingroup$ It's on pages $36$ and $37$ of this pdf: damtp.cam.ac.uk/user/tong/relativity/seven.pdf which correspond to the numbered pages $142$ and $143$. $\endgroup$ Commented Jun 15, 2022 at 19:38
  • $\begingroup$ "Of course, this observer is not in an inertial frame, but at any time t we can consider the inertial frame that is momentarily at rest with respect to the accelerated particle. This allows us to simply use the Lorentz contraction formula." What don't you understand of this explanation given in the paper? $\endgroup$
    – Mattia
    Commented Jun 16, 2022 at 17:58
  • $\begingroup$ Another notable quantity for characterizing "the path of the accelerating observer" would be (calculated by the fomula) $$ \int {\rm d}x' := \int \frac{1}{\gamma} \, {\rm d}x := \int \frac{1}{\gamma} \, \frac{\rm d}{{\rm d}t}\big[ \, x \, \big] \, {\rm d}t := \int \frac{\alpha \, t}{1 + (\alpha^2 / c^2) \, t^2} \, {\rm d}t= \frac{c^2}{\alpha} \, \text{Ln}\left[ \, \sqrt{1 + (\alpha^2 / c^2) \, t^2} \, \right]$$ (which diverges, as the duration $t$ diverges). $\endgroup$
    – user12262
    Commented Jun 17, 2022 at 20:28
  • $\begingroup$ @Mattia: "[...] consider the inertial frame that is momentarily at rest wrt. the accelerated particle." -- Arguable better understandable is to call and identify such a relevant inertial frame as "momentarily co-moving wrt. the accelerated particle". After all, a (constantly, uniformly, hyperbolically) accelerating particle is never itself a member of any inertial frame and thereby "at rest wrt." other members of an inertial frame in the sense those members are at rest wrt. each other; nor is the accelerating particle ever "individually at rest" a.k.a. "inertial". $\endgroup$
    – user12262
    Commented Jun 17, 2022 at 20:32

2 Answers 2

0
$\begingroup$

Let's say we line up million marbles in empty space, distance between two consecutive marbles being 1 light year. We call this line of marbles a "ruler".

Now when observer is at rest relative to the ruler, he observes the length of the ruler to be million ly.

When observer moves at speed 0.87 relative to the ruler, he observes the length of the ruler to be half million ly.

When observer accelerates from zero to speed 0.87 relative to the ruler, he observes the length of the ruler to change by half million ly.

When observer accelerates so that the gamma of the ruler goes from 10 to 20, according to said observer, he observes the length of the ruler to go from 1/10 million ly to 1/20 million ly.

This contracting "ruler" must not be used to measure distances by the observer, maybe the observer carries a good ruler with himself, so that he can do valid distance measurements.

Now if observer starts acceleration from one end of the ruler, which we call "start", at first the observer observes his distance from the start increasing, but after some time he observes that his distance to the start is decreasing.

Because when the "ruler" has become very short, according to the observer, and observer has not passed the end of the "ruler" yet, then the start of the "ruler" must be close to the observer, according to the observer.

$\endgroup$
0
$\begingroup$

That calculation of traveled distance consist of two parts:

  1. calculate the traveled distance in the launchpad frame
  2. transform the result to the traveler frame

The second part has nothing to do with acceleration, it's a basic transformation of length.

The traveling person agrees that the calculated distance is his traveled distance.

The method to visit a distant place is:

  1. go to a frame where the place is a nearby place
  2. travel the short distance
  3. go back to the frame where the place was a distant place
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.