Relativity - Oh dear, here we go again!

[Ziggurat]

OK, let’s try this if it hasn’t been done all ready.

Not explicitly, but the key difference here is the distinction between what you see and what you observe, and if you understand that distinction, these questions are easy.

We have two spacemen and three clocks and they are big clocks that can be seen from a 100 feet with no problem. The speed of light for this experiment is 100 ft. per second. They are all somewhere in deep space, who cares where. The clocks are all reading the same time.

I am going to assume furthermore that the clocks and the spacemen also start at the same place and at rest with respect to some specified reference frame (which I'll refer to as the start frame).

1. Now one of the spacemen takes one of the clocks and travels at 99.9999% speed of light for one second and stops and looks back at the other spaceman and the two clocks, he is 100 ft away.

The one second time must be refering to the time in the start frame, not the time he experiences. I will refer to this spaceman as spaceman 1, and his clock as clock 1.

Is the time on his clock is the same as the other two clocks.
Is the time on his clock is one second behind the other two clocks.
Is the time on his clock is two seconds behind the other two clocks.
Is the time on his clock is one second ahead of the other two clocks.
Is the time on his clock is two seconds ahead of the other two clocks.

The time on clock 1 is observed to be different than the other two clocks. The proper time he experiences while moving with respect to the start frame is very short, so clock 1 will be observed to be almost 1 second behind the other clocks. However, spaceman 1 will see the other clocks as being about the same time as clock 1, while spaceman 2 will see clock 1 being two seconds behind the other clocks, since there's an approximately 1 second delay for the signal to get transmitted from one location to another. They don't see the same thing, but they both observe the same thing: clock 1 is ~1 second behind the other clocks.

2. Now the other spaceman takes one of the two remaining clocks

I'll call this spaceman 2 and clock 2, with the clock left behind being clock 3.

and travels very slowly, about one inch a second to the other spaceman.

You can go slower than that, the point is that we take the limiting behavior of slow travel time.

When he gets to the other spaceman he looks at the other spaceman’s clock.

Clock 1 and clock 2 are now in the same location. This is important.

Is the time on his clock is the same as the other spaceman’s clock.
Is the time on his clock is one second behind the other spaceman’s clock.
Is the time on his clock is two seconds behind the other spaceman’s clock.
Is the time on his clock is one second ahead of the other spaceman’s clock.
Is the time on his clock is two seconds ahead of the other spaceman’s clock.

Clock 1 will be observed to be about 1 second behind clock 2, as before. Because they are now in the same location, all observers will also see this same ~1 second difference, since any time delay in signal propagation to any location is the same from both clocks.

3. Now the other spaceman looks back at the other clock left behind.

Is the time on his clock is the same as the other clock.
Is the time on his clock is one second behind the other clock.
Is the time on his clock is two seconds behind the other clock.
Is the time on his clock is one second ahead of the other clock.
Is the time on his clock is two seconds ahead of the other clock.

Spaceman 1 and 2 both observe that clock 2 is reading approximately the same time as clock 3. Spaceman 1 and 2 both see clock 2 reading approximately 1 second ahead of clock 3 because of the time delay for the signal from clock 3 to reach them.

 

[Thabiguy]

When the two twins meet, they do have different ages, but it is not because of SR and its time dilation. The effect is a result of GR: the change occurs when the traveling twin decelerates, changes direction and accelerates back towards the earth.

Not quite. The twin paradox is indeed because of SR and time dilation, and nothing else. GR needn't be invoked at all.
Explanation.

 

[ponderingturtle]

The rock only exists in its own rest frame and falls in a straight line toward the force of gravity when dropped. That it is perceived to be moving on an angle when observed from another rest frame is an illusion. To use the movie analogy used earlier, the single event is being separated in to a strip of many events and this creates the illusion.

Is it me, or does this seem to disprove orbital mechanics?

 

[Gertrude]

Not quite. The twin paradox is indeed because of SR and time dilation, and nothing else. GR needn't be invoked at all.
Explanation.

Actually there had been a debate among writers in the late 50's regarding the need to introduce GR to explain the twin paradox but nowadays, most physicists agree that SR is sufficient. And yes, SR can deal with acceleration. But we must be clear. SR deals only with acceleration as measured in inertial reference frames. This is not quite the case in the twin paradox, where one of the twins himself undergoes acceleration. In this sense, it is a problem of GR. We are appealing to the reality of this acceleration to explain the paradox.

The reason most writers agree that SR is sufficient in explaining the paradox is that we can use methods such as the Doppler effect (ie. pretend the twins send each other signals and count these) as described by SR and obtain the same result. But this is not to say that SR can handle acceleration as experienced by an observer.

 

[Ziggurat]

Do not confuse this with the twin paradox. When the two twins meet, they do have different ages, but it is not because of SR and its time dilation. The effect is a result of GR: the change occurs when the traveling twin decelerates, changes direction and accelerates back towards the earth. The twin paradox imho is the worst example in teaching SR.

This is simply wrong. Let me expand upon Thai's response: special relativity can accomodate acceleration. It would be a crap theory if it couldn't. The math can get ugly and so most introductory treatments don't address it, but there is absolutely NO impediment to using acceleration within special relativity. It's mathematically exactly equivalent to the problem of curved lines in Euclidean space. You don't have to abandon Euclidean geometry or the pythagorean theorem in order to calculate the length of a curved line, and you don't need to go to general relativity to handle acceleration.

But it is a common misconception that you need it, for two reasons: first, because it's often not taught (because you don't need to do it to understand the basics of SR), and second, because of a misunderstanding of the equivalence principle (which I could go into in more detail but isn't really important at the moment). So you're far from alone in your mistake, but the twin paradox really can be handled completely within the confines of special relativity, including the bits where one twin (or even both) accelerates. The ONLY thing you cannot handle correctly within the confines of special relativity is gravity. If you think the twin paradox gets resolved because of some GR effect, then you don't understand the problem.

 

[Ziggurat]

Actually there had been a debate among writers in the late 50's regarding the need to introduce GR to explain the twin paradox but nowadays, most physicists agree that SR is sufficient. And yes, SR can deal with acceleration. But we must be clear. SR deals only with acceleration as measured in inertial reference frames. This is not quite the case in the twin paradox, where one of the twins himself undergoes acceleration. In this sense, it is a problem of GR. We are appealing to the reality of this acceleration to explain the paradox.

The reason most writers agree that SR is sufficient in explaining the paradox is that we can use methods such as the Doppler effect (ie. pretend the twins send each other signals and count these) as described by SR and obtain the same result. But this is not to say that SR can handle acceleration as experienced by an observer.

This is also incorrect. There's generally little reason to treat special relativity problems from within a non-inertial frames (and plenty of reason not to - the math can get ugly), but there are no actual impediments. All the mathematical framework is there already. It's no different than Newtonian physics in that respect - using a rotating coordinate system introduces coriolis and centripidal pseudo-forces, but that's it. The reason that GR is different is not that it can handle non-inertial reference frames and SR can't, it's that GR can handle non-uniformity of that non-inertiality. That requires a whole new (and much more complicated) mathematical framework, but just non-inertiality can be handled within special relativity alone.

 

[Gertrude]

This is simply wrong. Let me expand upon Thai's response: special relativity can accomodate acceleration. It would be a crap theory if it couldn't. The math can get ugly and so most introductory treatments don't address it, but there is absolutely NO impediment to using acceleration within special relativity. It's mathematically exactly equivalent to the problem of curved lines in Euclidean space. You don't have to abandon Euclidean geometry or the pythagorean theorem in order to calculate the length of a curved line, and you don't need to go to general relativity to handle acceleration.

But it is a common misconception that you need it, for two reasons: first, because it's often not taught (because you don't need to do it to understand the basics of SR), and second, because of a misunderstanding of the equivalence principle (which I could go into in more detail but isn't really important at the moment). So you're far from alone in your mistake, but the twin paradox really can be handled completely within the confines of special relativity, including the bits where one twin (or even both) accelerates. The ONLY thing you cannot handle correctly within the confines of special relativity is gravity. If you think the twin paradox gets resolved because of some GR effect, then you don't understand the problem.

Like I said, SR can very well deal with accelerated motion as measured in inertial frames. But to deal with accelerated frames, we must use approximations (and ingenuity). There is no rigorous treatment (ie. without the need for approximations) that I know of. If there is, please do point me to some reference.

The reason I stated that it was a problem of GR (and you are right that I should have weighed my words more as it does come accross as a 'miracle' of GR) is that we must appeal to the reality of this acceleration to explain the paradox. Some treatment that does not involve GR can very well explain the paradox (for instance, the relativistic Doppler effect), but the turnaround is the key.

I stated this because it appears that many people take the example of the twin paradox to illustrate and understand SR. In this sense, they will think that time dilation/length contraction is still valid when the observers meet. But the twin paradox appears at first sight to be in contradiction with the tenets of SR and imho should not be used to understand the basics of SR.

 

[Gertrude]

This is also incorrect. There's generally little reason to treat special relativity problems from within a non-inertial frames (and plenty of reason not to - the math can get ugly), but there are no actual impediments. All the mathematical framework is there already. It's no different than Newtonian physics in that respect - using a rotating coordinate system introduces coriolis and centripidal pseudo-forces, but that's it. The reason that GR is different is not that it can handle non-inertial reference frames and SR can't, it's that GR can handle non-uniformity of that non-inertiality. That requires a whole new (and much more complicated) mathematical framework, but just non-inertiality can be handled within special relativity alone.

When I mention 'acceleration', I certainly refer to "all kinds" of accelerations, not only uniform acceleration...

But I'd be curious to see a rigorous treatment of (uniform) accelerating frames within SR. If you have any reference, I'd be highly interested.

 

[boooeee]

This is simply wrong. Let me expand upon Thai's response: special relativity can accomodate acceleration. It would be a crap theory if it couldn't. The math can get ugly and so most introductory treatments don't address it, but there is absolutely NO impediment to using acceleration within special relativity. It's mathematically exactly equivalent to the problem of curved lines in Euclidean space. You don't have to abandon Euclidean geometry or the pythagorean theorem in order to calculate the length of a curved line, and you don't need to go to general relativity to handle acceleration.

But it is a common misconception that you need it, for two reasons: first, because it's often not taught (because you don't need to do it to understand the basics of SR), and second, because of a misunderstanding of the equivalence principle (which I could go into in more detail but isn't really important at the moment). So you're far from alone in your mistake, but the twin paradox really can be handled completely within the confines of special relativity, including the bits where one twin (or even both) accelerates. The ONLY thing you cannot handle correctly within the confines of special relativity is gravity. If you think the twin paradox gets resolved because of some GR effect, then you don't understand the problem.

I know this issue has probably been covered ad nauseum in this forum, but can somebody provide an SR interpretation of the travelling twin's point of view?

In other words, the travelling twin is the one that accelerates away from his twin and then accelerates back. The travelling twin is sitting in his spaceship and observing the age of his twin. How does the travelling twin use Special Relativity to explain that time is moving faster for his twin?

 

[Ziggurat]

When I mention 'acceleration', I certainly refer to "all kinds" of accelerations, not only uniform acceleration...

That's not what I mean (though I phrased it poorly). I don't mean that your acceleration is uniform over time (there's no requirement for that, just as Euclidean geometry can handle curves with varying curvature), I mean that you're adopting a frame in which all of space-time adapts to your single acceleration. This goes back to the equivalence principle: standing still in a gravitational field is locally equivalent to accelerating on a flat space-time background. If you can handle acceleration with special relativity, why can't you handle gravity? Because gravity isn't local, and it isn't uniform. If we look at two people sitting on either side of a planet, their "accelerating" frames are accelerating away from each other (because gravity is not uniform, but pulls them in different directions), and yet the distance between them isn't changing. So any attempt to simply use a special relativity non-inertial reference frame to describe these two guys is going to fail, but not because special relativity can't handle accelerated frames.

But I'd be curious to see a rigorous treatment of (uniform) accelerating frames within SR. If you have any reference, I'd be highly interested.

Not at my fingertips, sorry.

 

[Ziggurat]

The reason I stated that it was a problem of GR (and you are right that I should have weighed my words more as it does come accross as a 'miracle' of GR) is that we must appeal to the reality of this acceleration to explain the paradox. Some treatment that does not involve GR can very well explain the paradox (for instance, the relativistic Doppler effect), but the turnaround is the key.

Depends what you mean by that.

Let's look at the metric:
ds² = dx² + dy² + dz² - (ct)²To find the proper time of any path, we integrate this quantity along that path. This is EXACTLY like taking the integral of a Euclidean metric:
ds² = dx² + dy² + dz²to find the length of a curve, it's just a different metric. Now, we can make our traveling twin's trajectory curved (finite acceleration), or we can make it kinked (infinite acceleration, instant velocity change), it doesn't matter: the process of taking that path integral is the same either way. Saying that it's the turnaround that matters is equivalent to saying that (say) a right angle in a Euclidean path is what makes it longer than the straight line. That may be true in one sense, but the extra length in the bent path isn't found in the bend itself. Likewise, the "missing" time isn't subtracted from the turnaround of the twin either. Each leg of the traveling twin's path really is shorter than half the stationary twin's path.

I stated this because it appears that many people take the example of the twin paradox to illustrate and understand SR. In this sense, they will think that time dilation/length contraction is still valid when the observers meet.

I'm not even sure what you mean by this anymore.

But the twin paradox appears at first sight to be in contradiction with the tenets of SR and imho should not be used to understand the basics of SR.

That's the whole point, though: it's only an apparent contradiction if you don't really understand what those tenets actually are.

 

[Ziggurat]

I know this issue has probably been covered ad nauseum in this forum, but can somebody provide an SR interpretation of the travelling twin's point of view?

In other words, the travelling twin is the one that accelerates away from his twin and then accelerates back. The travelling twin is sitting in his spaceship and observing the age of his twin. How does the travelling twin use Special Relativity to explain that time is moving faster for his twin?

Recall that your space axis (which marks all points in space at the same time for your reference frame) tilts when you change reference frame. If you're accelerating, that axis is continually tilting. So as the traveling twin accelerates back towards the stationary twin, the intersection of his space axis with the stationary twin's world-line move upwards (in time). During the acceleration phase, then, the traveling twin will indeed observe the stationary twin's clock to be moving faster, in his accelerated frame. If he changes velocity instantaneously, then his space axis will tilt instantly, and he will observe that the stationary twin's current time will have instantaneously jumped forward. An important caveat, though: this is what he oberves, not what he sees. Even in the case of instantaneous velocity change, there's no discontinuity in what the twin sees.

If you want to ask what the twins see (which is also interesting, BTW), that's actually pretty straight forward. While moving away from each other, each twin sees the other twin's clock slowed by time dilation, plus additional slowing from doppler shift. When the traveling twin turns around and heads back, he will see the stationary twin's clock slowed by time dilation, but also speeded up by the doppler shift. So for half of his journey, the doppler shift increases the clock rate, and for half of it the doppler shift decreases the clock rate.

The stationary twin, however, doesn't see the traveling twin turn around at the moment he turns around, because the signal is time-delayed. But when he does see the traveling twin heading back towards him, he will see the clock slowed by time dilation, plus speeded up because of the doppler shift. Unlike the traveling twin, however, he sees the doppler shift slowing the clock for MORE than half the time interval, and speeding up the clock for LESS than half the time interval (because it took time for the light from the turnaround to get back to him). Which is how the twins end up with different times on their clocks: what they see over the length of the whole journey isn't the same, even if instantaneous snapshots on the outward and return journeys look superficially symmetric.

 

[Gertrude]

Depends what you mean by that.

Let's look at the metric:
ds² = dx² + dy² + dz² - (ct)²To find the proper time of any path, we integrate this quantity along that path. This is EXACTLY like taking the integral of a Euclidean metric:
ds² = dx² + dy² + dz²to find the length of a curve, it's just a different metric. Now, we can make our traveling twin's trajectory curved (finite acceleration), or we can make it kinked (infinite acceleration, instant velocity change), it doesn't matter: the process of taking that path integral is the same either way. Saying that it's the turnaround that matters is equivalent to saying that (say) a right angle in a Euclidean path is what makes it longer than the straight line. That may be true in one sense, but the extra length in the bent path isn't found in the bend itself. Likewise, the "missing" time isn't subtracted from the turnaround of the twin either. Each leg of the traveling twin's path really is shorter than half the stationary twin's path.

Absolutely. The point here is that, neglecting accelerations, two inertial frames cannot be brought together and still experience different durations. I could put it differently: the lack of symmetry in the paradox is key. The fact that one twin definately feels an acceleration (be it a constant acceleration as in a circle or a sudden turnaround) while the other one doesn't makes up for the difference.

In SR, time dilation works both ways. In all cases we can say: "A is travelling at X m/s relative to A" in the same way as "B is travelling at X m/s relative to A". If the only effect that we considered was velocity, then there indeed would be a paradox, as the effects of SR are symmetric for both observers, ie. both observers would see the other aging slower. It is the idea that the situation is not symmetrical that resolves the paradox.

In this sense, I think the subtlety may lead to an incorrect understanding that time dilation is not only apparent and occurs to the 'fastest traveler' only(whatever that means).

 

[Yllanes]

But I'd be curious to see a rigorous treatment of (uniform) accelerating frames within SR. If you have any reference, I'd be highly interested.

Do you mean references as in web pages or books? Because if it is the latter, most textbooks discuss them. For example, let us see how uniform accelerated motion looks in SR.

First of all we must define it, because acceleration is relative. I'm going to find the trajectory of a particle with constant acceleration in its instantaneous rest frame. This rest frame is the inertial reference frame that happens to have the same velocity as the particle at each moment, and so it is different at each point of the movement. Let a be the ordinary acceleration of Newtonian physics, and v the ordinary velocity. As you may now, we can define a four velocity if we differentiate the position x with respect to the proper time

[latex]\footnotesize
\begin{align*}
x^\mu&=(ct,\ x,\ y,\ z),\\
u^\mu&=\frac{\mathrm{d} x^\mu}{\mathrm{d} \tau} = \left(\gamma c,\ \gamma {\vec v}\right),\qquad \tau =t \sqrt{1-v^2/c^2} = t/\gamma,\\
w^\mu&=\frac{\mathrm{d} x^\mu}{\mathrm{d} \tau^2}
\end{align*}
[/latex]

The important thing about four vectors is that their magnitude is the same in all reference frames. We write that magnitude as u² and we have to remember it is not calculated with the Euclidean metric, but that the time component is subtracted. For example

[latex]\footnotesize
\[u^2 = (u^1)^2+(u^2)^2+(u^3)^2-(u^0)^2 = -c^2
\]
[/latex]

In our accelerated motion, we now that w^m = (0, a, 0,0) in the rest frame. So w² = a² in every reference frame. We are interested in the acceleration as seen from a 'fixed' frame. So we go to the value of w^m in that system and force it to have w²=a². This leads to the equation (v is the velocity of the particle in the fixed frame).

[latex]\footnotesize
\[\frac{\mathrm{d}}{\mathrm{d}t} \frac{v}{\sqrt{1-\frac{v^2}{c^2}}} = a
\]
[/latex]

It's just a matter of integrating now. If we set x = 0 at t = 0 the movement is

[latex]\footnotesize
\begin{align*}
v&=\frac{at}{\sqrt{1+\frac{a^2t^2}{c^4}}}, &
x &= \frac{c^2}{a}\left(\sqrt{1+\frac{a^2t^2}{c^2}}-1\right)
\end{align*}
[/latex]

And it is inmediate to see that in the limit at << c we recover the Newtonian equations for the uniform accelerated movement ( v ~ at, x ~ at²/2). As you see, it is very simple to handle accelerated motion in SR.

 

[Ziggurat]

It is the idea that the situation is not symmetrical that resolves the paradox.

Correct.

In this sense, I think the subtlety may lead to an incorrect understanding that time dilation is not only apparent and occurs to the 'fastest traveler' only(whatever that means).

I'm confused by what you mean here. If you're claiming that time dilation is only an apparent effect, that is wrong. It's a very real effect. But it can be wrongly applied to the traveling twin paradox if the issue of simultaneity is not handled correctly as well.

 

[Gertrude]

First of all we must define it, because acceleration is relative. I'm going to find the trajectory of a particle with constant acceleration in its instantaneous rest frame. This rest frame is the inertial reference frame that happens to have the same velocity as the particle at each moment, and so it is different at each point of the movement.

I lost you right there! If the rest frame has the same velocity as the particle, it cannot be inertial since the particle undergoes acceleration!?

In our accelerated motion, we now that wm = (0, a, 0,0) in the rest frame. So w2 = a2 in every reference frame. We are interested in the acceleration as seen from a 'fixed' frame. So we go to the value of wm in that system and force it to have w2=a2. This leads to the equation (v is the velocity of the particle in the fixed frame).

Ok. So you are analysing the motion as an observer in an inertial frame. Actually, I have seen this treatment before. What interests me is looking at events from an accelerating frame. If you have any suggestion...

 

[Ziggurat]

I lost you right there! If the rest frame has the same velocity as the particle, it cannot be inertial since the particle undergoes acceleration!?

No. Rather, acceleration is always measured with respect to some specific reference frame. This doesn't sound significant until you realize that 1) the same accelerating trajectory will have a different acceleration when measured from different reference frames, and 2) this is not the same as classical mechanics where velocity is relative but acceleration is absolute. To adopt constant proper acceleration, you basically find a trajectory where the instantaneous acceleration in the inertial reference frame defined by the instantaneous velocity is always constant. Which frame you use has to keep changing, which means that in any fixed reference frame the acceleration will not appear constant, but each of those tangent frames is still inertial.

 

[Gertrude]

I'm confused by what you mean here. If you're claiming that time dilation is only an apparent effect, that is wrong. It's a very real effect. But it can be wrongly applied to the traveling twin paradox if the issue of simultaneity is not handled correctly as well.

I am quite good at expressing things in the most confusing possible manner. I'll give an example that may be clearer. The statement "moving clocks run slower" is confusing because it does imply that time slows down with speed (relative to any observer). The reality is that moving clocks measure events occuring in a slower inertial frame as having longer duration than what a clock at rest relative to said event would measure.

The confusion is only increased with the twin paradox. It could be interpreted that velocity indeed slows time down for every observer (as 'confirmed' by the fact that when they meet, the two twins no longer have the same age). But this is not the case. If A and B are in two different inertial frames, B would see the clock of A running slower because it is moving relative to A and A would see the clock of B running slower because it is moving relative to B. There is no 'absolute' time dilation and it is in this sense that I say time dilation is apparent.

 

[Paulhoff]

OK, let’s try this again.

We have two spacemen and three clocks and the clocks are big so that they can be seen and read from a 100 feet with no problem. The speed of light for this experiment is 100 ft. per second. The spacemen and clocks are all together somewhere in deep space, who cares where, and who cares about their speed. The clocks are all reading the same time.

1. Now spaceman (1) takes one of the clocks and travels in any direction at 99,9999999% speed of light for one second and stops. Spaceman (1) now is 100 ft away from spaceman (2) and the two clocks remanding clocks, and spaceman (1) looks at the two clocks with spaceman (2).

A. Is the time on spaceman's (1) clock is the same as the other two clocks.
B. Is the time on spaceman's (1) clock is one second behind the other two clocks.
C. Is the time on spaceman's (1) clock is two seconds behind the other two clocks.
D. Is the time on spaceman's (1) clock is one second ahead of the other two clocks.
E. Is the time on spaceman's (1) clock is two seconds ahead of the other two clocks.

2. Now spaceman (2) takes one of the two remaining clocks and travels very slowly, about one inch a second to spaceman (1). When he gets to spaceman (1) he looks at spaceman's (2) clock.

A. Is the time on spaceman's (2) clock is the same as spaceman’s (1) clock.
B. Is the time on spaceman's (2) clock is one second behind spaceman’s (1) clock.
C. Is the time on spaceman's (2) clock is two seconds behind spaceman’s (1) clock.
D. Is the time on spaceman's (2) clock is one second ahead of spaceman’s (1) clock.
E. Is the time on spaceman's (2) clock is two seconds ahead of spaceman’s (1) clock.

3. Now spaceman (2) looks back at the clock left behind.

A. Is the time on his clock is the same as the clock left behind.
B. Is the time on his clock is one second behind the left clock.
C. Is the time on his clock is two seconds behind the left clock.
D. Is the time on his clock is one second ahead of the left clock.
E. Is the time on his clock is two seconds ahead of the left clock.

Paul

 

[Yllanes]

I lost you right there! If the rest frame has the same velocity as the particle, it cannot be inertial since the particle undergoes acceleration!?

Ziggurat answered this. There is a different rest frame at each moment of time.

Ok. So you are analysing the motion as an observer in an inertial frame. Actually, I have seen this treatment before. What interests me is looking at events from an accelerating frame. If you have any suggestion...

Ah, well it's just the same. For example, t is time as measured by the stationary observer. To switch to the accelerated observer we have to use his proper time, which is just a matter of integrating

[latex]\footnotesize
\[
\tau=\int_0^t \sqrt{1-\tfrac{v^2}{c^2}}= \frac{c}{a} \mathrm{arcsinh}\left(\frac{at}{c}\right)\]
[/latex]

so the distance from its starting point, as measured by the accelerated particle, is

[latex]\footnotesize
\[
x(\tau)=\left(\cosh \tfrac{at}{c} - 1\right)\frac{c^2}{a}
\]
[/latex]

Incidentally, such a uniformly accelerated observer can outrun a photon, even though his velocity never reaches c.