# Clock on constantly accelerating object approaches Gudermannian limit?

If an object is moving away from X with a constant acceleration of a, its velocity at time t (relative to X and accounting for relativity) is given by:

$v(t)=\tanh (a t)$

Also, if an object is traveling at velocity v (measured as a fraction of the speed of light) relative to X, the time dilation factor is:

$\sqrt{1-v^2}$

For example, if an object is traveling at .99c relative to X, the time dilation factor is approximately 0.14, meaning that for every second X measures on its own clock, it sees 0.14 seconds ticked off on the moving object's clock.

Combining these two equations, I find the time dilation factor for an object with constant acceleration a is:

$\sqrt{1-v(t)^2}=\sqrt{1-\tanh ^2(a t)}=\text{sech}(a t)$

In other words, at time t for X, the object's clock is ticking at $\text{sech}(a t)$ seconds for every second on X's clock.

To find the total elapsed time, I should be able to just integrate:

$\int \text{sech}(a t) \, dt = \frac{2 \tan ^{-1}\left(\tanh \left(\frac{a t}{2}\right)\right)}{a} = \frac{\text{gd}(a t)}{a}$

where gd is the Gudermannian function.

The problem: as t approaches infinity...

$\lim_{t\to \infty } \, \frac{\text{gd}(a t)}{a} = \frac{\pi }{2 a}$

If true, this means X will never see the object's clock pass $\frac{\pi }{2 a}$.

This seems incorrect. What am I doing wrong? Am I incorrectly applying special relativity formulas to a constantly accelerating object?

One reason I think this is incorrect. According to:

http://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html

it should take me 1.94 arccosh (n/1.94 + 1) years (object time) to travel n light years at constant acceleration (of 1g). This quantity is unbounded as n increases, so either my calculations or the page's calculations are incorrect, and I'm pretty sure I'm more likely to be wrong than the page above.

Note: I came across this while attempting to answer https://astronomy.stackexchange.com/questions/13817/

Here's the discrete analog of what I (the moving object) am doing. Every second:

• I drop a beacon that has zero relative velocity to me.

• I accelerate until I'm traveling at 10m/s (or whatever a I choose) with respect to the beacon.

I believe:

• When smoothed out to be continuous, I will feel a constant acceleration.

• As viewed from X (the stationery observer), my velocity is tanh(a*t)

You mention in your answer "for every unit of your time the object accelerates to the speed of an object that it currently thinks is moving at a certain speed. But this means it has to accelerate at a faster rate according to its own clock", but I'm not sure I understand this.

As I see it, the moving object can be seen as accelerating with respect to a beacon it just dropped, and the small 10m/s velocity increase shouldn't have significant time dilation. In the continuous case, there should be no time dilation at all.

I believe your answer is correct, but think I'm still missing something.

Response:

"If they say you need to do in 1s of beacon time, then this will feel constant since you are always comoving to the beacon you just dropped.". That's what I'm doing. I'm not doing this: "But if instead you have to do it in time that is 1s of that first beacon you dropped way way way back and which you are now moving very fast with respect to this is a problem". I agree that's a problem especially since that would mean v(t) = 10*t, and once you're at c - 10m/s, you can't accelerate 10m/s at all. But I still don't see how you got your formula. I'll try to add to my question re how I got my formula (relativistic adding of velocities) shortly.

EDIT (this is the discrete case, just for fun, but with my misunderstanding corrected):

NOTE: for the below "my" refers to the accelerating frame of reference and "fixed" refers to non-accelerating frame of reference.

The formula for adding relativistic velocities (when both are given as a fraction of light speed) is:

$\text{add}(u,v)=\frac{u+v}{u v+1}$

If I start at zero velocity (with respect to some fixed X), and follow the process above (drop beacons and accelerate by a where a is much smaller than c) every second of my time, my speed as viewed by the fixed observer is given by:

$\text{speed}(a,0)=0$

$\text{speed}(a,n+1)=\frac{a+\text{speed}(a,n)}{a * \text{speed}(a,n)+1}$

The closed-form solution (simplest form Mathematica could find):

$\text{speed}(a,n)=\frac{2}{\left(\frac{2}{a+1}-1\right)^n+1}-1$

Although I'm dropping beacons every 1 second in my own time frame, I am dropping them slower and slower to the fixed observer X. This was the crux of my misunderstanding

For the fixed observer, how much time passes between my dropping beacon $n$ and beacon $n+1$?

When 1 second passes on my clock, time dilation tells me $\frac{1}{\sqrt{1-\text{speed}(a,n)^2}}$ passes on the fixed observer's clock. Plugging in $\text{speed}(a,n)$ and simplifying:

$\text{dilation}(a,n) = \frac{1}{2} \left(1-a^2\right)^{-n/2} \left((1-a)^n+(a+1)^n\right)$

For the continuous case, we accelerate $\frac{a}{k}$ (k times slower) $k n$ (k times as often) and take the limit as $k\to \infty$. This yields:

$\text{contspeed}(n)=\lim_{k\to \infty } \, \text{speed}\left(\frac{a}{k}, nk\right) = \tanh (a n)$

Note this is the same value I had earlier, but that it refers to seconds elapsed in my frame of reference.

How many seconds have elapsed in the fixed frame of reference in the continuous case? The instantaneous time dilation is:

$\text{contdilation}(a,n) = lim_{k\to \infty } \text{dilation}\left(\frac{a}{k}, nk\right) = lim_{k\to \infty } \frac{1}{2} \left(k^2-a^2\right)^{-\frac{k n}{2}} \left((k-a)^{k n}+(a+k)^{k n}\right) = \cosh (a n)$

Thus, when the nth infinitesimal time unit ticks off in my reference frame, $\cosh (a n)$ infinitesimal time units tick off in the fixed frame. The total elapsed time for the fixed frame is simply the integral of this or $\frac{\sinh (a n)}{a}$

So, when $\frac{\sinh (a n)}{a}$ seconds have elapsed in the fixed reference frame, my speed (relative to the fixed reference frame) is $\tanh (a n)$:

$\text{fixedspeed}\left(\frac{\sinh (a n)}{a}\right)=\tanh (a n)$

To solve, we apply this change of variable to both sides:

$n\to \frac{\sinh ^{-1}(a x)}{a}$

to yield:

$\text{fixedspeed}(n)=\frac{a n}{\sqrt{a^2 n^2+1}}$

which is equivalent to @timeaus' answer for v(t).

The total distance is just the integral of this, or:

$\text{fixeddist}(a,n)=\int \text{fixedspeed}(a,n) \, dn = \frac{\sqrt{a^2 n^2+1}}{a}$

which, as expected, is equivalent to timeaus' answer for x(t).

• Why does your intuition tell you that it's incorrect? I don't feel all that surprised, at all. A constant acceleration seems to act like an event horizon. The object just redshifts away in a finite time. – CuriousOne Feb 28 '16 at 4:12
• OK, but when the object's clock reads Pi/2a, how far away will it be from X? An infinite distance? And how about the second after that? If my calculations are correct, it means we can travel anywhere in the universe in Pi/2a units of time (relative to our vessel). – barrycarter Feb 28 '16 at 4:28
• I think the relativistic rocket equations predict something like that, yes... but at an insane expense of energy. OTOH, why would you expend an insane amount of energy on traveling in a finite time when you can travel with a constant and reasonably low amount of energy in no time using light? – CuriousOne Feb 28 '16 at 4:36
• Relativistic addition of velocities has nothing to do with it. Everyone can compute relative velocities, they are a frame invariant quantity because you are just asking what the most recent beacon sees. Everyone agrees on what the most recent beacon sees. So we all agree when you get to your target speed for that discrete part of your journey. The question is when do you drop a new beacon and accelerate again. The FAQ has it happen every unit of time for the recent beacon. You are doing it for every unit of time of the first beacon. Which is way way more quickly. – Timaeus Mar 3 '16 at 18:56

You do not need general relativity to solve for motion in a flat spacetime. Using general relativity for a flat spacetime problem would be like using Riemannian geometry (the geometry of curved surfaces) to solve for the length of a curve in a flat Euclidean space. If your space is flat, regular flat geometry works.

So your accelerating curve is treated the same as in flat Euclidean geometry. Your space is flat so Minkowksi geometry is fine and you don't need General Relativity. You compute the length of a curve the same way in Minkowski geometry and in Euclidean geometry. You break the curve into pieces, approximate the curve with a line segment, compute the length of that segment and add up the lengths for each segment (and in Euclidean geometry that would be an underestimate of the final length, and in Minkowski geometry if the curve always has a time-like tangent it's an overestimate of the final time) and then you repeat for a finer mesh. And the sums should approach the actual length of the curve as a limit.

What am I doing wrong?

The reason you don't agree with the physics FAQ is that it is solving a different problem than you are solving.

Your formula is for an object that accelerates at a rate that increases the rapidity by the same amount per unit of inertial frame time, $t$. To move along your curve, the object needs to accelerate with faster and faster proper acceleration. Because your formula is saying that for every unit of your time the object accelerates to the speed of an object that it currently thinks is moving at a certain speed. But this means it has to accelerate at a faster rate according to its own clock, which measures the proper time along the curve.

Whereas the problem in the physics FAQ has the proper acceleration be the same at every moment. So for instance if you pick an inertial frame with regular $t,$ $x,$ $y,$ and $z$ coordinates and take off with a constant proper acceleration, $a,$ then you undergo hyperbolic motion and then your object is on a curve like $$x(t)=\sqrt{c^4/a^2+c^2t^2}.$$ In which case your velocity as a function of time is $$v(t)=\frac{c^2t}{\sqrt{c^4/a^2+c^2t^2}}.$$

So to put it simply, the first equation you wrote requires that it feel like it is accelerating faster and faster, and the equations I wrote have it feel like it is accelerating just as hard at every moment. And these are different problems.

Here's the discrete analog of what I (the moving object) am doing. Every second:

If you do this every second as measured by your most recently just dropped beacon, you get the formula I gave. If you do it every second of the clock in the first frame then you accelerate much faster. It comes down to whether you accelerate in time $\Delta t/\gamma$ or in time $\Delta t.$

I believe: When smoothed out to be continuous, I will feel a constant acceleration.

If you accelerated to 10 m/s relative to your beacon in 1s of beacon time then that would feel like a constant acceleration.

As I see it, the moving object can be seen as accelerating with respect to a beacon it just dropped, and the small 10m/s velocity increase shouldn't have significant time dilation.

The question isn't about the time dilation of the 10 m/s relative to the beacon or the 0 m/s relative to the beacon. The question is someone tells you to fire your rockets to go faster than the comoving beacon you just dropped and you ask how quickly you need to get up to that speed. If they say you need to do in 1s of beacon time, then this will feel constant since you are always comoving to the beacon you just dropped.

But if instead you have to do it in time that is 1s of that first beacon you dropped way way way back and which you are now moving very fast with respect to this is a problem. And in fact you want to do it super fast, so fast that the first beacon ticks 1s while you get up to speed relative to your most recent beacon.

• So, just to confirm, you're saying v(t) = tanh(at) is wrong for an object that has fixed proper acceleration? I believe all of my calculations derive from this equation, so, if it's wrong, that would explain my mistake. What would be the correct formula here? – barrycarter Mar 3 '16 at 13:35
• Still a little confused, edited question to reflect. – barrycarter Mar 3 '16 at 15:33
• @barrycarter Response to edit. – Timaeus Mar 3 '16 at 17:47
• @barrycarter You are misreading me. It's not a question of how fast to go. Your discrete version always gets up to a speed that the most recent beacon say is 10 m/s. The question is how quickly you get up to that speed. Does the most recent beacon think you did it in 1s (that's the proper acceleration) or does the first beacon way back long ago compute how fast that happened. Let's say you fire thrusters really fast and spend most of your time coasting at 10 m/s relative to the recent beacon. Your formula has the first beacon think you fire your rockets once a second. – Timaeus Mar 3 '16 at 18:53
• Got it at last. If I'm dropping beacons every 1 second my time, the stationery observer sees me dropping beacons slower and slower (because my clock appears slow to him). – barrycarter Mar 3 '16 at 19:13