# Proper Time with a non-zero Initial Velocity

I have a question regarding how to find the proper time for a body with an initial velocity to slow down to 0. For example, the equation I have been working with looks like: $$\int^\tau_0 d\tau=\int^t_0\frac{1}{\sqrt{1+\frac{(\alpha t+v)^2}{c^2}}}dt$$ Where $t$ is equal to the coordinate time for the rocket to slow down, $\alpha$ is the proper acceleration and $v$ is the intial velocity. Integrating gives: $$\tau=\frac{c}{a}arcsinh(\frac{\alpha t+v}{c})$$ I believe, however, that this is incorrect as the values it returns are far too small to be a legitimate value for proper time. I do believe that this is on the right track to deriving the correct value.

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Your formula is almost correct. If $\alpha$ is the (constant) proper acceleration and $v_0$ the initial coordinate velocity, then $$v(t) = \frac{\alpha t+w_0}{\sqrt{1+(\alpha t+w_0)^2/c^2}},$$ with $$w_0 = \frac{v_0}{\sqrt{1-v_0^2/c^2}}.$$ See this post for the derivation. The proper time is then found by integrating $$\text{d}\tau = \sqrt{1 - v^2/c^2}\text{d}t = \frac{\text{d}t}{\sqrt{1+(\alpha t+w_0)^2/c^2}},$$ thus $$\tau = \int_0^t\frac{\text{d}t'}{\sqrt{1+(\alpha t'+w_0)^2/c^2}}.$$ If we use the substitution $y = (\alpha t'+w_0)/c$, we get $$\tau = \frac{c}{\alpha}\int_{w_0^{\phantom{1}}/c}^{(\alpha t+w^{\phantom{1}}_0)/c}\frac{\text{d}y}{\sqrt{1+y^2}},$$ and that's where the problem was: the lower bound is not zero. The result is \begin{align} \tau &= \frac{c}{\alpha}\left[\text{arcsinh}\left(\frac{\alpha t+w_0}{c}\right) - \text{arcsinh}\left(\frac{w_0}{c}\right)\right]\\[0.4cm] &= \frac{c}{\alpha}\left[\ln\left(\frac{(\alpha t+w_0)/c + \sqrt{1+(\alpha t+w_0)^2/c^2}}{w_0/c + \sqrt{1+ w_0^2/c^2}}\right)\right]. \end{align}

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Hmmmm, alright I believe I understand it for the most part. Is there any distinction between the $\tau$ and $t'$ you use in the integral? – Hattar Wu Sep 10 '13 at 2:24
@HattarWu $t'$ is just a name for the coordinate time in the integration, to avoid confusion with $t$ in the upper bound. – Pulsar Sep 10 '13 at 2:30
Ah, I see. For the coordinate time used in the final equation, is that equivalent to the coordinate time used in the first equation for $v(t)$? Wouldn't that imply that, because the rocket is slowing to a stop, $t=\frac{-\omega_0}{\alpha}$ where $\alpha = -(2*9.81)$? That feels like it does not take into account relativistic effects. – Hattar Wu Sep 10 '13 at 2:44
@HattarWu Yes. If you insert that value, the expression simplifies a lot. – Pulsar Sep 10 '13 at 2:47

Yeah, something there is incorrect. I imagine $t\rightarrow\infty$, and see that $\alpha t+v$ will $\rightarrow\infty$ too. But, unless you're using some weird definitions, always $v<c$.

Given proper acceleration $\alpha$, the way to make use of it is from its definition, and of the four-velocity, proper time, and Lorentz transforms. The rocket's velocity at some time $t+\delta t$ should be $\alpha\delta\tau$ as seen in the rocket's own frame at time $t$. Its own change in velocity over a short duration of time as it knows it.

Write out the relativistic four-velocity, think about the relation between $\delta\tau$ and $\delta t$, and of course we're thinking of small time intervals, $\delta t \rightarrow 0$ - it just seems natural to write a differential equation. Have you done that? Where did that integral of yours come from?

There's the concept of "rapidity" which seems to come up more in high energy physics than in other areas, you might find enlightening to read about.

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The integral comes from the definition of the Lorentz factor: $$\gamma = \frac{dt}{d\tau}$$ $$d\tau = \frac{dt}{\gamma}$$ and then substituting the version of the Loretnz factor with its definition in terms of $\alpha$ and integrating. I'm sorry, but I don't quite understand exactly what you are saying....what exactly is the problem with the equations? I know there is a problem, but I can't figure out where it is! – Hattar Wu Sep 10 '13 at 1:13