User:L33th4x0r/twin

Dear Wikipedians:

I've been trying to work out the details of some of the calculations found in arXiv:physics/0411233, I got very very close to one of the final results but seem to have gotten stuck on a small bit of detail that didn't add up, which I was hoping to get your help with punching through.

The general formula for calculating time elapsed (T) for the stay-at home twin purely from the proper time elapsed ($$\bar{\tau}$$) for the traveling twin, and purely from the traveling twin's non-inertial frame of reference, was given in the paper as:

In the extreme (and classic when talking about any traveling twin problems) case of the infinite acceleration happening in an instant mid-trip for the traveling twin, the paper modeled it using Dirac's Delta as:

where $$\operatorname{artanh} (-v)$$ is the rapidity of the ending half of the journey, $$\operatorname{artanh} (v)$$ is the rapidity of the starting half of the journey, and $$\delta(\tau - \tilde{\tau})$$ is Dirac's Delta centered at $$\tilde{\tau}$$, where $$\tilde{\tau}$$ is the instant the infinite acceleration (so-called "turn-around") happened for the traveling twin.

According to the paper, plugging ($$) into ($$), and after some work with hyperbolic functions, one finds $$ T = \bar{\tau}/\sqrt{1-v^2} $$, which is one of the classical twin paradox results. I did get this result, but only when starting from the following expression:

Plugging ($$) into ($$), we have:

$$\begin{align} T^2 &= \left[\int_0^\bar{\tau} e^{\int_0^\tau -2\operatorname{artanh} v\delta\left( \tau' - \frac{\bar{\tau}}{2}\right) \text{d}\tau'} \text{d}\tau \right] \left[\int_0^\bar{\tau}  e^{-\int_0^\tau -2\operatorname{artanh} v\delta\left( \tau' - \frac{\bar{\tau}}{2}\right) \text{d}\tau'} \text{d}\tau \right] \\

&= \left[\int_0^\frac{\bar{\tau}}{2} e^{\int_0^\tau -2\operatorname{artanh} v\delta\left( \tau' - \frac{\bar{\tau}}{2}\right) \text{d}\tau'} \text{d}\tau + \int_\frac{\bar{\tau}}{2}^\bar{\tau} e^{\int_0^\tau -2\operatorname{artanh} v\delta\left( \tau' - \frac{\bar{\tau}}{2}\right) \text{d}\tau'} \text{d}\tau \right] \left[\int_0^\frac{\bar{\tau}}{2} e^{-\int_0^\tau -2\operatorname{artanh} v\delta\left( \tau' - \frac{\bar{\tau}}{2}\right) \text{d}\tau'} \text{d}\tau + \int_\frac{\bar{\tau}}{2}^\bar{\tau} e^{-\int_0^\tau -2\operatorname{artanh} v\delta\left( \tau' - \frac{\bar{\tau}}{2}\right) \text{d}\tau'} \text{d}\tau \right] \\

&= \left[\int_0^\frac{\bar{\tau}}{2} e^{-2\operatorname{artanh} v\int_0^\tau \delta\left( \tau' - \frac{\bar{\tau}}{2}\right) \text{d}\tau'} \text{d}\tau + \int_\frac{\bar{\tau}}{2}^\bar{\tau} e^{\int_0^\tau -2\operatorname{artanh} v\delta\left( \tau' - \frac{\bar{\tau}}{2}\right) \text{d}\tau'} \text{d}\tau \right] \left[\int_0^\frac{\bar{\tau}}{2} e^{-(-2\operatorname{artanh} v) \int_0^\tau \delta\left( \tau' - \frac{\bar{\tau}}{2}\right) \text{d}\tau'} \text{d}\tau + \int_\frac{\bar{\tau}}{2}^\bar{\tau} e^{-\int_0^\tau -2\operatorname{artanh} v\delta\left( \tau' - \frac{\bar{\tau}}{2}\right) \text{d}\tau'} \text{d}\tau \right] \\

&= \left[\int_0^\frac{\bar{\tau}}{2} e^{-2\operatorname{artanh} v(0)} \text{d}\tau + \int_\frac{\bar{\tau}}{2}^\bar{\tau} e^{\int_0^\tau -2\operatorname{artanh} v\delta\left( \tau' - \frac{\bar{\tau}}{2}\right) \text{d}\tau'} \text{d}\tau \right] \left[\int_0^\frac{\bar{\tau}}{2} e^{-(-2\operatorname{artanh} v) (0)} \text{d}\tau + \int_\frac{\bar{\tau}}{2}^\bar{\tau} e^{-\int_0^\tau -2\operatorname{artanh} v\delta\left( \tau' - \frac{\bar{\tau}}{2}\right) \text{d}\tau'} \text{d}\tau \right] \\

&= \left[\int_0^\frac{\bar{\tau}}{2} e^0 \text{d}\tau + \int_\frac{\bar{\tau}}{2}^\bar{\tau} e^{\int_0^\tau -2\operatorname{artanh} v\delta\left( \tau' - \frac{\bar{\tau}}{2}\right) \text{d}\tau'} \text{d}\tau \right] \left[\int_0^\frac{\bar{\tau}}{2} e^0 \text{d}\tau + \int_\frac{\bar{\tau}}{2}^\bar{\tau} e^{-\int_0^\tau -2\operatorname{artanh} v\delta\left( \tau' - \frac{\bar{\tau}}{2}\right) \text{d}\tau'} \text{d}\tau \right] \\

&= \left[\int_0^\frac{\bar{\tau}}{2} (1) \text{d}\tau + \int_\frac{\bar{\tau}}{2}^\bar{\tau} e^{-2\operatorname{artanh} v \int_0^\tau \delta\left( \tau' - \frac{\bar{\tau}}{2}\right) \text{d}\tau'} \text{d}\tau \right] \left[\int_0^\frac{\bar{\tau}}{2} (1) \text{d}\tau + \int_\frac{\bar{\tau}}{2}^\bar{\tau} e^{-(-2\operatorname{artanh} v) \int_0^\tau \delta\left( \tau' - \frac{\bar{\tau}}{2}\right) \text{d}\tau'} \text{d}\tau \right] \\

&= \left[\int_0^\frac{\bar{\tau}}{2} \text{d}\tau + \int_\frac{\bar{\tau}}{2}^\bar{\tau} e^{-2\operatorname{artanh} v (1)} \text{d}\tau \right] \left[\int_0^\frac{\bar{\tau}}{2} \text{d}\tau + \int_\frac{\bar{\tau}}{2}^\bar{\tau} e^{ 2\operatorname{artanh} v (1) } \text{d}\tau \right] \\

&= \left[\int_0^\frac{\bar{\tau}}{2} \text{d}\tau + \int_\frac{\bar{\tau}}{2}^\bar{\tau} e^{-2\operatorname{artanh} v} \text{d}\tau \right] \left[\int_0^\frac{\bar{\tau}}{2} \text{d}\tau + \int_\frac{\bar{\tau}}{2}^\bar{\tau} e^{ 2\operatorname{artanh} v} \text{d}\tau \right] \\

&= \left[ \tau\Bigg|_0^\frac{\bar{\tau}}{2} + \int_\frac{\bar{\tau}}{2}^\bar{\tau} e^{-2 \left( \frac{1}{2} \ln \frac{1+v}{1-v} \right) } \text{d}\tau \right] \left[ \tau\Bigg|_0^\frac{\bar{\tau}}{2} + \int_\frac{\bar{\tau}}{2}^\bar{\tau} e^{ 2 \left( \frac{1}{2} \ln \frac{1+v}{1-v} \right) } \text{d}\tau \right] \\

&= \left[ \frac{\bar{\tau}}{2} + \int_\frac{\bar{\tau}}{2}^\bar{\tau} e^{-\ln\frac{1+v}{1-v}} \text{d}\tau \right] \left[ \frac{\bar{\tau}}{2} + \int_\frac{\bar{\tau}}{2}^\bar{\tau} e^{\ln\frac{1+v}{1-v}} \text{d}\tau \right] \\

&= \left[ \frac{\bar{\tau}}{2} + \int_\frac{\bar{\tau}}{2}^\bar{\tau} \left( e^{ \ln \frac{1+v}{1-v} } \right) ^{-1} \text{d}\tau \right] \left[ \frac{\bar{\tau}}{2} + \int_\frac{\bar{\tau}}{2}^\bar{\tau} e^{ \ln \frac{1+v}{1-v} } \text{d}\tau \right] \\

&= \left[ \frac{\bar{\tau}}{2} + \int_\frac{\bar{\tau}}{2}^\bar{\tau} \left( \frac{1+v}{1-v} \right) ^{-1} \text{d}\tau \right] \left[ \frac{\bar{\tau}}{2} + \int_\frac{\bar{\tau}}{2}^\bar{\tau} \frac{1+v}{1-v} \text{d}\tau \right] \\

&= \left[ \frac{\bar{\tau}}{2} + \int_\frac{\bar{\tau}}{2}^\bar{\tau} \frac{1-v}{1+v} \text{d}\tau \right] \left[ \frac{\bar{\tau}}{2} + \int_\frac{\bar{\tau}}{2}^\bar{\tau} \frac{1+v}{1-v} \text{d}\tau \right] \\

&= \left[ \frac{\bar{\tau}}{2} + \frac{1-v}{1+v} \int_\frac{\bar{\tau}}{2}^\bar{\tau} \text{d}\tau \right] \left[ \frac{\bar{\tau}}{2} + \frac{1+v}{1-v} \int_\frac{\bar{\tau}}{2}^\bar{\tau} \text{d}\tau \right] \\

&= \left[ \frac{\bar{\tau}}{2} + \frac{1-v}{1+v} \tau\Bigg|_\frac{\bar{\tau}}{2}^\bar{\tau} \right] \left[ \frac{\bar{\tau}}{2} + \frac{1+v}{1-v} \tau\Bigg|_\frac{\bar{\tau}}{2}^\bar{\tau} \right] \\

&= \left[ \frac{\bar{\tau}}{2} + \frac{1-v}{1+v} \frac{\bar{\tau}}{2} \right] \left[ \frac{\bar{\tau}}{2} + \frac{1+v}{1-v} \frac{\bar{\tau}}{2} \right] \\

&= \frac{\bar{\tau}}{2} \left[ 1 + \frac{1-v}{1+v} \right] \frac{\bar{\tau}}{2} \left[ 1 + \frac{1+v}{1-v} \right] \\

&= \frac{\bar{\tau}}{2} \left[ \frac{1+v}{1+v} + \frac{1-v}{1+v} \right] \frac{\bar{\tau}}{2} \left[ \frac{1-v}{1-v} + \frac{1+v}{1-v} \right] \\

&= \frac{\bar{\tau}}{2} \frac{\bar{\tau}}{2} \frac{1+v+1-v}{1+v} \frac{1-v+1+v}{1-v} \\

&= \frac{\bar{\tau}}{2} \frac{\bar{\tau}}{2} \frac{2}{1+v} \frac{2}{1-v} \\

&= \frac{\bar{\tau}}{\cancel{2}} \frac{\bar{\tau}}{\cancel{2}} \frac{\cancel{2}}{1+v} \frac{\cancel{2}}{1-v} \\

&= \frac{\bar{\tau}\bar{\tau}}{(1+v)(1-v)} \\

\end{align}$$

This is based on the fact that

$$ \left\{ \begin{array}{lll} \int_0^\tau \delta\left( \tau' - \frac{\bar{\tau}}{2}\right) \textrm{d}\tau' = 0 & \text{if} & 0 \leq \tau < \frac{\bar{\tau}}{2} \\ \int_0^\tau \delta\left( \tau' - \frac{\bar{\tau}}{2}\right) \textrm{d}\tau' = 1 & \text{if} & \frac{\bar{\tau}}{2} \leq \tau \leq \bar{\tau} \\ \end{array} \right.$$