Special relativity with acceleration: Mathematically solving the twin paradox (1 Viewer)

blyatman · Sep 17, 2019

To help fix the misconception that general relativity is required to tackle accelerations in special relativity, I've provided an example to demonstrate how to solve the twin paradox in special relativity. This example also illustrates how mathematics is utilised in physics (the language of physics is, after all, partial differential equations). This example will only involve simple calculus from Adv Math, so it should be straightforward to follow.

Suppose we have 2 twins, A and B, both starting on Earth. Twin A remains on Earth, whilst twin B accelerates off in a rocketship before returning sometime later when 10 years have elapsed on earth. The objective is to find the proper time, $\tau$ , that has elapsed for twin B upon their return to Earth.

From the perspective of twin B, the differential form of the kinematic time dilation equation is:
$dt=\frac{d\tau}{\sqrt{1-\frac{v^2}{c^2}}}$
where $dt$ is the elapsed coordinate time for twin A (stationary frame), and $d\tau$ is the elapsed proper time for twin B (moving frame).

For twin B to make a roundtrip, we must vary their velocity such that they speedup to move away from Earth, slow down and stop at some point, and then return back to Earth in a similar manner. A simple velocity function that achieves this is a sine function: $v(t)=V_0\sin nt$ (note that the velocity $v$ must be a function of the coordinate time $t$ rather than the proper time $\tau$ to ensure that the velocity is symmetric). To arrive back at Earth after $t=10~\mathrm{yrs}$ , we can solve for $n$ based on this period:

$T=\frac{2\pi}{n}\Rightarrow n=\frac{2\pi}{T}=\frac{\pi}{5}~\mathrm{yr}^{-1}$

For simplicity, we'll set $V_0 = c$ so that the maximum velocity is $c$ (I'll justify this later*).

Hence,
$v(t)=c\sin\frac{\pi}{5}t$
$\Rightarrow dt=\frac{d\tau}{\sqrt{1-\sin^2\frac{\pi}{5}t}}=\frac{d\tau}{\sqrt{\cos^2\frac{\pi}{5}t}}=\frac{d\tau}{\left|\cos\frac{\pi}{5}t\right|}$
$\Rightarrow d\tau=\left|\cos\frac{\pi}{5}t\right|dt$

The goal is to calculate the elapsed coordinate time $\tau$ for twin B once 10yrs after elapsed on Earth. Hence:
$\int_0^{\tau}d\tau=\int_0^{10}\left|\cos\frac{\pi}{5}t\right|dt=4\int_0^{5/2}\cos\frac{\pi}{5}t\,dt$
$\Rightarrow\tau=\frac{20}{\pi}\left[\sin\frac{\pi}{5}t\right]_0^{5/2}$
$\Rightarrow\tau=\frac{20}{\pi}\approx6.4~\mathrm{yrs}$

Hence, only 6.4 yrs have elapsed for twin B, whereas 10 years have elapsed for twin A!

Hope this helps gives some insight into how to treat simple accelerations in special relativity, as well as illustrate how calculus is utilised in physics. Note that no general relativity was required!

*Justification for $V_0=c$ : From above, this value was chosen to simplify the integration. We could have set $V_0=\lim_{v\rightarrow c} v$ , which would mean that in the limit as $v\rightarrow c$ , twin B would experience a proper time of 6.4 yrs.

Special relativity with acceleration: Mathematically solving the twin paradox (1 Viewer)

blyatman

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