36-4

In this problem, the father wants take a very speedy trip, so that his daughter left on earth will age 40 more years than he does. But he ages 4 years during the trip, so she must age 44 years total. More explicitly, if we let $n$ be the daughters initial age and $m$ her final age, we have

age	d	f
i)	$n$	$n+20$
f)	$m$	$m-20$

The we have $(m-20)-(n+20)=4$, that is, the father ages $4$ years. Simplifying the expression yields $m-n=44$, so that the daughter ages 44 years.

On the diagram below (fig.4) the daughter stays on earth, so her line is just the $ct$ axis. She is at $x=0$ at every time $t$. The father moves with constant speed away from earth for 2 years in his frame ($t'=2 {\rm yr}$) and then abruptly turns around and heads back to earth at the same constant speed for another 2 years (in his frame). The daughter ages 44 years in her frame during the whole trip, or 22 years on the father's way out and 22 more on his way back.

Figure 4: Space-time diagram for 37-4.

So, our event of interest is when the father turns around in his trip. We know both the $t$ and $t'$ coordinates of this point, so lets see if we can find the corresponding $\beta$. The Lorentz time transformation says that

$$ct'=\gamma(ct-\beta x),$$

so that we need to know $x$, that is, how far the daughter measures the father to have traveled before turning around. Well, if the father moves with constant speed $v=\beta c$ and for a time $t$, then the distance traveled is

$$x=vt=\beta ct.$$

We could also determine this from the second Lorentz transform equation noting that the $x'$ coordinate of when the father turns around is exactly 0 because one is always at rest in their own frame of reference (the origin of his coordinate system is always centered on him). Then we have

$$0=x'=\gamma(x-\beta ct),$$

which immediately gives $x=\beta ct$ since $\gamma\neq 0$.

So, now we can substitute for $x$ giving

$$\begin{eqnarray*} ct' &=& \gamma(ct-\beta(\beta ct))\\ &=& \gamma ct(1-\beta^2)\\ &=& \gamma ct(\gamma^{-2})\\ &=& ct/\gamma, \end{eqnarray*}$$

or

$$t'=\frac{t}{\gamma},$$

which is the equation for time dilation: since $\gamma>1$ the time $t$ measured in the stationary frame is larger than the time $t'$ measured in the moving frame by the factor $\gamma$. Note that we made the substitution $1-\beta^2=\gamma^{-2}$. (Show this is true.)

Now we just substitute in for gamma and solve for $\beta$

$$\begin{eqnarray*} t' &=& \sqrt{1-\beta^2}t\\ (t')^2 &=& (1-\beta^2)t^2\\ \left... ...right)^2}\\ &=&\sqrt{1-\left(\frac{2}{22}\right)^2}\\ &=&.996. \end{eqnarray*}$$