# Proper Time Along a Trajectory with Changing Velocity

I'm relatively new to SR and just encountered the Twin Paradox. I don't think I had trouble understanding the resolution to the paradox, but I was curious how an observer in an inertial reference frame during the first half of the accelerating trajectory would calculate proper time.

Letting the observer be at rest along DE, and setting the total time travelled along the stationary trajectory to $$\Delta t$$, the time undergone during that section should just be $$\tau_1 = \frac{1}{2}\Delta t.$$

Keeping with that reference frame though, I calculated that the velocity they measure along EF would be $$v_2= \frac{-2v}{1+v^2/c^2},$$ corresponding to a time interval of $$\tau_2= \frac{\Delta t}{2\gamma} = \frac{1}{2}\Delta t \sqrt{1-(\frac{-2v}{1+v^2/c^2})^2},$$

which clearly doesn't add up to $$\tau = \tau_1+\tau_2 = \Delta t \sqrt{1-v^2}$$. Is the problem just assuming a discrete change in velocity rather than continuous acceleration, or some other part of the setup?

Edit: just fixed a minus sign in my initial velocity formula that I mistyped - the issue still stands.

In Alice's frame: Betty leaves at time $$0$$, arrives at Alpha Centauri at time $$5$$, having aged $$3$$ years, and returns to earth at time $$10$$, having aged another $$3$$ years.

In Outbound Betty's frame: Betty separates from earth at time $$0$$, touches Alpha Centauri at time $$3$$ having aged $$3$$ years, and rejoins earth at time $$16.667$$, having aged another $$3$$ years.

In Inbound Betty's frame: Betty separates from earth at time $$-10.667$$, touches Alpha Centauri at time $$3$$ having aged $$3$$ years, and rejoins earth at time $$6$$, having aged another $$3$$ years.

How to see this: Use the Lorentz transformation to calculate the coordinates of events $$E$$ and $$F$$ in Outbound Betty's frame. To see what's happening in Inbound Betty's frame, you can do the same, remembering that the relevant velocity is now $$-v$$, not $$+v$$. Or alternatively, save yourself some work by using the symmetry between Outbound and Inbound Betty, so that if Outbound says that Inbound's journey takes $$16.667$$ years, then Inbound must say the same of Outbound's journey.

• Thanks for commenting: I'm glad instantaneous acceleration isn't an issue, but I still don't understand the error in my calculation. Could you give a brief summary of how you calculated the 13.667 time interval for the inbound segment from outbound betty's frame? Commented Jul 8 at 3:07
• According to Inbound-Betty when her clock reads “-10.66”, Outbound-Betty separates from Earth. When the Bettys meet at AlphaCentauri, both Betty-clocks will read “3” (as if they had arranged a time-calibration so that non-inertial Betty continues as Inbound-Betty). Of course, the elapsed time of DE on inboundBetty’s clock is 13.66 (from reading “-10.66” to reading “3”). See the spacetime diagram in my answer. Commented Jul 10 at 16:57
• @robphy: I've corrected my arithmetic. Thanks for catching this. Commented Jul 12 at 11:21
• @Lambda : Outbound Betty says that the time interval from D to E is three years and the time interval from E to F is 16.667 years. By symmetry, Inbound Betty must say the time interval from D to E is 16.667 years and the time interval from E to F is 3 years. I assume that Inbound Betty does not reset her clock at event E , so that at E the clock says 3. In order for the time interval from D to E to be 16.6667, she must say that E took place at −10.6667 . You could get the same answer by applying Lorentz transformations, but this is quicker and simpler. Commented Jul 12 at 11:22

Below my original answer, find UPDATED calculations of some features of the problem and an attempt to address your calculations.

This spacetime diagram with the ticks drawn might help you visualize the calculation given by @WillO, as well as analyze your attempted calculation. (The rotated graph paper makes drawing the ticks easier when using nice speeds like (4/5)c, which has a rational Doppler factor [here,3].)

I've drawn the lines of simultaneity for the two piecewise-inertial portions of Betty's non-inertial worldline. (The lines are parallel to the spacelike diagonals of the corresponding "clock diamonds", which are traced out by a light-clock. All clock diamonds have the same area.)

I've included clock readings for the Outbound and Inbound worldlines so that their clocks agree when they meet at event E [on Alpha Centauri]. Both read "3" at E (in this setup).

Note that

• Outbound Betty (who visits D and E, but not F) says
DE took 3 ticks (reading 0 to 3),
EF took 13.66 (reading 3 to 13.66), and thus
DF took 16.66 ticks (reading 0 to 16.66).
• Inbound Betty (who visits E and F, but not D) says
DE took 13.66 ticks (reading -10.66 to 3),
EF took 3 ticks (reading 3 to 6) and thus
DF took 16.66 (reading -10.66 to 6).

## UPDATE

Here's some spacetime geometrical calculations, which underlie and clarify standard textbook formula [when they are used correctly]. Hopefully this will help you sort out your attempted calculation.

Minkowski right-triangle DME,
with Minkowski-right-angle at M and where DM and ME are the time and space components of DE according to the observer ([Outbound-]Alice) along DM.

Using the Minkowski-angle (called the rapidity) $$\theta_{out}$$ between timelike rays DM and DE, think of

• DM is the adjacent side (here, 5)
• ME is the opposite side (here, 4)
• DE is the hypotenuse (since it is opposite the right-angle) (here, 3)

We check that \begin{align} (HYP)^2 &=(ADJ)^2-(OPP)^2 \\ (DE)^2 &=(DM)^2-(ME)^2\\ (3)^2 &=(5)^2-(4)^2 \end{align}

According to [Outbound-]Alice, the velocity of DE is the slope (with t running upwards). $$\beta_{out}=(v_{out}/c)=\tanh\theta_{out}=\frac{OPP}{ADJ}=\frac{ME}{DM}=\frac{4}{5}$$ and the time-dilation factor $$\gamma_{out}=\cosh\theta_{out}=\frac{ADJ}{HYP}=\frac{DM}{DE}=\frac{5}{3}$$

Since there are other Minkowski-right triangles, it might be good to adopt a notation:
for Minkowski right-triangle DME (right-angle at M, ADJ = DM, OPP = ME, HYP =DE)

• $$\displaystyle v_{DME}=\mbox{tangent}=\frac{OPP}{ADJ}=\frac{ME}{DM}=\frac{4}{5}$$
• $$\displaystyle \gamma_{DME}=\mbox{cosine}=\frac{ADJ}{HYP}=\frac{DM}{DE}=\frac{5}{3}$$

Note the letter orderings and positions!

If you didn't know DE but do know the time-dilation factor formula (essentially the identity for hyperbolic-cosine in terms of hyperbolic-tangent), you could use that formula:

$$DE=\frac{DM}{\gamma_{DME}}=\frac{DM}{\frac{1}{\sqrt{1-\beta^2_{DME}}}}=(5)\sqrt{1-(4/5)^2}=3$$

Minkowski right-triangle FME (for [Inbound-]Alice)

• $$\displaystyle v_{FME}=\mbox{tangent}=\frac{OPP}{ADJ}=\frac{ME}{FM}=\frac{4}{-5}=-\frac{4}{5}$$
• $$\displaystyle \gamma_{FME}=\mbox{cosine}=\frac{ADJ}{HYP}=\frac{FM}{FE}=\frac{-5}{-3}=\frac{5}{3}$$

Minkowski right-triangle DGF (for [Outbound-]Betty)

• $$\displaystyle v_{DGF}=\mbox{tangent}=\frac{OPP}{ADJ}=\frac{GF}{DG}$$
• $$\displaystyle \gamma_{DGF}=\mbox{cosine}=\frac{ADJ}{HYP}=\frac{DG}{DF}$$

I drew the diamonds for DG (16.66) but not for GF.
Let's suppose we know neither.

Let me suggest a spacetime-geometrical argument motivated by the physics. By the Relativity Principle, Minkowski right-triangles DME and DGF are "similar" Minkowski-right-triangles with the same rapidity-magnitude $$\theta_{out}$$. We expect that corresponding sides are in proportion.

Thus, $$\displaystyle \frac{DM}{DE}=\gamma_{DME}\stackrel{RP}{=}\gamma_{DGF}=\frac{\stackrel{?}{DG}}{DF}$$ implies $$\stackrel{?}{DG}=(DF)\gamma_{DGF}\stackrel{RP}{=}(DF)\gamma_{DME}=(10)(5/3)=50/3 \approx 16.66$$ We expect the velocities to be equal in magnitude but opposite in sign. So, $$\displaystyle \frac{ME}{DM}=\beta_{DME}\stackrel{RP}{=} -\beta_{DGF}= -\frac{\stackrel{?}{GF}}{DG}$$ implies $$\stackrel{?}{GF}= -(DG)\beta_{DGF}\stackrel{RP}{=} -(DG)\beta_{DME}=-(50/3)(4/5)=-40/3 \approx -13.33.$$ So, $$FG=(-GF)=13.33$$.

## Now to your attempted calculation

It seems to want to:
use the inertial-segments DE and EF of non-inertial Betty's round trip and
some standard textbook formulas that would be used by Outbound-Betty (along DEG) and maybe Inbound-Betty (along EF).
to write DF as the sum of two elapsed times $$DF=DX+XF$$
using an event X between D and F on that inertial worldline (of Alice).

But which event X is not clear to me.

Assumptions: we know

• DE=3, EG=13.66 and that, according to Outbound-Betty, G is simultaneous with F.
• according to Outbound-Betty, that Alice has velocity (-4/5)c
• according to Alice, the Bettys have the same outbound and inbound speeds of (4/5)c.

From the diagram, I see that one can proceed in three ways

1. Outbound-Betty alone.
Outbound-Betty would use event H which she says is simultaneous with event E.
Knowing DE and $$v_{Alice}=(-4/5)$$, DH is gotten from right-triangle DEH (where $$\gamma_{DEH}=(DE)/(DH)$$). $$DH=(DE)/\gamma_{DEH}=(DE)/\gamma_{DME}=(3)/(5/3)=(9/5)=1.8.$$ Since DEH and DGF are similar right-triangles (where $$\gamma_{DGF}=(DG)/(DF)$$), then DF is determined. $$DF=(DG)/\gamma_{DGF}=(DG)/\gamma_{DEH}=(50/3)/(5/3)=10.$$ So, get HF as (DF-DH). $$HF=(DF)-(DH)= 10 - 1.8 = 8.2.$$ (No need for Inbound-Betty or Alice, and no need to know EF.)

2. Outbound-Betty and Inbound-Betty, but no Alice.
Outbound-Betty would use event H which she says is simultaneous with event E to find DH, but not directly find HF (as above).
Outbound-Betty would determine EF (along Inbound-Betty's worldline) using right-triangle EGF with rapidity $$\phi=\theta_{out}+\theta_{in}$$ (where $$\gamma_{EGF}=(EG)/(EF)$$),
but then, the rest is up to Inbound-Betty. (One can use the addition-formula then calculate the corresponding time-dilation factor. Using the Doppler factor for $$(v/c)=\tanh(\theta)=(4/5)c$$ (which is $$k=\exp(\theta)=3$$), one can use a fancy method to get $$\gamma_{EGF}=\cosh(\ln 9)=(41/9)\approx 4.55$$ and $$v_{EGF}=\tanh(\ln 9)=(40/41)\approx 0.9756c$$.)

Inbound-Betty would use event Q which she says is simultaneous with event E. Then Inbound-Betty can find QF.
So, $$DF=DH+\stackrel{?}{HQ}+QF.$$
However, Inbound-Betty can't easily use EF to find HQ (sometimes called the "missing time").
Inbound-Betty would need to use an event on her worldline simultaneous with H (which would be before event E).
So, there is no easy method for Inbound-Alice (using standard textbook methods). (Somehow "appealing to symmetry" because of an equal return speed would be "a trick" and would not be a general method.)

3. Outbound-Betty, Inbound-Betty, and Alice.
We get EF as above in attempt 2.
We ask Alice to use event M which she says is simultaneous with E.
Thus, with Alice's measurements (via right-triangles DME and FME), we write $$DF=DM+MF$$.

(Although it took me a lot of time, it was helpful for me to write this all down.)

I think an important lesson is that "equations" and "formulas" can easily be misused. Draw a good spacetime diagram with labels and use formulas with extra notation to specify what one is referring to.

• I've been out on vacation and forgot to come back to this until now: thanks so much for the detailed answer! Commented Jul 23 at 3:16

Your picture says v=0.8c so time for Betty is 0,6 times time of Alice. what you calculated for v I don't understand. the velocity of Betty is always ±0.8c from E to F its 4 Ly in 5y same as from D to E

There is another way of looking at the problem: we replace the coordinates by the speeds in the Lorentz transformations (LT), i.e.

$$v'=\gamma(v-at)\;\;,\;\;t'=\gamma(t-\frac{ac}{a_{l}^{2}})$$

we assume for the light : $$t=\frac{c}{a_{l}}$$

with the condition: $$(a_{l}t')^{2}-v'^{2}=0 \;\;,\;\; (a_{l}t)^{2}-v^{2}=0$$

$$a_{l}$$ : the acceleration limit.

by analogy with the LT: $$\gamma=\frac{1}{\sqrt{1-\frac{a^{2}}{a^{2}_{l}}}}=\frac{1}{\sqrt{1-\frac{(at)^{2}}{(a_{l}t)^{2}}}}=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$$

so : $$\Delta t'=\gamma(\Delta t-\frac{c}{a_{l}^{2}}\Delta a)$$

with $$j=\Delta a=0$$, $$\;\;\;\Delta t'=\frac{\Delta t}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$$

The question:"... but I was curious how an observer in an inertial reference frame during the first half of the accelerating trajectory would calculate proper time."

Éric Gourgoulhon, $$\mathbf{Special\ Relativity\ in\ General\ Frames,\ From\ Particles\ to\ Astrophysics}$$, page 41, 42, ISBN 978-3-642-37275-9, 2013 has the solution.

Important points to consider. Alice is not accelerating. Betty is an accelerated traveler.

• $$A$$ Alice and Betty are in the same inertial reference frame.
• $$AC_{1}$$ Betty accelerates.
• $$C_{1}P$$ Betty decelerates.
• $$PC_{2}$$ Betty accelerates.
• $$C_{2}B$$ Betty decelerates.
• $$P$$ and $$B$$ Alice and Betty are in the same inertial reference frame.
• The top speed at $$C_{1}$$ is bigger than $$(4/5)c$$ so the average between $$AP$$ can be $$(4/5)c$$. The proper time delta at $$C_{1}$$ for a very short space-time interval along the worldlines between Alice and Betty is much bigger compared to time when Betty just left at the event $$A$$.
• Invariance of space-time interval gives us the difference in aging, Betty's proper time of $$AP + PB$$ space-time interval is shorter than Alice's proper time of $$AB$$ space-time interval.

In short, the calculation of time-like space-time intervals gives us an observer's proper time.