SR time dilation discussion meaningless when there is gravity?

Let us say you have three clocks fixed to the spinning earth at the equator. If you set one of this clocks on a plane going eastwards around the globe it will have measured a lesser amount of time passed than the earth-bound clock when it gets back. The faster the plane travels the larger the difference will be. If you set the third clock on a plane going westwards around the globe it will have measured a larger amount of time passed than the earthbound clock when it gets back. This is of course because of the rotation of the earth. For the sake of simplicity we are ignoring gravitational time dilation.

Now if you send clocks out in the solar system around some arbitrary path but not to close to Jupiter or any other planet and back again the difference in time passed or "aging" of clocks can basically be calculated by knowing the velocity of the clock as compared to the Sun or the solar system barycentre along the paths of the clocks. This is again ignoring gravitational time dilation.

Is the discussion of the twin paradox meaningless in a real world scenario where you have gravity? Should the question of which twin that has been aging the most really be answered by something along the line of "as long as you are close to the Earth the twin that moves faster as compared to the centre of the earth ages less, but out in the solar system the twin that moves faster as compared to the solar system barycentre ages the less and out in the galaxy the twin that moves faster with respect to the center of the galaxy ages less".

Do you always have to pay respect to gravitational fields in a way that makes the classical twin-paradox discussion that ignores gravitational fields meaningless?

• The twins paradox has not much to do with relative motion, it is that one twin accelerates (or is in a gravitational field). – Rob Jeffries Mar 23 at 16:01

$$\let\g=\gamma \let\d=\delta \let\De=\Delta \def\rA{{\rm A}} \def\rB{{\rm B}} \def\rC{{\rm C}} \def\rD{{\rm D}} \def\rE{{\rm E}} \def\rF{{\rm F}} \def\rG{{\rm G}} \def\rU{{\rm U}} \def\tA{t_\rA} \def\tB{t_\rB} \def\tC{t_\rC} \def\tD{t_\rD} \def\tE{t_\rE} \def\tF{t_\rF} \def\tG{t_\rG} \def\xA{x_\rA} \def\xB{x_\rB} \def\xC{x_\rC} \def\xD{x_\rD} \def\xE{x_\rE} \def\xF{x_\rF} \def\xG{x_\rG} \def\tauA{\tau_\rA^\phz} \def\tauB{\tau_\rB^\phz} \def\tauC{\tau_\rC^\phz} \def\tauD{\tau_\rD^\phz} \def\tauE{\tau_\rE^\phz} \def\tauF{\tau_\rF^\phz} \def\tauG{\tau_\rG^\phz} \def\phz{{\phantom0}} \def\cH{{\cal H}} \def\cS{{\cal S}} \def\dt{\d t} \def\Dt{\De t} \def\D#1#2{{d#1\over d#2}}$$ I think the question shouldn't be closed leaving the last word to an erroneous answer. Up to now I've seen only words and no equations but IMO physics without numbers and equations is just chat. I want to fill the gap giving a solution, complete as well as standard - a homework exercise, I could say. Whoever disagrees is asked to show what's wrong in my answer.

We are in flat spacetime. No stars, planets, galaxies nearby. In other words, only SR comes into play. From a space station $$\cS$$ a spaceship $$\cH$$ is started, which for a time $$\dt$$ is moving with a constant proper acceleration $$a$$ (hyperbolic motion) until it reaches a cruise velocity $$v$$ (wrt $$\cS$$). A cruise phase follows when $$\cH$$ keeps its velocity constant for a time $$\Dt$$ (as measured by $$\cS$$ clocks). Now $$\cH$$ accelerates backwards (with proper acceleration $$-a$$) for a time $$2\,\dt$$, thus reaching velocity $$-v$$. The return journey begins, lasting $$\Dt$$. $$\cH$$'s mission ends with a braking phase: acceleration $$a$$, duration $$\dt$$. At this point $$\cH$$ is still near $$\cS$$.

The total journey time, as measured by $$\cS$$, is $$2\,\Dt+4\,\dt$$. How long did the journey last as measured by spaceships clock?

Events

I'll mark 7 notable events:

• A: start of $$\cH$$
• B: end of acceleration
• C: end of cruise phase - counter-acceleration begins
• D: max distance from start point is reached
• E: return journey begins
• F: end of return journey - brake phase begins
• G: end of braking and stop.

Only one inertial frame is used: $$\cS$$ rest frame, with coordinates $$(t,x)$$. The whole mission takes place at $$x\ge0$$. Each event, say $$U$$, has its own coordinates $$t_\rU,x_\rU$$. Time and space origins are fixed at $$\rA$$, so that $$\tA=0$$, $$\xA=0$$. We already know all $$t$$-coordinates: \eqalign{ \tB &= \dt \cr \tC &= \dt + \Dt \cr \tD &= 2\,\dt + \Dt \cr \tE &= 3\,\dt + \Dt \cr \tF &= 3\,\dt + 2\,\Dt \cr \tG &= 4\,\dt + 2\,\Dt.\cr} $$x$$-coordinates instead are to be calculated, but $$\xB$$, $$\xC$$ will suffice as the others follow by symmetry. The same holds for proper times. I'll only fix proper time origin by setting $$\tauA=0$$.

The $$\rB$$-event

Equations of hyperbolic motion are \eqalign{ \tB &= t_0\,\sinh {\tauB \over t_0}\cr \xB &= x_0 \left(\cosh {\tauB \over t_0} - 1\right) \cr} \tag1 where $$t_0 = {c \over a} \qquad x_0 = c\,t_0 = {c^2 \over a}.$$ By eliminating $$\tauB$$ in eqs. (1) $$\xB = c\,\sqrt{t_0^2 + \tB^2} - x_0 = c\,\sqrt{t_0^2 + \dt^2} - x_0 \tag2$$ and from (1) $$\tauB = t_0 \sinh^{-1} {\tB \over t_0} = t_0 \sinh^{-1} {\dt \over t_0}.\tag3$$

The $$\rC$$-event

From (2) and from definition of $$v$$ $$v = \D{\xB}{\tB} = {c\,\tB \over \sqrt{t_0^2 + \tB^2}} = {c\,\dt \over \sqrt{t_0^2 + \dt^2}}$$ $$\sqrt{1 - {v^2 \over c^2}} = {t_0 \over \sqrt{t_0^2 + \dt^2}}$$ $$\xC - \xB = v\,\Dt$$ $$\tauC - \tauB = \Dt\,\sqrt{1 - {v^2 \over c^2}} = {t_0\,\Dt \over \sqrt{t_0^2 + \dt^2}}.\tag4$$

Total proper time was requested, i.e. $$\tauG = 4\,\tauB + 2\,(\tauC - \tauB) = 4\,t_0 \sinh^{-1} {\dt \over t_0} + {2\,\Dt \over \sqrt{1 + (\dt/t_0)^2}}.$$ (eqs. (3), (4) were used). To be compared with $$2\,\Dt+4\,\dt$$.

Let's keep fixed the acceleration phase parameters: $$a$$, $$\dt$$ and then $$v$$. It can be seen that by increasing $$\Dt$$ the difference between both times grows as large as we like, i.e. twins effect is proportional to journey's duration whereas acceleration phases give a constant contribution.

• "We are in flat spacetime. No stars, planets, galaxies nearby". You are imagining a universe without celestial bodies? Sure you are going to find places in the universe where the gradient of the gravitational potential is very small and the difference in "gravitational time dilation" between the two twins is small even when one of them accelerates away from the other. – Agerhell Mar 28 at 13:10

You are correct to say that the twin paradox is solvable with gravitational time dilation.

It is a misconception to think that the twins age differently just because they move at a constant speed relative to each other. Now speed is symmetrically relative, what that means is both twins could say, that the other one is moving compared to them so they should age less. The twin on Earth could say, that the other twin is moving relative to him, and the twin on the spaceship could say that the twin on Earth is moving relative to him so both could say that the other one should age less.

The solution is gravitational time dilation. It is because acceleration is absolute in the universe. The twin in the spaceship is moving with constant speed, so non if the twins ages differently, until the point of return. At the point of return, the spaceship has to slow and accelerate, that is it has to turn around. Now you are right to think that as per the equivalence principle, this is the same as the effects of gravity, which slow down the spaceship in the time dimension.

Nothing happens until the point of return, because the spaceship is moving with constant speed and both twins age the same way. Now at the point of return, the spaceship accelerates, and that has the same effect as gravity, and the spaceship slows down in the time dimension. An object with rest mass (like the spaceship) moves in the time dimension with speed c, and in the spatial dimensions with speed less then c.

Now when the spaceship turns around, it accelerates, and this causes it to slow down in the time dimension. The universe is built up so, and the four vector is built up so that the magnitude of the four vector has to be c. Now if the spaceship accelerates, its spatial speed changes, so in the spatial dimensions its speed changes, but the four vector's magnitude has to stay c. To compensate, the spaceship will slow down in the time dimension, and that means that the twin on the spaceship will age slower compared to the twin on Earth. This is for the period of turning only, when the spaceship turned around and starts to move with constant speed again towards Earth, the twins will age the same way again.

But there was a period, at the turn, when they aged differently, and that difference is now there. The twin in the spaceship aged slower, relative to the twin on Earth. So when they meet again on Earth, they will see that the twin in the spaceship is younger. There is a distance now between them in the time dimension, and that stays so (as long as non of them accelerates).

Now this is only true if we disregard the difference between the gravitational potentials in the region of space where the spacehip is moving and the Earth's potential. You are right that the difference between the gravitational potential where the spaceship is and the Earth will too affect the way the two twins age.

• I tried to read the Nasa documentation but it is not a super-easy document. It seems to me that if you send a spacecraft from Earth and to for instance Jupiter and you want the spacecraft to turn on its engines at exactly the right time to do some maneuver when it gets to Jupiter it has do do two things: Compensate for how the gravitational potential of the Earth, the Sun and Jupiter will influence clocks. Compensate for how the velocity of the spaceship in relation to the same celestial objects will influence clocks. descanso.jpl.nasa.gov/monograph/series2/Descanso2_all.pdf – Agerhell Mar 23 at 17:12
• @Árpád Szendrei: If your argument were correct, then the twins' difference in age would be the same regardless of how long the trip lasted, as long as the periods of acceleration & deceleration were the same. Are you claiming that the length of the trip is unimportant? – D. Halsey Mar 23 at 18:52
• @D.Halsey yes, as long as the trip is with constant velocity. The only thing that matters is the period of acceleration as you state it correctly. – Árpád Szendrei Mar 23 at 22:21
• @Árpád Szendrei: But that is incorrect. The difference in the twins' ages gets larger as the duration of the trip increases. – D. Halsey Mar 23 at 22:46
• @D.Halsey false, that is a misconception. If the spaceship travels with constant velocity, there is no difference how they age. speed is symmetrically relative. It is only when the spaceship accelerates (or is in a different gravitational zone). In my answer I only take acceleration, and I disregard the difference between Earth's stress-energy and empty space's (if that is where the spaceship travels) stress-energy. The only thing that matters is when the spacehip accelerates. That is equal to as if the spaceship would be in a different gravitational zone then Earth's gravitational zone. – Árpád Szendrei Mar 23 at 22:50