# Does the twin paradox require both twins to be far away from any gravity field?

If one twin is on earth at 1 g and the other twin accelerates away from earth following a great big elliptical counterclockwise trajectory. He travels at .9 g for 20 years(according to earth time) as well as some small amount of left acceleration (.44 g since $$\sqrt{0.9^2+0.44^2}=1$$) to make the first semicircle . He then turns around and decelerates at .9 g in the opposite direction for 20 years(according to earth time) and now experiences some small amount of rightward acceleration (.44 g) to complete the semicircle. He then arrives at earth. Will they both be the same age?

Although the twin paradox has been discussed on this site on 42 pages of questions, there are only 4 pages that address the question of gravitational acceleration compared to motion acceleration. My intuition suggests that gravitational acceleration should have the same effect as motion acceleration so they should be the same age. In reading some of these questions I find contradictory answers. For example, this question

Gravitational Time Dilation vs Acceleration Time Dilation

suggests

a higher acceleration would yield the same results as more gravity

Why does only one twin travel in the twin paradox?

says

"that the earth twin experiences the same RELATIVE acceleration as the space twin (in the opposite direction) this is incorrect."

So which is the correct interpretation?

My twist to this question is the elliptical orbit. The direction of the principle component of acceleration (from the rear of the space ship) remains unchanged yet the lateral acceleration does change. That the lateral acceleration changes direction should not affect time dilation because that is a scalar quantity. Am I overlooking something when I make that statement?

Aside from the focus of the question in the title, there is a slight difference in the geometry. Gravitational acceleration gives a tidal effect whereas motion acceleration does not.This distinction though does not seem to enter into the calculation of time dilation though.

• I assumed you intended the magnitude of the acceleration to be 1 g, so I made some edits – Dale Jun 27 at 11:19
• Your question's title doesn't seem to correspond well with the body of your question. – PM 2Ring Jun 27 at 11:46
• please see edit – aquagremlin Jun 27 at 14:57
• Please do not make edits that invalidate responses already received. – Dale Jun 27 at 16:28
• The edit only clarified the objection. I will post again a CLEARER NEW question. – aquagremlin Jun 28 at 22:44

Will they both be the same age?

No, they will not. The traveling twin will be substantially younger. To calculate the age of each twin simply integrate the metric over their worldline: $$\tau = \int d\tau = \int \sqrt{g_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}}d\lambda$$ (in units where c=1). This procedure is general, it works for any twin motion and any spacetime, with or without gravity.

For example, this question Gravitational Time Dilation vs Acceleration Time Dilation suggests a higher acceleration would yield the same results as more gravity

Unfortunately, this one is worded a little poorly in a way that appears to be contributing to your confusion. Gravitational acceleration does not cause time dilation. Gravitational time dilation is caused by the gravitational potential. Also, the equivalence principle only applies over small enough regions of spacetime that spacetime curvature can be neglected.

So what would be true is that a clock on the ground on earth would tick slower than a clock raised 1 m off the ground on earth, and a clock on the back of the rocket would tick slower than a clock raised 1 m off the back of the rocket, and the difference in tick rate would be the same for both cases. That is how the equivalence principle would apply in this scenario.

That the lateral acceleration changes direction should not affect time dilation because that is a scalar quantity. Am I overlooking something when I make that statement?

Since the change in lateral acceleration is experimentally detectable it is in itself sufficient to break the symmetry. However, in this case it is rather irrelevant since acceleration does not cause time dilation anyway. But the two twins are in no way symmetric in this version of the problem.

Gravitational acceleration gives a tidal effect whereas motion acceleration does not.

This is correct. In fact, the tidal effect you mention is spacetime curvature. So the equivalence principle only is valid over regions of spacetime small enough that tidal effects are negligible.

• thank you for a direct answer. Though I am still troubled by your statement that " a clock on the back of the rocket would tick slower than a clock raised 1 m off the back of the rocket". Whether the clock is on the floor of the rocket or on a table on the floor of the rocket should make no difference. It is accelerating the same in both instances. – aquagremlin Jun 27 at 15:01
• My conclusion from your statement " Gravitational acceleration does not cause time dilation. Gravitational time dilation is caused by the gravitational potential. " is that gravity may feel the same as acceleration due to motion, but the effect on spacetime is very different. That we feel they are the same is an illusion caused by the limits of our senses. – aquagremlin Jun 27 at 15:04
• @aquagremlin as you said “It is accelerating the same in both instances.” That is indeed the whole point of the example. The acceleration is the same but the (pseudo) gravitational time dilation is different because time dilation is based on potential and not acceleration. – Dale Jun 27 at 15:46
• @aquagremlin you said “gravity may feel the same as acceleration due to motion, but the effect on spacetime is very different”. Locally they are identical. The problem I was addressing was not an incorrect application of the equivalence principle, but a misunderstanding of how gravitational time dilation itself works. It is a common mistake to think that gravitational time dilation depends on the gravitational acceleration when in fact it depends on the gravitational potential – Dale Jun 27 at 16:22
• So if you understand the no-gravity twin paradox 1st, it's the change in velocity times the distance to Earth that causes the Earth clock to jump forward on turn around. That is exactly why it's the potential, and not $g$-force alone that matters. – JEB Jun 27 at 22:56

Gravitational time dilation on the surface of the Earth vs. life at infinity is minuscule and has no impact on the Twin Paradox.

Time dilation due to acceleration does not play a role in the age difference for the traveling twin: all that matters is that he changes direction after traveling at high speed. Including linear acceleration only confuses the matter, and adding a big elliptical loop just adds another dimension.

It's best to understand the idealized twin paradox 1st: that is the one with instant acceleration. If Earth twin says each leg lasts $$T$$, then he ages $$2T$$ for whole trip.

Meanwhile he sees space-twin age $$T/\gamma$$ on each leg of the trip.

The paradox arises because space twin also sees himself age $$T/\gamma$$ on each leg of the trip, but he sees Earth twin age $$T/\gamma^2$$ on each leg of the trip.

Note that:

$$2T/\gamma^2 \ne 2T$$

so that space twin has a discrepancy of $$\Delta T = 2T(1-1/\gamma^2)$$.

Once can look too gravitational time dilation during the acceleration, but the problem is that space twin took $$0$$ seconds to turn around in his reference frame. He also to $$0$$ seconds to turn around Earth's reference frame.

Note that:

$$0 - 0 \ne \Delta T$$

However, at the turn around event, the Earth's clock is both at $$T/\gamma$$ (for the outgoing twin) and at $$T/\gamma + \Delta T$$ (for the ingoing twin) at the same time.

When the space twin switches reference frames, the Earth clock advances by $$\Delta T$$. You can work that a gravitational time dilation back on Earth would correspond to $$\Delta T$$, but that has the unfortunate property of being reversible, if the twin turns around again...and nobody wants to accept time going backwards, so considering it as gravitational time dilation is tricky.

I think it's better to remain in flat space time and be cognizant of the fact that the time on Earth at the turn event is not well defined, and depends on the velocity of the space twin. If he turns around, Earth time jumps forward; however if he decides to accelerate away from Earth even faster, then time on Earth can jump backwards.

• Thank you for answering but I’m afraid your answer has too much information. Especially when you say “ Time dilation due to acceleration does not play a role in the age difference for the traveling twin: all that matters is that he changes direction after traveling at high speed.“ This seems distracting and confusing. So you are saying that being in 1 g at earth gravity is not the same as accelerating at 1 g in a space ship? – aquagremlin Jun 27 at 2:58
• @aquagremlin You seem to have a misconception that acceleration resolves the twin paradox. It doesn’t. “Time dilation due to acceleration does not play a role in the age difference for the traveling twin: all that matters is that he changes direction after traveling at high speed.” - Is exactly the correct resolution of the paradox. If you find this “distracting and confusing”, then you don’t understand how the paradox is resolved and should give it more thought. This answer is correct +1 – safesphere Jun 27 at 4:52
• @aquagremlin I think that your confusion stems from your belief that acceleration is the key to the twin paradox. In the simplest version of the twin paradox, the age difference happens because the traveling twin changes inertial reference frame, but the stay at home twin doesn't change frames. True, acceleration is involved in the process of changing frames, but it's the frame change which is crucial. – PM 2Ring Jun 27 at 12:07
• please see edit – aquagremlin Jun 27 at 14:58
• It may help to consider the space twin to be a photon going say 5 light years and hitting a mirror. The whole journey out, T_earth = 0y. The nanosecond after reflection, T_earth = 10y, and it holds there the whole way back. On arrival, photon's age is 0, Earth is 10. All the 'missing' time occurs at the instant of reflection because "now" on Earth (from 5 light years out) can be any value within a 10 year span, depending only on your speed (see: the Andromeda Paradox). – JEB Jun 27 at 17:08