This question has been asked before in the form of the 'Twin paradox' , and there are 42 pages of questions on this site alone when I search for 'twin paradox'.

For example

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

Clarification regarding Special Relativity

The counterargument has been 'that the traveling twin has to accelerate and therefore is not in an inertial frame of reference.'

OK, I believe that; it is consistent with the lay explanation of general relativity. This effect of gravitational time dilation was dramatically portrayed in the film 'Interstellar' when the crew goes to a planet near a black hole and everyone else gets a lot older.

So what happens if we use the 'equivalence principle' and provide a gravitational acceleration to the non traveling twin? I have read many scenarios described on this site where the observer will not be able to tell if he is on earth or on the accelerating spaceship, yet the many of the questions here on physics stack exchange are answered with the statement 'the spaceship traveler will age more slowly'.

If I can slow aging by 'dialing up gravity', or I can slow aging by traveling fast, why is there always the statement that 'the spaceship traveler will age more slowly'? Why does 1 g on a spaceship age you more slowly than 1 g on earth?


Suppose that the traveller twin in $t = -t_0$ is passing by the Earth with $v = 0.9 c$, and turns its engines on, in order to meet once again his brother at Earth. And to achieve this, the ship keeps the same acceleration $g$ until the moment of return.

At the moment $t_0$:

$$\frac{dt}{d\tau} = \frac{1}{{(1-0.9^2)}^{1/2}} = 2.294$$

For a frame at uniform acceleration $g$:

$$t = \frac{c}{g}senh\left(\frac{g\tau}{c}\right)$$ $$\frac{dt}{d\tau} = cosh\left(\frac{g\tau}{c}\right)$$

When $\tau = (+/-)45067942,31s$ => $t = (+/-)63206372,8s$ and $\frac{dt}{d\tau} = 2.294$

So, the total time until the twins meet again is: $t' = 2*45067942,31s = 2.86$ years for the traveller twin and $t = 2*63206372,8s = 4$ years for the Earth twin.

While both are under the same acceleration $g$, the gravitational potentials are very different.

In the Earth, it is enough to send a test mass faster than 11.4km/s for it escapes our well and never more return.

In an uniformly accelerated frame, there is no escape. If at any moment during the trip, a test mass is send to the Earth direction, with any velocity, one day the ship will meet it again, if the acceleration $g$ is never turned off.

It is almost a one directional black hole, in the meaning that there is no escape from any mass in the direction of the acceleration.

| cite | improve this answer | |
  • $\begingroup$ thank you! I cannot tell you how many times (as well as the above answers) well-meaning people have told me there is no difference. Yet when I calculate gravitational time dilation using GR and motion accelerated time dilation using SR, I get different ages. $\endgroup$ – aquagremlin Jun 30 at 16:12

Why does 1 g on a spaceship age you more slowly than 1 g on earth?

It doesn't. The gravitational time dilation at 1 g is rather small, approximately 2.19 seconds per century, as I mentioned in this answer. It doesn't matter whether that 1 g is due to gravity or a spaceship undergoing constant acceleration.

Such time dilation is negligible in the usual Twin Paradox scenarios.

The traveling twin in the Twin Paradox ages less than than the earthbound twin because the traveling twin has to switch inertial frames in order to return home. It's that change of reference frame that makes the big difference, not the acceleration. The acceleration is merely the mechanism whereby the change of reference frame is performed.

| cite | improve this answer | |
  • $\begingroup$ However, I think your answer does not include time dilation due to special relativity and the increasing velocity of the space ship. Thank you though. $\endgroup$ – aquagremlin Jun 30 at 16:13
  • $\begingroup$ @aquagremlin Yes, I'm only talking about the time dilation directly caused by gravitational potential or acceleration. My point is that that time dilation is negligible when discussing the Twin Paradox. $\endgroup$ – PM 2Ring Jun 30 at 16:32

One G ages you the same way whether you are sitting on a beach on earth with a mai tai in your hand or in deep space, accelerating your rocket ship at 9.8 meters per second squared. Increasing your G force by moving to a more massive planet or by jazzing up your rocket engines will slow down your clocks in the same way.

Should you choose to perform experiments to verify this I would advise substituting the mai tai with Herradura Gold tequila mixed with fresh-squeezed pink grapefruit juice. Remember to repeat the experiment enough times to achieve statistical significance.

One more point: If you are looking for experimental proof that biological activity is slowed down in a speeding reference frame, note this: If you are sitting in that speeding frame looking at brownian motion through a microscope, your own internal processes will be slowed down in exactly the same manner as the brownian motion, and you will observe no slowdown of the brownian motion because YOU are slowing down too.

| cite | improve this answer | |
  • $\begingroup$ I cannot find a SINGLE experiment where aging was investigated. Everyone assumes that because time is affected that probabilistic processes (the collision of atoms in chemical reactions, the catalysis of biochemical reactions by proteins, the effect of mass action) will be symmetrically affected. Yet all experimental proofs of special relativity only involve single particles and their lifetimes(in experiments involving decay). I would like to see an experiment at least where Brownian motion is examined under relativistic conditions. $\endgroup$ – aquagremlin Jun 30 at 16:17
  • $\begingroup$ @aquagremlin According to SR, all good clocks react the same way to velocity. If you have a special clock that doesn't behave in accordance with the Lorentz transformation, then an inertial observer could use that clock, in conjunction with a normal clock, to calculate their velocity without making an external distance measurement. But what would such a velocity even mean? $\endgroup$ – PM 2Ring Jun 30 at 16:57
  • $\begingroup$ It's not so much the idea of a 'clock' but rather the interaction of particles when I ask about mass action, etc. As all 'proofs' of SR look at individual particles or their motions as determined by EM waves, I have not found empirical or experimental info on the effect of SR on interactions mediated by forces. I hope my thinking is not too disordered, but I am imagining that SR affects the electric field lines that govern particle interactions differently than the particles that generate them. $\endgroup$ – aquagremlin Jun 30 at 18:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.