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A common thought experiment illustrating gravitational time dilation involves Alice and Bob in an accelerated rocket. Alice is above Bob. Bob fires periodic signals upwards. Alice observes the signals at a lower rate due to kinematic Doppler and a link is made between acceleration and gravitational time dilation (Bob is at lower gravitational potential).

After some time has elapsed, Alice would presumably observe Bob a little younger than herself - they are twins. If she then moved (slowly) down the rocket to meet Bob, will Bob be younger? If so, how can this be understood from the perspective of someone observing the accelerating ship. If not, how does this fit in with the equivalence principle.

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    $\begingroup$ To go and meet Alice after some time Bob has to move quite fast for a quite long time. To go and meet Alice after a long time Bob has to move very fast for a very long time. $\endgroup$ – stuffu Jan 16 '17 at 13:41
  • $\begingroup$ @stuffu thanks - very interesting comment. Are you able to be expand on this (as an answer)? It would be really good if I could reconcile this with the familiar gravitational time dilation result. $\endgroup$ – Big AL Jan 16 '17 at 19:28
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    $\begingroup$ In an uniform gravity field g is proportional to time dilation. If we observe an accelerating rocket, rapidity of its rear is at, rapidity of its bow is redshift_factora*t. Instantaneous kinetic time dilation can be calculated from instantaneous rapidity. Bob's "gravitational time dilation" consists of two things: difference of kinetic time dilations and a Doppler shift. When Bob is approaching Alice, the non-real age difference that was caused by the Doppler shift time dilation is being converted to real age difference caused by kinetic time dilation. $\endgroup$ – stuffu Jan 17 '17 at 0:42
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Yes, Alice will find she has aged more than Bob has.

The equivalence principle tells us that acceleration is locally indistinguishable from a gravitational field. We use the qualifier locally because gravitational fields always have tidal forces and these are not present in an accelerating frame. However if the setup is small enough that the tidal forces are undetectable then acceleration and gravity are equivalent.

Suppose the acceleration of the rocket is $a$, then the force on a unit mass is also $a$, and the work needed to lift that unit mass from Bob to Alice is just:

$$ W = ah $$

where $h$ is the vertical spacing between Bob and Alice. The gravitational potential energy per unit mass in a gravitational field with gravitational acceleration $g$ is:

$$ U = gh $$

And if $g = a$ then the two energies are the same. I mention this because for a weak gravitational field there is a simple equation that relates the relative time dilation of two observers to their difference in gravitational potential energy:

$$ \frac{dt_B}{dt_A} = \sqrt{1 - \frac{2(U_A - U_B)}{c^2}} $$

In this case $U_A - U_B = +ga$, because Alice is above Bob, so we get:

$$ \frac{dt_B}{dt_A} = \sqrt{1 - \frac{2ga}{c^2}} \lt 1$$

And that means that Bob's time measurements, $t_B$, are always less than Alice's time measurements $t_A$. In other words Bob ages more slowly.

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  • $\begingroup$ Thanks for the response. Sorry, but I don't believe this answers my question. I am trying to understand a result of grav time dilation - specifically how two synchronised clocks can fall out of sync - by considering accelerated motion. Invoking EEP and quoting the grav time dilation formula doesn't achieve that. I am trying to see how one can arrive at the very same result, but by different means. $\endgroup$ – Big AL Jan 16 '17 at 19:20

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