# How exactly to calculate the age difference in Twin paradox

How exactly to calculate the age difference in "Twin paradox"

In so much of the text I have seen the "twin paradox" is resolved by simply showing how the condition is asymmetrical, they make no calculation on how this asymmetry results in age difference. I have seen some claiming that its complete resolution requires General Relativity too, whereas other claim to resolve it only through special relativity?

I would like to know its resolution using only S.R if possible.
(consider the typical case, when twin B, first accelerates away from twin A, then travels at constant speed, then deaccelerates, and accelerates in the returning direction then constant speed and finally deaccelerates to stop near his twin A.)

mgb_phys
Homework Helper

If you ignore the acceleration/deceleration time it's simple ( just special relativity)
The time slowing factor is 1/sqrt( 1 - v^2/c^2)
(this is the same factor that their mass increases and their length decreases)

Then you just multiply this by the 'time' for the stationary twin.

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Nabeshin

The asymmetry arises due to the fact that the observer in the rocket changes reference frames (accelerates) at some point during the voyage. Considering the two parts of the journey separately, as mgb_phys notes, allows the problem to be solved in special relativity.

Janus
Staff Emeritus
Gold Member

and for the acceleration period you use:

$$\frac{c}{a} \sinh^{-1}\left (\frac{at}{c} \right )$$

Where a is the acceleration felt by the traveling twin.

I have got to know that if the acceleration time for the startup , round-turn and the stoping are all small i.e. they happen almost instantly, then the age difference caused within this time are all negligible.
Is this correct?

Dale
Mentor
2020 Award

In so much of the text I have seen the "twin paradox" is resolved by simply showing how the condition is asymmetrical, they make no calculation on how this asymmetry results in age difference. I have seen some claiming that its complete resolution requires General Relativity too, whereas other claim to resolve it only through special relativity?
The most general way to calculate the difference is to integrate the metric along each twin's worldline. As long as the metric is the Minkowski metric then you are using special relativity.

George Jones
Staff Emeritus
Gold Member

I have got to know that if the acceleration time for the startup , round-turn and the stoping are all small i.e. they happen almost instantly, then the age difference caused within this time are all negligible.
Is this correct?
Does "within this time" refer to the whole trip, or just the part during which there is non-zero acceleration. If the former, then the answer in "no"; if the latter, then, "yes" is a possible answer.

Does "within this time" refer to the whole trip, or just the part during which there is non-zero acceleration. If the former, then the answer in "no"; if the latter, then, "yes" is a possible answer.
I think if by "within this time" you mean time wrt the twin in the inertial frame of reference then whatever you say is right , but what about within this time wrt to the non inertial twin ?

DrGreg
Gold Member

I have got to know that if the acceleration time for the startup , round-turn and the stoping are all small i.e. they happen almost instantly, then the age difference caused within this time are all negligible.
Is this correct?
If you are calculating everything in the inertial frame of the twin who stays at home, then, yes if whatever happens during the acceleration is a very short period of time, then it won't make much difference to the final answer.

If, however, the accelerating twin wants to keep track of his twin's age continuously during the acceleration periods, then it can't be ignored. During acceleration, the twin's definition of simultaneity rapidly changes so his assessment of the other twin's age also rapidly changes. This is known as "gravitational time dilation". (During the acceleration, the twin feels a "pseudo-gravitational" G-force.)

After Einstein laid out the several examples in his 1905 paper, there were many varied articles dealing with the subject ... If you want to include the acceleration phases, you can use the technique of post 6...but a simple solution doesn't require more than special relativity as you can reformulate the problem by having the traveling twin already in motion has he passes the stay at home twin and they synchronize their clocks as the traveler passes by(e.g. the traveler could have left from San Fancisco and gets up to full speed as he passes his brother in NY. Likewise, there need be no turn around time factor as one can place a clock at that point and compare the time during flyby. To get the total age difference you simply double the one way time difference.

The problem was made more complicated by an article Einstein wrote in 1918 where he attempted to assign a physical cause to the age difference by analogizing to a pseudo G field - at the turn around point the acceleration gives the same dilation that would result from simply using SR (because the distace R to turn around result in a long time difference as seen from the other twin) - this led a number of authors to claim the twin thing was in reality a General relativity problem (Sciama, Born, Lederman to mame but a few). So you will find several ways of arriving at the correct answer - but it is really a simple one way SR problem that has been confused by too many solutions

I think that the absolute distance travelled, compared to the acceleration would be a big factor. As inferred above by George Jones.

Accelerating slowly would mean that the dilation effects become more significant for a shorter part of the journey. Just how significant the effect on the twins ages this acceleration is, is therefore largely dependant on the acceleration in terms of the overall distance travelled.

A longer distance (hopefully more travelled AFTER the maximum velocity has been reached) indicates that there is more distance travelled at velocities where the effects of dilation are at maximum.

Both acceleration and deceleration would need consideration, especially if at different rates, although the same principle applies.

Al68

In so much of the text I have seen the "twin paradox" is resolved by simply showing how the condition is asymmetrical, they make no calculation on how this asymmetry results in age difference. I have seen some claiming that its complete resolution requires General Relativity too, whereas other claim to resolve it only through special relativity?

I would like to know its resolution using only S.R if possible.
(consider the typical case, when twin B, first accelerates away from twin A, then travels at constant speed, then deaccelerates, and accelerates in the returning direction then constant speed and finally deaccelerates to stop near his twin A.)
The simplest way, if the duration of the acceleration is small, is to just divide the distance (according to each twin) by the relative velocity. Then double the results to get the answer for a two way trip.

For example, if a ship travels to a star 10 light yrs away from earth at 0.8c, the elapsed time on earth will be 10ly/0.8c= 12.5 yrs. And the elapsed time on the ship will be 6ly/0.8c= 7.5 yrs (the distance is 6 ly in the ship's frame due to length contraction).

So the difference in elapsed time (between earth departure and star arrival) between the twins for a one way trip is 5 yrs. For a two way trip, it's 10 yrs.

The asymmetry in this "resolution" is the simple fact that the distance between the earth and the turnaround point is defined in earth's frame (un-contracted) and contracted in the ship's (inertial) frame. The distance "expands" back from 6 ly to 10 ly in the ship's accelerated frame during the deceleration.

thanks to you all