### Statement

The so-called 'Twin Paradox' is one of the most famous problems in all of science. Fortunately for relativity it is not a paradox at all. As has been mentioned, Special and General Relativity are both self-consistent within themselves and within physics. We will state the twin paradox here and then describe some of the ways in which the paradox can be resolved.

The usual statement of the paradox is that one twin (call her A) remains at rest on the earth relative to
another twin who flies from the earth to a distant star at a high velocity (compared to *c*). Call the flying
twin B. B reaches the star and turns around and returns to earth. The twin on earth (A) will see B's clock
running slowly due to time dilation. So if
the twins compare ages back on earth, twin B should
be younger. However, from B's point of view (in her reference
frame) A is moving away at high speed
as B moves towards the distant star and later A is moving towards B at high speed as B moves back towards
the earth. According to B, then, time should run slowly for A on both legs of the trip; thus A should be
younger than B! It is not possible that both twins can be right-the twins can compare clocks back on earth
and either A's must show more time than B's or vice-versa (or perhaps they are the same). Who is right?
Which twin is younger?

### Resolution

The reasoning from A's frame is correct: twin B is younger. The simplest way to explain this is to say that in
order for twin B to leave the earth and travel to a distance star she must accelerate to speed *v*. Then
when she reaches the star she must slow down and eventually turn around and accelerate in the other
direction. Finally, when B reaches the earth again she must decelerate from *v* to land once more on the
earth. Since B's route involves acceleration, her frame cannot be considered an inertial reference frame
and thus none of the reasoning applied above (such as time dilation) can be applied. To deal
with the situation in B's frame we must enter into a much more complicated analysis involving accelerating
frames of reference; this is the subject of General Relativity. It turns out that while the B is moving with
speed *v* A's clock does run comparatively slow, but when B is accelerating the A's clocks run faster to
such an extent that the overall elapsed time is measured as being shorter in B's frame. Thus the reasoning in
A's frame is correct and B is younger.

However, we can also resolve the paradox without resorting to General Relativity. Consider B's path to the
distant star lined with many lamps. The lamps flash on and off simultaneously in twin A's frame. Let the
time measured between successive flashes of the lamps in A's frame be *t*_{A}. What is the time between
flashes in B's frame? As we learned in Heading
the flashes cannot occur
simultaneously in B's frame; in fact B measures the flashes ahead of him to occur earlier than the flashes behind him (B is
moving towards those lamps ahead of him). Since B is always moving towards the flashes which happen
earlier the time between flashes is less in B's frame. In B's frame the distance between flash-events is zero
(B is at rest) so *Δx*_{B} = 0, thus *Δt*_{A} = *γ*(*Δt*_{B} - *vΔx*_{B}/*c*^{2}) gives:

Δt_{B} = |

Thus the time between flashes is less in B's frame than in A's frame. N is the total number of flashes that B sees during her entire journey. Both twins must agree on the number of flashes seen during the journey. Thus the total time of the journey in A's frame is

*T*

_{A}=

*NΔt*

_{A}, and the total time in B's frame is

*T*

_{B}=

*NΔt*

_{B}=

*N*(

*Δt*

_{A}/

*γ*). Thus:

T_{B} = |

Thus the total journey time is less in B's frame and hence she is the younger twin.

All this is fine. But what about in B's frame? Why can't we employ the same analysis of A moving past flashing lamps to show that in fact A is younger? The simple answer is that the concept of 'B's frame' is ambiguous; B in fact is in two different frame depending on her direction of travel. This can be seen on the Minkowski diagram in :

Here is lines of simultaneity in B's frame are sloped one way for the outward journey and the other way for the trip back; this leaves a gap in the middle where A observes no flashes, leading to an overall measurement of more time in A's frame. If the distant star is distance*d*from the earth in A's frame and the flashes occur at intervals

*Δt*

_{B}in B's frame, then they occur at intervals

*Δt*

_{B}/

*γ*=

*Δt*

_{A}in A's frame, as per the usual time dilation effect (this is the same for inward and outward journeys). Again let the twins agree that there are N total flashes during the journey. The total time is B's frame is then

*T*

_{B}=

*NΔt*

_{B}and for A,

*T*

_{A}=

*N*(

*Δt*

_{B}/

*γ*) +

*τ*where

*τ*is the time where A observes no flashes (see the Minkowski diagram). In B's frame the distance between the earth and the star is (half the total journey time times the speed) which is also equal to

*d*/

*γ*due to the usual length contraction. Thus:

T_{B} = | |||

T_{A} = + τ = + t |

What is

*τ*? We can see from that the slopes of the lines are ±

*v*/

*c*so the time in which A observes no flashes is

*ct*= 2

*d*tan

*θ*= 2

*dv*/

*c*. Thus:

T_{A} = + = frac2dv |

Comparing

*T*

_{A}and

*T*

_{B}we see

*T*

_{B}=

*T*

_{A}/

*γ*which is the same result we arrived at above. A measures more time and B is younger.