THE RELATIVITY TWIN PARADOX

THE TWIN PARADOX

 

 

 

 

The Twin Paradox, sometimes referred to as the Clock Paradox, is frequently used as an attempt to illustrate an error with special relativity.  In SR, two frames in motion relative to each other are in a sense equivalent.  Viewing the other from either one requires the same transformation of time and length.  Thus a dimension on B viewed from A should be exactly the same as the corresponding dimension on A viewed from B.  Furthermore,  B’s velocity measured in A using A’s instruments is exactly the same as A’s velocity measured in B.

 

Therefore, because of this symmetry, if one member of a set of twins takes a ride on a spaceship then he/she should not gain or lose any time resulting from the trip.

 

Not so.  The very short reason is that the space-time paths taken by the twins are not symmetrical with respect to each other as would be required for the elapsed time to remain the same for both frames.  Twin A, the traveler, starts on Earth, boards a moving frame for the trip out, then stops and boards another frame in the opposite direction, then finally stops moving (relative to the Earth) and debarks.  Twin B, on the other hand, does none of these movements.

 

In the following, a twin will travel out for 150 days at 2/10 of the speed of light and return at the same speed.   

 

DETAILS OF THE JOURNEY

 

Twin B jumps on a space ship travelling at 0.2c, two-tenths of the speed of light.  We can use a version of the “Relativity Window” to visualize the transformations involved.  Fig. 1 shows the scene at time zero just after twin B boarded the ship and is travelling to the right at 0.2c.  The clocks in our, A’s, frame of reference are seen all synchronized at t = 0.  The clocks In B’s frame are seen from our frame with the readings shown.  Remember that in B’s frame, B will see synchronized clocks.  If a photograph is made in B’s frame at a great distance along an axis perpendicular to the two axes shown with a camera stationary in B’s frame, the clocks in B’s frame will all have the same readings and our clocks will differ. 

 

Next, allow twin B to travel for 150 time units.   See Fig. 2.   As described in the Fermi’s Question section, these views are normalized so that time can be any chosen units.  Let’s say one TU is a day so the twin travels for 150 days at 0.2c.  At that time we observe that twin B is located at a distance of 30 light-days away and his clock is seen to read 146.97.  (146.969)

 

Red notes below are from twin appendix

 

     From (1)  t’ = beta (150 - .2 * 30) = 146.969

             Where beta = 1.0206

 

Twin B however, is not concerned with the clock opposite him.  He is concerned with the clock of his twin, back at x = 0.  He will conduct an observation which is exactly symmetrical to the one described in the previous paragraph.  See Fig. 3 in which we adjusted time in the stationary frame to the value 146.97.  In Fig. 3 one of B’s clocks reads 150 (days) and is directly opposite x = 0 which is, of course, the location of twin A.  This occurs when the clock at x = 0 reads 146.97.  From twin B’s viewpoint, all clocks in his frame read the same and so he concludes that his twin’s clock is reading lower than his clock.  Note carefully the symmetry here:  twin A observes twin B’s clock to be lower than his by exactly the same amount twin B notes A’s clock to be less than his!

 

But the symmetry stops here.

 

Let us return to the conditions of Fig. 2 and direct the twin to return to Earth.  He does this by jumping from his frame onto another frame moving left at the same speed, 0.2 of the speed of light.  We will assume the transition takes place in a very brief period of time so that we may neglect any small transitional time accumulation (or loss) and we will also assume he can miraculously survive the acceleration.  Fig. 4 shows the scene just after the “jump”.  In Fig 4 we use the same transformations we’ve used all along but we assume, for simplicity of calculation we establish a set of auxiliary frames of reference:  One frame is stationary and has its origin at twice the distance of the “jump” point and another frame, a moving frame which becomes the return frame for the traveler.  Two conditions are required for this frame;  (1) its velocity with respect to the home frame be the same as twin B’s velocity was going out and (2) the clock at the jump point in the new frame read exactly what twin B’s clock read just before the jump.  (See Twin Appendix for details)

 

We can now advance the time to t=300, the time we expect the traveler to arrive home.  See Fig. 5. 

 

Using the last equation in the twin appendix;  t’’’ = beta * 300 – 2 * 150 * 0.2 * beta = 293.94

 

Thus, if  the Lorentz transformation equations are believed,  it’s hard to escape the conclusion that the twin returns younger.  It might still be argued by some that despite the mathematics, the twins remain the same age regardless of the behavior of the clocks.  This would be a dangerous proposal because:

 

(1)  The observant traveler will then notice that his clock seems to be running slowly with respect to his metabolism.

 

(2)  It will seem to run slowly regardless of his direction or speed.

 

(3)  Therefore, it will always appear to run fastest only when he is stationary with the Earth.

 

(4)  Therefore, there is something special about the velocity of the Earth.

 

Not a very good line of reasoning. 

 

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TWIN PARADOX APPENDIX

 

We can simplify the picture so as to make the solution more intuitive.

 

For the first half of the journey, the outgoing half, we will simply employ the methods discussed earlier.  The return trip is a little more difficult and will be described here. 

 

The approach will be to establish a frame of reference moving from right to left which will be the frame the traveler twin moves into when he/she reverses direction to return home.  The reversed frame, O’’’, will “start” from a point on the x axis twice as far as the reversal point.  We will also define another frame, an auxiliary frame, O’’, which is stationary and also has its origin at the same point along the x axis.  The two frames, O’’’ and O’’ are related through simple Lorentz equations.  Then converting x’’ values from O’’ back to x in the home frame, O, requires nothing more than a simple translation. 

 

 

We start with the simplified, normalized Lorentz equations from the “Fermi’s Question” section.  (Refer to that section for the special initial conditions required for the simplifications)

 

(1)  t’ = beta(t – vx)

 

(2)  x’ = beta(x – vt)

 

(3)  x = (x’/beta) + vt

 

where beta = 1/sqr(1 – vv)

 

 

To find x’, t’ opposite x = 0:

 

     x’(at x=0) = -beta v t

     t’(at x=0)  = beta t

 

What is the travelers frame immediately after reversal when his/her velocity has become –v? 

 

(Reversal time is tr, reversal x is xr, the return speed will be the same as the outgoing speed, v).

 

The reversal frame is equivalent to a frame, O’’’, which had a t’’’ = x’’’ = 0 crossover at a point x = 2(xr) and with a velocity of -v.  We view the frame O’’’ as moving relative to another, (stationary), frame O’’ whose origin is also located at 2(xr).  See Fig. App 1.

 

We again use the normalized Lorentz transforms, following equations (1), (2), and (3), after reversal, to convert from O’’’ to O’’:

 

Note that v is now reversed,

 

(4)  t’’’ = beta(t’’ + v x’’)

 

(5)  x’’’ = beta(x’’ + v t’’)

 

We now convert O’’ values to O values, x, t after reversal time, tr:

 

          Converting from O’’ to O:       x = x’’ + 2 v tr

 

                   Where tr is the reversal time in the O frame.

                    Note that the last term in the above equation is just double xr; 2xr.

 

                    x = x’’ + 2xr ,       x’’ = x – 2xr

 

                   Furthermore, since O’’ is stationary relative to O,   t’’ = t.

 

          Converting from O’’’ to O’’:    

 

                   If x’’’ = 0, and since t’’ = t, then from (5):    x’’ = -v t

                   from (4), also if x’’’=0:   t’’’ = beta (t – v*v t) = t/beta = t’’/beta

 

 t’’’ (at x’’’=0) has the same value as the traveler had before the reversal.

 

                   (    t’(x’=0) = t / beta    (obtained from equation (1) above.)    )

 

 

What is t’, x’  opposite  x=0 before reversal?   t’ (opp x=0)  = beta t    (from (1))

                 x’ (opp x=0) = -beta v t   (from (2))

 

 

Next, some general results after the velocity reversal, after tr.

 

What are x’’’ and t’’’ opposite any x at any time t  > tr ?  (note that v is –v after t=tr)

 

           Since        x’’ = x – 2xr

           and            t’’ = t

 

         using (5)     x’’’ = beta(x – 2 xr + v t)

            thus          x = (x’’’/beta)  + 2xr - vt

             and          x = 2xr - vt   opposite x’’’=0

 

Next, using (4)    t’’’ = beta(t + v(x – 2 xr))          (note,  (x-2 xr) is x’’ opposite x

            Then        t’’’ = beta(t + vx – v 2xr)

 

Next, what are the values in O’’’ opposite x=0?

            x’’’ = -2xr beta  + vt beta          and          t’’’ = t beta - 2xr v beta