In a message dated 12/18/2010 8:37:25 PM Eastern Standard Time,
zubarsky@gmail.com writes:
I have always wondered something about the twin paradox. If one twin is
moving away from the Earth at high speed time moves slower for him compared
to his brother. If the brother on Earth were able to see his brother on
the
fast moving ship everything would appear in slow motion. Here is my
question. If the on the fast moving space ship were to look at the Earth
is
it moving away from him at a very high speed too. So doesn't everything on
Earth appear to be moving in slow motion compared to time on the space
ship?
So, why upon return is the traveling twin young and the twin that stayed on
Earth old? I don't get it.
Thanks,
Mike Barr
I wrote this up a few years ago to explain this. Since this is stored in
word the server is going to crunch the text. Hopefully it will still be
readable.
*********************
Twin Paradox Defeated
Travel to star and back at .8c Star 20 LY away.
In a message dated 7/14/2003 2:48:37 PM Eastern Standard Time,
particlephysics@juno.com writes:
Complete the drawing for the deceleration phase--which completes
all of the ticks for the second line-- and when the drawing is finally
done have them hand COUNT the total NUMBER of clock ticks for your
stay-at-home clock and compare this count with the total NUMBER of ticks
on the accelerated/decelerated spaceship clock. Of course, they will
discover that the stay-at-home clock has recorded many MORE clock ticks
than the go-for-a-ride in a spaceship clock (as heard on the "walkie
talkie"). So, when the spaceship captain steps out of the spaceship to
shake your hand he will be YOUNGER (he will have aged slower) than you,
the stay-at-home observer.
This is the key to really understanding the so called twins paradox. This
is also the point that Lawrence made in his post. Let's see if by combining
this with what I posted we can get a even better understanding of this.
Remembering our twins and their measure of the major events.
Event 1- departure
Jack's Clock =2000 Mack's clock=2000 (These are annual clocks)
Event 2- Arrival at the star.
Jack's clock =2015 Mack's clock=2045
Jack due to time dilation experiences 15 years to travel 20 LY. Mack
observes Jack arrival 25 years after departure plus an additional 20 years for
this information to travel from the star to the earth.
Event 3- Arrival Home
Jack's Clock=2030 Mack's clock=2050
However this time we will send a radio pulse every year from the telescope
on earth to the spaceship and from the spaceship to the telescope so that
the twins can have a "real time" picture of what is happening. Now we can
no longer ignore Doppler so we need the equation
f=f_0*[sqrt[(1+v)/(1-v)] (c=1)
What Jack Sees
As Jack travels toward the star he based on the equation above he receives
a pulse once every three year from earth. (t=1/f) So since it takes 15
years his time
to reach the star Jack counts 5 pulses.
As Jack travels back to earth we must reverse the sign of v so that Jack
receives 45 pulses. Since 5+45=50 Jack knows his brother has aged 50 years.
What Mack sees.
As Jack travels away from the earth and back again he sees the same two
rates of annual pulses. So is the situation symmetric. NO! Remember Mack
doesn't see Jack return until 2045 and he arrives back on earth in 2050. So
from 2000 to 2045 Mack receives 1 pulse every 3 years (15 pulses) and from
2045 to 2050 3 pulses per year. (15 pulses)
Therefore 15+15=30. Mack knows his brother has aged 30 years. The twin's
Paradox is defeated.