Hey Jorrie, it looks like you're pretty alone on this one! Maybe it's too much hard thinking, or you have bowled over the regular responders to your challenges!

Not fully knowing what I'm doing, I'll give it a shot.

Using equation 2 of your tips to get the relativistic difference velocity between the ships (after 2008) I get (0.96c-0.75c)/(1+(0.96c x -0.75c)/c² = 0.75c.

Jim has a 3 year head start on Pam, so it will take her 3/0.75 = 4 years to catch up. During this time Jim has flown 4 x 0.75 = 3 light years.

So the answers to questions 1 and 2 are: 3 light years distance and the end date is 2008 + 4 = 2012.

Using your tips equation 1 to get the time ratios:

Jim: dtau/dt = √[1-0.75^2] = 0.66 for the whole 12 years and also for Pam for the 1st 4 years.

So Jim's date must be 2000 + 0.66*12 ~ 2008.

Pam: dtau/dt = √[1-0.96^2] = 0.28 for the last 4 years.

So Pam's date: 2000+0.66*4 + 1*4 + 0.28*4 ~ 2007.76.

Hence, despite being stationary for 4 years, Pam just beats Jim in being younger at the end!

How was that?

Hi Scruffy, nice try - you have done a good job, except right at the start!

Do read the tips again, esp. the use and example of equation 2.

Have fun...

Hey Jorrie, thanx.

If I understand correctly then, I can simply subtract Jim's speed from Pam's speed because they were not given relative to each other, right?

That leaves 0.21c, so updating my sums:

Jim has a 3 year head start on Pam, so it will take her 3/0.21 = 14.29 years to catch up. During this time Jim has flown 14.29 x 0.75 = 10.7 light years.

So the answers to questions 1 and 2 are: 10.7 light years distance and the end date is the year 2008 + 14.29 = 2022.29 (mid April 2022?)

Using your tips equation 1 to get the time ratios:

Jim: dtau/dt = √[1-0.75^2] = 0.66 for the whole 22.29 years and also for Pam for the 1st 4 years.

So Jim's date must be 2000 + 0.66*22.29 ~ 2014.29.

Pam: dtau/dt = √[1-0.96^2] = 0.28 for the last 14.29 years.

So Pam's date: 2000+0.66*4 + 1*4 + 0.28*14.29 ~ 2010.65.

This puts Pam very much younger than Jim, which is not what I expected. After all, she were stationary in the reference frame for some time. So where did I go wrong?

SL

Hi Scruffy, let's leave it for a while and see if there are any other opinions...

I have never done any relativity stuff so I'm going to have a try. But, the main problem (if I've correctly understood that the relationship between speed and distance is unchanged) doesn't seem to involve any relativity.

J d3 = 0.75*t3
P d3+3 = 0.96*t3

solving d3 = 75/7 and t3 = 100/7

Now for the relative calendars

First 4 real years.

dtau/dt = SQRT(1-.75^2)= SQRT(7)/4

So P = SQRT(7) and J = SQRT(7)

2nd 4 real years P = 4+ SQRT(7) and J = 2*SQRT(7)

third period

P: dtau/dt = SQRT(1-.96^2) = 7/25

So P = 8 + SQRT(7) and J = 2*SQRT(7) + 25/SQRT(7)

P = 10.65 J = 14.74

d3 = 10.71

real time = 2022.29

Q1) 10.71

Q2) 2022.29

Q3) P = 2010.65 and J = 2014.74

I see that I have the same answers as scruffy except for a small transcription error. We didn't need the second equation, unless I've completely misunderstood.

hi Randall, glad we got the same answers!

I still have a problem with the validity of the sums. I recall Jorrie writing on this blog that the inertial observer being present at both events will always record a shorter time than the inertial observer not present at both.

In this case, Jim is an inertial observer present at both events and should record the shortest time. According to the calcs Pam records the least time. How come?

SL

Hi Scruffy,

...........Don't know, I don't think I understand any of it. If we have got the right answers then it seems incredible that Pam can squeeze all that last time period into 4 years of her own life.

Hi Randall.

You guys understand it better than what you make out here!

If it was possible for Pam to achieve the speed of light (which isn't), she could have done the last stint in the blink of her eyes! Hard to wrap one's head around, but all indications are that it's true!

Regards.

Hi Scruffy, you wrote: "In this case, Jim is an inertial observer present at both events and should record the shortest time. According to the calcs Pam records the least time. How come?"

Recall that Pam was not an inertial observer for the whole trip, while Jim was, at least since his initial acceleration stopped. Pam did two more stints of acceleration.

Jorrie

Hi Jorrie, OK, I see what you mean.

So acceleration is what makes the difference. Does it mean that the larger the acceleration the more severe the effect would be? Or is it perhaps the level of acceleration together with for how long it continues?

SL

Hi Scruffy, you asked: "Or is it perhaps the level of acceleration together with for how long it continues?"

No - acceleration level multiplied by time does give you a velocity change, which by itself does not affect clocks. It is for how long the velocity is maintained that makes all the difference.

In the teaser we are taking the velocity change to happen fast enough to be ignored. If you go and do the sums for different distances but with identical velocities, you will get different answers.

If we make the acceleration small and hence the acceleration time significant, it will change the results, but only because the speed builds up slowly and cannot be considered constant over the whole period.

Regards.

Hi Jorrie,

I have given some calculations of mine, but too late for you to consider perhaps.

Pam & Jim will pass Station yellow at their own time of

4*√(1-9/16) = 2.645 years after 2000 = 2002.65 date

Jim will pass station green at his own time of

2002.645 + 2.645 = 2005.29 date

Pam will meet Jim at a time of T years (own time) after reaching station yellow such that

T = 4 + t √(1-0.96*0.96) = 4 + 0.28 t where t is the assumed station travel time of Pam

The distance travelled by Pam during the time = 0.96ct

The distance travelled by Jim during the same station time = (4+t)*075c

Equating the distance travelled by Jim and Pam at the same station tim

(4+t)0.75c = 0.96ct or t = 14.28 years

The distance travelled by Jim from station yellow = (4 + 14.28)0.75c = 13.71 light years

The distance between station green and blue = 3 light years

Therefore The distance between station green and blue = 13.71 - 3 = 10.71 light years

The own time spent by Pam after reaching station yellow at the time of meeting Jim = 4 + 0.28t = 7.998 years

The time spent by Jim after reaching station yellow before meeting Pam =

(4+14.28)*√(1-9/16) = 12.09 years

Therefore the difference in time between the times of Pam and Jim = 12.09 - 8 = 4.09 years

Jorrie,

In my 6th line from bottom (comment #12) I put down

Distance between station green to blue = 3 light years. It should read as below

Distance between station yellow and green = 3 light years station time.

kasharma

Hi kasharma, I'm afraid you have it wrong.

Read replies #3 and #5, they have it right.

Jorrie