Undergrad Handshake paradox involving the Starship Enterprise

[Nugatory]

I expected the last two diagrams to be just rotated, but they follow the Lorentz transformation so do not agree space/time measurements

They do. A Minkowski diagram is drawn on the two-dimensional surface of a computer screen or a sheet of paper which obeys Euclidean geometry so the Pythagorean theorem works: if we have two points $(x_1,y_1)$ and $(x_2,y_2)$ the distance $s$ between them is given by $s^2=(x_2-x_1)^2+(y_2-y_1)^2$. However, spacetime is non-Euclidean - $s^2=(x_2-x_1)^2-(t_2-t_1)^2$ - so when we plot our $(x,t)$ points on the sheet of paper the distance between the points on the sheet of paper doesn't correspond to the actual spacetime distance.

 

[Ibix]

I expected the last two diagrams to be just rotated,

They are Lorentz boosted, which turns out to be the Minkowski spacetime analog of a Euclidean rotation, but they are not rotated in the Euclidean sense, no. The axes of the boosted frame "scissor" together towards the $ct=x$ line - at least, as they are drawn on the first diagram. They remain orthogonal in the Minkowski sense, which is why they are drawn orthogonal in the last diagram.

they follow the Lorentz transformation so do not agree space/time measurements

They have different definitions of space and time, yes, but as Nugatory notes they do share measurements like $c^2\Delta t^2-\Delta x^2$.

 

[ilario980]

this question puzzled me, so now i'm on vacation from my job and can give this answer

if spoke back to earth instead of mars this becomes similar to twin paradox and at start time both frame agree [x=0, t=0]

while ship is moving measure a slower earth clock but the two frames becomes equals if ship stop "immediatly" on spock landing and both
clocks start ticking at same rate starting from [ x=0, t=t_landing] and [ x=x_sheeplanding
, t=t^'_landing]; in a sense time dilation can be
thought as some sort of delay other than that given by finite speed of light

at start time both frames agree [x=0, t=0] and ship start moving at c/2 (γ=1.15)
after 2 days spock leaves ship (in a flat space time)

after 4 days spock back to earth
to keep speed of light constant Lorentz transformations shrink to whole universe in x direction of ship frame, so earth frame measure ship at [ x=2 daysLight , t=4days] moving at speed c/2;
at t^'=0 ship measure distance of 2 daysLight as 1.732 daysLight, it reach this equivalent distance (in a shrunken universe) in just 3.464 days moving at speed c/2

in conclusion 3 seconds of handskake from earth frame is measured as 3*γ seconds in ship frame, but this is a "delayed" measure because if ship stops immediatly when spock land on earth, it measures a shorter time than earth clock

 

[cianfa72]

Mars started their clocks at the same time (by their frame's definition of "the same time") as Enterprise left Earth - but here you can see that (by Enterprise's definition of "at the same time") they started early but Earth didn't.

I read your very illustrative post, what about this last sentence ? In the first part by their you mean clocks Einstein's synchronized in Mars's rest frame and their common/shared definition of "the same time", I believe.

In the last diagram, which is the "zero" Mars's clock tick ? It is the one red on the bottom right that is on "the same time" straight line that joins it to the green/blue/grey tick at the origin?

 

[PeterDonis]

I read your very illustrative post, what about this last sentence ? In the first part by their you mean clocks Einstein's synchronized in Mars's rest frame and their common/shared definition of "the same time", I believe.

In the last diagram, which is the "zero" Mars's clock tick ? It is the one red on the bottom right that is on "the same time" straight line that joins it to the green/blue/grey tick at the origin?

The answers to these questions are obvious to anyone who knows how to read spacetime diagrams. If you don't, please learn how offline. You are hijacking someone else's thread and you have now been banned from further posts in this thread.