I Handshake paradox involving the Starship Enterprise

[ilario980]

TL;DR Summary

How long does the handshake between martian presidents and dr. Spoke last?

suppose the enterprise departs from planet earth on a mission the the other side of the milky way at 90% of the speed of light; since time is dilated while cpt Kirk drinks coffe on planet earth some time passes (let's say 1h). Approaching mars captain Kirk orders dt. Spock to land on Mars with his spacecracft; to do this dr. Spock vehicle must move at the opposite direction of enterprise at about 90% of speed of light; so again time is dilated, while dr. Spock drinks a coffe on enterprise about 1h passes. On planet Mars dr. Spoke shake hands with the martian president: how long does the handshake last from enterprise reference system?
since mars clock is almost aligned with earth clock it should last a few seconds of earth/mars clock, but from enterprise dr. Spoke 5 seconds should last 1h and 5 seconds of earth/mars clock should last a few microseconds

 

[Ibix]

First off, it's Mr Spock - he's an officer in an organisation patterned loosely after the US Navy, so is referred to as Mister or by rank. Dr Spock was a child psychologist.

The Enterprise's inertial reference system sees Mars moving at 0.9c, which is a time dilation factor of about 2.3. So the 5s handshake would last about 11.5s by the Enterprise clocks.

 

[ilario980]

ok, but Mars clock should be aligned whith earth clock so Enterprise should see events of Mars faster, not slower; what's wrong?

 

[Ibix]

so Enterprise should see events of Mars faster

No. Time dilation is a symmetrical effect - if I measure your clocks ticking slow, you measure mine ticking slow. So Earth and Mars clocks tick the same rate and find Enterprise's clocks lagging further and further behind; Enterprise determines that Earth and Mars clocks are lagging further and further behind.

This is not paradoxical; the two frames of reference don't agree on what "at the same time as my clock reads zero" means, which is a phenomenon called "the relativity of simultaneity". This doesn't get as much limelight as time dilation but is arguably a much more important effect. I'd suggest googling the term and asking more questions after that.

 

[Dale]

ok, but Mars clock should be aligned whith earth clock so Enterprise should see events of Mars faster, not slower; what's wrong?

Earth clocks are irrelevant to the question. The handshake takes place on Mars and the question is the duration in the Enterprise frame.

Edit: I see that @Ibix is assuming that Earth and Mars are at rest in the same frame. It isn’t a bad assumption, but I assumed they were moving with respect to each other. Any discrepancies are due to that.

 

[Nugatory]

ok, but Mars clock should be aligned whith earth clock so Enterprise should see events of Mars faster, not slower; what's wrong?

We can think of this problem as either:
a) The ship is at rest while the two planets are moving away from it at .9c so the planet clocks are slow compared to the ship clock.
b) The planets are at rest while the ship is moving away from them at .9c so the ship clock is slow compared to the planet clocks.

There is no paradox here - when we allow for the relativity of simultaneity we see that both descriptions are accurate and do not contradict one another.

 

[ilario980]

Thank you very much for your replies; my doubt is that Mr. Spock clock cannot be aligned with Mars clock since it should be slower than ship. I will read about relativity of simultaneity, at the moment I'm busy

Ilario M.

 

[Dale]

my doubt is that Mr. Spock clock cannot be aligned with Mars clock since it should be slower than ship

Both Spock and Mars clocks tick at the same rate. As you said:

dr. Spock vehicle must move at the opposite direction of enterprise at about 90% of speed of light; so again time is dilated,

So Spock is at rest with respect to Mars and both are moving at 0.9 c with respect to Enterprise.

 

[ilario980]

just a bit of math:

t_spock< t_ship< t_mars

t_spock = t_mars

 

[Dale]

just a bit of math:

t_spock< t_ship< t_mars

t_spock = t_mars

Ok, there is your problem. $t_{spock}=t_{mars}$ is true, but none of your inequalities are true.

By the Lorentz transform $$t_{mars}= \frac{1}{\sqrt{1-v^2/c^2}} \left(t_{ship}-\frac{v\ x_{ship}}{c^2}\right)$$ Whether this is greater than or less than $t_{ship}$ depends on $x_{ship}$. Similarly for the other inequalities. You should study the Lorentz transform.

 

[PeroK]

just a bit of math:

t_spock< t_ship< t_mars

Whether ##t_spock< t_ship## or not depends on your frame of reference. The first postulate of SR says that all motion is relative. You've gone wrong from the very beginning by assuming that Spock is "really moving" and "really time dilated". When, in fact, time dilation is relative.

 

[ilario980]

at t0 ship starts and all clock signs 00:00:00; when the ship is near mars

t_earth=t_mars = 00:30:00
t _ship = t _spock =00:29:00


then spock leaves the ship

t_earth=t_mars = 00:31:00
t _ship =~ 00:29:30
t _spock = ?

t_spock should tick at the same rate of mars clock and a slower tick than ship clock, wich is not possible

 

[PeroK]

at t0 ship starts and all clock signs 00:00:00; when the ship is near mars

t_earth=t_mars = 00:30:00
t _ship = t _spock =00:29:00


then spock leaves the ship

t_earth=t_mars = 00:31:00
t _ship =~ 00:29:30
t _spock = ?

t_spock should tick at the same rate of mars clock and a slower tick than ship clock, wich is not possible

I thought this was about a handshake?

In any case, there is no universal time which can be used to set all clocks to zero simultaneously. You've invented your own theory of relativity, which is clearly wrong. That has nothing to do with special relativity.

 

[ilario980]

nope, just to share some thinking...

God bless you all

 

[Dale]

tspock should tick at the same rate of mars clock and a slower tick than ship clock, wich is not possible

It is possible, and abundant evidence shows that is in fact the way the world works.

However the numbers you made up above aren’t matching your words. When you are talking about the rate of Spock’s clock the appropriate quantity is not $t_{spock}$ but $d\tau_{spock}/dt_{ship}$ etc.

$\tau_{spock}$ is time on Spock’s clock. $t_{spock}$ is his coordinate time in his reference frame, not time on his clock. And a rate is a derivative.

So $d\tau_{spock}/dt_{mars}=1$ says that Spock’s clock ticks at the same rate as Mars’ time. And $d\tau_{spock}/dt_{ship}<1$ says that Spock’s clock ticks slowly according to the Enterprise’s time.

 

[Ibix]

nope, just to share some thinking...

God bless you all

I have the impression you're bowing out here, but just in case some further explanation might help, here are a few Minkowski diagrams. If you came across displacement-time diagrams in school physics, these are more or less the same - we just tend to draw time up the page instead of across. You can read off "where everything is at one instant" by finding a time on the vertical axis and finding where each line is at that level. We take these diagrams slightly more literally in relativity, though, because it models everything as taking place in spacetime - so these diagrams become maps of spacetime.

Anyway - here's the first one.

Things that have constant positions in space appear as vertical lines here, because their x position is the same at all times. The green line represents Earth and the red line represents Mars. Things that are moving are slanted lines, and the Enterprise appears as a grey line, initially in the same place as Earth at the bottom of the diagram and, as time goes on, moving towards Mars and arriving there after about eight time units. Mr Spock's path is marked as a blue dotted line, travelling with the Enterprise until they reach Mars, where he stays. Finally, there are some fine horizontal grid lines to help with reading this frame's time.

Now let's draw the same diagram, but add markers on each line when everybody's clock ticks.

The Earth's clocks and Mars' clocks tick once per time unit - always on the grid lines. The Enterprise's clocks tick slow - the first tick is above the first grid line, and by the fifth grid line it's only ticked four times. Mr Spock's watch ticks at the same time as the Enterprise's clocks, until he reaches Mars. After Mars the Enterprise's clock continues ticking slow - you can just see its eighth tick on the tenth gridline. But Mr Spock's watch now ticks at the same rate as the Mars clocks. It's not synchronised with them - it does not tick at the same time - but it ticks at the same rate. That's perhaps easier to see if we drop everybody else's clock ticks and Mr Spock's line:

So it seems that the Enterprise's clocks tick slowly. How can the Enterprise say that Mars and Earth clocks tick slow? Well, the point about spacetime is that there isn't a unique definition of "time", just as there isn't a unique definition of "left" in space - if we aren't facing the same way we don't share the same meaning of "left". Similarly, in spacetime objects that aren't travelling at the same speed don't share a meaning of "space" or "time". Those grey horizontal lines mark all the events that Earth and Mars say happen at the same time as their clocks tick 0, 1, 2, 3 etc. But the Enterprise doesn't agree. Here are the grid lines Enterprise would draw:

The grey lines go through each of Enterprise's clock ticks, just as the ones in the earlier diagrams go through each of Earth's clock ticks. But, in this frame, those lines are slanted. Enterprise does not divide spacetime into space and time in the same way as Earth does. And notice how the lines compare to the green markers of Earth's clock ticks. The first Earth tick is after the first line, and the fourth Earth tick is on the fifth grey line - just like the Enterprise's ticks were on the Earth's grey lines. The slant of those lines is what I called "the relativity of simultaneity" in an earlier post - the two frames don't agree what parts of spacetime are "at the same time" - one frame uses the horizontal lines and one the slanted lines.

Enterprise would say that this is a valid map of spacetime, but not the most convenient one. She'd prefer a map with her grey lines horizontal. The Lorentz transforms provide the maths for this, and Enterprise would draw this map:

This time, Enterprise is stationary and appears as a vertical line. Earth and Mars are moving to the left and appear as slanted lines. Just like before, the clock ticks on the moving objects are slower, with only four ticks every five grid lines. You can see the relativity of simultaneity in action again by looking at Mar's clock ticks. 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.

Relativity has some strange concepts in it (the one most people seem to struggle with is the idea that "at the same time" doesn't mean the same thing to different frames), but it is not paradoxical.

 

[ilario980]

It is possible, and abundant evidence shows that is in fact the way the world works.

However the numbers you made up above aren’t matching your words. When you are talking about the rate of Spock’s clock the appropriate quantity is not $t_{spock}$ but $d\tau_{spock}/dt_{ship}$ etc.

$\tau_{spock}$ is time on Spock’s clock. $t_{spock}$ is his coordinate time in his reference frame, not time on his clock. And a rate is a derivative.

So $d\tau_{spock}/dt_{mars}=1$ says that Spock’s clock ticks at the same rate as Mars’ time. And $d\tau_{spock}/dt_{ship}<1$ says that Spock’s clock ticks slowly according to the Enterprise’s time.

yes, i used $t_{actor}$ as time on actor clock , $d\tau_{actor}$ is the wright notation
let's put this way, suppose that enterprise has on board 2 ultra wide camera; at departure time all clocks signs 00:00:00 and camera 1 focuses on a big clock on mars signing 00:00:00

when enterprise reaches a fixed point near mars the video recorded on ship from camera 1 is shorter than the time signed on mars clock, $d\tau_{ship}/d\tau_{mars}<1$
$d\tau_{ship}=00:29:00$

from mars frame:
$d\tau_{mars}=d\tau_{earth}=00:30:00$

so if captain Kirk replay the video recorded from camera1 he sees that 1 second signed on mars big clock is shorter than 1 second on the ship

then spock leaves the ship and camera 2 focuses on a clock visible on his spacecraft ; since spock time is dilated 1 second from the clock on spacecraft is longer from 1 second on ship $d\tau_{ship}/d\tau_{spacecraft}>1$


when spock reaches mars president camera1 is recording a video where $d\tau_{ship}/d\tau_{mars}<1$ and camera 2 is recording a video where $d\tau_{ship}/d\tau_{spock}>1$ but of course all cameras should record the same video and spock cannot shake hands of mars president
what's wrong?

 

[Ibix]

Introducing cameras often makes things messier because they introduce light speed delay - what they see nearby is what is happening now (or near enough) but what they see far away is happening in the past. And clocks moving with respect to the camera are not recorded ticking at the "real" rate because the delay correction is changing. I'm not sure to what extent you are trying to correct for these issues.

What actually happens is that Spock shakes hands with the president, but the president's watch shows 00:30:00 and Spock's watch shows 00:29:00 at the start of the handshake. At the end of the handshake (say 5s long) their watches have advanced to 00:30:05 and 00:29:05. Depending on who defines "at the same time", the Enterprise's clocks either read 00:29:11.5 or 00:29:02.2 at the same time as the handshake finishes.

 

[Sagittarius A-Star]

spock cannot shake hands of mars president
what's wrong?

In addition to what others wrote, Spock doesn't shake hands. He uses the following salutation:

Source:
https://memory-alpha.fandom.com/de/wiki/Vulkanischer_Gruß

 

[Ibix]

Now I've had time to do a re-read, you're not including for either the distance change nor the relativity of simultaneity. In a line, that's what's wrong - you're applying your own incorrect model of the theory.

Some quick numbers: you are assuming a time dilation factor of 29/30, which corresponds to a speed a bit above 0.25c. The journey time is 30 minutes, so you're assuming an Earth-Mars distance of about 7.5 light minutes.

at departure time all clocks signs 00:00:00 and camera 1 focuses on a big clock on mars signing 00:00:00

No - Mars is 7.5 light minutes away, so the camera would see the Mars clock reading 23:52:30, or thereabouts.

when enterprise reaches a fixed point near mars the video recorded on ship from camera 1 is shorter than the time signed on mars clock, $d\tau_{ship}/d\tau_{mars}<1$

This can't be directly concluded - you need to correct for the changing distance from Enterprise to Mars in order to derive that the Mars clocks were ticking fast or slow. Depending on whether you choose to use the Earth/Mars frame's definition of simultaneity and distance or the Enterprise's frame, you will either obtain that the Mars clock was ticking slow or the Enterprise clock was ticking slow.

$d\tau_{ship}=00:29:00$

from mars frame:
$d\tau_{mars}=d\tau_{earth}=00:30:00$

Here you are using the times measured by the frames, where you have corrected for any light speed delay. But...

so if captain Kirk replay the video recorded from camera1 he sees that 1 second signed on mars big clock is shorter than 1 second on the ship

...here you're using what the camera actually shows, which is uncorrected. And you haven't allowed for that. It would have 29 minutes of footage showing the Mars clock ticking off 37.5 minutes of time. When you correct for the changing light speed delay, you would either conclude that the Mars clock had ticked off 30 minutes (if you use the Earth frame's correction) or a little over 28 minutes (if you use Enterprise's frame's correction).

then spock leaves the ship and camera 2 focuses on a clock visible on his spacecraft ; since spock time is dilated 1 second from the clock on spacecraft is longer from 1 second on ship $d\tau_{ship}/d\tau_{spacecraft}>1$

Yes, but the camera will show both his and Mars' clocks ticking much slower than just time dilation due to the now increasing light speed delay - both Spock's and the president's clocks will appear to tic about 23 times to every 30 of the Enterprise's clock ticks. Again, once you correct for the lag, Enterprise will conclude that Spock and the president's watches are ticking slow.

what's wrong?

The consistent error you have is that you are not allowing for the relativity of simultaneity, which means that the clocks on Earth and Mars do not read the same time according to the Enterprise. That means that simply taking the difference between a reading on an Earth clock and the reading on a Mars clock does not give the Enterprise the correct answer about clock rates.

This is a very common misunderstanding in beginners in the topic - we have all been there. Failing to understand it is one of the most common reasons for people thinking relativity is wrong - but actually it's their own mental model of relativity that's wrong, not the theory itself. The best solution is actually a text book, but at absolute minimum you need to learn the Lorentz transforms (not time dilation and length contraction, which are special cases that are popular but not generally applicable) and I strongly recommend learning to draw and interpret Minkowski diagrams.

 

[Ibix]

In addition to what others wrote, Spock doesn't shake hands. He uses the following salutation:

View attachment 343829​

Source:
https://memory-alpha.fandom.com/de/wiki/Vulkanischer_Gruß

We can time how long it takes them to say "Live Long and Prosper" to each other.

 

[Dale]

yes, i used $t_{actor}$ as time on actor clock , $d\tau_{actor}$ is the wright notation
let's put this way, suppose that enterprise has on board 2 ultra wide camera; at departure time all clocks signs 00:00:00 and camera 1 focuses on a big clock on mars signing 00:00:00

when enterprise reaches a fixed point near mars the video recorded on ship from camera 1 is shorter than the time signed on mars clock, $d\tau_{ship}/d\tau_{mars}<1$
$d\tau_{ship}=00:29:00$

from mars frame:
$d\tau_{mars}=d\tau_{earth}=00:30:00$

so if captain Kirk replay the video recorded from camera1 he sees that 1 second signed on mars big clock is shorter than 1 second on the ship

then spock leaves the ship and camera 2 focuses on a clock visible on his spacecraft ; since spock time is dilated 1 second from the clock on spacecraft is longer from 1 second on ship $d\tau_{ship}/d\tau_{spacecraft}>1$


when spock reaches mars president camera1 is recording a video where $d\tau_{ship}/d\tau_{mars}<1$ and camera 2 is recording a video where $d\tau_{ship}/d\tau_{spock}>1$ but of course all cameras should record the same video and spock cannot shake hands of mars president
what's wrong?

Oh so much is wrong. You really should study some of the basics. If you actually work through the real math here there are no conflicts.

First, a notation like $d\tau_{ship}/d\tau_{mars}$ makes no sense. Two clocks moving with respect to each other are only co-located at one event. Comparing rates requires that they be colocated over some interval. So this sort of quantity doesn’t make sense. When talking about the rate of a clock it will always be compared to some coordinate time, as the coordinate time covers all of spacetime. So $d\tau_{ship}/d\tau_{mars}$ is meaningless and $d\tau_{ship}/dt_{mars}$ is the rate of the ship’s clock with respect to Mars’ coordinate time.

Also, to describe the reading on a clock would be $\tau_{ship}=00:29:00$ rather than $d\tau_{ship}=00:29:00$. If you wanted to be more explicit you could indicate $\tau_{ship}\big|_{event}=00:29:00$ to indicate what event had that time.

Now, another big issue is that you are just randomly throwing out numbers with no basis. There are two relevant formulas that you need to use here. The first formula is the Lorentz transform, which relates the coordinates in two inertial frames.
$$t_{mars}= \frac{1}{\sqrt{1-v^2/c^2}} \left(t_{ship}-\frac{v\ x_{ship}}{c^2}\right)$$
$$x_{mars}= \frac{1}{\sqrt{1-v^2/c^2}} \left(x_{ship}-v\ t_{ship}\right)$$

Note that in this scenario Spock’s frame is non inertial. So you will need a more complicated method if you wish to use $t_{spock}$ anywhere. I would recommend sticking with $\tau_{spock}$.

The other relevant formula is the spacetime interval. It relates the time on a clock to the coordinates in an inertial frame. The clock does not need to be inertial so
$$c^2 d\tau_{spock}^2=c^2 dt_{mars}^2-dx_{mars}^2$$

And similarly for other clocks and other inertial frames.

Finally, the idea that all the cameras should record the same video is nonsense. Determining what a camera records is a complicated matter of ray optics. And in a relativistic scenario it gets exceptionally complicated. You need to get the basics down first before you are ready to approach something like that. The cameras are not going to be helpful for you at the level you need to work.

Focus on using the right concepts in the right formulas. If you use the actual formulas you will find that it all works. If you just make up stuff it will continue to cause problems for you.

 

[ilario980]

Okay, I need to study the theory, what seems like a paradox to me is that both cameras show that time flows at the same speed after mars landing

 

[ilario980]

or it could be a flight recorder instead of a camera to avoid ligth delays

 

[Dale]

Okay, I need to study the theory, what seems like a paradox to me is that both cameras show that time flows at the same speed after mars landing

Forget the cameras for now. Focus on the clocks.

When Spock is on Mars then $$c^2 d\tau_{spock}^2=c^2 dt_{mars}^2-dx_{mars}^2\bigg|_{spock}$$$$\left(\frac{d\tau_{spock}}{dt_{mars} }\right)^2=1-\left(\frac{dx_{mars}}{c \ dt_{mars}}\right)^2 \bigg|_{spock}=1$$ and $$c^2 d\tau_{spock}^2=c^2 dt_{ship}^2-dx_{ship}^2\bigg|_{spock}$$$$\left(\frac{d\tau_{spock}}{dt_{ship} }\right)^2=1-\left(\frac{dx_{ship}}{c \ dt_{ship}}\right)^2\bigg|_{spock}=1-\left(\frac{v}{c}\right)^2$$

 

[PeroK]

Okay, I need to study the theory, what seems like a paradox to me is that both cameras show that time flows at the same speed after mars landing

Good idea. At the moment, you are stumbling about in the dark without understanding the nature of space, time, motion and reference frames.

 

[ilario980]

at first glance the Lorentz transformations say that in order to agree on the constant c we must disagree on the measure of length and time; I will study this in more detail later, thank you very much for your answers

 

[PeroK]

at first glance the Lorentz transformations say that in order to agree on the constant c we must disagree on the measure of length and time; I will study this in more detail later, thank you very much for your answers

That's absolutely correct!

Newtonian physics: absolute space and time; but no invariant speed (of light).

Special Relativity: invariant speed (of light), but the Lorentz Transformation between inertial reference frames; and, spacetime as a single four-dimensional entity.

Both these possibilities arise naturally from basic considerations of space and time. And, experiments show that our universe is relativistic and not Newtonian. E.g. all the particle experiments at CERN or other particle accelerators confirm SR every day of the week.

 

[jbriggs444]

at first glance the Lorentz transformations say that in order to agree on the constant c we must disagree on the measure of length and time; I will study this in more detail later, thank you very much for your answers

We disagree about three things.

We disagree about length (length contraction).
We disagree about duration (time dilation).
We also disagree about simultaneity (the relativity of simultaneity).

If you only account for time dilation and length contraction, you get a wonky theory that does not work out at all. Which is why you see contradictions.

 

[ilario980]

I have the impression you're bowing out here, but just in case some further explanation might help, here are a few Minkowski diagrams. If you came across displacement-time diagrams in school physics, these are more or less the same - we just tend to draw time up the page instead of across. You can read off "where everything is at one instant" by finding a time on the vertical axis and finding where each line is at that level. We take these diagrams slightly more literally in relativity, though, because it models everything as taking place in spacetime - so these diagrams become maps of spacetime.

Anyway - here's the first one.
View attachment 343821
Things that have constant positions in space appear as vertical lines here, because their x position is the same at all times. The green line represents Earth and the red line represents Mars. Things that are moving are slanted lines, and the Enterprise appears as a grey line, initially in the same place as Earth at the bottom of the diagram and, as time goes on, moving towards Mars and arriving there after about eight time units. Mr Spock's path is marked as a blue dotted line, travelling with the Enterprise until they reach Mars, where he stays. Finally, there are some fine horizontal grid lines to help with reading this frame's time.

Now let's draw the same diagram, but add markers on each line when everybody's clock ticks.
View attachment 343822
The Earth's clocks and Mars' clocks tick once per time unit - always on the grid lines. The Enterprise's clocks tick slow - the first tick is above the first grid line, and by the fifth grid line it's only ticked four times. Mr Spock's watch ticks at the same time as the Enterprise's clocks, until he reaches Mars. After Mars the Enterprise's clock continues ticking slow - you can just see its eighth tick on the tenth gridline. But Mr Spock's watch now ticks at the same rate as the Mars clocks. It's not synchronised with them - it does not tick at the same time - but it ticks at the same rate. That's perhaps easier to see if we drop everybody else's clock ticks and Mr Spock's line:
View attachment 343823
So it seems that the Enterprise's clocks tick slowly. How can the Enterprise say that Mars and Earth clocks tick slow? Well, the point about spacetime is that there isn't a unique definition of "time", just as there isn't a unique definition of "left" in space - if we aren't facing the same way we don't share the same meaning of "left". Similarly, in spacetime objects that aren't travelling at the same speed don't share a meaning of "space" or "time". Those grey horizontal lines mark all the events that Earth and Mars say happen at the same time as their clocks tick 0, 1, 2, 3 etc. But the Enterprise doesn't agree. Here are the grid lines Enterprise would draw:
View attachment 343824
The grey lines go through each of Enterprise's clock ticks, just as the ones in the earlier diagrams go through each of Earth's clock ticks. But, in this frame, those lines are slanted. Enterprise does not divide spacetime into space and time in the same way as Earth does. And notice how the lines compare to the green markers of Earth's clock ticks. The first Earth tick is after the first line, and the fourth Earth tick is on the fifth grey line - just like the Enterprise's ticks were on the Earth's grey lines. The slant of those lines is what I called "the relativity of simultaneity" in an earlier post - the two frames don't agree what parts of spacetime are "at the same time" - one frame uses the horizontal lines and one the slanted lines.

Enterprise would say that this is a valid map of spacetime, but not the most convenient one. She'd prefer a map with her grey lines horizontal. The Lorentz transforms provide the maths for this, and Enterprise would draw this map:
View attachment 343825
This time, Enterprise is stationary and appears as a vertical line. Earth and Mars are moving to the left and appear as slanted lines. Just like before, the clock ticks on the moving objects are slower, with only four ticks every five grid lines. You can see the relativity of simultaneity in action again by looking at Mar's clock ticks. 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.

Relativity has some strange concepts in it (the one most people seem to struggle with is the idea that "at the same time" doesn't mean the same thing to different frames), but it is not paradoxical.

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

 

[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.