# How to plot correctly spacetime diagrams

• RiccardoVen

#### RiccardoVen

Hello,
after having done a bit of exercices on Taylor & Wheeler ( just for self-study ), I felt
the need to go on a bit differently, i.e. trying to solve qualitative problems instead of
quantitative ones, i.e. using a bit of diagrams ( all in all, as Susskind always suggest
"when you face a SR problem, plot a diagram first ).

That said, I was trying to solve this Trekker oriented problem:

http://www.phys.vt.edu/~takeuchi/relativity/practice/problem07.html

The overall diagram is quite clear to me, since I was able to answer to all
questions "almost" without problems. But then I thought "wait a minute: it's easy looking
at a already well cooked diagram...let's try to do it by myself from scratch, now!"

( Incidentally this is needed, for example, if you want to to solve the T & W 3.7 exercice,
called Space War, which is exactly this problem without Kirk & friends ).

But here start having problem, highligthing probably my real doubts on it.

So I started to draw a simple x/t diagram, i.e. the enterprise frame, adding a
light beam from origin heading to the right ( x > 0 ).

The I plotted the klingon's primed frame, obviously making it a simmetric around
the light beam, with a certain rapidity angle with x/t axes.

So far, so good. Now I decided to add the Enterprise at rest in its frame ( the unprimed ).
This actually leads to 2 vertical lines ( parallel to t time axis ). This also result in deciding
where to put the event of klingon front reaching the enterprise rear.

Looking at the original plot, the author decided to put it onto the light beam line, i.e.
C point. It's not really clear from this diagram, but if you look at its book ( I own it ), you
can see the enterprise rear wordline intersects the lightbeam EXACTLY in C point.

So my point is: is this a licky strike, or a specific design? Why choosing the C event
to be light-like?

From C, you can draw a line parallel to t', which intersect the enterprise x-axis in point
V ( let's imagine to call the intersection of x-axis on the right of the O origin respectively
V, W and Z ).

So the distance OV is actually the contracted distance of the klingon ship as the enterprise sees it ( this is because the V point is got as intersection between the front klingon wordline with the
enterprise x-axis ).

So, the way contraction is actually a f() in which you choose the C point.

This puzzles me a bit: imagine to keep the enterprise with the same length at rest, i.e. WZ length
as constant. Now we decide to move the W point shifting on its right. This shifts the Z point as well, leading to a new C point with the beam : this result in a klingon ship contracted a bit wider than before.

So the point is: how the contraction, which a f() of just the relative beta ( v/c ), may vary
changing the point in which we choose to draw the enterprise ship? ( i.e. V point )

2) I cannot figure how to relate the VW distance with the OV one. From the picture it seems
exactly the same distance. I mean, I'm able to derive the contracted X from the uncontracted
one ( at t=0, for instance ), but I'd like to desume it from the plot, not with formulas...

This is also because let's imagine to keep the W point as costant, i.e. getting a given OV contracted klingon ship. where is the constrain for the Z point, i.e. for the enterprise front point?
I'm asking this, because if we'll be able to move the W on its left without constrains, i.e. to change the enterprise size, we could lead to a situation in which the enterprise front it's on the left of firing E points, leading to a hit, instead of a miss.

This doesn't make sense, of course. So my reasonign is wrong in somewhere.
Let me know where you find my error, please, eventually.

Regards

Ricky

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Hello,
after having done a bit of exercices on Taylor & Wheeler ( just for self-study ), I felt
the need to go on a bit differently, i.e. trying to solve qualitative problems instead of
quantitative ones, i.e. using a bit of diagrams ( all in all, as Susskind always suggest
"when you face a SR problem, plot a diagram first ).

That said, I was trying to solve this Trekker oriented problem:

http://www.phys.vt.edu/~takeuchi/relativity/practice/problem07.html

The overall diagram is quite clear to me, since I was able to answer to all
questions "almost" without problems.
I haven't looked at the answers but if I were giving the answers, I would say that the whole problem is a trick question. It's an impossible scenario. It's simply not possible for there to be "a standard Klingon tactic...to fly by its enemy and fire this cannon as the front end of the Battle Cruiser passes the enemy ship's rear end." That would take an impossible instantaneous communication which I will explain shortly.

But then I thought "wait a minute: it's easy looking
at a already well cooked diagram...let's try to do it by myself from scratch, now!"
...
Looking at the original plot, the author decided to put it onto the light beam line, i.e.
C point. It's not really clear from this diagram, but if you look at its book ( I own it ), you
can see the enterprise rear wordline intersects the lightbeam EXACTLY in C point.

So my point is: is this a licky strike, or a specific design?
It's pure coincidence having no significance whatsoever.

Why choosing the C event
to be light-like?
First off, a single event isn't light-like. The spacetime interval between two events, the origin and event C just happens to be light-like but I don't think the author was even aware of this.

From C, you can draw a line parallel to t', which intersect the enterprise x-axis in point
...
Without seeing your diagram, it's a little challenging to follow the rest of your dialog and I would rather show you how I would draw the spacetime diagrams.

First, I will replicate as close as I can the diagram from your link showing just the two spaceships (refer to comments on the diagram for detailed information):

Now looking at the diagram, it would appear that when the front of the Klingon Cruiser (red) reaches the rear of the Enterprise (green) we can draw a horizontal line to the rear of the Klingon Cruiser (blue) and see that it is within the front boundary of the Enterprise (black) and so McCoy would be right and the Enterprise destroyed:

However, if we draw a similar line in the rest frame of the Klingon Cruiser, we see that the rear of the Klingon Cruiser (blue) is outside the front boundary of the Enterprise (black) and so Kirk would be right and the Enterprise saved:

Incidentally, we can transform this diagram back to the rest frame of the Enterprise to see how that line of simultaneity for the Cruiser's rest frame appears in the Enterprise's rest frame (as depicted in the linked diagram):

Personally, I don't see the point in doing this. To me, simultaneity should be shown in the frame in which it applies and if you draw in grid lines, then you don't even need special lines of simultaneity, the horizontal grid lines are already there for that purpose.

Now here's where I want to explain the trickery of this scenario. The only way the Klingon Cruiser could have their "standard tactic" that relies on simultaneity, is to analyze the approaching spaceship long before they get near each other and preprogram the cannon to fire at the correct time. But the implication of the scenario is that a detector in the front of the Klingon Cruiser is what causes the cannon to fire and if that's the case, then the whole story changes. The fastest that a signal could be propagated from the front of the cruiser to the rear would be at the speed of light such as:

And now we see that the Enterprise will be destroyed. In fact, it doesn't matter what frame we analyze this in. Here is the rest frame for the Enterprise:

I would encourage you to draw your spacetime diagrams on graph paper and to focus on just one frame. When you get done, use the Lorentz Transformation to convert the coordinates to a new frame and draw another diagram and another piece of graph paper.

Alternatively, you could copy my drawings and add in your labels and maybe I could follow the rest of your explanation.

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Hi,
thanks for your answer. the problem of simultaneity is clear, of course, I mean this problem is fake,
since it assumes the signal from front to rear travels istantaniously, which could not be.
If you look to other problem in the list:

http://www.phys.vt.edu/~takeuchi/relativity/practice/problem08.html

this is actually a variation in which the author explains this situation is unrealistic, for the reason you are mentioning.
Nevertheless, my problem wasn't that. It was clear this was unreal, etc. etc.

My problem, if you read better to my post, is different. The diagram I'm talking about is exactly
the same as the one in the link, BUT ( I wrote it in my post above ) you have to label 3 points
on the x axis, like I did, i.e. V, W and Z. The V is the intersection between the front contracted
klingon ship with the x axis, W is the front of Enterprise at t=0 and Z it's its rear at t=0.

All the remaining diagram is unchanged. So, again, my problem is in how to relate the uncontracted
enterprise with the contracted klingon.
Let me know if now you are able to follow my post now, otherwise I could rearrange my post a bit.

Ricky

Hi,
thanks for your answer. the problem of simultaneity is clear, of course, I mean this problem is fake,
since it assumes the signal from front to rear travels istantaniously, which could not be.
If you look to other problem in the list:

http://www.phys.vt.edu/~takeuchi/relativity/practice/problem08.html

this is actually a variation in which the author explains this situation is unrealistic, for the reason you are mentioning.
Nevertheless, my problem wasn't that. It was clear this was unreal, etc. etc.

My problem, if you read better to my post, is different. The diagram I'm talking about is exactly
the same as the one in the link, BUT ( I wrote it in my post above ) you have to label 3 points
on the x axis, like I did, i.e. V, W and Z. The V is the intersection between the front contracted
klingon ship with the x axis, W is the front of Enterprise at t=0 and Z it's its rear at t=0.

All the remaining diagram is unchanged. So, again, my problem is in how to relate the uncontracted
enterprise with the contracted klingon.
Let me know if now you are able to follow my post now, otherwise I could rearrange my post a bit.

Ricky
I've gone back and studied your first post some more and as I said earlier, you're giving importance to point C where there is none.

The contracted length of the Klingon ship can be measured or seen on the diagram at any time, not just at the Coordinate Time of 0. It's length in my first diagram is always 400 feet. You can see this easily at the Coordinate Time 500 nsecs or 1000 nsecs where the blue and red dots line up on vertical grid lines spaced 400 feet apart but it's also true everywhere else. You just have to remember to measure it horizontally.

The contracted length of the Klingon ship has nothing to do with the location of the Enterprise or even if there is an Enterprise. It only has to do with the fact that the Klingon ship is traveling at some speed in a particular frame. It's speed determines the amount of contraction--nothing else.

Don't get confused by the common terminology that the Enterprise "sees" the Klingon ship as contracted because it doesn't. That terminology is short-cut for saying, "in the frame in which the Enterprise is at rest, the Klingon ship is length contracted" but we could also say, "in a frame in which the Klingon ship is traveling, it is length contracted".

I see your point, but this unfortunately doesn't help me too much.
I've used the word "see" but I know'd better using "measure". But again, this is
not the point.
And I know as well the contracted length can me measured at any horizontal line
in which Deltat is 0 in Enterprise frame, i.e. where simultaneous events occur
to Enterprise.
The real point is the ship should have the same length when BOTH are at rest.
This is fine. So I can choose the Enterprise's length and the contracted one "born"
naturally from the rapidity by which the axes a inclined, i.e. is just a f(v/c).

My point is: how can I measure the uncontracted length length of Klinkgon's ship
directly on the x't' diagram? Is this possible? It seems not to me.

And again, I know you are saying point C is nothing special. This makes a lot of sense to me.

But, if I have to plot the contracted length from scratch, I would you do it?
I mean, let's imagine to have the bare two frames axes, with nothing more.
You add, for instance, the uncontracted Enterprise first, with 2 vertical lines.

The how you's add the contractred one? Is seems the original author is
using the C point, making a line parallel to t' and then keeping my V intersection.

This makes sense, again, since the line parallel to t' passing through C is actually
the end of contracted ship.

But here's my doubt: if you move to the right the C point, since it's aribtrarely chosen,
you'd land to a wider contracted length.

So probably, since you are so gentle and you are able to plot so fine diagrams, may you explain
your point simply drawing it from scratch?

I hope my point is clear, now.

Regards

Ricky

I see your point, but this unfortunately doesn't help me too much.
I've used the word "see" but I know'd better using "measure". But again, this is
not the point.
And I know as well the contracted length can me measured at any horizontal line
in which Deltat is 0 in Enterprise frame, i.e. where simultaneous events occur
to Enterprise.
The real point is the ship should have the same length when BOTH are at rest.
This is fine. So I can choose the Enterprise's length and the contracted one "born"
naturally from the rapidity by which the axes a inclined, i.e. is just a f(v/c).

My point is: how can I measure the uncontracted length length of Klinkgon's ship
directly on the x't' diagram? Is this possible? It seems not to me.

And again, I know you are saying point C is nothing special. This makes a lot of sense to me.

But, if I have to plot the contracted length from scratch, I would you do it?
I mean, let's imagine to have the bare two frames axes, with nothing more.
You add, for instance, the uncontracted Enterprise first, with 2 vertical lines.

The how you's add the contractred one? Is seems the original author is
using the C point, making a line parallel to t' and then keeping my V intersection.

This makes sense, again, since the line parallel to t' passing through C is actually
the end of contracted ship.

But here's my doubt: if you move to the right the C point, since it's aribtrarely chosen,
you'd land to a wider contracted length.

So probably, since you are so gentle and you are able to plot so fine diagrams, may you explain
your point simply drawing it from scratch?

I hope my point is clear, now.

Regards

Ricky
Let me start by saying that we have a special term for the uncontracted length. It's called the Proper Length.

OK, you want to start with the Proper Length of the Enterprise in its own rest frame showing as two vertical lines:

We can make this any width we want but we'll stick with the same width we had earlier, 500 feet. (The scaling on these diagrams is somewhat smaller than before.)

Now we want to add in the contracted Klingon ship traveling to the right at 0.6c. If we don't know how to do that, we can start by transforming to the rest frame of the Klingon ship. The speed parameter we want to use in the Lorentz Transformation is 0.6c because want the new frame to be moving to the right with respect to the original one which will make the Enterprise move to the left:

Now we can add in the Klingon ship anywhere we want. I have arbitrarily chosen as shown here:

Finally, we transform back to the original frame using a speed parameter of -0.6c and we get:

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Yes sorry for the proper distance miss. Sometimes I use the terms properly, someother I forgot to use them.
I see your point, and this is good to me, I mean now I can understand a way of doing it.
But, this not answer to my doubts completely. Let me try to explain me better.
First of all, I'd like to do just in a unique original spacetime diagram, WITHOUT the need to jump back and forth
between 2 diagrams as you did.
Also, I'd like to do it desuming the contracted length from the proper one WITHOUT the need of measuring with 0.6c or so. All in all, the "inclined" primed system should already takes care by the rapidity about
the contraction, since its inclination is a f(v/c). So, there MUST be a way to plot the contracted length
from the proper just by geometric construction.
Hope my point is clear about that.
So let's try to change a bit the way in which I would plot it from scratch. this should explain my point a bit better ( just to point it: I know my reasoning is wrong and C point is almost meaningless in choice, but I cannot
see where it's wrong ).

Let's start again in a different way: let's start always from the 2 frames, the Enterprise unprimed, and the
Klingon primed.
Now, don't put yet the Enterprise in the game. Let's just choose the C point somewhere, i.e. let's put the
meeting event on the plot. As you stated, it doesn't need to lie on the light beam.
This point represent the event in which the Enterprise's rear meets the Klingon front. This mean, we could plot
a vertical line in unprimed frame, representing the Enterprise rear world line.
At the same time, we could draw a line parallel to t' axis passing through C: this represents the Klingon's ship
front world line.
This is correct to me, and I cannot see error so far. Now, we can take the intersection of the Klingon's front world line with any of horizontal line in the unprimed frame, i.e. with any line at the same time in Enterprise frame.
We could, for instance, take the intersection with x axis, leading to "my" V point. Or we could take the one with
the horizontal line passing through C, i.e. the line of event at the same time o C in Enterprise frame.

Let's choose this second option. If we look at the original diagram, we can take the B intersection with the t' axis, which is also, of course, the klingon's rear word line.
this means, we can interpret the BC segment as the klingon's contracted length measured at time tC in Enterprise frame.
So, with this reasoning, we are able to fix the klingon's contracted length, WITHOUT any need to
plot the Enterprise proper length. Hope this is clear.
So far, it makes a lot of sense to me.

So now we should be able to plot the Enterprise wordlines using the proper Enterprise length. this means adding
a vertical line passing through the A point. So AC is actually the proper Enterprise length.
Comparing AC and BC we can BC is contracted, which is good.

But let's try to redo from scratch choosing a C point a bit on its right ( towards increasing xs ), but lying on the
same horizontal line. We now lead to a bigger klingon's contracted length, since the C point can be chosen arbitrarely.
If now we put back the vertical line keeping the Enterprise proper length, we could lead to something
which seems ( to me ) to be unrelated with the contracted one.

Just to resemble:

1) we choose the meeting event point C
2) from it we can desume directly the klingon's contracted length

then, there must be a way of getting the Enterprise proper length by a
PURE GEOMETRICAL BASIS, desuming it from the BC segment.
this is roughly, desuming the AC segment from the BC segment just from the plot.

So you are basically draw arbitrarely spaces ships world lines, and the you get the C
point as intersection, which is good. I'm starting from an arbitrarely chosen C point
and desuming from it the contracted length. I can do it since this is a pure qualitative
problem without numbers. Of course, in a real problem I'd start like yuo did in your case.

This is my real problem. Probably I'm wrong in doing it. May be I cannot start putting the C point and then
desuming from the rest from it.

But my feeling there must be a way to desume lengths without neither measuring them nor being forced to
swap between frames for doing.

Let me know where's my reasoning is failing, please.

Thanks

Ricky

EDIT: Reading deeply you first explanation makes a lot of sense and I will try it for sure, in future.
But my point is just for understanding if it's possible to gather all info from the same diagram.
I mean: all in all there's anything which actually force me to avoid to start from a "meeting point
event", and then desuming all rest of data from it.
I know, this is a bit of an "odd" and couterintuitive way of proceeding, but, since the info are already
all there embedded in the axes inclination, there must be a way of doing it in the way I-m depicting.

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I see now what you are asking: you want to know how to put two frames on one drawing. I have no idea how people make that kind of drawing. I would struggle with the same confusion you're struggling with. Maybe someone else can help you with that.

No problem, you have been really really helpful, and now, at least, I know how to face
the problem somehow.
I'm a bit obsessed by information content, sometimes. So I'm almost sure all the needed data are already there.
But I'm not able to retrieve them properly.

Just a final question: your way of building the 4 wordlines intersecting is very good.

That said, I think the original diagram is actually making the klingon's rear world line coincident with the t' axis,
is just a handy way of doing it, just to have them coincident. I mean, we could take the rear wordline simply
parallel to the t' axis ( like we do with the enterprise front, which is quite far from unprimed origin ).

So again, this is just how the author decided to put its frames origin, I guess.

If you look at my first diagram in post #2 and my last diagram in post #6, you will see that the relationships between all the worldlines are identical, even though their relationship to the origins is different and the dots representing one-nsec ticks are different. I don't know if the author's diagram allows for such freedom or whether there needs to be a specific relationship of the worldlines to the origin. Hopefully, someone who knows how to put two frames on one diagram will provide the answer.

I don't think there's any need to synchro the wordline with the origin at all, neither in the primed frame, not in the unprimed one. The worldline, as far as I know, are actually uncoupled from any choice of spacetime diagram.
The only requirement, since they represent rest frames, is of course the enterprise world lines are parallel to t axis, while the klingon's ones are parallel to t'.
But, as for plot of classical mechanics dynamics, the point in which you choose the origin is totally arbitrary, since they are just equivalent canvas in which represent the same reality.

All in all, we wouldn't see any different hit or miss if we choose the enterprise to be 100 feet from the origin of unprimed frame, while the klingon's ship is 10000 feet far from the primed origin.

This wouldn't simply make sense. So, again, I think the author has done probably something like you, for getting
the 4 canvas wordlines, then he decided to put the klingon's rear wordline passing through the origin, just for getting a "small" or a more handy diagram, or so.

Thanks for your help. we will see if somehow else will eventually help us further. I know the problem is tricky,
and also it's not the common way for doing it.

Regards

Ricky

... So, again, I think the author has done probably something like you, for getting
the 4 canvas wordlines, then he decided to put the klingon's rear wordline passing through the origin, just for getting a "small" or a more handy diagram, or so.

Thanks for your help. we will see if somehow else will eventually help us further. I know the problem is tricky,
and also it's not the common way for doing it.

Regards

Ricky

I don't see anything wrong with the statement of the problem. Perhaps the resolution will become more apparent to all if the statement of the problem is inverted.

Given:
A Clingon vessel pointing in the positive x-direction and 1 light-second long in proper length has a disruptor attached to its rear with a trigger accessible to the captain at the front.

The Enterprise pointing in the negative x-direction and 1 light-second long in proper length is approaching the Clingon vessel from ahead at v = -.866c on course to just miss a head-on collision.

As the Enterprise passes the Clingon vessel, the Clingon captain waits till the rear of the Enterprise aligns with the front of his vessel and then fires the disruptor. But the Clingon captain does not know about relativity and does not realize the Enterprise is length contracted by 50%. The disruptor misses its target.

From the point of view of those on the Enterprise, it is the Clingon vessel that is length contracted so that Dr. McCoy, with only a superficial understanding of relativity, thinks the Enterprise is going to get hit. Captain Kirk, however, knows SR quite well and knows the disruptor will miss.

Show how Captain Kirk comes to his conclusion. (Presented this way many of you will already recognize the resolution. It has to do with Relativity of Simultaneity.)

Put one clock with the Clingon captain, call it t1, and one with the disruptor, call it t2. From the reference frame of the Clingon captain the trigger is pulled and the disruptor fires simultaneously. Let’s say both clocks record the events at time t1=t2=0. That’s 4 events, all of which occur simultaneously in the reference frame of the Clingon captain.

From the reference frame of the Enterprise the pulling of the trigger and the front clock recording t1=0 still occur at the same time. And the firing of the disruptor and the rear clock recording t2=0 also occur at the same time. But the two pairs of events do not occur at the same time. As far as Captain Kirk is concerned, the disruptor clock, t2, reads .866seconds ahead of the Clingon captain’s clock.

So, according to Captain Kirk, when the Clingon captain pulls the trigger at t1=0 the disruptor clock, t2, already reads .866seconds. Since the disruptor fired at t2=0 the firing took place .866 seconds before the Clingon captain pulled the trigger. Where was the disruptor then? A little Newtonian Physics tells us. Distance = Velocity x Time = .866c(m/s) x .866(s) = .75 light-seconds. Captain Kirk knows the disruptor pulse will pass .25 light-seconds in front of the Enterprise.

I don't see anything wrong with the statement of the problem.
If you can't see what's wrong, then I'll point it out later when you get to it.

Perhaps the resolution will become more apparent to all if the statement of the problem is inverted.

Given:
A Clingon vessel pointing in the positive x-direction and 1 light-second long in proper length has a disruptor attached to its rear with a trigger accessible to the captain at the front.

The Enterprise pointing in the negative x-direction and 1 light-second long in proper length is approaching the Clingon vessel from ahead at v = -.866c on course to just miss a head-on collision.

As the Enterprise passes the Clingon vessel, the Clingon captain waits till the rear of the Enterprise aligns with the front of his vessel and then fires the disruptor. But the Clingon captain does not know about relativity and does not realize the Enterprise is length contracted by 50%. The disruptor misses its target.

From the point of view of those on the Enterprise, it is the Clingon vessel that is length contracted so that Dr. McCoy, with only a superficial understanding of relativity, thinks the Enterprise is going to get hit. Captain Kirk, however, knows SR quite well and knows the disruptor will miss.

Show how Captain Kirk comes to his conclusion. (Presented this way many of you will already recognize the resolution. It has to do with Relativity of Simultaneity.)

Put one clock with the Clingon captain, call it t1, and one with the disruptor, call it t2. From the reference frame of the Clingon captain the trigger is pulled and the disruptor fires simultaneously.
This is what's wrong. The Klingon captain cannot pull the trigger and cause the distant disruptor to fire simultaneously. That would be action at a distance or having a signal travel faster than the speed of light.

Let’s say both clocks record the events at time t1=t2=0. That’s 4 events, all of which occur simultaneously in the reference frame of the Clingon captain.

From the reference frame of the Enterprise the pulling of the trigger and the front clock recording t1=0 still occur at the same time. And the firing of the disruptor and the rear clock recording t2=0 also occur at the same time. But the two pairs of events do not occur at the same time. As far as Captain Kirk is concerned, the disruptor clock, t2, reads .866seconds ahead of the Clingon captain’s clock.

So, according to Captain Kirk, when the Clingon captain pulls the trigger at t1=0 the disruptor clock, t2, already reads .866seconds. Since the disruptor fired at t2=0 the firing took place .866 seconds before the Clingon captain pulled the trigger.
Now you've got the effect before the cause! Doesn't this bother you?

Where was the disruptor then? A little Newtonian Physics tells us. Distance = Velocity x Time = .866c(m/s) x .866(s) = .75 light-seconds. Captain Kirk knows the disruptor pulse will pass .25 light-seconds in front of the Enterprise.
Your Newtonian Physics calculation is wrong. You have incorrectly applied the Proper Time for the disruptor clock to the Enterprise's frame. The correct answer is 1 light-second, not .25 light-seconds.

Here is a spacetime diagram for the rest frame of the Klingon spaceship using the same colors as described in post #2:

And here is the same scenario transformed to the rest frame of the Enterprise:

You can see that the firing of the disruptor occurs in Kirk's rest frame one light-second in front of (to the left of) the front of the Enterprise (the black worldline).

As I said in post #2, the fastest that a signal could be propagated from the trigger to the disruptor is at the speed of light, in which case Kirk is wrong and the Enterprise is destroyed. Here's how it looks in the Klingon's rest frame:

And as I also said in post #2, it doesn't matter which frame is used. Here's how it looks in the Enterprise's rest frame:

But if the propagation of the signal from the trigger to the disruptor is at less than the speed of light, as is common with electrical cables, the Enterprise could be saved. Specifically, if it is less than 0.866c in the Klingon's rest frame, then the signal will propagate behind the Enterprise and the disruptor will fire after the Enterprise has passed it.

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Thanks for your help MikeLizzi,
but we didn't talk about the was correct or not. My problem is how the initial diagram in the link I've posted
was plotted.
Read carefully the post above, and you will notice I'm talking about how to retrieve contracted klingon's length from enterprise contracted one, but:

1) using the same spacetime diagram and not jumping to different frame views
2) not using any direct measurement, but just retrieving data in a pure geometrical way.

Thanks

Ricky

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Hi ghwellsjr,
I was trying to apply the overlapping of 4 wordlines as you did in post 6. So, first you plot the enterpise in its rest frame and then you get its transformation in the klingon's rest frame.
I was wondering how to get these plots on a paper in a quick way. So, I guess you have may be an applet or so which is actually drawing and inclining them or so?

I mean, the lines can be actually got from applying lorentz transform to couple of points per wordline, isn'it?

So, when transforming from enterpise to klingon's frame I can take 2 points for each line, applying the lorentz
transformation getting 2 new points per line, and then draw the 2 new lines, is this correct?

The same from transforming back with the same trick.

I mean, I checked the transformed worldlines of enterprise form correctly an slant angle related to 0.6c ( I checked for instance some points, getting 9/15 squares to the left, i.e. 0.6 ( since we are assuming c=1 to have
PI/4 light inclination as usual ).
But this is just the inclination. It seems you diagram correctly transforms back and forth, i.e. why I guess you are applying effectively lorentz transformations to couple of points per line, or at least transforming one point + 0.6 of slant.

Am I correct?

thanks

Ricky

Hi ghwellsjr,
I was trying to apply the overlapping of 4 wordlines as you did in post 6. So, first you plot the enterpise in its rest frame and then you get its transformation in the klingon's rest frame.
I was wondering how to get these plots on a paper in a quick way. So, I guess you have may be an applet or so which is actually drawing and inclining them or so?

I mean, the lines can be actually got from applying lorentz transform to couple of points per wordline, isn'it?

So, when transforming from enterpise to klingon's frame I can take 2 points for each line, applying the lorentz
transformation getting 2 new points per line, and then draw the 2 new lines, is this correct?

The same from transforming back with the same trick.

I mean, I checked the transformed worldlines of enterprise form correctly an slant angle related to 0.6c ( I checked for instance some points, getting 9/15 squares to the left, i.e. 0.6 ( since we are assuming c=1 to have
PI/4 light inclination as usual ).
But this is just the inclination. It seems you diagram correctly transforms back and forth, i.e. why I guess you are applying effectively lorentz transformations to couple of points per line, or at least transforming one point + 0.6 of slant.

Am I correct?

thanks

Ricky
If you are making your own plots on graph paper, you only have to transform the two endpoints of each inertial worldline segment, as you surmised, and then you can space the dots proportionally. After you get the hang of it, you could just transform one endpoint and apply the correct inclination, but this will require you to also apply the correct time dilation. For simple cases, you can probably do this correctly but the safest way is to transform the endpoints. Just remember to apply the negative sign to the speed when going back from a transformed frame to its original frame.

I wrote my own computer program to make these diagrams. It transforms the coordinates of each dot.

Yes I realized it must be an application fro doing that transformation.
thanks for the hints about end points, I realized the same by myself doing some trials.

I've tried to transform back the red/black wordlines meeting point at (-2, 1) in klingon rest frame getting correctly the transformed point (-1, 1) in Enterprise's frame. In that case, you land to a rear klingon's ship wordline ( blue one ) which doesn't go through (0, 0), but the hit/miss reasoning would hold the same ( I'm going to show this in a further post ). In your first post, incidentally, you chose the blue world line pass through (0, 0) in Klingon's rest frame, which is of course transformed as (0, 0) in enterprise's frame. In that way you got exactly the same diagram as done by the author in the link.

Thanks

Regards

Ricky

Hi MikeLizzi,
I would not enter in any discussion between you and ghwellsjr. I'm here just to have fun with physics,
and usually I don't like forums, just because of situation like these.

But I'm here to talk about physics, so I have to support ghwellsjr theory completely. I mean, don't let me be misunderstood, I'm not talking about "the way in which things are said", I'm talking about physics.

And ghwellsjr is right in what he said: the original problem is not "real". I mean, it stills remains a problem
to be solved, but it assumes the signal from the front to the rear of klingon' ship is sent without any delay
due to finite speed of light. You simply cannot have signals running without delay and transmitted simultaneously
from A to B.
ghwellsjr was just saying that, and he was right in stating it.
To prove this further, consider the author next problem, which I posted above, i.e.:

http://www.phys.vt.edu/~takeuchi/rel...problem08.html [Broken]

In this new variant, the author states:

"The tactic assumed above is actually unrealistic since there is no way for the gunner at the rear end of the Battle Cruiser to know when the Cruiser's front end has reached the rear end of the Enterprise. So instead, consider the situation in which a light signal is sent from the front end of the Battle Cruiser to its rear when a sensor detects the rear end of the Enterprise pass by, and the cannon fires when it receives this signal."

So ghwellsjr has landed to the same conclusions, even without knowing about the second corrected problem.

Regards

Ricky

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OK, I had found some time to hand plotting some spacetime diagrams to proof the origin choice
is not important, i.e. we can easily choose the klingon's rear wordline different from t' axis ( but of course parallel to it ).

So, transforming back and forth as we talked about I was finally able to draw correctly the worldlines. I've analysed directly the realistic case with the signal from front to rear.
Here's the first diagram:

The F point is the firing event, i.e. when the light ray reaches the klingon's rear. At this point is easy to evaluate situations from enterprise and from klingon's ship firing to lines parallel to x and x' axis.
With the first we can see the F point occurs exactly spatially in between the enterprise ship, hitting it completely. This is because of course the F point lies between the ED segment, which is the enterprise proper length. At the same time we can see from klingon POV the F lies between the HL segments, which represents again enterprise's length.

As we can see, there's not problem at all in having the klingon's rear worldline far from the t' axis: all the reasoning still rule. This makes sense since the origin is just a choice and doesn't influence the physics of the system.

But at this point I was interested in seeing better the Klingon's rest frame POV, just to understand better about your sentence about the speed of the signal, for having a hit or a miss.

Here the picture, hence:

This is "almost" exactly the same picture you can obtain with your app ( a bit less precise, of course ), i.e. the wordlines are correctly transformed using the Lorentz transformations.
I kept the same point naming and meaning, just to be able to easily compare both POVs.
Here we can see the "same reality" saw from enterprise POV, i.e. the F point lying in between HL and ED segments, reasulting in a plain hit.
I've also shadowed by a solid hatching the interesting triangle FLC. this is the area in which signals sent back to the klingon's ship rear still hit the enterprise. The fastest signal is, of course, the light ray, which result in the line passing through CF, hitting at F. If now we slow down the signal, we get all lines to light ray right side until we reach the line CL. This is the other extreme signal speed which leads to a hit, having the klingong's front facing at last the enterprise rear.
This line, as you depicted in your post, is actually the same as the ships and frames speed, i.e. 0.6c in our case.
Less than this, the signal is too slow to reach the klingon's rear just in time to get a hit.
This makes a lot sense to me: if the signal is slower than the ship speeds, this latter are so fast
to paste each other completely before the klingon's rear reaches the signal and fires.

Regards

Ricky

EDIT: if you click on images you can see them zoomed. I don't know why
they are so little in the post.

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EDIT: if you click on images you can see them zoomed. I don't know why they are so little in the post.
I don't know how you posted the images but the way I do it is I go down and click on Manage Attachments and then I browse to the images and upload them and close the window. That adds links to the images above the Manage Attachments button. Then I right-click on one of the links and select Copy Link Location, go up to the Insert Image button, and when it says Please enter the URL of your image, I just paste and click OK.

Transforming back and forth between two diagrams for each frame is the only way I know to end up with both frames on one diagram. You did a great job! Are you satisfied with that method or are still hoping for someone to show how you can do it without actually drawing two diagrams?

I don't know how you posted the images but the way I do it is I go down and click on Manage Attachments and then I browse to the images and upload them and close the window. That adds links to the images above the Manage Attachments button. Then I right-click on one of the links and select Copy Link Location, go up to the Insert Image button, and when it says Please enter the URL of your image, I just paste and click OK.
I uploaded the images as you are telling and this make them appear under the attachement combo box ( sorry for the naming, I'm a programmer ). After this, when I need to insert one of them, I simply put the cursor where I need it and then choose the image from the combo. But probably this trick is actually making them so little. I will try yours, in future.

Transforming back and forth between two diagrams for each frame is the only way I know to end up with both frames on one diagram. You did a great job! Are you satisfied with that method or are still hoping for someone to show how you can do it without actually drawing two diagrams?

Thanks. I'd be able, I guess, to write a similar app as yours, but, despite I'm working every day with latest technology, I'm a bit old fashioned and I like to write, draw etc etc. This is my only way to "carve" the concepts in my mind.

Yes I'm fully satisfied and I'm a bit convinced there's no direct way to plot correctly wordlines without your method. All in all, I've realized the proper length is meaningful only in the frame which is drawn with the "normal" perpendicular xt axis couple. That's why you have to transform back and forth from xt to x't' and viceversa.
So I cannot see any other method than yours, after having use it.

Thanks for your support

Ricky

All in all, I've realized the proper length is meaningful only in the frame which is drawn with the "normal" perpendicular xt axis couple. That's why you have to transform back and forth from xt to x't' and viceversa.

I noticed this sentence is a bit funny, a may be foggy. First of all, by "frame which is drawn with the "normal" perpendicular xt axis couple" I meant the rest frame of course. Also, I really meant "the metric is not euclidean on the diagram", so you cannot measure with a ruler on your plot the distances separating points not simultaneous in a given frame. ( I know you will think I have reinvented the wheel, but just breaking your head on this kind of things you can REALLY realize the big structure underlying this diagrams ).
Also, the unit of measure on "inclined" frame axes are somewhat scaled compared to the unit of measure on orthogonal axes ( I realized as well why you are probably plotting equidistant points on the wordline: you can easily see how inclined wordlines lines are actually stretching the points ). This is of course due to the particular properties of Lorentz transformation.
So, for example, let's take the plot in first diagram ( Enterprise's rest ). We know the segment FM actually represents
the klingon's ship width, taken at same time in Klingon's primed frame. We know, by construction, the proper width
is the same as the enterprise's rest one. In my plot this represented by six unit.
Now, we could be tempted to measure by a ruler with the same unit the distance between points F and M, but this would be completely wrong, because THIS IS not euclidean space. The metric is different. So if you measure that distance
you get a totally wrong value. The trick is to notice the x' axis has a different unit than the one in unprimed Enterprise's one. If I had plotted the equidistants transformed points as you did by your app, I would notice that FM segment
is wide exactly as 6 transformed units. So it's correct and coherent: that distance is the klingon's ship width at rest ( taken at two simultaneous times in klingon's rest frame ) BUT evaluated using the unit of measure on x'.
The ratio between the two units of measure is, of course, a f() of the relative speed between the 2 frames.

I know these are trivial things for the most of you here on the forum. But realizing on the diagram is a different thing to me.

That said, I think it's really hard to retrieve all distances just from the diagram as I thought initially.
The metric is different as using geometry is a bit hard ( you can land to segment which seems bigger than other
ones if measured with euclidean metric, but are shorter using the Minkowsky metric ).
May be, using the hyperbolic geometry, we could try to do something more in that sense ( i.e. retrieve in a native
way the proper distances from contracted ones using pure geometrical tools ).

Thanks for all the support and patience

Regards

Ricky

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I see now what you are asking: you want to know how to put two frames on one drawing. I have no idea how people make that kind of drawing. I would struggle with the same confusion you're struggling with. Maybe someone else can help you with that.

Never heard of Loedel diagrams?
On a few occasions Loedel diagrams have been posted on this forum (BobC2 posted a lot of them) , but apparently you (and many others) never took the effort to study them...?
Loedel diagrams show same time and space units in both frames. A LOT easier to read reciprocal contration and time dilation than from a Minkowski diagram.
No offence, your charts are O.K. and correct, but you will not get the bigger 4D spacetime picture if you can not read them from one 4D spacetime diagram.

Never heard of Loedel diagrams?
On a few occasions Loedel diagrams have been posted on this forum (BobC2 posted a lot of them) , but apparently you (and many others) never took the effort to study them...?
You're just saying that because you don't remember this post or your response.

Loedel diagrams show same time and space units in both frames. A LOT easier to read reciprocal contration and time dilation than from a Minkowski diagram.
No offence, your charts are O.K. and correct, but you will not get the bigger 4D spacetime picture if you can not read them from one 4D spacetime diagram.
You always acknowledge that my charts are O.K. and correct, thanks for that.

The OP was wondering how to draw a diagram of the type in his link in his first post. I admitted that I don't know how to do that, ie, put two frames on one diagram. If you know, why don't you teach him?

I've looked to Loedel diagrams and they seem really interesting.
I mean, they finally remove that asymmetry between scaling from the
rest frame and the moving one, leaving the two frames to their simmetric
importance, since no one of them is more important than the other.

I will try to replot "my" problem with them for sure.

But, again, this does not answer to my original doubts, i.e how much info I could
retrieve from a single diagram. Now that I finally was able to draw it using the
jump between frames ( which is very rich in contents, to me ), I can try to restate
my problem a bit better, if some of you if still interested in.

So, let's start from a bare diagram in which is present just the enterprise xt frame
and the klingon's primed one.

Now, as told initially I imagined to do these steps:

1) I could fix, like the original author did, the klingon's rear wl ( wordline since now )
to be coincident with the prime t' axis. this is somehow really handy, as we'll
see, but it's not mandatory as I demoed in my plots above.

2) instead of starting in adding the enterprise wls in its rest frame ( vertical lines ),
I imagined simply to know in advance where the C point as been measured
from both frames. This is the spacetime point in which the klingon's front met
the enterprise's rear. We could imagine a mental problem in which we know this
data, instead of ships' width in their rest frame.

Let's plot it in this way, for instance:

( still inserted too little, sorry for that, but if you click on them you can zoom them ).

3) After having fixed the C point, we could for sure run through it 2 other wordlines:
the blue Enterprise's rear wl and the Klingon's red front wl. These are correct for
sure, since the C point is a spacetime event involving E's rear and K's front wls.

4) now we can imagine to sit at rest on the Enterprise, and to measure the Klingon's
ship width ( I took really really care to avoid the word "to see" ). The measurement
is done at the same time, so this leads to draw a horizontal line ( parallel to x-axis )
running through the C point.

5) this line meets the t' axis, which is the K's rear wl as well, in the B point. This is actually
the second point measured by the Enterprise, so we could state for sure, the Klingon's
ship length as measured by the Enterprise is the BC segment length, measured
using the unit of measure belonging to the Enterprise's rest frame, of course.

This point IS very important: using this simple technique, and using the choice of K's rear wl
coincident with the t', we was able to gather the measured contracted Klingon's ship length
just starting from the C meeting point.

Now, just a wl is still missing in action: the Enterprise's front wl. And this is really the core
of all this post. As you can from the diagrama above, I just sketched a hypothetical wl
in green, representing this E's front wl. We can be sure, from a pure qualitative point of view,
it must be where it is, i.e. on the left of the B point. This is because the segment AC is actually
the proper Enterprise length in its frame, and we can compare it with the contracted BC length
( which is contracted for sure due to length contraction ).
So, the real point is this: how could I add that green word line, just using the info about
how the BC segment is wide? All in all, we know the 2 ships share the same length when both at rest, so the AC length must a f() of the BC one.
As stated, I'd like to be able to gather that segment just by geometric construction.

We have basically 2 way for doing it without getting crazy with my approach:

1) we can measure the BC segment using E's rest frame unit of length. We know from the theory,
since we are measuring at the same time in the E's frame, we could use the length contraction
formula, i.e. AC = BC * γ, where γ is the Lorentz factor which is > 1

2) we could use a different info. Look at the following diagram:

As you can see, I've added the segment DC here. Since it's parallel to the x' axis, this is
actually the Klingon's witdth as measured in Klingon's rest frame. since this frame is
moving its length it's actually underlying to Lorentz transformations. As I stated in my
previous post, the "inclined" moving frame has a different unit of measure, a bit streched
than the unprimed enterprise's frame.
I've already "demonstrated" this length is actually the Klingon's proper length, BUT MEASURED
using streched primed unit of measure. So, if we were able to measure it with a ruler
sharing the same stretched unit of measure, we could read the Klingon's width, which should
be the Enterprise as well.
So getting it, we could use taking the same value but expressed in unprimed unit of measure,
and applying it directly in drawing the AC segment.
To explain this better in my not really clear english, let's imagine the unit of measure of the
primed x' is actually twice as the unprimed one. If we draw the DC line getting 12 suing unprimed units, we could get the actual length is 12 / 2, so 6 units of measure in primed frame.
This also means the enterprise length is 6 but using unprimed units of measure.

This is very effective, but also a bit tricky since involves measuring being able to use
as many meaningful digits as possible.

So, actually, my doubt still rules: is there a geometric exact way ( by construction, intersection or so ) to retrieve the AC segment from the BC one without using any of the 2 points above?

I really think this could be got by construction using the BC segment somehow altogether
the hyperbolic geometric underlying the diagram.

Hope to have been more clear, now.

Thanks

Ricky

EDIT: looking better as the 2nd diagram, be warned the vertical hypothetical
green wl, must actually lie between point D and B, but I drawn it quite
far just for sake of clearity.

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...
4) now we can imagine to sit at rest on the Enterprise, and to measure the Klingon's
ship width ( I took really really care to avoid the word "to see" ). The measurement
is done at the same time, so this leads to draw a horizontal line ( parallel to x-axis )
running through the C point.
You haven't describe exactly how we can make this measurement "at the same time". I can think of two (equivalent) ways:

The first way is that we could have a series of previously synchronized (using light signals) clocks all along the width of the Enterprise and a ruler all along the width the Enterprise (or distance markings on the floor adjacent to each clock). Then we either have a bunch of people stationed at each clock or a bunch of recording devices (video cameras). Eventually the rear of the Klingon travels along the entire width of the Enterprise and we get a record of when it passed each location. Now we note the time on the synchronized clock at the rear of the Enterprise when the front of the Klingon reached it. Then we interrogate all the people or recording devices and find out which one noted the same time for the rear of the Klingon to have passed it and the ruler or marking on the floor tells us the measured length of the Klingon.

The other way we can make the measurement "at the same time" is for us, sitting at the rear of the Enterprise, to keep a record of radar measurements and examine them after the fact. This involves us continually sending signals that bounce off the rear of the Klingon. We keep track of when each signal was sent and when it was received. Then we assume that each signal took the same amount of time to get to the Klingon as it took for it to return to us. Then we can look at our record and find the one that matches the time the front of the Klingon reached us and by knowing how fast the light traveled, we can calculate "where" the rear of the Klingon was "when" its front reached us.

5) this line meets the t' axis, which is the K's rear wl as well, in the B point. This is actually
the second point measured by the Enterprise, so we could state for sure, the Klingon's
ship length as measured by the Enterprise is the BC segment length, measured
using the unit of measure belonging to the Enterprise's rest frame, of course.

This point IS very important: using this simple technique, and using the choice of K's rear wl
coincident with the t', we was able to gather the measured contracted Klingon's ship length
just starting from the C meeting point.
Well I hope you can see that it's not quite as simple as you make it out to be if you imagine us to be sitting on the Enterprise. Of course it's easy if we are sitting at our desk looking at a diagram that we have constructed and declared what's going on, but that's not what you said you want to imagine.

Now, just a wl is still missing in action: the Enterprise's front wl. And this is really the core
of all this post. As you can from the diagrama above, I just sketched a hypothetical wl
in green, representing this E's front wl. We can be sure, from a pure qualitative point of view,
it must be where it is, i.e. on the left of the B point. This is because the segment AC is actually
the proper Enterprise length in its frame, and we can compare it with the contracted BC length
( which is contracted for sure due to length contraction ).
So, the real point is this: how could I add that green word line, just using the info about
how the BC segment is wide? All in all, we know the 2 ships share the same length when both at rest, so the AC length must a f() of the BC one.
As stated, I'd like to be able to gather that segment just by geometric construction.
If you are continuing to imagine that we are on the Enterprise, we can measure its width with either of the two methods I described earlier for measuring the width of the Klingon:

The first method involves just looking at the end of the ruler or the last marking on the floor.

The second method involves bouncing a radar signal off the front of the Enterprise and measuring the round trip time, dividing by two and calculating how far light traveled during that amount of time.

We have basically 2 way for doing it without getting crazy with my approach:

1) we can measure the BC segment using E's rest frame unit of length. We know from the theory,
since we are measuring at the same time in the E's frame, we could use the length contraction
formula, i.e. AC = BC * γ, where γ is the Lorentz factor which is > 1
Sure we could make this calculation, if we knew how fast the Klingon ship was traveling relative to us and that would require making at least two distance measurements of its position versus time. We can use our previous measurement and as the second one we could use when the rear of the Klingon passed the rear of the Enterprise and from that we could calculate its speed.

2) we could use a different info. Look at the following diagram:

View attachment 61644

As you can see, I've added the segment DC here. Since it's parallel to the x' axis, this is
actually the Klingon's witdth as measured in Klingon's rest frame. since this frame is
moving its length it's actually underlying to Lorentz transformations. As I stated in my
previous post, the "inclined" moving frame has a different unit of measure, a bit streched
than the unprimed enterprise's frame.
I've already "demonstrated" this length is actually the Klingon's proper length, BUT MEASURED
using streched primed unit of measure. So, if we were able to measure it with a ruler
sharing the same stretched unit of measure, we could read the Klingon's width, which should
be the Enterprise as well.
So getting it, we could use taking the same value but expressed in unprimed unit of measure,
and applying it directly in drawing the AC segment.
To explain this better in my not really clear english, let's imagine the unit of measure of the
primed x' is actually twice as the unprimed one. If we draw the DC line getting 12 suing unprimed units, we could get the actual length is 12 / 2, so 6 units of measure in primed frame.
This also means the enterprise length is 6 but using unprimed units of measure.

This is very effective, but also a bit tricky since involves measuring being able to use
as many meaningful digits as possible.
This is very confusing to me. Why do you say "stretched"? It seems to me that you should be saying "contracted".

So, actually, my doubt still rules: is there a geometric exact way ( by construction, intersection or so ) to retrieve the AC segment from the BC one without using any of the 2 points above?

I really think this could be got by construction using the BC segment somehow altogether
the hyperbolic geometric underlying the diagram.

Hope to have been more clear, now.

Thanks

Ricky

EDIT: looking better as the 2nd diagram, be warned the vertical hypothetical
green wl, must actually lie between point D and B, but I drawn it quite
far just for sake of clearity.
You have to decide whether this question is from the point of view of us riding on the Enterprise and making actual measurements, making assumptions and doing calculations or if it's from the point of view of us sitting at our desks and specifying everything about the scenario and simply drawing spacetime diagrams consistently. If it's the later, you should recognize that any single spacetime diagram for a single frame contains all the information we need. All we have to do is use the Lorentz Transformation process to transform to another frame. I'm assuming that we have well-defined coordinates for events on diagrams with clearly marked grid lines.

OK,
of course probably I can restrict my above speech only about "consistently drawing space-time diagram on my desktop". I was just trying to figure out how the enterprise would measure the contracted Klingon's ship, but it seems to me this make the discussion deviating from my real problem.

that said, you are stating a spacetime contains all the information for a single frame. This is the key point. I'm not sure this is fully true, but this is just a feeling. I will eventually work it out better, if I'll have time.
All in all, I need a way to draw properly spacetime diagrams, and now I got it.

Just a final point. Looking to your spacetime you draw in these posts, and also if you do it manually as I did,
you can see those points you are regularly sampling on the axes, are transformed with a wider gap in the other frame. This is because the Lorentz transformation actually stretches the unit of measure. So a segment sampled between 2 points in the rest frame, is actually streched to a longer segment in its transformed dual.
Of course, since the unit of measure is different, if you started with a 1 as unit in the rest frame, you still land to
1 as in the transformed moving frame. But the segment between samples are streched.
This is occurs, of course when you transform from rest to moving. If you do the opposite, i.e. from moving to rest, you land to a contracted segment.
Hope now it's clear what I meant.

Thanks

Ricky

EDIT: trying to refine better what I wrote above about stretching. As we see in out previous posts, let's take two parallel vertical wordlines, as you did in your post. You can, for instance, choosing them 6 unit apart in your rest frame. Then you transform them applying Lorentz and getting 2 inclined parallel lines, as we depicted.
Now, looking at your plots, we can easily the distance ( Euclidean distance, of course ) between to transformed points is actually wider then the corresponding at rest.
But: if you count the number of units the distance between the 2 transformed lines you can see it's piercing exactly 6 new units. So, from a Euclidean point of view, those trasformed segment is bigger, but since the unit of measure is now bigger as well, the length measured in the new unit of measure is actually 6, as it was at rest.

So the streching is from a Eclidean point of view, not from a Lorentz metric one.

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that said, you are stating a spacetime contains all the information for a single frame.
You rearranged what I said and made it sound like I was saying something different than I actually said. Here's what I actually said: "...any single spacetime diagram for a single frame contains all the information we need." By that I meant that our spacetime diagram will have only a single frame in it with a single set of axes and a single set of horizontal and vertical grid lines. In other words, it can be for x,t or x',t' but not both. Whichever one it is for will have the axes at right angles to each other. Then you show all the events of interest and connect them with the worldlines of interest and you can throw in some labels to identify anything else of interest.

The way you misquoted me makes it sound like different frames contain different information which might be the implication you get from your linked drawing in your first post. It makes it look like the x,t frame contains information only about the Enterprise and the x',t' frame includes information only about the Klingon ship and so to get them both on the same drawing, you have to draw two frames, including two sets of axes. If they were really consistent, they should have also drawn in two sets of grid lines along with grid markings. The whole idea of doing it this way is so that you can see both sets of frame coordinates for each event, but since most people don't include the grid lines or the markings, it leads some people to think that each ship exists only in its own rest frame.

This is the key point. I'm not sure this is fully true, but this is just a feeling.
You should be sure that it's fully true (that each frame contains all the information regarding a given scenario)--especially since you're a programmer and you know that I wrote a program to perform the Lorentz Transformation process. Once I put all the worldline information for a scenario into the original frame, I just change the speed setting applied by the LT and I get a new frame--there is no other source of information.

...
Just a final point. Looking to your spacetime you draw in these posts, and also if you do it manually as I did,
you can see those points you are regularly sampling on the axes, are transformed with a wider gap in the other frame. This is because the Lorentz transformation actually stretches the unit of measure. So a segment sampled between 2 points in the rest frame, is actually streched to a longer segment in its transformed dual.
Of course, since the unit of measure is different, if you started with a 1 as unit in the rest frame, you still land to
1 as in the transformed moving frame. But the segment between samples are streched.
This is occurs, of course when you transform from rest to moving. If you do the opposite, i.e. from moving to rest, you land to a contracted segment.
Hope now it's clear what I meant.
There are two things going on when you transform to a new frame. First off, just because the speed is different, pairs of events will appear to get stretched out or compressed but that has nothing to do with any relativistic effect. The second thing that is going on is Time Dilation. Two events that were at the same location in one frame (one above the other in the drawing) will get stretched out (dilated) along the time axis. That combined with the first effect might lead you to the conclusion that there is a stretching along the spatial axis but you should discard that notion. What matters when discussing how the length of an object is depicted in a second frame is most easily understood by looking at the spacing of the worldlines and not the same pair of events. For example, look at these two drawings for the Enterprise from post #6:

In the first drawing, you can pick any black event and any green event and take the difference between their x coordinates and you will get the same answer for the length of the Enterprise, 500 feet. But you can't do that with the second drawing because you could get just about any length you want. I think what you are saying you do is, for example, you take the bottom two events and conclude that the Enterprise has stretched out to 625 feet. Instead, if you look at the two events at the coordinate time of 2 nsec, you will see that the length is 400 feet. In fact, you can use any two events that occur at the same time and get the same answer of 400 feet.

EDIT: trying to refine better what I wrote above about stretching. As we see in out previous posts, let's take two parallel vertical wordlines, as you did in your post. You can, for instance, choosing them 6 unit apart in your rest frame. Then you transform them applying Lorentz and getting 2 inclined parallel lines, as we depicted.
Now, looking at your plots, we can easily the distance ( Euclidean distance, of course ) between to transformed points is actually wider then the corresponding at rest.
But: if you count the number of units the distance between the 2 transformed lines you can see it's piercing exactly 6 new units. So, from a Euclidean point of view, those trasformed segment is bigger, but since the unit of measure is now bigger as well, the length measured in the new unit of measure is actually 6, as it was at rest.

So the streching is from a Eclidean point of view, not from a Lorentz metric one.
It sounds like you are saying that, for example, in the top drawing, since the bottom two events are 5 units apart, if you go to the bottom drawing and measure how far apart they are using the spacing of the dots along the black or the green lines, you'll also get 5 units. While that is an interesting observation, I think it is leading to the wrong conclusion or at least to an unconventional one that will only lead to confusion.

I think a better observation would be to add a series of events spaced one unit apart in the first drawing that "connect" the bottom two dots. These would represent the markings on a ruler placed on the floor of the Enterprise from one end to the other. Then you could say that the ruler continues to measure the same 500-foot distance between the front and the rear of the Enterprise but it would not be correct to say that the ruler is stretched out just like the Enterprise is stretched out.

The reason is that it's not just a series of dots placed along the bottom of the scenario but rather it's a series of additional worldlines that parallel the two end worldlines and now the spacing of all the worldlines is contracted, not stretched.

It sounds like you are saying that, for example, in the top drawing, since the bottom two events are 5 units apart, if you go to the bottom drawing and measure how far apart they are using the spacing of the dots along the black or the green lines, you'll also get 5 units. While that is an interesting observation, I think it is leading to the wrong conclusion or at least to an unconventional one that will only lead to confusion.

This is exactly what I meant. You always get the same "number" of unit, and this makes perfect sense to me. The value is the same but the "base" unit has changed.
Probably you have understood I usually face things without taking too much respect to the convention or so. Don't let me be misunderstood: I'm not proud of it.
But I don't really see how this could lead to confusions, to me it's perfectly clear, even this is
not the usual in which diagrams are analysed.
I mean, despite there are a lot of years I'm studying physics, it's just few months I really started
to do problems or so on relativity ( I focused first on quantum mech, mainly due to my studies
in nuclear engineering in Turin ). So, as usual, I don't know how to "conventional" approach problems. Sorry for that.
The only thing I can see, and this is really embedded in Lorentz transformation, the number of units are actually conserved. I've done a lot of trials, computing different cases, and always I found the units on axes matches this rule.

I think a better observation would be to add a series of events spaced one unit apart in the first drawing that "connect" the bottom two dots. These would represent the markings on a ruler placed on the floor of the Enterprise from one end to the other. Then you could say that the ruler continues to measure the same 500-foot distance between the front and the rear of the Enterprise but it would not be correct to say that the ruler is stretched out just like the Enterprise is stretched out.

The reason is that it's not just a series of dots placed along the bottom of the scenario but rather it's a series of additional worldlines that parallel the two end worldlines and now the spacing of all the worldlines is contracted, not stretched.

I really think this is just a dual way to see what I'm saying, as far as I can see.
I mean, it cannot be a lucky strike to have the number of units equals in both frames.
And, I'm wondering why no one else has never seen this before. I don't mean I'm right.
Don't let me be misunderstood. I'm just trying to understanding why the convention you
are using is more "logic" than mine.

Try to think it in this way: let's take the klingon's ship length, expressed in its primed orthogonal rest frame.
The length is 6 units here. Now we apply the tranformation, going to unprimed enterprise's frame.
Here, if we take a generic line parallel to x' frame, and intersecting the klingon's inclined wls, we
are getting a segment which represents the length of the klingon's ship in its frame.
I'm stating, if you use the "streched" units on the x' axis, you can easily get the same
number of units as in its orthogonal version, i.e. 6 units.
Of course the units are different in length, in order to accommodate for the same number.
So I don't state there is any kind of contraction ( as intended as length contraction in relativity ).
We are just talking about the same ship length, expressed in the same x't' frame but from orthogonal
to inclined.
So, this is "just" a mathematical trick we can use to get the length of something ( length inteded
at same t', in thi case ) even when we are dealing with a segment parallel to inclined x' axis.
This doesn't actually means there a physics in it.
Notice, I'm not trying fooling stating there's a "real" physical space dilation ( stretching )
instead of a contraction. I'm aware we can experience only space contractions and time dilations.
This is straightforward.
I'm stating It's just a match trick.
But this trick, since I'm a programmer as you stated, is puzzling me about saying we have more info to be eventually used and we are not forced to jump back and forth between frames.

So, sorry I have eventually mis-quoted you in some way. My real intention and meaning was
the one just depicted, indeed.

Hoep I was clear.

Thanks, regards

Ricky

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And, I'm wondering why no one else has never seen this before.
Someone else has seen this before, namely, Minkowski, over a century ago. The reason why it's a new discovery for you is that people don't usually put grid lines or markings in their Minkowski diagrams with two frames. If they had, you would obviously see that the grid lines for the spatial separation between the two events at the bottoms of the two world lines are the same size as the grid lines for the time axis. I had to search for quite some time before I found a blank set of Minkowski grid lines:

Note that the slanted yellow grid lines are spaced slightly farther apart than the normal black grid lines. Unfortunately, there are no yellow grid markings so you'll have to imagine that they are there.

I've created another spacetime diagram with two worldlines depicting the ends of an object that is 5 feet wide:

We would show this on the Minkowski diagram using the black grid lines and markings:

Would it ever occur to you to say that the spatial separation of the two worldlines (or even the bottom two events) would be measured using the vertical axis for 0 to 5 nsecs? It's the same physical distance on the diagram, isn't it? Wasn't that your argument for doing that earlier?

Now we transform to a speed of -0.5:

And for this one we use the yellow grid lines and markings:

Similarly, I ask you, would it make sense to use the yellow markings along the t1 axis to measure the x1 distance between the bottom two events? Again, they are the same physical distance on the drawing.

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Similarly, I ask you, would it make sense to use the yellow markings along the t1 axis to measure the x1 distance between the bottom two events? Again, they are the same physical distance on the drawing.

English is not my preferred main language, even I use it every day for my job.
So I cannot understand the "mood" of your sentence just above. It seems a bit
you are doing some irony, but I'm not sure ( notice there's not any polemic in my words, I'm just trying
to figure out if you are confirming what I stated or you are trying to disprove it, somehow ).
I guess you are suggesting to use t1 marking to measure x'1 just because the x' inclined axis
has not any mark on it. But, looking better to your last plot, you can count the grid lines parallel
to t' and you get the red wordline intersecting the x' axis pretty after 5 of them. So you can lead
to 5 as it was in your orthogonal version. So, adding some marks on x' axis as well, would fix this ambiguity, I think.

So the answer is: yes. That logic can be applied both for times and for space axis, of course, since the symmetry of Lorentz transformations are actually "stretching" the unit of measure in the same way( this mainly because the Lorentz transformation IS linear ). Drawing the grids really helps a lot in explaining better what I meant.
And if Minkowski has seen it as well, this is a good point for sure :-).

I was just trying to explain this is not adding any new physics to it. It's straightforward, when you notice a distance in inclined x't' frame is 6, and this is also 6 when the x't' is orthogonal at rest, you are stating nothing really new. You are just noticing a distance is the same when measured in the same frame ( primed in this case ). Of course, if you use euclidean metric to measure that segment when the x' is inclined, you'd get a longer segment. But this, of course, doesn't mean that distance is measured as stretched from xt frame. that distance makes sense only at the same t' time, and cannot be interpreted as a distance in xt, since that would be no more simultaneous in xt frame.
So that's why I'm stating the unit of measure is stretched when we transform it form orthogonal to inclined.

If you look to this article on Loedel diagrams, for instance:

http://www.einsteins-theory-of-relativity-4engineers.com/loedel-diagrams.html

One of the benefits of the diagram is that unlike in Minkowski diagrams, the scales of both axes of both frames are identical. In effect, the worldlines of both observers progress along their respective time axes at the same rate. Hence, the Loedel diagram does not give apparent preference to one of the inertial frames

so "different scaling" is also another way to me to tell "units are scaled ( so contracted or stretched, according to one you are referring to ).

Anyhow, it seems ( I hope ) you got my point, so yes using grid my point ismore noticable. So, coming back to my latest plot, I was just saying the DC segment is actually the klingon's ship length since it lasts from the extremes on klingon's wordlines.
So, if you had an hypothetical ruler owning the stretched units of primed inclined frame, you would be able to measure ( on the plot I mean ) the DC length and you got the length expressed as that distance was still in the dual orthogonal x't' frame.
If you don't have that ruler ( of course is difficult to own one of that kind :-) ), you could store how the unit segment in orthogonal x't' primed is stretched inclining it in xt frame. Let's say you are jumping from 1 to 1.45.
So, for example, if you measure by a ruler the DC segment would be 14.5, you could state for sure the klingon's proper length is 10. So you could use that proper length, and use it for choosing the A point: since the Enterprise and the Klingon's ship share the same length when both at rest, you could use that to choose an A point which x coordinate is 10 units far from C.x ( this means of course forcing the AC segment to be 10 units in length ).

So, theoretically, you would be able to sketch and gather the Enterprise length's without the need to transform and jump between 2 frames. This was an explanation on what I meant as "more information".

Of course this is not a trick can be used "really" to sketch wordlines, this is because is really to measure distances by a ruler and use that reading to add new segments based on that measurement, without loosing in precision. Your way is of course by far more reliable and precise, but for a "rough" analysis would be enough, I think.

Ricky

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English is not my preferred main language, even I use it every day for my job.
So I cannot understand the "mood" of your sentence just above. It seems a bit
you are doing some irony, but I'm not sure ( notice there's not any polemic in my words, I'm just trying
to figure out if you are confirming what I stated or you are trying to disprove it, somehow ).
I guess you are suggesting to use t1 marking to measure x'1 just because the x' inclined axis
has not any mark on it. But, looking better to your last plot, you can count the grid lines parallel
to t' and you get the red wordline intersecting the x' axis pretty after 5 of them. So you can lead
to 5 as it was in your orthogonal version. So, adding some marks on x' axis as well, would fix this ambiguity, I think.
I was trying to get you to agree that if you would not use the t spacings to measure x distances in the normal frame, then you shouldn't use the t' spacings to measure the x' distances in the slanted frame, even if it works. It makes more sense to put in the grid lines and markings in both frames and use the t or t' markings to measure time intervals and the x or x' markings to measure the distance spacings. One applies exclusively to time and the other applies exclusively to distances. I'm not denying that your trick works, just that it's not a good idea to use it.

So the answer is: yes. That logic can be applied both for times and for space axis, of course, since the symmetry of Lorentz transformations are actually "stretching" the unit of measure in the same way( this mainly because the Lorentz transformation IS linear ). Drawing the grids really helps a lot in explaining better what I meant.
And if Minkowski has seen it as well, this is a good point for sure :-).

I was just trying to explain this is not adding any new physics to it. It's straightforward, when you notice a distance in inclined x't' frame is 6, and this is also 6 when the x't' is orthogonal at rest, you are stating nothing really new. You are just noticing a distance is the same when measured in the same frame ( primed in this case ). Of course, if you use euclidean metric to measure that segment when the x' is inclined, you'd get a longer segment. But this, of course, doesn't mean that distance is measured as stretched from xt frame. that distance makes sense only at the same t' time, and cannot be interpreted as a distance in xt, since that would be no more simultaneous in xt frame.
So that's why I'm stating the unit of measure is stretched when we transform it form orthogonal to inclined.

There's no stretching going on. Let me try another argument. Here is a diagram from my previous post:

Now here's another diagram that depicts exactly the same scenario:

Now would you say this second diagram depicts a stretched scenario compared to the first one? I hope not and I hope you understand that it is the grid lines and the markings that establish that the two diagrams are exactly the same. This is why the slanted frame is not stretched from the normal frame.

If you look to this article on Loedel diagrams, for instance:

http://www.einsteins-theory-of-relat...-diagrams.html [Broken]

One of the benefits of the diagram is that unlike in Minkowski diagrams, the scales of both axes of both frames are identical. In effect, the worldlines of both observers progress along their respective time axes at the same rate. Hence, the Loedel diagram does not give apparent preference to one of the inertial frames

so "different scaling" is also another way to me to tell "units are scaled ( so contracted or stretched, according to one you are referring to ).
I looked at the article and there's no mention of any stretching (or contracting) going on.

If you read carefully, they are saying that the Loedel diagram has three frames, the two that are drawn and the reference frame which is "imaginary", meaning it is not drawn. But this is the same thing that you would get if you had two of the Minkowski diagrams with one skewed one way and the other skewed the other way and then you eliminated the axes, grid lines and markings for the normal frame. There's nothing special about the Loedel diagram that other diagrams can't also show.

By the way, do you understand Fig. 2?

Anyhow, it seems ( I hope ) you got my point, so yes using grid my point ismore noticable. So, coming back to my latest plot, I was just saying the DC segment is actually the klingon's ship length since it lasts from the extremes on klingon's wordlines.
So, if you had an hypothetical ruler owning the stretched units of primed inclined frame, you would be able to measure ( on the plot I mean ) the DC length and you got the length expressed as that distance was still in the dual orthogonal x't' frame.
If you don't have that ruler ( of course is difficult to own one of that kind :-) ), you could store how the unit segment in orthogonal x't' primed is stretched inclining it in xt frame. Let's say you are jumping from 1 to 1.45.
So, for example, if you measure by a ruler the DC segment would be 14.5, you could state for sure the klingon's proper length is 10. So you could use that proper length, and use it for choosing the A point: since the Enterprise and the Klingon's ship share the same length when both at rest, you could use that to choose an A point which x coordinate is 10 units far from C.x ( this means of course forcing the AC segment to be 10 units in length ).
I got your point, I just don't think it's a good one. And I don't understand this paragraph. I don't know why you can't just follow the normal processes using the Lorentz Transformation and draw your diagram based on the results you get from it.

So, theoretically, you would be able to sketch and gather the Enterprise length's without the need to transform and jump between 2 frames. This was an explanation on what I meant as "more information".
But you are still jumping between 2 frames if you are using a diagram like the one in your link:

Do you see how they specify two frames in the drawing? They call the normal one the Enterprise Frame and they call the slanted one the Klingon Frame. They actually show the t and x axes for both frames. Unfortunately, they don't show any grid lines or markings for either frame so it isn't quite a obvious that there are two frames and two sets of implied coordinates for each event.

Furthermore, you should not think of the Enterprise Frame as containing only the Enterprise spaceship. It also contains the Klingon spaceship. Same with the Klingon frame--it also contains the Enterprise space ship. All frames contain all objects.

I have redrawn the Minkowski diagram to move the markings for the normal frame to the left and bottom edge and I have added markings for the slanted frame at the top and right:

Note that I have put in three events.

The coordinates for these three events in the normal frame and the primed frame are:
Code:
Blue   t=-3, x=-4   t'=-1, x'=-3
Red    t=4,  x=1    t'=4,  x'=-1
Green  t=-3, x=3    t'=-5, x'=5

Note that I can read these coordinates right off the diagram without having to do any Lorentz Transformation calculations. If you look up Minkowski diagram in wikipedia, you'll see that that is the point of a Minkowski diagram--you don't have to do any calculation to perform the Lorentz Transformation between two frames.

But now that we have computers that can draw the transformed diagrams for us, I don't know why we want to continue using the graphical techniques, especially sense they tend to be rather confusing to people who don't realize what they are actually depicting.

So here is a diagram for the three events in the normal frame:

And here is a diagram for the three events in the slanted (primed) frame:

Do you understand that just because I drew two separated diagrams for these two frames, the one Minkowski diagram also has two frames on the one diagram?

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I'm not denying that your trick works, just that it's not a good idea to use it.
This is not a good point to me. you are expressing a personal opinion, which is
not actually really meaningful to me. If my idea is working ( and I can assure it's working )
that's all no opinion is needed.

There's no stretching going on.

Now would you say this second diagram depicts a stretched scenario compared to the first one?
Of course not. you are comparing two diagrams related to the same rest frame. I'm comparing
the frame at rest with the SAME frame slanted. Here there's a changing in scale for sure.

I looked at the article and there's no mention of any stretching (or contracting) going on.
This is not true. the article tells there's a different scaling between rest frame and slanted ones
on "normal" overlapped Minkowsky diagram. Probably you lost that part, and you forgot to quote it as well, since I posted it in my above post, i.e.:

One of the benefits of the diagram is that unlike in Minkowski diagrams, the scales of both axes of both frames are identical. In effect, the worldlines of both observers progress along their respective time axes at the same rate. Hence, the Loedel diagram does not give apparent preference to one of the inertial frames

This is just taken from that article. I've fully understood Loedel diagrams and they are using an intermediate "fake" inertial frame in order to make the 2 in analysis to be symmetric. And you can see from the above sentence that this remove the different scaling between rest and moving frame. I know my English is poor, but different means to me not equal, which again leads to dfferent units.
this is written in the quoted part above.

I got your point, I just don't think it's a good one. And I don't understand this paragraph.
I think this is the real problem. Since you don't understand it, you don't find a good one
( I was wondering how you can state to having got my point but at the same time stating
you didn't understand it ).
I cannot find any other words to explain you better. Sorry.

I don't know why you can't just follow the normal processes using the Lorentz Transformation and draw your diagram based on the results you get from it.
this is the other problem. It seems you are traying to tell me not to use my method, almost just
because you didn't understood it. And this is clear since you are telling: "using Lorentz transformation". I'm actually using them. Your way is not just the unique way to use.
You started to express personal opinion. I've used your method and I will use it for sure.
So, why are you trying to convince myself mine is wrong?
I think trying to disprove it, as you did it from the very beginning, it's a not a good
way to help understand it you better.

Do you see how they specify two frames in the drawing? They call the normal one the Enterprise Frame and they call the slanted one the Klingon Frame. They actually show the t and x axes for both frames. Unfortunately, they don't show any grid lines or markings for either frame so it isn't quite a obvious that there are two frames and two sets of implied coordinates for each event.

Furthermore, you should not think of the Enterprise Frame as containing only the Enterprise spaceship. It also contains the Klingon spaceship. Same with the Klingon frame--it also contains the Enterprise space ship. All frames contain all objects.

All these things are straightforward to me. When I say "jumping between 2 frames" I really meant "your" method
about building the wordlines in primed rest frame and then "press" your app button to get them in slanted version.
So "jumping" meant this to me.
So, fir the last time I hope, I was just stating you can use the info the slanted klingon's ship width in Enterprise rest frame ( let's say 6 ) could be used to deduce the Enterprise width, as explained in the paragraph it seems you don't understand.
We are using different approaches: you started plotting the 4 wordlines, while as explained as above, I started putting the C event point and to choose the rear klingon's wl to be coincident with the slanted t' ( which is
the choice used by the author as well ).
Doing that, you got directly the contracted klingon's width, so I was looking for a method to deduce the Enterprise's width WITHOUT jumping as you did. This led to my "not good" approach ( at least to you, of course ).

That said, just to let egos in peace, I will use your method for sure ( without any computer, of course). Mine was just a trick to understand better diagrams, but, AS STATED, it cannot be used regularly.

But now that we have computers that can draw the transformed diagrams for us, I don't know why we want to continue using the graphical techniques, especially sense they tend to be rather confusing to people who don't realize what they are actually depicting.
I totally disagree with you. Using computers for doing for us things it's like to use calculators for doing divisions. This is actually making minds sleeping.
I do them manually and I will do my plots as well for sure. So agan, since you are expressing personal
opinions which actually don't concern at all with physics, let's try to avoid them/

that said, sorry, but I have not too much time to convince you I'm correct and to explain you better
my point, and I'm not interested in convince you neither, indeed.

Thanks for your support, Regards

Ricky

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I understand your trick. Let me express it. You want to use the distance of the Proper Time ticks of a moving object on a diagram to measure the distance of the Proper Length for the moving object on the same diagram.

But that trick only works because we usually draw our spacetime diagrams with the spacing of the horizontal and vertical coordinates so that it will work. I had to do a lot of work on my computer program to make that happen. It's not trivial. You should try to write you own program to make it work and you'll see what I mean. But there is nothing wrong with having the horizontal and vertical scaling be arbitrary, as a computer program will normally do without extra work. For example, here's a repeat of one of the diagrams from my first post:

Now here is another version of the same diagram but with the horizontal and vertical scaling changed:

This is a perfectly legitimate diagram conveying exactly the same information as the previous one. But your trick won't work, will it?

If you always use the time coordinates to establish time information and you use the spatial coordinates to establish length information, and you use the Lorentz Transformation equations to draw a new diagram with any coordinates you choose, then you'll never get tricked. This always works, even in situations where you trick would also work.

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I understand your trick. Let me express it. You want to use the distance of the Proper Time ticks of a moving object on a diagram to measure the distance of the Proper Length for the moving object on the same diagram.

Not true. I'm using proper distance to measure the proper length. time is not involved at all.
Try to tell it better: look to my diagram I've already posted:

I'm talking about measuring the DC segment and using it to apply it to the AC one.
The DC segment is actually the proper distance of the klingon's ship evaluated at the
same t' in klingon' slanted frame. We can say it's a proper distance since it's measured
at the same t' and it's the segment between the klingon's front and end wls. So it's
actually the klingon's proper length.
Now, we can try to count the streched dots pierced by this DC distance ( which represent to me
the stretched unit in primed slanted system ). Let's say we count them to be 8 ( just for sake of simplicity ).
Then, since this is the proper distance and it's the same of the Enterprise's proper distance,
we can use it to fix the A point: this will be 8 units horizontally from the left of C, but evaluated using the x unit of measure belonging to the Enterprise's frame ( i.e. the distance between 2 dots
on X axis, which are unstretched ).
And all of this can be done on the same plot, without the need to put temporarily the x't' frame
in orthogonal view ( i.e. putting x't' at rest ).

So, actually the time is involved at all and there's not ambiguity with any different scaling
between t' and x' axes.

DC is actually representing a proper length, and not a proper time at all.

I can see this approach is unconventional, but it works.

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