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How to plot correctly spacetime diagrams

  1. Aug 31, 2013 #1
    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:


    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 wanna 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 lenght 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.


    Last edited: Sep 1, 2013
  2. jcsd
  3. Sep 1, 2013 #2


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    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.

    It's pure coincidence having no significance whatsoever.

    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.

    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.

    Attached Files:

  4. Sep 1, 2013 #3
    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:


    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.

    Thanks for your help and your answer.

  5. Sep 1, 2013 #4


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    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".
  6. Sep 2, 2013 #5
    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.


  7. Sep 2, 2013 #6


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    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:


    Does this address your questions?

    Attached Files:

  8. Sep 2, 2013 #7
    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.



    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.
    Last edited: Sep 2, 2013
  9. Sep 2, 2013 #8


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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.
  10. Sep 2, 2013 #9
    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.
  11. Sep 2, 2013 #10


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    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.
  12. Sep 2, 2013 #11
    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.


  13. Sep 2, 2013 #12
    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.

    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.
  14. Sep 3, 2013 #13


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    If you can't see what's wrong, then I'll point it out later when you get to it.

    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.

    Now you've got the effect before the cause!! Doesn't this bother you?

    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.

    Attached Files:

  15. Sep 3, 2013 #14
    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.


    Last edited: Sep 3, 2013
  16. Sep 3, 2013 #15
    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?


  17. Sep 3, 2013 #16


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    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.
  18. Sep 3, 2013 #17
    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.



  19. Sep 3, 2013 #18
    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.


    Last edited by a moderator: May 6, 2017
  20. Sep 4, 2013 #19
    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 :smile: ), 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.



    EDIT: if you click on images you can see them zoomed. I don't know why
    they are so little in the post.
    Last edited: Sep 4, 2013
  21. Sep 5, 2013 #20


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    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?
  22. Sep 5, 2013 #21
    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.

    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

  23. Sep 5, 2013 #22
    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


    Last edited: Sep 5, 2013
  24. Sep 7, 2013 #23
    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.
  25. Sep 8, 2013 #24


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    You're just saying that because you don't remember this post or your response.

    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?
  26. Sep 9, 2013 #25
    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.



    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.
    Last edited: Sep 9, 2013
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