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Spacetime Diagram

  1. Mar 19, 2004 #1
    I would apprecite it if someone could please check my spacetime diagram for the following scenario. I appreciate any help. I'm tutoring myself based on texts and have no instructor to ask.

    S frame of reference with S' superimposed. S' is moving at velocity [tex]v=3/4 c [/tex] in the +x direction with respect to S. Therefore
    [tex]\Delta y= \Delta y' = 0[/tex], and [tex]\Delta z= \Delta z' = 0[/tex]

    I calculated [tex]\gamma=1.5[/tex] (with a little approximation).
    I calculated angle [tex]\theta=33.75[/tex] by taking 45 degrees times 3/4 (Beta of S').


    Is length l correct? If I instead superimposed S on S' would the distance of [tex]\Delta ct[/tex] along the superimposed ct axis be 0.66 and the length of [tex]\Delta ct'[/tex] be 1?

    Thanks for any help!!
  2. jcsd
  3. Mar 20, 2004 #2
    I believe that you did this incorrectly. Notice that each point on the ct' axis represents the origin of the S' frame at different values of ct. i.e. the ct' axis is the worldline traced out buy the origin of the S' frame. So the Abscissa of each event on this line is x and the ordinate is ct where

    x = vt = (v/c)(ct) = beta*(ct)

    If you rotated the spacetime diagram and plotted x as a function of ct then the ct' axis would have a slope of v/c. If theta is the angle the ct' axis makes with the ct axis then tan (theta) = v/c and therefore

    theta = arctan(v/c) = arctan (3/4) = 36.87 degrees
  4. Mar 22, 2004 #3
    pmb_phy your willingness to help me learn this topic is truly a blessing. I've created a brand new diagram and thrown super-imposing spacetime diagrams to the wind. I've instead taken the more conservative route to help me learn the relationships between S and S'.

    Here's a link to the page...use IE if possible. If you don't get an image you'll have to look at the less attractive second link:

    (don't neglect the Zoom button in the lower right of the first link, it's extremely nice)

    The problem was a bit more difficult than a more basic "event X at the origin" problem. In this problem I placed some time prior to the event A, and therefore had to plot all of the events using [tex]\gamma[/tex] converstions for length and time. If my graphs are correct, the event order changes for S' from A/D simultaneous and B/C simultaneous to D, A, B, C in that order. Is that right!? And if so is this because S' is moving toward planet L and observes planet L's time speeding up?

    If anyone could confirm or deny it's validity I'd appreciate it :)
    Last edited: Mar 22, 2004
  5. Mar 25, 2004 #4
    In continuing my research in SR I found that this is indeed NOT correct.

    You cannot determine a sequence of events just by assuming the lowest (ct) coordinate values happen first. You must accound for "light lag" and trace a 45 degree line from the event to the worldline of the reference frame to find their proper order! It's most succintly put this way: "we cannot make any pair of events change their order simply by changing frames!"
  6. Mar 25, 2004 #5
    That is incorrect. I'm being lazy and not taking a close look at your diagrams (too much to think about and my eyes are straining to see it).

    But I think I know what you're getting at. And you can determine the sequence of events by assuming the lowest values happen first. Whether there is light signals flying around is irrelevant in regards to determining the sequence of events.

    This sounds like a nice idea for a web page on my site. Let me try and whip one up tommorow.
  7. Mar 26, 2004 #6
    Please notice the button in the VERY far lower right corner that says "Fit In Window." This offers zooms to the image of up to 1200%, that should be big enough for ya!

    Your statement is very interesting, pmb_phy. It has me wondering what it means for a sequence of events to occur. For a boring example let's have A throw a ball to B at event C, and B catches the ball at event D. Due to causality the events can NEVER change order in ANY frame of reference, am I right? Otherwise we could manipulate a theoretical traveler's velocity so that event D happens before event C.

    But I'm reading as i'm posting this...

    I've found something that clears up the statement I made in my last post. I think it can be summed up like this (straight from the text):
    timelike separation (positive interval) events must occur in the same order in all frames.
    spacelike separation (negative interval) events can occur in different orders in different frames because they're causally disconnected.
    null separation events must have a null separation in all frames.
  8. Mar 26, 2004 #7
    I did. It was still hard to read. I probably have my screen settings on different settings than you do
    Cool. By the way, you can call me Pete.
    For me it means something like this. At t = -1 event occurred at x = 6, at t = +1 even B occured at x = 12. So event A happened before event B.
    For it to be possible for the sequence of events to be different in different frames of reference the events must have a spacelike spacetime seperation. you gave an example where the events have a timelike spacetime seperation.
    Me too. [:-)]
    Note that for the sequence of events to be frame dependant they cannot occur at the same place.
    Exactly! If you try drawing this on a spacetime diagram then it will become clear. I'lll try to make one later today or this weekend.

  9. Mar 26, 2004 #8
    Sweet, thanks Pete. I'm brad btw. Pleeeeased to meet you!
  10. Mar 26, 2004 #9
    Hi Brad. Pleased to meet you to! :-)

    I'll try to get to this in the weekend but I pulled my back out and am having problems sitting in this chair. In the meantime I suggest that you work with the spacetime diagrams and try to convince yourself of all this by working with some examples that you create. Don't do any math. Just draw pictures. Note that lines of simultaneity are lines which are parallel to the x-axis. So events above such a line come later and below these lines events are "before"

    Working with spacetime diagrams is nice since when you master them you can solve problems of a qualitative nature such as this one in a speedier fashion. You just picture the spacetime diagram and you solve it like that. Mind you its not easy but its worth doing. I don't think I'll ever master these because just when I thought I knew all aspects of them it turns out there was something I didn't know! :-)

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