Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Quick question on Minkowski space diagram

  1. Sep 19, 2014 #1
    Consider the lower right plot in this picture (or many similar ones).

    I interpret the angle of the t' axis with respect to the t axis as: From the point of view of the stationary observer, all progress in time for the moving observer will be accompanied by the latter's spatial progress (to the right in this case), so the 'spatial baseline' for the moving observer is angled with respect to that of the stationary observer, as shown. Right? This seems straightforward.

    I find it harder to interpret the angle of x' to x. I get that it 'works out' as in how the lightning flash at the carriage's front occurs in the past for the moving observer (red dotted lines). (If the angle of x' to x were the negative of the actual one, then the flash would occur in the future, so it would not be a good representation.) But is there an intuitive equivalent to the interpretation for the t' axis? "From the point of view of the stationary observer, all progress in space for the moving observer will be accompanied by the latter's temporal ..." Somehow I get confused here. Something about the latter's temporal slowing down...? I don't quite see how that speaks from the diagram.

    Thanks!
     
  2. jcsd
  3. Sep 19, 2014 #2

    ghwellsjr

    User Avatar
    Science Advisor
    Gold Member

    I agree with you. Those kinds of diagrams are not intuitive and hard to read. One of the main reasons is that there are two sets of coordinate grid lines (which often don't get drawn in and they just have one drawing) but at least in this case, they have two drawings with the events and worldlines located physically on the two drawings in the same locations.

    When I make my spacetime drawings, I use the same orthogonal grid lines and move the events and worldlines around which I think makes the drawings intuitive and much easier to read. If I have time, I'll make some equivalent drawings to the ones you linked to.

    Maybe after I do that, your drawings will make more sense.
     
  4. Sep 19, 2014 #3

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The x' axis being tilted relative to the x axis is the phenomenon known as "relativity of simultaneity". Every observer describes spacetime as consisting of many copies of space labeled by the time coordinate. Lines parallel to the x axis are said to be simultaneity lines for the observer using the unprimed coordinates. Each of those lines is what that observer describes as "space", at some specific time. Lines parallel to the x' axis are said to be simultaneity lines for the observer using the primed coordinates. Each of those lines is what that observer describes as "space", at some specific time.
     
  5. Sep 19, 2014 #4

    Nugatory

    User Avatar

    Staff: Mentor


    All events on lines parallel to the x-axis (including the x-axis itself) happen at the same time but different places in the unprimed frame. All events on lines parallel to the x' axis (including the x' axis itself) happen at the same time but different places in the primed frame.

    All events on lines parallel to the t-axis (including the t-axis itself) happen at the same place but different times in the unprimed frame. All events on lines parallel to the t' axis (including the t' axis itself) happen at the same place but different times in the primed frame.

    The fact that the angle between the axes is not 90 degrees is unimportant. We are in the habit of drawing cartesian coordinates with the axes at right angles for two reasons:
    1) That's how preprinted graph paper comes
    2) It makes it easy to see how the pythagorean theorem (##\Delta{s}^2=\Delta{x}^2+\Delta{y}^2##) gives us the distance between any two points in the cartesian plane.

    However, #1 isn't physics, and #2 doesn't apply to Minkowski space, so there's no particular reason to care whether the axes are perpendicular in our space-time diagrams, nor to attach any particular significance to the angle that the primed axes make with one another. We generally choose 90 degrees for whichever frame we're calling the rest frame, but that's just a convention.
     
  6. Sep 20, 2014 #5
    Many thanks, all. Those are helpful answers.

    Please do, if you can. I had a quick stab at it just now, but not quite getting there.

    But suppose that in the picture mentioned, the t' axis was drawn at a 90° angle to the x' axis. Would the result not suggest that, from the point of view of the unprimed observer, future events for the primed observer happen spatially in the opposite direction in which the primed observer is moving? Which would be wrong. In other words: should the t' and x' axis not always lie in quadrant I and III for a right-moving primed observer (and in II and IV for a left-moving primed observer)? Which implies that the angle between the primed axes needs to be <90° (assuming the axes are chosen orthogonal in the fixed frame), so it is not quite arbitrary.

    Is it arbitrary that the angles between t and t', and between x and x', look identical in magnitude? If it is, would the diagram nonetheless be harder to interpret if they were not identical?
     
  7. Sep 20, 2014 #6

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The slope of the t' axis is 1/v. The slope of the x' axis is v. So the angle between the t' axis and the t axis is the same as the angle between the x' axis and the x axis. The angle between the t' axis and the t axis is always less than 45 degrees.

    The t' axis is the world line of an object that's stationary at position coordinate 0 in the primed coordinates. The x' axis is the set of events that are all assigned time coordinate 0 by the primed coordinate system.
     
    Last edited: Sep 20, 2014
  8. Sep 20, 2014 #7

    Nugatory

    User Avatar

    Staff: Mentor

    Yes, you are right. The primed axes need to be "inside" the unprimed ones, so the angles are not completely arbitrary.

    Somewhat surprisingly, you could (if you are willing to risk further confusing yourself just to see how twisted the on-paper representation can be) draw the picture with the primed axes at 90 degrees and the unprimed axes at some angle greater than 90 degrees, and as long as you remember that on a line parallel to the x axis means "events at the same time" and on a line parallel to the t-axis means "events at the same place" you'll get the right relationships.

    They angles will be the same, for the reason that Fredrik gave above.
     
  9. Sep 20, 2014 #8
    Thanks - again both enlightening answers. I think I understand everything that is being said. And yet: my original brain-block remains. It may have to do with understanding slopes more than relativity. :confused: I'm still trying to put into words what I'm seeing, and my intuition (which may be wrong) wants the formulations for t' and x' to mirror each other.

    Let's say the moving observer moves at 1/3c, i.e. at 1/3 lightyear / year, then:

    slope of t' = 3 = 1/v = 3y / ly
    "From the point of view of the stationary observer, every 1 lightyear of progress in space for the moving observer will be accompanied by the latter's 3 years of progress in time. (However, from the point of view of the moving observer, those 3 years did not bring him anywhere further than 'here'.)"

    No problems there. I move up one unit on the x-axis from the origin, and see that t'(that point) gives t=3.

    slope of x' = 1/3 = v = 1/3 ly / y
    "From the point of view of the stationary observer, every 1 year of progress in time for the moving observer will be accompanied by the latter's 1/3 lightyear of progress in space. [That's really just restating the premise.] (However, from the point of view of the moving observer, that 1/3 lightyear of travel still leaves him in 'now'.)"

    Now... How should I visualize this, in parallel with the case for t'? If I move up one unit on the t-axis from the origin, x'(that point) gives x=3, not 1/3 !

    Alternatively, to characterize x' verbally, one could say something like: "An event happening 'now' at 1 lightyear of spatial distance to the moving observer, is not 'now' but at 1/3 year of temporal distance from the point of view of the stationary observer." Fine. But that somehow gets the numbers or axes mixed up again, because it talks about 1/3 y and 1 ly, whereas the slope of x' = 1/3 ly / 1 y.

    Feels like a neuron in my brain is placed the wrong way around.
     
  10. Sep 20, 2014 #9

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    What I had in mind was that you should use the same definition of "slope" in both cases: ##\Delta t/\Delta x##.

    If you put your pen down on a point on the t' axis, and move it 1 unit to the right, and then move it 1/v=3 units up, you end up back on the t' axis.

    If you put your pen down on a point on the x' axis, and move it 1 unit to the right, and then move it v=1/3 units up, you end up back on the x' axis.
     
  11. Sep 20, 2014 #10
    Right, thanks...

    ... and I notice I made a mistake in my previous post: the slope is just a number so all my statements about slopes with units are wrong.

    Still can't quite pinpoint what about this is throwing me off. Something about the slope expressing ##\Delta t/\Delta x## for both t' and x', while I 'want' one to speak about time (and the effect on space) and the other about space (and the effect on time) - and this want arising from the question: whence these angles?
     
  12. Sep 20, 2014 #11

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Quick question on Minkowski space diagram
  1. Minkowski space (Replies: 28)

Loading...