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

Quick check as to space-like and time-like

  1. Oct 26, 2013 #1
    I keep getting the terms confused:

    1) if I have the equation
    -Δr2+c2Δt2=s2
    does one say that I am using the (-,-,-,+) signature?

    2) Given the above , if s2 > 0 then the interval is space-like, and if s2 < 0 it is time-like, or is it vice-versa?

    3) If I now use the proper time τ , is it the same (as far as signature, space-like and time-like) for -Δr2+Δτ2=R2?

    4) If R is the space-time radius of an arc in the above, what would that arc be?

    Thanks for any help.
     
  2. jcsd
  3. Oct 26, 2013 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi nomadreid! :smile:
    don't worry … so do i ! :rolleyes:

    (000,t1) and (000,t2) have time-like separation: two events at the same position but separated in time

    (000,t) and (x00,t) have space-like separation: two events at the same time but separated in space

    yes
    s2 > 0 means it's mostly a difference in t, so that's time-like :wink:
    not following you :confused:
     
  4. Oct 26, 2013 #3

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    Assuming that by ##\Delta r^2## you mean ##\Delta x^2 + \Delta y^2 + \Delta z^2##, then yes, although it's more customary to list the timelike component first, so that the signature would be (+,-,-,-).

    (Note: coordinate differentials are usually written with a ##d##, like this: ##ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2##. Note that I put the timelike component first.)

    Vice versa. You have one plus and three minuses in the coordinate differentials, so positive ##dt^2## must be a timelike interval; since ##ds^2## has the same sign as ##dt^2##, a positive ##ds^2## must also be a timelike interval.

    What you'e written here isn't a spacetime interval; in fact I'm not sure what it's supposed to mean, physically. If you're writing the spacetime interval in terms of the metric, you use coordinate differentials, as above. If you write the interval along a particle's worldline in terms of its proper time, the interval is just ##\Delta \tau##; that's the definition of proper time. If you're trying to write other intervals using the proper time, you would have to set up coordinates such that the particle's proper time was the time coordinate; but in those coordinates the metric probably wouldn't look as simple.
     
  5. Oct 27, 2013 #4
    Many thanks, tiny-time and PeterDonis. This was a huge help. As far as my last two questions, it turns out that you are both right, that those questions didn't make sense: I had come across the equation using tau in a context of Minkowski space, and automatically stupidly assumed that the author was referring to proper time, which is what tau usually means in this context. However, going over the paper again, I noticed tucked away in a paragraph that the author defined tau as ct. OK, my fault for not noticing this. So questions (3) and (4) were, as you both pointed out, senseless. But your answers to questions (1) and (2) were supremely helpful. Thanks again!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook