1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Timelike separated events cannot occur at the same time

  1. Jan 25, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove that for a timelike interval, two events can never be considered to occur simultaneously.

    2. Relevant equations

    Δs'2=∆s'2

    (∆s is invariant)

    s2=x2 - (ct)2
    s'2=x'2 - (ct')2


    3. The attempt at a solution

    I first imagined a reference frame K in which two events happened at (x1, t1) and (x2, t2)

    Here, ∆s2= x22 - x12 - c2t22 + c2t12 = s'2=∆x'2 - ∆(ct')2 where ∆(ct')2 = 0 (if I'm trying to prove that this CAN'T happen I thought it would be best to show what would happen if it were indeed zero)

    ...which can be rewritten as:

    ∆s2= x22 - x12 - c2t22 + c2t12 = s'2=∆x'2 = x2'2 - x1'2< 0 (my book says that for events with timelike separation that ∆s2<0)

    I guess I'm not really sure where to head from here. Should I substitute in x'=x-vt...(the lorentz transformations for velocity) and try and solve for some value of v such that (ideally) it might be v>c and thus false? I tried doing that but saw that the algebra seemed pretty long and so I thought that I might not be on the right track...I'm not exactly sure what I'm looking for here, I'm quite new to proof-based math (or physics or really anything). Any guidance would be appreciated.
     
  2. jcsd
  3. Jan 25, 2013 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You are thinking way too complicated. If the separation between two events is timelike then (∆s)^2<0. If the two events can be seen as happening at the same time in some frame then what's the sign of (∆s)^2? Now remember it is supposed to be invariant.
     
  4. Jan 25, 2013 #3
    Ah I see what you're saying--the sign would have to be ∆s^2>0 and the sign flip would be inconsistent with ∆s^2 also being invariant. Would it be feasible for me to show that the sign would have to be positive? Should I just state that it is known? I guess I'm not really sure how to prove that s^2 has to be positive in this case....
     
  5. Jan 25, 2013 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You'd better show it because you've got some misunderstanding about what ∆s^2 means. If (x1,t1) and (x2,t2) are your two events then ∆s^2=(x1-x2)^2-c^2*(t1-t2)^2. Not (x1^2-c^2*t1^2)-(x2^2-c^2*t2^2) or whatever you seem to think it is. Now suppose t1=t2?
     
  6. Jan 26, 2013 #5

    Why is ∆s^2=(x1-x2)^2-c^2*(t1-t2)^2? My book kind of led me to believe that it is possible to find s^2 for one event, which would look like s^2= (x1)^2 - (ct1)^2, and so I kind of assumed from this that ∆s^2 would have to have the form I listed earlier. Did I just straight misunderstand my book? Is s^2 a quantity that relates TWO events within one system (do I need to have both (x1, t1) AND (x2, t2) in order to get a value for s^2??)
     
  7. Jan 26, 2013 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, you are misunderstanding or the book in unclear. The interval ∆s^2 is measured between two events. Just like when you say 'distance' it implies there are two points. If you write ∆s^2= (x1)^2 - (ct1)^2, that's the interval between (x1,t1) and (0,0). The interval is usually written ∆s^2=(∆x)^2-(c∆x)^2, with the ∆'s indicating a difference between the two points. Once you figure this out it will be easy to figure out the sign if t1=t2.
     
    Last edited: Jan 26, 2013
  8. Jan 29, 2013 #7
    Can someone clarify this for me? If we are in a timelike interval, X^2-(ct)^2>0, but why exactly can't events occur at the same time? Say they are in the same time and same place?
     
  9. Jan 29, 2013 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Timelike means x^2-(ct)^2<0. If they occur at the same time then t=0. x^2 can't be less than zero. If they occur at the same time AND place then the interval is 0. It's not timelike. It's null.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook