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Time dilation

  1. Aug 1, 2009 #1
    A/to special relativity if one is moving faster, he sees that the clocks w.r.t which he is having relative motion are ticking faster.It increases my curiosity to know what would an observer w.r.t whom the person is having relative motion see, if he observes the
    watch of the traveller.Will he see that the traveller's watch is ticking slower or ticking faster?Logically he should observe that the traveller's watch is ticking slower.But it's confusing for me since both are having relative motion w.r.t each other and hence in both the frame of references we can use the same relative speed 'v'(suppose) to calculate the time interval between two events as observed by an observer in the other frame of reference, which tells both should see each other's watch ticking faster.
     
  2. jcsd
  3. Aug 1, 2009 #2

    CompuChip

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    That is correct. This is, as you say, precisely what the concept of relativity tells you.
     
  4. Aug 1, 2009 #3
    Both see the other person's clocks ticking slower, not faster. However, the spacetime interval will be the same when calculated in either frame: [itex]-t^2 + x^2 = -(t')^2 + (x')^2 [/itex].

    This is analogous to Euclidean space. Consider two Cartesian coordinate systems with the same origin but rotated with respect to each other by an angle [itex]\theta[/itex]. The [itex]x[/itex]-coordinates of the [itex]x'[/itex]-axis will be [itex]x'\cos\theta[/itex]. At same time, the [itex]x'[/itex]-coordinates of the [itex]x[/itex]-axis will similarly be [itex]x\cos\theta[/itex]. In other words, the tick marks on each system's [itex]x[/itex]-axis will be squeezed together on the other system's [itex]x[/itex]-axis. That's analogous to time dilation, except that each observer's [itex]t[/itex]-axis is stretched out on the other person's [itex]t[/itex]-axis due to the difference between Euclidean and Minkowski geometries (the minus sign in the spacetime interval).
     
    Last edited: Aug 1, 2009
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