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How to get time dilation & length contraction from Lorentz transform?

  1. Jun 10, 2012 #1
    OK, I've found a great explanation of the derivation of the Lorentz transformation, with

    x' = γ [ x - v t ]

    t' = γ [ t - ( v / c2 ) x ]

    so if I take the other term as 0, there is

    x'( t = 0 ) = γ x

    t'( x = 0 ) = γ t

    but the problem is that the time dilation & length contraction use reciprocals of each other, but these transforms use the same

    0 → proper

    v → relativistic

    Δtv = γ Δt0

    Lv = ( 1 / γ ) L0

    What am I interpreting wrong here?
     
  2. jcsd
  3. Jun 11, 2012 #2

    PeterDonis

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    2016 Award

    Staff: Mentor

    Try writing out both transform equations for each of the cases above:

    Case 1: (x, 0) transforms to (x', t') as follows:

    x' = γ x

    t' = - γ v x

    where v is the relative velocity.

    Case 2: (0, t) transforms to (x', t') as follows:

    x' = - γ v t

    t' = γ t

    You will see why this is useful in a moment.

    The common formulas for "length contraction" and "time dilation" actually play a trick on you behind the scenes: they change the pairs of events that they refer to behind your back, so to speak. The key is that the transformation equations above transform *intervals*, i.e., *pairs* of events, not single events; but which particular pair of events describes an object or its motion *changes* when we change frames.

    To see how this works, consider the two cases above. Case 1 might describe, for example, a ruler at rest in the unprimed frame, with proper length x. The ruler is moving in the primed frame, so we expect it to be "length contracted", but as you note, when we transform the interval describing its proper length in the unprimed frame (from the origin 0, 0 to x, 0) to the primed frame, the difference in x' coordinates comes out *larger* by the gamma factor, instead of smaller. What gives?

    What gives is that we are looking at the wrong interval; the ruler's "length" in the primed frame is described by a *different* interval, one which is purely along the x' axis instead of the unprimed x axis. In other words, the correct interval in the primed frame is the one between the origin of the primed frame and whatever event on the worldline of the ruler's other end is simultaneous with the origin in that frame. That is, we need to find the event on the worldline of the ruler's other end that has t' = 0. The easiest way to find that event is to go back to the unprimed frame and write the equation for the x' axis in that frame: it is simply t = vx. The worldline of the ruler's other end in the unprimed frame is just x = x (the other end is at rest at the same x coordinate for all time), so the point we are looking for is the intersection of that worldline with the x' axis, i.e., (x, t) = (x, vx). So we just have to transform the interval (0, 0) - (x, vx) to the primed frame:

    x' = γ (x - vt) = γ (x - v^2 x) = γ x (1 - v^2) = x / γ

    t' = γ (t - vx) = γ (vx - vx) = 0

    Voila! Now we have an interval that is purely along the x' axis, since t' = 0, and as you can see, this interval has a length that is *shorter* than x by the factor gamma, exactly as desired.

    A similar procedure can be done for case 2 to show how the correct "time dilation" can be determined by finding the right interval along the t' axis, instead of the interval transformed above, which is along the unprimed t axis.
     
  4. Jun 11, 2012 #3
    A very good "clock" for time dilation measurements is the muon lifetime, which is about 2.2 microseconds at rest. When the muon is stored in a storage ring (magnetic ring, similar to a synchrotron accelerator) at a gamma of 29.3, the measured lifetime in the lab was about 64 microseconds.
    The distance the muon in the ring travels during one lifetime at v=βc is βγcτ.
     
  5. Jun 11, 2012 #4
    OK, I've got it licked - it's a matter of the problem semantics.

    There are the time & length intervals - Δt & Δx - such that the intervals are scaled

    Δt' = γ Δt

    Δx' = γ Δx

    For time dilation, the proper time is that observed in the clock in the observer's IRF, with the dilated time being that observed in the clock that is in the IRF that is moving relative to the observer's IRF; hence the time dilation is the Lorentz factor.

    Δt' → { time interval observed in the clock that is in the IRF that is moving relative to the observer's IRF }

    Δt → { time interval observed in the clock that is in the observer's IRF }

    time dilation = Δt' / Δt = γ

    For the length contraction, the proper length would be that observed by an observer in the object's IRF – i.e., observed by an observer in the IRF that is moving relative to the IRF of the observer observing the length contraction - and the contracted length being that observed by an observer in the IRF that is moving relative to the object's IRF - i.e., observed by an observer in the IRF of the observer observing the length contraction - hence the length contraction is the reciprocal of the Lorentz factor.

    Δx → { length observed by an observer in the IRF that is moving relative to the object's IRF – i.e., observed by an observer in the IRF of the observer observing the length contraction }

    Δx' → { length observed by an observer in the IRF that is moving relative to the object's IRF – i.e., observed by an observer in the IRF of the observer observing the length contraction }

    length contraction = Δx / Δx' = 1 / γ

    Basically, the time observation is observer based, while the length observation is length based, hence the swapping!

    Talk about tricky!
     
    Last edited: Jun 11, 2012
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