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Lorentz-Transformation and time

  1. Mar 9, 2007 #1
    Well, I understand the Lorentz-Transformation and comprehend the principles of special relativity but I'm confused about something very basic.

    Let's assume S is a system (inertia) and S' is relatively moving to S with velocity v. Now I can take the Lorentz-transformation to get from S into S':

    x'=gamma(x-vt) and t'=gamma(t-beta*x/c)

    Now we take an event in S starting at t1 and ending at t2, thus taking the time dt=t2-t1 in S. I want to know how an observer in S' thinks about the time of the evening. Ok, using the Lorentz-transformation we yield:

    dt'=t2-t1=gamma(t2-beta*x/c)-gamma(t1-beta*x/c)=gamma(t2-t1)

    BUT SINCE GAMMA>1 IT FOLLOWS THAT dt'>dt and thus if the event in S takes 5s the event may take (if v is sufficient large) 10s in S'. But that would mean more time in S' has passed than in S. BUT WE KNOW, THAT IS SIMPLY NOT TRUE. All books tell one should use the following transformation:

    dt'=gamma^-1*dt, but WHY (I mean the Lorentz-transformation is defined in another way)???

    Thanks for your answer
     
  2. jcsd
  3. Mar 9, 2007 #2
    Why is not true? Observers moving relative to one another will disagree on the space and time intervals taken separately, but they will agree on one thing:the interval ds2 is invariant.
     
  4. Mar 9, 2007 #3

    Doc Al

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    Careful here! Events have both time and position: you forgot to consider x1 and x2. Here's how I would write it:

    [tex]\Delta t' = \gamma(\Delta t - v\Delta x/c^2)[/tex]

    Clearly the time between those events according to S' depends on where they took place according to S.


    That last formula, the so-called "time dilation" formula, applies to events that happen at the same place in the moving frame (where [itex]\Delta x' = 0[/itex]):

    [tex]\Delta t = \gamma(\Delta t' + v\Delta x'/c^2)[/tex]

    setting [itex]\Delta x' = 0[/itex] gives you:

    [tex]\Delta t = \gamma \Delta t'[/tex]

    An application of this would be observations of a clock at rest in S'. Since it stays in one place in S', according to S clocks it will appear to run slowly.

    Of course, the effect is symmetric: A clock at rest in S will appear to run slow according to S' clocks:

    Starting with:

    [tex]\Delta t' = \gamma(\Delta t - v\Delta x/c^2)[/tex]

    setting [itex]\Delta x = 0[/itex] gives you:

    [tex]\Delta t' = \gamma \Delta t[/tex]
     
    Last edited: Mar 9, 2007
  5. Mar 9, 2007 #4
    time transformation

    I used to tell my students that t and t' represent the readings of two clocks C(x)and C'(x') synchronized in accordance with Einstein's clock synchronization procedure in I and in I respectively located in the simplest case at the same point of the overlapped OX(O'X') axes. dt and dt' represent small changes in the readings of the two clock i.e. very small time intervals. Speaking about them we should make a net distinction between proper time intervals and coordinate time interval .If in I' a proper time interval is measured (dx'=0) whereas in I we measure a coordinate time interval (dx different from zero) then the relationship is dt=g(V)dt'. If in both reference frames coordinate time intervals are measured then they are related by
    dt=g(V)(dt'+Vdx'/cc) whereas if in both reference frames proper time intervals are measured then they are related by the Doppler shift formula.
    As I see Authors use different names for time intervals.
     
  6. Mar 9, 2007 #5
    Mhh, ok, let think me about your answers. I'll get a further response later ... (thanks)
     
  7. Mar 10, 2007 #6
    Again, something is missleading me. Let's take the following systems: S' system of a muon moving with v, S system earth.

    Question How long is the length h (system S) in S':
    ==> take Lorentz-transformation: dx'=gamma*(dx-v*t)
    ==> h'=gamma*h which is wrong.
    Why should it be h'=gamma^-1*h (explanation out of the definition of the Lorentz-transformation)?
     
  8. Mar 10, 2007 #7
    Another question: Let's say I'm an observer in S and there is a system S' moving with speed v relatively to S. Then take an event in S at x=0 that takes 5 ticks. If I transform this with Lorentz-transformation in S' then I get:

    dt'=gamma*dt=gamma*5s > dt (For me in S)

    Does that mean, that if I look on a clock in S' then I see this very 5 ticks take there longer then in my system S? And further. Let's say there's another observer in S'. This S' observer looks in my system S. So the person in S' observes the same effect. He thinks my clock is moving slower:

    dt=gamma*dt' (for the observer in S')
     
  9. Mar 10, 2007 #8
    Modeling student thinking: An example from special relativity

    I just read a paper entitled "Modeling student thinking: An example from relativity" Am.J.Phys.75 272 2007. In the introduction the Auithor mentions:
    "Science education research and physics education research in particular are concerned with the nature of students ideas."
    Not being very familiar with the subject of statistical analysis in the teaching process, and taking into account the typical question in that thread which reflects a possible student ideea (taken by chance) my question is: What should we analyse, the students idea or the way in which he was tought? My oppinion is: the second.
     
  10. Mar 10, 2007 #9

    HallsofIvy

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    So as long as you approve of the teaching method, it doesn't matter whether the student learns or not?

    And what does this have to do with the subject of this thread?
     
  11. Mar 10, 2007 #10
    Well, forget the whole thing about Lorentz-transformation. I was driving car today and thought about the very basic principle of a "transformation" and what it really means and applied my insight to special relativity and my question above.

    At the very moment I understand the concept. I think my problem was in the interpretation of a "transformation" from one system to another one. Forget about my last two posts. I'm sure my previous post includes false interpretations.
     
  12. Mar 10, 2007 #11

    Doc Al

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    length contraction from the LT

    To compare length measurements in two different frames, make sure the measurement of each end is taken at the same time!

    To find out what S' will measure for a given distance, we must make sure that [itex]\Delta t' = 0[/itex] (not [itex]\Delta t = 0[/itex] !). So start with this:

    [tex]\Delta x = \gamma(\Delta x' + v\Delta t')[/tex]

    And let [itex]\Delta t' = 0[/itex] to get:

    [tex]\Delta x = \gamma \Delta x'[/tex]

    Which is the expected length contraction formula.

    Realize that S frame observers will disagree that S' properly measured the distance--they will say that the position measurements were taken at different times since S measures the clocks in S' to be out of synch. (That's where the fun begins!)
     
  13. Mar 10, 2007 #12

    Doc Al

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    Yes. To an observer in S', the number of ticks will be greater. According to S', the moving clock in S runs slow.
    All observers in S' will measure your clock in S as running slowly according to the same formula:

    [tex]\Delta t' = \gamma \Delta t[/tex]
     
  14. Mar 12, 2007 #13
    teaching statistics

    The starting thread is a typical example for the fact that a learner has not correctly understood the subject he was tought. The Am.J,Paper I have quoted makes a statistics of the concepts of students about simultaneity.
    As not familiar with the use of statistics in the evaluation of the quality of teaching, my problem is what should we analyse: the students ideeas or the way in which he was tought. IMHO the second variant is the correct one. Thanks for your answer and of others.
     
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