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Proper length

  1. Jan 5, 2013 #1
    Proper length is given by $$ L = \sqrt{\Delta x^2 + \Delta y^2 + \Delta z^2 - c^2\Delta t^2 }$$
    So, when $$ \Delta x = \Delta y = \Delta z = 0 $$ there is no motion and $$ L = ic\Delta t $$ What does that mean, if anything?
     
    Last edited: Jan 5, 2013
  2. jcsd
  3. Jan 5, 2013 #2
    It means that the curve in spacetime connecting event A to event B is timelike, so it makes more sense to talk about proper time than proper length. Proper time is the time a clock carried along said curve would read, and is given by:

    [tex]\tau =\sqrt{c^2\Delta t^2- \Delta x^2 - \Delta y^2 - \Delta z^2}[/tex]
     
  4. Jan 5, 2013 #3

    pervect

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    It means that that you're computing the "length" of a time interval. Which is of course, [itex]\Delta t[/itex]
    The usual name for this (minus the multiplicative factors) is "proper time" rather than proper length.

    Are you worried about the factor of i? The factor of c?
     
  5. Jan 5, 2013 #4

    tiny-tim

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    proper length is the distance between two events, as measured by an observer who regards them as being at the same time

    if you regard the two events as being at the same place, then only an observer going infinitely fast (as measured by you) could regard them as being at the same time

    since such an observer can't exist, it would be very surprising if the proper length was real! :smile:

    (alternatively, if you insist on allowing observers who move faster than light, then you have to allow that their times and their distances can be imaginary)
     
  6. Jan 5, 2013 #5
    To get an answer to your question that you would be satisfied with, you would probably need to convey to us in what sense you are searching for a "meaning." One way to begin a discussion that could be carried out from a number of different points of view would be to first establish some kind of context.

    One context would be to just ask what your final equation is saying at face value. That's easy, you just put the equation into words: An incremental distance, dL, is traversed by moving at the speed of light over a time increment of dt. Beyond that you may be expecting a comment about the meaning of the incremental distance, dL, in this example.

    If you are searching for some physical meaning about dL, you might need to look for it in the context of a particular universe model. For example you might assume a 4-dimensional spacetime model of physical reality. Then, at face value, the equation could be interpreted as giving the incremental distance, dL, that an observer at rest in his own "rest frame" moves during the time increment, dt. In this case the dL is interpreted as distance along the rest frame 4th dimension, i.e., the observer moves along his own 4th dimension at the speed of light. This picture of course raises more questions--for example, what is the meaning of "observer", pictured here as moving along the 4th dimension? And what exactly is the 4th dimension? However, this forum is not intended for discussions of these types of questions, so you should not expect to pursue these kinds of ideas here.

    Others could provide other models providing a context for interpreting the "meaning" of dL. Logical positivists would caution you to avoid assigning physical meaning beyond the observation of measurement results and performing the indicated mathematical calculations.
     
    Last edited: Jan 5, 2013
  7. Jan 6, 2013 #6

    stevendaryl

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    In SR, suppose you have two events (point in space and time) [itex]e_1[/itex] and [itex]e_2[/itex], with a separation vector in some inertial frame [itex](\Delta x, \Delta y, \Delta z, \Delta t)[/itex]. For definiteness, let's assume [itex]\Delta t \geq 0[/itex]. Then there are three possible types of relationship between the events
    1. Timelike separation. This is the case in which [itex]\Delta t > \dfrac{1}{c}\sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}[/itex]. In this case, there is some inertial frame in which the two events take place at the same spatial location, but at different times. Then
      [itex]\tau = \sqrt{(\Delta t)^2 - \dfrac{1}{c^2}((\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2)}[/itex] is the time between the events in this frame.
    2. Spacelike separation. This is the case in which [itex]\Delta t < \sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}[/itex]. In this case, there is some inertial frame in which the two events take place at the same time, but at different locations. Then
      [itex]L = \sqrt{-c^2 (\Delta t)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}[/itex] is the distance between the events in this frame.
    3. Lightlike separation. This is the case in which [itex]\Delta t = \sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}[/itex]. In this case, all frames agree that the events do not take place at the same time or the same location, but that it is possible to send a light signal from [itex]e_1[/itex] to [itex]e_2[/itex].
     
  8. Jan 7, 2013 #7

    tiny-tim

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    on second thoughts, i don't think one should talk about the proper length (or proper distance) between events, but only about the proper length of a rigid body

    there's no such thing as the proper time between two events … the proper time depends on the path taken

    (wikipedia (http://en.wikipedia.org/wiki/Proper_length) defines the proper length of any path, and then is forced to define the proper length between two events as the proper length of the straight path between them)

    so by analogy the proper length or distance between two events should also depend on the path taken

    however, i dont see any point in defining the proper length of a path …

    although the proper time of a path has physical significance … it's the time shown on a clock that follows that path … i can't see any physical significance for the proper length of a path (unless that path is a same-time-for-some-inertial-observer path between two points of a rigid body) …​

    why would anyone want to know it? :confused:
     
  9. Jan 8, 2013 #8
    OK, with Δx = Δy = Δz = 0, then τ = cΔt, which is a length. What does that mean?
     
  10. Jan 8, 2013 #9
    This is great, thanks. Ignore the previous post.
     
  11. Jan 9, 2013 #10
    Whoops! I'm so used to working in units where c=1 that I forgot the extra factor of c:

    [tex]c \tau =\sqrt{c^2\Delta t^2- \Delta x^2 - \Delta y^2 - \Delta z^2}[/tex]
     
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