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Lorentz transformations, time intervals and lengths

  1. Dec 31, 2008 #1
    Prtesent the Lorentz transformations as
    dx=g(dx'+Vdt')
    dt=g(dx-Vdt)
    In my oppinion dx and dx' represent proper lengths measured in I and in I', dt and dt' representing coordinate time intervals. Do you aggree.
    Happy new year to all participamts on the Forum
     
  2. jcsd
  3. Dec 31, 2008 #2
    it should be
    dx=g(dx'+Vdt')
    dt=g(dt'+Vdx'/cc)
     
  4. Dec 31, 2008 #3

    Fredrik

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    I don't agree. I don't see a reason to call one of variables "proper something", and I think that calling one of them "proper something" and the other "coordinate something" is the worst option. You might as well choose the opposite convention: dt is the proper time measured by an observer whose world line coincides with the t axis, and dx is a spatial coordinate in the coordinate system that we'd normally associate with this guy.

    I'd just describe them both as coordinates.
     
  5. Dec 31, 2008 #4

    tiny-tim

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    Happy new year to you too, bernhard :smile:

    I agree with Fredrik :smile:

    there's no reason whatever to distinguish between x and t.

    "proper" means independent of the observer …

    the "proper length" of a rod (if anyone used the phrase) would mean its length as measured in a frame in which it is stationary.
     
  6. Dec 31, 2008 #5
    Thanks for your answer. As I see we do not read the same textbooks. I like the definitions given in :A Travelers Guide to Sapcetime Mc.Graw-Hill,Inc. 1995
    Proper time: The time between two events as measured by a single clock that is present at both events (its value depends on the world-line that the clock follows in getting from one event to the other.
    Coordinate time interval: The time between two events in an inertial reference frame by a pair of synchronized clocks one present at one event, the other present at the other event.
    The definitions of proper length and distorted length are well known.
    So my question is how to qualify dx (dx') and dt (dt') in accordance with the definitions given above. As long as we do not use standard definitons...
     
  7. Dec 31, 2008 #6
    I'd just describe them both as coordinates.
    dt and dt' are time intervals or coordinates?
     
  8. Dec 31, 2008 #7

    tiny-tim

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    t and t' are coordinates,

    dt and dt' are coordinate intervals. :wink:
     
  9. Dec 31, 2008 #8
    Thanks again. But that does not answer my question?
     
  10. Dec 31, 2008 #9

    tiny-tim

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    "dt and dt' are time intervals or coordinates"? …

    ok … they can't be coordinates, they are intervals, so they're time coordinate intervals. :smile:
     
  11. Dec 31, 2008 #10

    JesseM

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    What does it mean to put intervals like dx and dt in the Lorentz transformation rather than coordinates like x and t? Intervals imply you're talking about a pair of events, but normally the Lorentz transformation just deals with the coordinates of a single event E. Your notation only seems to make sense if dx is the distance along the x-axis between the event E you're looking at and an event at the origin at x=0, t=0 (or equivalently x'=0, t'=0). But if that's the case, dx cannot be a "proper length" unless E has a time-coordinate of t=0 (since proper length always refers to the spatial distance between two events in the frame where they are simultaneous).
     
  12. Dec 31, 2008 #11

    Fredrik

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    A global coordinate system in SR is just a smooth function [itex]x:M\rightarrow\mathbb R^4[/itex], where M is Minkowski space. (Yes, M and [itex]\mathbb R^4[/itex] is the same set. Minkowski space is [itex]\mathbb R^4[/itex] with a manifold structure and a metric). A Lorentz transformation is just a transition function [itex]x\circ y^{-1}:\mathbb R^4\rightarrow\mathbb R^4[/itex], such that both x and y are inertial frames. It clearly doesn't matter if you express the 4-tuple that a Lorentz transformation acts on as (t,x,y,z) or (dt,dx,dy,dz).
     
  13. Dec 31, 2008 #12
    Thanks. Would you present all that to freshmans?
     
  14. Dec 31, 2008 #13
    Because the problem presents importance in my activity please read the following lines with the definitions given above in mind.
    Consider a clock K"(0,0) located at the origin O" of the I" inertial reference frame. I" moves with constant speed u relative to a I inertial reference frame and with speed u' relative to I', I' moving with speed V relative to I. At the origin of time in the three mentioned frames their origins are located at the same point in space. After a given time of motion K"(0,0) arrives in front of a clock K(x,0) of I and in front of a clock K'(x',0) of I'. The clocks of I and I' are Einstein synchronized. When the three clocks are located at the same point in space they read t, t' and t" respectively. Detected from I" the trip of the moving clock lasts a proper time interval (t"-0). Detected from I it lasts a coordinate time interval (t-0) whereas detected from I' it lasts the coordinate time interval (t'-0). In accordance with the formula which accounts for the time dilation effect we have
    (t-0)=g(u)(t"-0) (1)
    (t'-0)=g(u')(t"-0) (2)
    Combining (1) and (2) we obtain that the coordinate time intervals are related by
    (t-0)=(t'-0)g(u)/g(u') (3)
    Expressing the right side of (3) as a function of physical quantities measured from I' via the addition law of parallel speeds the result is
    (t-0)=g(V)(t'-0)[1+Vu'/cc] (4)
    During the time interval (t'-0) the origin O" generates in I' a "rod" of proper length
    (x'-0)=u'(t'-0) (5)
    with which (4) becomes
    (t-0)=g(V)[1+V(x'-0)/cc]. (6)
    Do you consider that the way in which the Lorentz-Einstein transformation (6) was derived proofs that (t-0) is a coordinate time interval whereas (x'-0) is a proper length.
    All comments are highly appreciated.
     
  15. Dec 31, 2008 #14

    JesseM

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    Your last sentence seems like a non sequitur to me--can you explain why what you said prior to that sentence implies "it clearly doesn't matter" if you transform between coordinates or intervals? Let's say v=0.6 (so gamma = 1.25), and we have two events A: (x=3 l.s., t=2 s.) and B: (x=5 l.s., t=7 s). In this case dx=2 l.s., dt=5 s. Now we can find the coordinates of A in the x',t' system:

    x' = 1.25*(3 - 0.6*2) = 2.25
    t' = 1.25*(2 - 0.6*3) = 0.25

    and of B:

    x' = 1.25*(5 - 0.6*7) = 1
    t' = 1.25*(7 - 0.6*5) = 5

    If we subtract the first from the second we get dx' = -1.25 and dt' = 4.75. This is the same as what we get if we take the original dx=2 and dt=5 and plug them into the equations:

    dx' = 1.25(dx - 0.6*dt)
    dt' = 1.25(dt - 0.6*dx)

    ...but I still don't understand why it works out this way. Suppose we instead used the following coordinate transformation:

    x' = x^2 + 3t
    t' = 4x + 2t^3

    In this case A would have coordinates (x'=15, t'=28) and B would have coordinates (x'=46, t'=706), so dx'=31 and dt'=678. But this is different than what we get if we plug in dx=2, dt=5 into the equation:

    dx' = dx^2 + 3dt
    dt' = 4dx + 2dt^3

    So why does it work for the Lorentz transformation but not the transformation above? Everything you said about the Lorentz transformation would seem to be true about it as well.

    Thinking about it some more, I think the answer probably has to do with the fact that the x' and t' coordinates of an event in the second frame are each linear functions of the x,t coordinates of the same event in the first frame (a function being linear if f(A + B) = f(A) + f(B)), so if you have two events A: (x1, t1) and B: (x2, t2) and subtract their coordinates to get a new event C: (x3 = [x2 - x1], t3 = [t2 - t1]) then for C, x'(C) = gamma*x3 - gamma*v*t3 = gamma*(x2 - x1) - gamma*v*(t2 - t1) = [gamma*x2 - gamma*v*t2] - [gamma*x1 - gamma*v*t1] = x'(B) - x'(A), and similarly for t' of C.
     
  16. Dec 31, 2008 #15

    Fredrik

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    I'm just saying that it doesn't matter what symbol you use for a variable. f(x) with x=3 is the same as f(y) with y=3.
     
  17. Dec 31, 2008 #16

    JesseM

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    But if dx and dt are understood to mean intervals between a pair of events rather than coordinates of a single event, then you can't assume that you can replace x and t in a coordinate transformation with dx and dt and get correct answers for that coordinate transformation, although as it turns out this does work for the Lorentz transformation because it's a linear transformation (but see the example of a nonlinear coordinate transformation I gave, where it wouldn't work).
     
  18. Dec 31, 2008 #17
    Feel free to ignore this comment as everyone knows my maths suck, but it seems to me Fredrik is describing a guage transformation of local symmetry which happens in this case to be synonymous with a Lorentz transformation, or vise aversa.

    Happy New Year
     
  19. Dec 31, 2008 #18

    JesseM

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    Does "in front of" mean they all meet at a negligible distance from one another, so we can treat the meeting as a single point in spacetime? Also, I'm confused by what the coordinates next to each K mean...I would have thought they were the coordinates in each frame of the point where the three clocks meet, but you said the clock K"(0,0) started at the origin of I'', so if the meeting occurred "after a given time of motion", wouldn't the K'' clock have a different t'' coordinate at the time of the meeting?
     
  20. Dec 31, 2008 #19
    I still think that dx is a "proper length". Consider the relative positions of the reference frames I and I' as detected from I when the standard synchronized clocks of I read t. Let E(x,t) and E'(x',t') be two events located on the overlapped OX(O'X') axes. The distance between the origins O and O' , V(t-0) , and (x-0) are proper lengths. The distance (x'-0) detected from I' is a proper length as well, (t-0) being a coordinate time interval.. Measured from I it is the contracted length
    (x'-0)/g. Adding distances measured from I we have
    (x'-0)/g=(x-0)-V(t-0) (1)
    where from
    (x'-0)=[(x-0)-V(t-0)]g (2)
    in which we recognise the Lorentz transformation relating proper lengths and coordinate time intervals.
    Let
    In what concerns the speed, the distance travelled by a particle and the time interval during which the distance is covered we have
    speed=proper distance covered/coordinate time interval
    resulting that
    proper distance covered=speedxcoordinate time interval.
    As a starter of the thread I consider that the colateral discussions do not elucidate my problem.
    Thanks for your help
     
  21. Jan 1, 2009 #20

    Fredrik

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    These are fine operational definitions of "proper time" and "coordinate time". I don't disagree with them, but I don't think of them as definitions, but rather as postulates about correspondences between things we measure and things in the mathematical model. For example, we define the "proper time" of a curve as the integral of [itex]\sqrt{-g_{\mu\nu}dx^\mu dx^\nu}[/itex] along that curve. The first of the "definitions" you quoted is the postulate that tells us how to measure proper time.

    Yes I would, and you're not a freshman. You have 25 articles about SR at arxiv.org.

    "Proper length" isn't a property of a number like dx. It's not a property of a pair of events. It's a property of a curve. That's why I wouldn't call dx a proper length even if it represents a number that agrees with some proper length measurement. (The definition of dx doesn't mention any curves. You could argue that it does implicitly, and that the relevant curve is the shortest possible curve that lies entirely in the plane of constant t that we're considering, but you could make a similar argument about dt).
     
    Last edited: Jan 1, 2009
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