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How To Understand This Lorentz Transform?

  1. Jan 29, 2015 #1
    Prof Ramamurti Shankar has this Youtube video 'Introduction to Relativity' at

    And in it he derives the Lorentz Transforms something like this, at about 58 minutes into it.

    |------------------------- t ----------------------| time
    |-----ut---------------|---------- x' -------------|
    |--------------------------- x --------------------|


    x' = (x - ut)Ɣ
    x = (x'+ ut')Ɣ

    Ɣ = 1/(√(1 - (u²/c²)))

    x'= (x - ut)/(√(1-u²/c²))

    I thought I understood but I now find I don't. Because when I look at that last equation it seems to me that x' is going to increase in length as speed increases.

    There's a failure in my understanding somewhere. Can someone please point it out to me?
     
  2. jcsd
  3. Jan 29, 2015 #2

    PAllen

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    x' is not a length, it is coordinate defined relative to an origin world line (which is completely different than the origin world line of the x,t coordinates). However, there is something partially valid about your observation. Per (x,t) observer, 'x rulers are smaller, leading to measurements that x thinks are too large.
     
  4. Jan 29, 2015 #3

    Ibix

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    x' is just a point, not a length. If you want to think about lengths, you need to be a little careful.

    Let's imagine two points at opposite ends of a rod of length 2L, at rest parallel to the x-axis and centered at the origin. We measure the positions of the ends of the rods at some time. Choosing t=0 makes the maths simpler, so let's choose that. End 1 of the rod is at x1=L and end 2 is at x2=-L. Now we transform those measurements into a frame moving at speed u in the +x direction using the Lorentz transform ##x'=\gamma(x-ut)##:
    [tex]
    \begin{eqnarray}
    x'_1&=&\gamma(x_1-0)\\
    &=&\gamma L\\
    x'_2&=&\gamma(x_2-0)\\
    &=&-\gamma L
    \end{eqnarray}
    [/tex]
    That makes the length in the new frame [itex]x'_1-x'_2=2γL[/itex] and, as you note, that is greater than 2L. We've apparently got length dilation, not length contraction. What's going on?

    We've been bitten by the relativity of simultaneity. It is very important to remember that two things that happen simultaneously in one frame don't generally happen simultaneously in another. You can see this here by Lorentz transforming the times of the measurements using ##t'=\gamma(t-ux/c^2)##:
    [tex]
    \begin{eqnarray}
    t'_1&=&\gamma(0-ux_1/c^2)\\
    &=&-\gamma uL/c^2\\
    t'_2&=&\gamma(0-ux_2/c^2)\\
    &=&\gamma uL/c^2
    \end{eqnarray}
    [/tex]
    Those measurements of the positions of the ends of the rod weren't made simultaneously in the new frame. This is significant because, in this frame, the rod is moving at speed -u. Imagine trying to measure the length of a beetle. If it sits still on the ruler, you can just note down the ruler measurement at each end and take the difference. It it's walking, however, you need to make certain that you note down the measurements at the same instant or you'll get nonsense because you conflate the beetle's motion with its size. In this case, we measured simultaneously in the rest frame of the rod (when it didn't actually matter), but not simultaneously in the frame where it's moving (when it did matter).

    We need to answer the question: in the moving frame, where is end 2 of the rod at the time ##t'_1##? This is easy enough to answer - we know ##x_2=-L## and we've just chosen ##t'_2=-\gamma uL/c^2##. We can just sub these into the Lorentz transforms to get what we don't know. It turns out to be easiest to use the inverse Lorentz transform for x, ##x=\gamma(x'+ut')##:
    [tex]
    \begin{eqnarray}
    x_2&=&\gamma(x'_2+ut'_2)\\
    -L&=&\gamma(x'_2-\gamma\frac{u^2}{c^2}L)
    \end{eqnarray}
    [/tex]
    which, with a bit of work, yields
    [tex]
    x'_2=\gamma L-2\frac{L}{\gamma}
    [/tex]
    That in turn makes the length (correctly measured with simultaneous measurements) in the moving frame:
    [tex]
    \begin{eqnarray}
    x'_1-x'_2&=&\gamma L-\left(\gamma L-2\frac{L}{\gamma}\right)\\
    &=&2\frac{L}{\gamma}
    \end{eqnarray}
    [/tex]
    ...which is length contraction.

    To summarise, then, you added some of the rod's velocity to its length because you didn't measure the length in one go. You can fix this by making measurements simultaneously in the frame you are interested in.
     
    Last edited: Jan 29, 2015
  5. Jan 29, 2015 #4
    Thanks for those. I'm going to have to think for a while. I'm a bit slow, you know, a bit slow.

    Is it right to say that in the first place I've misunderstood what the transforms are?

    I've thought we're transforming a length. I have thought that that gamma is the factor that defines the length change, the length contraction.

    But it is a transformation of co-ordinates. Is that it? And gamma is defining the change in the x co-ordinate in whichever frame?
     
  6. Jan 30, 2015 #5

    Ibix

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    Yes - it's just a change of coordinates. This particular change mixes together the x and t coordinates along with a scale change. The mixing together means that notions of "here" and "now" both get modified, while the gamma changes scales on your clocks and rulers.

    The idea of time and distance being mixed together is where the notion of time being a dimension arises. The Lorentz transforms amount to a funny kind of rotation (a hyperbolic rotation) in the x-t plane.
     
  7. Jan 30, 2015 #6

    PeroK

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    The Lorentz Transformation is a co-ordinate transformation. In this case, you have two co-ordinate systems whose origins coincide but where one (the primed one) is moving with velocity u in the +ve x direction with respect to the unprimed system.

    The Lorentz Transformation tells you how to convert co-ordinates of a point/event in spacetime from one system to the other.

    Note that the Lorentz Transformation encapsulates Time Dilation, Length Contraction and "leading clocks lag". That's why it's such a powerful tool.
     
  8. Jan 30, 2015 #7
    And so this equation that I started with

    defines a change in the co-ordinate x'. A negative change or diminution of the value bringing it closer to the origin. A positive change or addition taking it further away.

    'Taking it further away' implying a longer length? Or we simply have no lengths in question here? ( I remember that Prof Shankar was talking observers and events, not lengths).

    I was looking at http://en.wikipedia.org/wiki/Lorentz_transformation

    and got beat by 'matrix form'. Seems I'll have to learn what that's about. Would be very useful to know?
     
  9. Jan 30, 2015 #8

    PeroK

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    It's not about lengths. Imagine the non-relativistic equivalent:

    ##x' = x - ut, \ \ t' = t##

    There's no change to lengths here, simply a linear change in coordinates with time. Maybe look first at this (Galilean) transformation and understand it, then try the Lorentz again.

    The Lorentz transformation is more complicated because both t and x affect the x' and t' co-ordinates. It is relatively easy to recover time dilation for a moving clock and leading clocks lagging from Lorentz. But, recovering length contraction from the Lorentz is trickier. (Ibix did this above.)

    With the Lorentz, you have two things to think about:

    1) A co-ordinate transformation from S to S'

    2) What that does to clocks and rulers that are stationary in one frame (in terms of observing them in another).

    You must separate in your mind the concept of "an x coordinate" from "the length of something". An x-coordinate is not the length of anything.
     
  10. Jan 30, 2015 #9

    jtbell

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    The length of an object is the difference between the x-coordinates of the ends of the object, both taken at the same time. If (x1,t1) and (x2,t2) are spacetime coordinates of the two ends of the object, then L = x2 - x1 provided that t2 = t1.
     
  11. Jan 30, 2015 #10
    well of course i realise that an x coordinate is not a length. that's why i said 'taking it further away implying a longer length'.
    what i meant was assume a length along the x axis from origin to a position on that axis called 'x'.
    that's common.
    then x = 12 would mean there was a length of 12 units there.
    common usage. i was just doing that.
    then a positive change in x - from 12 to 14, say, would imply - in fact it would mean - a longer length.

    that's how i've been thinking all along. but realising now, as i said in that post, that there's no length at all meant in Shankar's presentation.
    he's simply talking coordinates of events, an event.

    Ahh.. but stop. I just thought to check back. I refer you to time 52:12 in Shankar's presentation where he says 'but i will not accept your length' and goes on to introduce the gamma factor.

    right there is where i got the length idea.
     
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