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Derivation Of Lorentz Transformation by Lawden

  1. Aug 23, 2007 #1
    I have been reading DF Lawden's nice book on tensor calculus and relativity. I keep going back to first chapter (p 9-11) as I did not 100% get his derivation of the "special" Lorentz transformation. He writes about a rotational transformation in one plane and then writes about a translational transformation, and then out pops the Lorentz transforms. The math in each step is easy but I am missing the logic of his argument and have never seen the Lorentz transformation derived in this fashion. How does the rotation lead to equations of translation? What exactly is he doing here?

    Thanks, Howard
     
  2. jcsd
  3. Aug 23, 2007 #2

    robphy

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    Could you scan the relevant pages and post them here as an attachment?
    (It's been a while since I've read Lawden's book. I used it as a text for an independent studies course.)
     
  4. Aug 23, 2007 #3
    Lawden and Lorentz Transform

    Great idea re the scans. I appreciate your time!

    Howard
     

    Attached Files:

  5. Aug 24, 2007 #4
    Hacky,

    It might help to compare it with this one.

    His first postulate amounts to "motion is relative", a'la Galileo.
    He uses the equation of motion of a particle (eqn 4.1) for the same purpose as Pal uses a (metaphorical) rigid rod: The transformation function preserves the uniformity. In Pal, a span "here" and a span "there" that are the same length (e.g. move a rod) will also be seen as the same (possibly different) length in S'. In Lawden, the distance traveled by a particle in a given duration is the same at different times. The choice of duration doesn't matter, so you don't have to worry about what it was in S' -- he choses his own. But in S', he also sees the particle move the same distance in any period of the same duration.

    He then says that a sphere of light maps from one coordinate system to the other. This is the "light cone" at one specific slice. That is pretty much the defining characteristic of Minkowski space-time (the "metric"), so you have enough information to derive the function to translate from S to S'.

    As for the rotation, the x4 axis is the t axis, but using imaginary numbers so as to flip the sign in the "metric" calculation and be able to use a plain summation in (eq 4.9). If you didn't do that, but defined the metric to be the square root of x squared minus t squared, in a normal Cartesian plane, then you can see that "acceleration" (change in velocity) is a hyperbolic rotation about the origin. No imaginary angles needed.

    I admit, I can't follow what he said between (5.5) and (5.6).

    For section 5 in general, he finished off 4 by saying it is, in general, a linear transform R4->R4. He begins section 5 by saying "suppose it is a rotation". That is indeed an example of a linear transform, but I don't see the point. His x4 axis is leaning the opposite way it is drawn in x-t diagrams. I think he ends up by saying it was an inspired guess; the result fits the previous constraints. Wouldn't that be more of a proof than a derivation?

    I think that part is saying that, given the proposed transform, and that an event has a different location in S and in S', back out the parameter for how far the rotation was.

    —John
     
  6. Aug 24, 2007 #5
    I agree that his explanations are somewhat confusing. What he says is this: consider three coordinate planes in the reference frame [itex]\overline{S}[/itex]. These are planes given by equations [itex] \overline{x}=0 [/itex], [itex]\overline{y}=0[/itex], and [itex]\overline{z}=0[/itex]. He then finds equations of the same three planes seen by the observer [itex]S[/itex]:

    [tex] x = -ict \tan \alpha [/tex]..............(1)
    [tex] y = 0 [/tex]
    [tex] z = 0 [/tex]

    Since [itex]\overline{S}[/itex] moves with respect to [itex] S[/itex] with velocity [itex] u [/itex], eq. (1) must also have the form [itex] x = ut [/itex]. From this he finds the relationship between the parameter [itex] \alpha[/itex] and the velocity [itex] u [/itex]:

    [tex] \tan \alpha = iu/c[/tex]

    (I am afraid he changes the sign of [itex] u [/itex] without mentioning it). From this, it is easy to find [tex] \sin \alpha [/tex] and [tex] \cos \alpha [/tex] in eq. (5.7) and Lorentz transformations (5.8).

    I think it can be rigorously proven that in the 4-dimensional space there exist linear transformations of coordinates which leave [itex]x_2 [/itex] and [itex]x_3 [/itex] untouched. And if these transformations preserve the combination [itex] x_1^2 + x_4^2 [/itex], then they must depend on a single parameter [itex] \alpha [/itex] and have the form (5.1). The proof is not very difficult.


    I think his drawing is fine. Note that [itex]x_i [/itex] and [itex] \overline{x}_i [/itex] are coordinates of the same point in two frame. To verify the consistency between (5.1) and the drawing, you can choose for example point [itex] x_1=1, x_4=1 [/itex] and see that after rotation of axes the coordinates of these points change in accordance with (5.1).

    Eugene.
     
  7. Aug 24, 2007 #6
    The thing about his argument I cannot follow is how he turned the rotation between S and [tex]\overline{S}[/tex] into a translation of -ict tan[tex]\alpha[/tex]. I feel like I get everything in the section except this and thus it seems like the Lorentz transformation magically appears out of nowhere,. He writes all about a rotation between the two frames (Fig. 5.1) but ends with a translation (Fig. 5.2)? Maybe this has something to do with rotation in complex plane? He made some difficult things clear as I read ahead but this I do not follow.
     
    Last edited: Aug 24, 2007
  8. Aug 24, 2007 #7

    jcsd

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    okay first of all do you understand his argument why the special Lorentz transformation should be viewed as a rotation in spacetime? We want to preserve the origin (we don't have to of course it's just simpler to consider only those transformations that presreve the origin), we want to preserve the use of 'Cartesian' cooridnates and preserve the scale too.

    The transformation that is being performed is purely the spacetime equiavlent of a rotation, no translation at all (the same event is the origin in both the unbarred and barred frames). However as a consequence of this 'rotation' the spatial axes of barred frame have translational motion through the spatial slices of the unbarred frame and vice versa.
     
  9. Aug 24, 2007 #8
    His explanations are tricky, because it is difficult for us to imagine a 4D spacetime. The best we can do is to visualize some lower-dimensional sections of it. Perhaps it would be more clear to you if you notice that Fig. 1 is a section of the 4D spacetime by the plane [itex] x_2 = x_3 = 0 [/itex] with explicitly shown axes [itex] x_1 [/itex] and [itex] x_4 [/itex], while Fig. 2 is a section of the same spacetime by the hyperplane [itex] x_4 = t = const [/itex] with explicitly shown axes [itex] x_1, x_2 [/itex] and [itex] x_3 [/itex]. You can also think that Fig. 2 is a section of Fig. 1 by a line (actually a 3D hyperplane) which goes parallel to the [itex] x_1 [/itex] axis and has [itex] x_4 =t [/itex]. What shows up as a "rotation" in Fig. 1 (in position-time coordinates) is represented as a shift of coordinate axes in Fig. 2 (in 3D space).

    Eugene.
     
  10. Aug 26, 2007 #9
    I think I am getting the gist of it, can anyone point to another reference that uses the same line of reasoning to obtain the Lorentz transform? Thanks.

    As an aside this really is a nice inexpensive book. It has not been revised since 1982 or so and thus the nomenclature might be a bit off but it is really good for the autodidact. You can also email Dover and they will promptly send a set of detailed solutions to the problems.

    Howard
     
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