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In the back of the popular Relativity book

  1. Oct 17, 2011 #1
    I'm looking at Einstein's "Simple Derivation of the Lorentz Transformation"

    http://www.bartleby.com/173/a1.html

    and am puzzled. I see how equation (3) is a sufficient condition for (1) implying (2) and vice versa, but...it also seems more than sufficient, doesn't it? He's assuming a linear relationship between x-ct and x'-ct' without saying why this should be so.
     
  2. jcsd
  3. Oct 18, 2011 #2
    Here is the relevant passage. Coordinate systems k and k' are moving relative to each other along their x and x' axes:

    "...Any such event is represented with respect to the co-ordinate system k by the abscissa x and the time t, and with respect to the system k' by the abscissa x' and the time t' when x and t are given.

    A light-signal, which is proceeding along the positive axis of x, is transmitted according to the equation

    x=ct

    or x-ct = 0 . . . . .(1)

    Since the same light-signal has to be transmitted relative to k' with the velocity c, the propagation relative to the system k' will be represented by the analogous formula

    x'-ct' = 0 . . . . .(2)

    Those space-time points (events) which satisfy (1) must also satisfy (2). Obviously this will be the case when the relation

    (x'-ct') = λ (x-ct) . . . . (3)

    is fulfilled in general, where λ indicates a constant; for, according to (3), the disappearance of (x – ct) involves the disappearance of (x' – ct')."
     
  4. Oct 19, 2011 #3
  5. Oct 19, 2011 #4
    Hmm..Well they seem well defined to me, and I understand why the transformation would be linear, and so look like


    [tex] x' = k_{1} x + k_{2} ct [/tex]
    [tex] ct' = k_{3} x + k_{4} ct [/tex]

    but Einstein in effect goes further than that by saying that

    [itex] k_{1} = k_{4} [/itex] and [itex] k_{2} = k_{3} [/itex].

    This is the step I don't understand. I believe that no other postulates are necessary, but could use help with that gap.

    Edit: Whoa, is something wrong with the LaTeX today or have I just forgotten how to use it already??
     
    Last edited: Oct 19, 2011
  6. Oct 19, 2011 #5

    Fredrik

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    They're not even mathematical statements, so in what sense are they well-defined? If you need me to point at a specific problem, then how about the fact that it's not clear what terms like "inertial frame", "light" and "law of physics" are referring to? For Einstein's postulates to make sense, the term "inertial frame" must be defined in advance. But the mathematical statements that best correspond to Einstein's postulates are part of the definition of that term. So it's all a circular mess. This is of course OK as long as we don't claim to be proving anything. We do these calculations in order to find a theory that makes predictions that are similar to the statements we started with.

    I did however say one thing that was wrong in that thread. in post #3, I said that you have to assume that coordinate transition functions are smooth to get linearity. This is wrong. Smoothness is sufficient to ensure linearity, but not necessary.

    Einstein's postulates suggests that you should assume that points on the line t=x are mapped to points on that same line. Similarly for the line t=-x. The equalities you're asking about follow from those assumptions.

    This is how I do these things:

    (Copied from this thread with minor edits to remove blank lines that were made unnecessary when the forum switched to MathJax. The "numbered list" I'm referring to contains Einstein's postulates in the following form: 1. The laws of physics are the same in all inertial coordinate systems. 2. The speed of light is the same in all inertial coordinate systems.)

    The explicit formula for a Lorentz transformation that I used above follows from a few very naural assumptions about how to make mathematical sense of the ideas in the numbered list in my previous post. Let's start by assuming that a coordinate change between inertial coordinate systems will be a linear function [itex]\Lambda[/itex]. (This is very natural. If these functions don't take straight lines to straight lines, an object that's moving with a constant velocity forever in one inertial coordinate system would be accelerating in another). A linear function can be represented by a matrix, so let's write [tex]\Lambda=\begin{pmatrix}a & b\\ c & d\end{pmatrix}.[/tex] The lines t=x and t=-x represent light rays. #2 on the numbered list in my previous post suggests that we should assume that any point on the line t=x is taken to a point on the same line, and similarly for t=-x. These assumptions imply that there exist numbers [itex]\alpha[/itex] and [itex]\beta[/itex], such that [tex]\Lambda\begin{pmatrix}1\\ 1\end{pmatrix}=\alpha\begin{pmatrix}1\\ 1\end{pmatrix}[/tex] [tex] \Lambda\begin{pmatrix}1\\ -1\end{pmatrix}=\beta\begin{pmatrix}1\\ -1\end{pmatrix}.[/tex] It's easy to show that these equations imply a=d, b=c, so that [tex]\Lambda=\begin{pmatrix}a & b\\ b & a\end{pmatrix}[/tex] Now note that the t' axis, expressed in the coordinates of S, is the line x=vt. This combined with the equivalence of different inertial coordinate systems alluded to in #1 on the numbered list, suggests that we should assume that the t axis expressed in the coordinates of S' is the line x'=-vt'. This means that there's a number [itex]\gamma[/itex] such that [tex] \Lambda\begin{pmatrix}1\\ 0\end{pmatrix}=\gamma\begin{pmatrix}1\\ -v\end{pmatrix} [/tex] This equation implies [itex]b=-av=-\gamma v[/itex]. We have [itex]a=\gamma[/itex], but this might be a different number for a different v, so I'm redefining [itex]\gamma[/itex] to be a function instead of a number, and I'll write [itex]\gamma(v)[/itex] where I would previously have written [itex]a[/itex] or [itex]\gamma[/itex]. [tex]\Lambda=\gamma(v)\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix}[/tex] #1 also suggests that we should assume that we can obtain the inverse just by changing the sign of the velocity. [tex]\Lambda^{-1}=\gamma(-v)\begin{pmatrix}1 & v\\ v & 1\end{pmatrix}[/tex] So [tex]I=\Lambda\Lambda^{-1}=\gamma(v)\gamma(-v)\begin{pmatrix}1-v^2 & 0\\ 0 & 1-v^2\end{pmatrix}[/tex] [tex]\gamma(v)\gamma(-v)(1-v^2)=1[/tex] But #1 also suggests that we should assume that [itex]\gamma[/itex] is an even function. (Without this assumption, statements about time dilation that two inertial observers make about each other won't be symmetrical). And that gives us the final result [tex]\Lambda=\frac{1}{\sqrt{1-v^2}}\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix}[/tex]
     
  7. Oct 19, 2011 #6
    Thanks Fredrik - will look this over later tonight when I'm free. :)
     
  8. Oct 20, 2011 #7
    Ok, makes sense to me. Thanks again.
     
  9. Oct 20, 2011 #8

    DrGreg

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    If you are already prepared to believe that the relationship between (t',x') and (t,x) is linear, you can perform a change of coordinates[tex]u = x - ct[/tex][tex]v = x + ct[/tex]and then the relationship between (u',v') and (u,v) has got to be linear, too, i.e.[tex](x'-ct') = \lambda (x-ct) + \alpha (x+ct)[/tex][tex](x'+ct') = \beta (x-ct) + \mu (x+ct) [/tex]Now consider the special case when x = ct and x' = ct'; this can be possible only if [itex]\alpha = 0[/itex]. Similarly when x = −ct and x' = −ct', this can be possible only if [itex]\beta = 0[/itex].
     
  10. Oct 20, 2011 #9
    Wow, that's a good one Dr. Greg - thanks!
     
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