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Showing that Lorentz transformations are the only ones possible 
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#19
Nov1312, 11:22 AM

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#20
Nov1312, 12:19 PM

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#21
Nov1312, 12:27 PM

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#22
Nov1312, 04:21 PM

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Well, that's certainly not true in one dimension, where the map f(x)=x^3 maps the entire line onto itself without being linear, or affine. But perhaps in higher dimensions...? Is there a theorem of this kind? 


#23
Nov1312, 04:26 PM

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#24
Nov1312, 05:05 PM

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Didn't the paper that Ben mentioned in another thread, http://arxiv.org/abs/physics/0302045, go through all this?
The assumptions that that paper made were (skimming) * replacing v with v must invert the transform * isotropy *homogeneity of space and time with a few tricks along the way: * adding a third frame * noting that x=vt implies x'=0 The result was pretty much that there must be some invariant velocity that was the same for all observers. (THere were some arguments about sign of a constant before this to establish that it was positive). The remaining step is to identify this with the speed of light. 


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Nov1312, 05:25 PM

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#26
Nov1312, 06:33 PM

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Personally, I prefer a different definition of spacetime homogeneity: i.e., that it should look the same wherever and whenever you are. IOW, it must be a space of constant curvature. This includes such things as deSitter spacetime, and admits a larger class of possibilities. But another way that various authors reach the linearity assumption is to start with the most general transformations preserving inertial motion, which are fractionallinear transformations. (These are the most general transformations which map straight lines to straight lines  see note #1.) They then demand that the transformations must be welldefined everywhere, which forces the denominator in the FL transformations to be restricted to a constant, leaving us with affine transformations. In the light of modern cosmology, these arbitrary restrictions are becoming questionable.  Note #1: a simpler version of Fock's proof can be found in Appendix B of this paper: http://arxiv.org/abs/grqc/0703078c/0703078 by Guo et al. An even simpler proof for the case of 1+1D can also be found in Appendix 1 of this paper: http://arxiv.org/abs/physics/9909009 by Stepanov. (Take the main body of this paper with a large grain of salt, but his Appendix 1 seems to be ok, though it still needs the reader to fill in some of the steps  speaking from personal experience. :) 


#27
Nov1312, 06:53 PM

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A linear transformation ##T:U\to V## is said to preserve collinearity if for all collinear x,y,z in U, Tx,Ty,Tz are collinear. It's trivial to prove that linear maps preserve collinearity. Since ##T(ax+by)=aTx+bTy## for all a,b, we have ##T((1a)x+ay)=(1a)Tx+aTy## for all a. I still haven't been able to prove that if T preserves collinearity, T is linear. Suppose that T preserves collinearity. Let x,y be arbitrary vectors and a,b arbitrary numbers. One idea I had was to rewrite ##T(ax+by)=T(ax+(1a)z)##. All I have to do is to define ##z=by/(1a)##. But this is a lot less rewarding than I hoped. All we can say now is that there's a number c such that $$T(ax+by)=cTx+(1c)Tz =cTx+(1c)T\left(\frac{by}{1a}\right).$$ The fact that we can't even carry the numbers a,b over to the righthand side is especially troubling. I don't know, maybe I've misunderstood a definition or something. The book I'm talking about is "Functional analysis: Spectral theory" by Sunder. It can be downloaded legally from the author's web page. Scroll down to the first horizontal line to find the download link. See exercise 1.3.1 (2) on page 9 (in the pdf, it may be on another page in the actual book). Edit: Direct link to the pdf. 


#28
Nov1312, 07:01 PM

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#29
Nov1312, 07:12 PM

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#30
Nov1312, 07:30 PM

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Edit: looking at his exercise, I think he means "##x,y,z## in ##V##", meaning that ##x,y,z## are vectors in ##V##. So the "straight line" also includes the origin. That makes his exercise almost trivial because "being on a straight line" means that the vectors are all simple multiples of each other (i.e., they're on the same ray), and linear transformations preserve this. But this is somewhat tangential to the current issue since in relativity we want something more general which preserves continuous inertial motion. 


#31
Nov1312, 07:44 PM

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I think he just meant that x,y,z are on the same line, not that they're all on the same line through the origin. 


#32
Nov1312, 08:20 PM

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In the second one, I apply the chain rule to ∂f/∂t' and there appears a factor of ∂x/∂t' that I don't see how to deal with, so I don't understand (35). I guess I need to refresh my memory about partial derivatives of multivariable inverses. 


#33
Nov1312, 08:49 PM

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The crucial idea here is that the straight line is being parameterized in terms of an arbitrary real ##\lambda##. Also think of ##x_0^i## as an arbitrary point on the line so that ##\lambda## and ##v^i## generate the whole line. Then they adopt a confusing notation that ##x## is an abbreviation for the 3vector with components ##x^i##. Using a bold font would have been more helpful. But persevering with their notation, ##x = x(\lambda) = x_0 + \lambda v##. Since we want the transformed ##x'^{\,i}## to be a straight line also, in general parameterized by a different ##\lambda'## and ##v'##, we can write $$ x'^{\,i}(x) ~=~ x'^{\,i}(x_0) ~+~ \lambda'(\lambda) \, v'^{\,i} $$ where the first term on the RHS is to be understood as what ##x_0## is mapped into. I.e., think of ##x'^{\,i}## as a mapping. It might have been more transparent if they'd written ##x'^{\,i}_0## and then explained why this can be expressed as ##x'^{\,i}(x_0)##. Confusing? Yes, I know that only too well. I guess it becomes second nature when one is working in this way all the time. Fock also does a lot of this sort of thing. $$ dx' = df = f_x dx + f_t dt = (f_x u + f_t) dt ~;~~~~~~ dt' = dg = g_x dx + g_t dt = (g_x u + g_t) dt ~; $$ and so Stepanov's (35) is obtained by $$ u' ~=~ dx'/dt' ~=~ \frac{f_x u + f_t}{g_x u + g_t} ~. $$ [Edit: I have more detailed writeups of both proofs where I try to fill in some of these gaps, but they're in ordinary latex, not PF latex. If you get stuck, I could maybe post a pdf.] 


#34
Nov1312, 10:10 PM

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It doesn't make a lot of sense to put a prime on the λ, but I guess they're doing it as a reminder that if the old straight line is the map B defined by ##B(\lambda)=x_0+\lambda v##, then the new straight line isn't necessarily ##\Lambda\circ B##. It could be ##\Lambda\circ B\circ f##, where f is a "reparametrization". I really don't like that they write v' for the vector I denoted by u, because it suggests that ##v'=\Lambda v##. I realized something interesting when I looked at the statement of the theorem they're proving. They're saying that if ##\Lambda## takes straight lines to straight lines, there's a 4×4 matrix A, two 4×1 matrices y,z, and a number c, such that $$\Lambda(x)=\frac{Ax+y}{z^Tx+c}.$$ If we just impose the requirement that ##\Lambda(0)=0##, we get y=0. And if z≠0, there's always an x such that the denominator is 0. So if we also require that ##\Lambda## must be defined on all of ##\mathbb R^4##, then the theorem says that ##\Lambda## must be linear. Both of these requirements are very natural if what we're trying to do is to explain e.g. what the principle of relativity suggests about theories of physics that use ##\mathbb R^4## as a model of space and time. 


#35
Nov1312, 11:54 PM

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#36
Nov1412, 07:36 AM

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