Showing that Lorentz transformations are the only ones possibleby bob900 Tags: lorentz, showing, transformations 

#1
Nov1212, 01:27 PM

P: 40

In a book ("The special theory of relativity by David Bohm") that I'm reading, it says that if (x,y,z,t) are coordinates in frame A, and (x',y',z',t') are coordinates in frame B moving with v in realtion to A, if we have (for a spherical wavefront)
[itex]c^2t^2  x^2  y^2  z^2 = 0[/itex] and we require that in frame B, [itex]c^2t'^2  x'^2  y'^2  z'^2 = 0[/itex] then it can be shown that the only possible transformations (x,y,z,t) > (x',y',z',t') which leave the above relationship invariant are the Lorentz transformations (aside from rotations and reflections). I'm wondering how exactly can this be shown? 



#2
Nov1212, 02:28 PM

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To show it for a general LorentzHerglotz transformation is really difficult, you should only consider a (lorentzian) boost along Ox, for example, i.e. equal y to 0 and z to 0.
You should consider then x(x',t') and t(x',t') to be linear functions. Place some unknown coefficients and then determine them from physical assumptions. 



#3
Nov1212, 02:42 PM

PF Gold
P: 4,081

Any transformation of the form
[tex]\left[ \begin{matrix} a & \sqrt{a^21}\\ \sqrt{a^21} & a \end{matrix} \right] \left[ \begin{matrix}dt\\ dx \end{matrix} \right] = \left[ \begin{matrix}dt'\\ dx' \end{matrix} \right] [/tex] will preserve dt'^{2} + dx'^{2} = dt^{2} + dx^{2}. More restraints than preserving the interval are needed. 



#4
Nov1212, 03:29 PM

P: 40

Showing that Lorentz transformations are the only ones possible[itex]x' = x  k[/itex] [itex]t' = \sqrt{t^22kx+k^2}[/itex] work (where k is some constant)? Seems like that, and any other similarly arbitrary transformation could work... 



#5
Nov1212, 04:06 PM

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#6
Nov1212, 05:12 PM

PF Gold
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Taking the Taylor expansion of the matrix and dropping terms of order a^{2} or greater gives the generator, I think. Exponentiating this gives a = cosh(something) but no idea what 'something' is. That's probably an illicit fudge, in any case. 



#7
Nov1212, 06:38 PM

P: 40

1. [itex]c^2 t^2  x^2  y^2  z^2 = 0[/itex] 2. [itex] c^2 t'^2  x'^2  y'^2  z'^2 = 0 [/itex] is it "difficult" or actually impossible to show that the Lorentz transformation is the only possibility (aside from rotation [itex]x^2+y^2+z^2=x'^2+y'^2+z'^2[/itex] and t=t', and reflection x=x', t=t', etc.)? 



#8
Nov1212, 08:01 PM

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I will use units such that c=1. I will also use the definition
$$\eta=\begin{pmatrix}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{pmatrix},$$ because I'm more used to this sign convention than the other one. The Minkowski form (pseudoinner product) on ##\mathbb R^4## is defined by ##g(x,x)=x^T\eta x## for all ##x\in\mathbb R^4##. Define ##P=\{x\in\mathbb R^4g(x,x)=0\}##. The OP is asking us to prove the following statement: If ##g(\Lambda(x),\Lambda(x))=g(x,x)## for all ##x\in P##, then ##\Lambda## is a Lorentz transformation.I don't know how to do that. I don't even know if it's possible. But I can prove a similar theorem that starts with a stronger assumption: If ##\Lambda## is linear and ##g(\Lambda x,\Lambda x)=g(x,x)## for all ##x\in\mathbb R^4##, then ##\Lambda## is a Lorentz transformation.Proof: Suppose that ##\Lambda## is linear and that ##g(\Lambda x,\Lambda x)=g(x,x)## for all ##x\in\mathbb R^4##. Let ##y,z\in\mathbb R^4## be arbitrary. We have $$g(\Lambda(yz),\Lambda(yz))=g(yz,yz).$$ If we expand this using the linearity of ##\Lambda## and the bilinearity of g, and use that ##g(\Lambda x,\Lambda x)=g(x,x)## for all ##x\in\mathbb R^4##, we see that ##g(\Lambda y,\Lambda z)=g(y,z)##. Since y,z are arbitrary, this means that we have proved the following statement: For all ##x,y\in\mathbb R^4##, ##g(\Lambda x,\Lambda y)=g(x,y)##. Now let ##x,y\in\mathbb R^4## be arbitrary. We have $$x^T\eta y=g(x,y)=g(\Lambda x,\Lambda y)=x^T\Lambda^T\eta\Lambda y.$$ Let ##\{e_\mu\}_{\mu=0}^3## be the standard basis for ##\mathbb R^4##. I will use the notation ##M_{\mu\nu}## for the component on row ##\mu##, column ##\nu##, of a matrix ##M##. For all ##\mu,\nu\in\{0,1,2,3\}##, we have $$\eta_{\mu\nu}=e_\mu{}^T\eta e_\nu=e_\mu{}^T\Lambda^T\eta\Lambda e_\nu=(\Lambda^T\eta\Lambda)_{\mu\nu}.$$ So ##\Lambda^T\eta\Lambda=\eta##, and this means that ##\Lambda## is a Lorentz transformation. (The definition of "Lorentz transformation" goes like this: A linear ##\Lambda:\mathbb R^4\to\mathbb R^4## is said to be a Lorentz transformation if ##\Lambda^T\eta\Lambda=\eta##). 



#9
Nov1212, 08:22 PM

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Hm, it looks like I can also prove the following variant:
If ##\Lambda## is surjective, and ##g(\Lambda(x),\Lambda(y))=g(x,y)## for all ##x,y\in\mathbb R^4##, then ##\Lambda## is a Lorentz transformation.With these assumptions, I can prove linearity by messing around with the expression $$g(x,ay+bz)=g(\Lambda(x),\Lambda(ay+bz)).$$ 



#10
Nov1212, 08:53 PM

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#11
Nov1212, 09:05 PM

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Alternatively, it is possible to find the conformal transformations by direct solution of the differential equations defining the transformation. The (messy, difficult) details can be found in Appendix A of this older text: V. Fock, N. Kemmer (translator), The theory of space, time and gravitation. 2nd revised edition. Pergamon Press, Oxford, London, New York, Paris (1964). You might be able to access a copy at Library Genesis. ;) Fock also shows that if you assume only the relativity principle (equivalence of inertial observers) then the most general transformations are of linearfractional form  which are not the same as conformal transformations since the latter involve a quadratic denominator in general. But if we add the light principle (which is what you used above), then the "intersection" between linearfractional and conformal transformations is indeed the Lorentz transformations. Regarding what the book said about requiring that the transformations be well behaved everywhere: although the more general transformations fail to be nonsingular in general everywhere, there have been recent attempts to use them to construct foundations that might account for the success of the LambdaCDM model in cosmology  the singular part of the transformation only occurs at the radius of the universe. But this is probably a subject for the BTSM forum. 



#12
Nov1312, 12:30 AM

P: 40

[itex]x' = \frac{xvt}{\sqrt{1v^2}}[/itex], [itex]t' = \frac{tvx}{\sqrt{1v^2}}[/itex], [itex]y'=y[/itex], [itex]z'=z[/itex] ? 



#13
Nov1312, 06:05 AM

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Frederik's calculation shows it's not trivial to get the LT from a few assumptions. 



#14
Nov1312, 08:17 AM

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Let's consider the 1+1, dimensional case. If we write $$\Lambda=\begin{pmatrix}a & b\\ c & d\end{pmatrix},$$ we can easily see that a≠0 and that c/a=v. To see this, first note that $$\Lambda\begin{pmatrix}1\\ 0\end{pmatrix}=\begin{pmatrix}a\\ c\end{pmatrix}.$$ In my notation, the upper component of a 2×1 matrix is the time coordinate. I will refer to it as "the 0 component", and the lower component, i.e. the spatial coordinate, as "the 1 component". I will also number the rows and columns of my 2×2 matrices from 0 to 1. For example, the 01 component of ##\Lambda## is b. If a=0, then the result above tells us that ##\Lambda## takes the time axis of the "old" coordinate system to the spatial axis of the "new" coordinate system. This corresponds to an infinite velocity difference, because the time axis is the (coordinate representation of) the world line of an object with velocity 0, and the spatial axis is the (coordinate representation of) the world line of an object with infinite velocity. This is why we can rule out a=0. This allows us to take a outside the coordinate matrix. $$\Lambda\begin{pmatrix}1\\ 0\end{pmatrix}=\begin{pmatrix}a\\ c\end{pmatrix} =a\begin{pmatrix}1\\ c/a\end{pmatrix}.$$ Now we can interpret c/a as v, because we know that ##\Lambda## maps the time axis to the line $$t\mapsto t\begin{pmatrix}1 \\ v\end{pmatrix},$$ i.e. the line with x'=vt'. Now consider the effect of ##\Lambda## on two coordinate matrices ##\begin{pmatrix}t_1\\ 0\end{pmatrix}## and ##\begin{pmatrix}t_2\\ 0\end{pmatrix}## with ##t_1<t_2##. \begin{align} \Lambda\begin{pmatrix}t_1\\ 0\end{pmatrix} =\begin{pmatrix}at_1\\ ct_1\end{pmatrix},\qquad \Lambda\begin{pmatrix}t_2\\ 0\end{pmatrix} =\begin{pmatrix}at_2\\ ct_2\end{pmatrix}\end{align} The 0 components of the new coordinate pairs are ##at_1## and ##at_2## respectively. If a>0, then ##t_1<t_2## implies that ##at_1<at_2##, but if a<0, then ##t_1<t_2## implies that ##at_1>at_2##. So ##\Lambda## preserves the temporal order of events on the 0 axis when a>0, and reverses them when a<0. To get the specific result you want, we need to assume that a>0. We are now dealing with an orthochronous Lorentz transformation. A similar argument shows that ##\Lambda## preserves the order of events on the spatial axis when d>0 and reverses them when d<0. So we also assume that d>0. We are now dealing with a proper Lorentz transformation. A Lorentz transformation that's both proper and orthochronous is sometimes called a restricted Lorentz transformation. Because of the above, we will write ##\Lambda## as $$\Lambda=\gamma\begin{pmatrix}1 & \alpha\\ v & \beta\end{pmatrix},$$ where ##\gamma,\beta>0##. \begin{align} \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix} &=\eta =\Lambda^T\eta\Lambda =\gamma^2\begin{pmatrix}1 & v\\ \alpha & \beta\end{pmatrix}\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}\begin{pmatrix}1 & \alpha\\ v & \beta\end{pmatrix}\\ &=\gamma^2\begin{pmatrix}1 & v\\ \alpha & \beta\end{pmatrix}\begin{pmatrix}1 & \alpha\\ v & \beta\end{pmatrix} =\gamma^2\begin{pmatrix}1+v^2 & \alphav\beta\\ \alphav\beta & \alpha^2+\beta^2\end{pmatrix} \end{align} The 00 component of this equality tells us that ##1=\gamma^2(1+v^2)##, which implies both that v<1 (because ##\gamma^2>0##) and that $$\gamma=\frac{1}{\sqrt{1v^2}}.$$ The 01 and 10 components both tell us that ##\gamma^2(\alphav\beta)=0##, which implies that ##\alpha=v\beta##. The 11 component tells us that ##\gamma^2(\alpha^2+\beta^2)=1##. So \begin{align}1v^2 &=\frac{1}{\gamma^2} =\alpha^2+\beta^2 =\beta^2v^2+\beta^2=\beta^2(1v^2)\\ \beta &=1\\ \alpha &= v\beta=v. \end{align} So our final result for ##\Lambda## is $$\Lambda=\frac{1}{\sqrt{1v^2}}\begin{pmatrix}1 & v\\ v & 1\end{pmatrix}.$$ If you prefer to write this out as a system of equations, $$\begin{pmatrix}t'\\ x'\end{pmatrix}=\gamma\begin{pmatrix}1 & v\\ v & 1\end{pmatrix}\begin{pmatrix}t\\ x\end{pmatrix}=\gamma\begin{pmatrix}tvx\\ vt+x\end{pmatrix}$$ \begin{align} t' &=\gamma(tvx)\\ x' &=\gamma(xvt). \end{align} 



#15
Nov1312, 10:27 AM

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#16
Nov1312, 10:45 AM

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#17
Nov1312, 10:53 AM

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#18
Nov1312, 11:08 AM

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It is always assumed that the transformation is linear (at least if the origin is mapped to the origin, otherwise affine). But what is the physical reason for this assumption?



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