By becko


2011-07-23 20:17:33 8 Comments

In almost all proofs I've seen of the Lorentz transformations one starts on the assumption that the required transformations are linear. I'm wondering if there is a way to prove the linearity:

Prove that any spacetime transformation $\left(y^0,y^1,y^2,y^3\right)\leftrightarrow \left(x^0,x^1,x^2,x^3\right)$ that preserves intervals, that is, such that

$$\left(dy^0\right)^2-\left(dy^1\right)^2-\left(dy^2\right)^2-\left(dy^3\right)^2=\left(dx^0\right)^2-\left(dx^1\right)^2-\left(dx^2\right)^2-\left(dx^3\right)^2$$

is linear (assuming that the origins of both coordinates coincide). That is, show that $\frac{\partial y^i}{\partial x^j}=L_j^i$ is constant throughout spacetime (that is, show that $\frac{\partial L_j^i}{\partial x^k}=0$).

Thus far all I've been able to prove is that $g_{ij}L_p^iL_q^j=g_{pq}$ (where $g_{ij}$ is the metric tensor of special relativity) and that $\frac{\partial L_j^i}{\partial x^k}=\frac{\partial L_k^i}{\partial x^j}$. Any further ideas?

8 comments

@Iván Mauricio Burbano 2019-03-14 14:15:36

The proof actually turns out to be a very simple exercise in linear algebra. I find this algebraic proof very satisfactory since it uses very little machinery. It also proofs that rotations and (with a slight rewording) unitary maps are linear.

Theorem: Let $U$ and $V$ be vector spaces over a field $F$ equipped with bilinear form $g$ and $h$ respectively. Further assume that $h$ is non-degenerate and we have a surjective map $f:U\rightarrow V$ such that $h(f(u),f(v))=g(u,v)$ for all $u,v\in V$. Then $f$ is linear.

Proof: Let $u,v,w\in V$ and $k\in F$. Then

$$h(f(ku+v)-kf(u)-f(v)v,f(w))=h(f(ku+v),f(w))-kh(f(u),f(w))-h(f(v),w)=g(ku+v,w)-kg(u,w)-g(v,w)=0.$$

Since $w$ was arbitrary and $f$ is surjective, non-degeneracy of $h$ guarantees that $f(ku+v)=kf(u)+f(v)$. Therefore, $f$ is linear.

By taking $U=V=M$ to be the vector space underlying Minkowski space and $g=h=\eta$ as its metric, we obtain that metric preserving transformations are linear. Lorentz transformations (distance preserving transformations) are metric preserving transformations due to the polarization formula in @Brian Moths answer. I think that one has to however include surjectiveness in the definition of a Lorentz transformation. Compare to the Mazur-Ulam theorem.

@Brian Moths 2016-12-21 02:43:00

First, notice that if $\Lambda$ is an isometry then it preserves dot products, since if $p'=\Lambda(p)$, $q'=\Lambda(q)$, and $r'=\Lambda(r)$ then $$\left( \langle r'-p' ,r'-p' \rangle -\langle r'-q' ,r'-q' \rangle - \langle q'-p' ,q'-p' \rangle \right)/2 = \langle r'-q' ,q'-p' \rangle, $$ and since the LHS is preserved so must the right hand side be.

Let's start with minkowski space and pick an origin $p$, and an orthnormal basis $\mathbf{e}_\mu$ satisfying $\langle \mathbf{e}_\mu , \mathbf{e}_\nu \rangle = \eta _{\mu \nu}$. Now any point $x$ in minkowski space can be written $p + x^\mu \mathbf{e}_\mu$, where $x_\mu = \langle x-p,\mathbf{e}_\mu\rangle$.

Now what about $x'=\Lambda(x)$? Well of course we want to say it has the same coordinates. So let's define the new basis. $\mathbf{e}'_\mu = \Lambda(p+\mathbf{e}_\mu)-\Lambda(p)$. Since $\Lambda$ perserves products of differences, we know that the $\mathbf{e}'_\mu$ are orthonormal and so $x'$ can be written $p' + x'^\mu \mathbf{e}'_\mu$, where $x'_\mu = \langle x'-p',\mathbf{e}'_\mu\rangle$.

But since $\Lambda$ preserves products, we have that $x'^\mu=x^\mu$. Therefore, since $$\Lambda(p + x^\mu \mathbf{e}_\mu)=\Lambda(p) +x^\mu\left(\Lambda(p + \mathbf{e}_\mu)-\Lambda(p)\right),$$ $\Lambda$ is affine.

@pppqqq 2016-12-21 08:34:35

Hi @NowIGetToLearnWhatAHeadIs, you are implicitly assuming that the infinitesimal condition $\text d s ^2 =\text d s' ^2$ (i.e. the condition on the tangent space) extends to a finite condition $s^2 = s'^2$ on the whole domain of the chart $x^\mu$. This seems to be the hard part of the proof, do you know any argument to show that this must be so?

@pppqqq 2016-12-21 08:41:36

(Actually this should be standard, it means that an isometry beetween (semi)-riemannian manifolds is also an isometry beetween (semi)-riemannian vector spaces if one of the two is such. However it doesn't look totally trivial to me)

@Brian Moths 2016-12-21 17:08:41

@pppqqq I see. The proof doesn't seem trivial to me either. To make things concrete, I am just considering two copies of minkowski space with a local isometry between them, and I am trying to prove it is linear. With euclidean space this is easy because straight lines minimize distance, so lines must get mapped to lines, but you can't use this fact I don't think. Either way, there is probably an easy answer that somebody knows. Should I open a new question?

@becko 2011-07-27 20:19:01

I had the feeling that a direct proof would be possible using only the relation $\eta _{ij}\frac{\partial y^i}{\partial x^p}\frac{\partial y^j}{\partial x^q}=\eta _{pq}$, assuming simple smoothness properties of the transformation and then using some algebra maneuvers. I found the following lovely argument in the book Gravitation and Cosmology by Steven Weinberg.

We start from the relation

$$\eta _{ij}\frac{\partial y^i}{\partial x^p}\frac{\partial y^j}{\partial x^q}=\eta _{pq}$$

Differentiating with respect to $x^k$ we obtain

$$\eta _{ij}\frac{\partial ^2y^i}{\partial x^p\partial x^k}\frac{\partial y^j}{\partial x^q}+\eta _{ij}\frac{\partial y^i}{\partial x^p}\frac{\partial ^2y^j}{\partial x^q\partial x^k}=0$$

We add to this the same equation with $p$ and $k$ interchanged, and subtract the same with $q$ and $k$ interchanged; that is,

$$\eta _{ij}\left(\frac{\partial ^2y^i}{\partial x^p\partial x^k}\frac{\partial y^j}{\partial x^q}+\frac{\partial y^i}{\partial x^p}\frac{\partial ^2y^j}{\partial x^q\partial x^k}+\frac{\partial ^2y^i}{\partial x^k\partial x^p}\frac{\partial y^j}{\partial x^q}+\frac{\partial y^i}{\partial x^k}\frac{\partial ^2y^j}{\partial x^q\partial x^p}-\frac{\partial ^2y^i}{\partial x^p\partial x^q}\frac{\partial y^j}{\partial x^k}-\frac{\partial y^i}{\partial x^p}\frac{\partial ^2y^j}{\partial x^k\partial x^q}\right)=0$$

This simplifies to

$$2\eta _{ij}\frac{\partial ^2y^i}{\partial x^p\partial x^k}\frac{\partial y^j}{\partial x^q}=0$$

Since the tensors $\frac{\partial y^i}{\partial x^j}$ and $\eta _{ij}$ are invertible, this implies that

$$\frac{\partial ^2y^i}{\partial x^p\partial x^k}=0$$

@Qmechanic 2011-07-27 21:13:28

In hindsight, here is a short proof.

The metric $g_{\mu\nu}$ is the flat constant metric $\eta_{\mu\nu}$ in both coordinate systems. Therefore, the corresponding (uniquely defined) Levi-Civita Christoffel symbols

$$ \Gamma^{\lambda}_{\mu\nu}~=~0$$

are zero in both coordinate systems. It is well-known that the Christoffel symbol does not transform as a tensor under a local coordinate transformation $x^{\mu} \to y^{\rho}=y^{\rho}(x)$, but rather with an inhomogeneous term, which is built from the second derivative of the coordinate transformation,

$$\frac{\partial y^{\tau}}{\partial x^{\lambda}} \Gamma^{(x)\lambda}_{\mu\nu} ~=~\frac{\partial y^{\rho}}{\partial x^{\mu}}\, \frac{\partial y^{\sigma}}{\partial x^{\nu}}\, \Gamma^{(y)\tau}_{\rho\sigma}+ \frac{\partial^2 y^{\tau}}{\partial x^{\mu} \partial x^{\nu}}. $$

Hence all the second derivatives are zero,

$$ \frac{\partial^2 y^{\tau}}{\partial x^{\mu} \partial x^{\nu}}~=~0, $$

i.e. the transformation $x^{\mu} \to y^{\rho}=y^{\rho}(x)$ is affine.

@Iván Mauricio Burbano 2019-02-13 03:10:56

This makes me feel uneasy. Can you conclude from $\frac{\partial^2y^\tau}{\partial x^\mu\partial x^\nu}=0$ that the transformation is affine? In general manifolds (such as in general relativity) the correct structure to even discuss affine transformation is lacking. Does $y^\mu-x^\mu$ have any physical meaning if $y^\mu$ are the coordinates of a point $p$ and $x^\mu$ the coordinates of a point $q$?

@Qmechanic 2011-07-26 16:44:31

Let us reformulate OP's question as follows:

Give a proof that a local coordinate transformation $x^{\mu} \to y^{\rho}=y^{\rho}(x)$ between two local coordinate systems (on a 3+1 dimensional Lorentzian manifold) must be affine if the metric $g_{\mu\nu}$ in both coordinate systems happen to be on constant flat Minkowski form $\eta_{\mu\nu}$.

Here we will present a proof that works both with Minkowski and Euclidean signature; in fact for any signature and for any finite non-zero number of dimensions, as long as the metric $g_{\mu\nu}$ is invertible.

1) Let us first recall the transformation property of the inverse metric $g^{\mu\nu}$, which is a contravariant $(2,0)$ symmetric tensor,

$$ \frac{\partial y^{\rho}}{\partial x^{\mu}} g^{\mu\nu}_{(x)}\frac{\partial y^{\sigma}}{\partial x^{\nu}}~=~g^{\rho\sigma}_{(y)}, $$

where $x^{\mu} \to y^{\rho}=y^{\rho}(x)$ is a local coordinate transformation. Recall that the metric $g_{\mu\nu}=\eta_{\mu\nu}$ is the flat constant metric in both coordinate systems. So we can write

$$ \frac{\partial y^{\rho}}{\partial x^{\mu}} \eta^{\mu\nu}\frac{\partial y^{\sigma}}{\partial x^{\nu}}~=~\eta^{\rho\sigma}. \qquad (1) $$

2) Let us assume that the local coordinate transformation is real analytic

$$y^{\rho} ~=~ a^{(0)\rho} + a^{(1)\rho}_{\mu} x^{\mu} + \frac{1}{2} a^{(2)\rho}_{\mu\nu}x^{\mu}x^{\nu} + \frac{1}{3!} a^{(3)\rho}_{\mu\nu\lambda}x^{\mu} x^{\nu} x^{\lambda} + \ldots. $$

By possibly performing an appropriate translation we will from now on assume without loss of generality that the constant shift $ a^{(0)\rho} =0 $ is zero.

3) To the zeroth order in $x$, the equation $(1)$ reads

$$ a^{(1)\rho}_{\mu} \eta^{\mu\nu}a^{(1)\sigma}_{\nu}~=~\eta^{\rho\sigma}, $$

which not surprisingly says that the matrix $a^{(1)\rho}_{\mu}$ is a Lorentz (or an orthogonal) matrix, respectively. By possibly performing an appropriate "rotation", we will from now on assume without loss of generality that the constant matrix

$$ a^{(1)\rho}_{\mu}~=~\delta^{\rho}_{\mu} $$

is the unit matrix.

4) In the following, it will be convenient to lower the index of the $y^{\sigma}$ coordinate as

$$y_{\rho}~:=~\eta_{\rho\sigma}y^{\sigma}.$$

Then the local coordinate transformation becomes

$$y_{\rho} ~=~ \eta_{\rho\mu} x^{\mu} + \frac{1}{2} a^{(2)}_{\rho,\mu\nu}x^{\mu}x^{\nu} + \frac{1}{3!} a^{(3)}_{\rho,\mu\nu\lambda}x^{\mu} x^{\nu} x^{\lambda}+ \ldots$$ $$+\frac{1}{n!} a^{(n)}_{\rho,\mu_1\ldots\mu_n}x^{\mu_1} \cdots x^{\mu_n}+ \ldots. $$

5) To the first order in $x$, the equation $(1)$ reads

$$ a^{(2)}_{\rho,\sigma\mu}+a^{(2)}_{\sigma,\rho\mu}~=~0.$$

That is, $a^{(2)}_{\rho,\mu\nu}$ is symmetric in $\mu\leftrightarrow \nu$, but antisymmetric in $\rho\leftrightarrow \mu$. It is not hard to see (by applying the symmetry and the antisymmetry property in alternating order three times each), that the second order coefficients $a^{(2)}_{\rho,\mu\nu}=0$ must vanish.

6) To the second order in $x$, the equation $(1)$ reads

$$ a^{(3)}_{\rho,\sigma\mu\nu}+a^{(3)}_{\sigma,\rho\mu\nu}~=~0.$$

That is, $a^{(3)}_{\rho,\mu\nu\lambda}$ is symmetric in $\mu\leftrightarrow \nu\leftrightarrow \lambda $, but antisymmetric in $\rho\leftrightarrow \mu$. For fixed $\lambda$, we can again reach the conclusion $a^{(3)}_{\rho,\mu\nu\lambda}=0$.

7) Similarly, we conclude inductively that the higher order coefficients $a^{(n)}_{\rho,\mu_1\ldots\mu_n}=0$ must vanish as well. So $y^{\mu}= x^{\mu}$. Q.E.D.

@Marek 2011-07-26 17:32:38

Very nice. I also concluded that my approach needs induction on Taylor order (I was trying to figure out a direct way but wasn't able to find it). This makes we wonder whether the direct attack is at all possible. After all, perhaps we need to exploit the antisymmetry and consequently the transformation being polynomial depends crucially on the fact that we preserve a quadratic form and not a cubic (say), or even something nonhomogenous. Indeed, it's probably easy to construct polynomial invariants preserved by non-polynomial transformations.

@becko 2011-07-27 17:13:43

@Qmechanic: I'm not clear on the 3rd step. Specifically, the part that says "By possibly performing an appropriate "rotation", we will from now on assume without loss of generality that the constant matrix ..." Can you elaborate? Thanks.

@Qmechanic 2011-07-27 17:40:17

@becko: Right, I'm being a bit brief. I'm trying to say that by introducing a third coordinate system $y'$, which is related by a standard linear Lorentz transformation (=a sort of "rotation") to the $y$ coordinate system, we may assume that the combined coordinate transformation $x\to y'$ is of the form claimed in the third step. The idea is that we can prove that the coordinate transformation $x\to y$ is linear, if we somehow can prove that that $x\to y'$ is linear. In the rest of the proof we remove the prime from the notation for convenience. Does that make sense?

@becko 2011-07-27 18:16:10

@Qmechanic: How does one prove that a tensor $a_{\mu }^{\rho }$ that satisfies $\eta ^{\mu \nu }a_{\mu }^{\rho }a_{\nu }^{\sigma }=\eta ^{\rho \sigma }$ can be put in the form $\delta _{\mu }^{\rho }$ through Lorentz transformations? If this is supposed to be elementary, at least point me to some website or something that explains it, since I'm self-studying this stuff. Thanks.

@Qmechanic 2011-07-27 19:23:39

@becko: The eq. you just wrote is the abstract/descriptive definition of a Lorentz matrix $a^{\rho}_{\mu}$. Does that help?

@becko 2011-07-27 20:32:10

@Qmechanic: Yes, it does help. The question I posed in my comment above now becomes: How do you prove that a Lorentz matrix can always be brought into the form $\delta _{\mu }^{\rho }$ using Lorentz transformations? This is equivalent to asking whether the Lorentz matrices are invertible, which they are, by definition. I hope I'm making more sense now.

@David Bar Moshe 2011-07-24 14:34:27

The first condition implies that the Jacobian matrix $L^i_j=\frac{\partial y^i}{\partial x^j}$ is a Lorentz transformation. By substitution of the definition of the Jacobian in this condition, we obtain:

$g^{ij}\frac{\partial y^k}{\partial x^i}\frac{\partial y^l}{\partial x^j} = g^{kl}$

In particular, taking the diagonal equations equating $l=k$, we have

$g^{ij}\frac{\partial y^k}{\partial x^i}\frac{\partial y^k}{\partial x^j} = g^{kk}= \pm 1 $

(The plus sign for the time coordinate and the minus sign for the space coordinates).

But this is just the Hamilton-Jacobi equation for a free relativistic particle, whose unique solution can be obtained by separation of variables:

$y^k = \sum_i f^{(k)}_i(x^i)$

By Substitution, we obtain:

$\frac {df^{(k)}_i(x^i)}{dx^i} = const$

Thus, the new coordinates are linear functions of the old coordinates. The constant coefficients are not independent, since the Jacobian matrix must be a Lorentz transformation.

Update:

Upon lurscher's suggestion, here are two references containing the Hamilton-jacobi equation of a relativistic particle. (Both references refer to a particle in an external electromagnetic field. In order to obtain the Hamilton-Jacobi equation for the free particle one needs the particualr case with a vanishing vector potential): reference-1 (by A. granik), reference-2

(The needed version appears in equation (33) of the first reference, the second reference contains the (proper) time dependent version).

In addition, I'll give here an other derivation based on the WKB approximation of the Klein- Gordon equation:

$\frac {1}{c^2}\frac {\partial^2\psi}{\partial t^2}-\nabla^2 \psi + \frac{m^2 C^2}{\hbar^2}\psi = 0$

The plane wave solutions are given by:

$\psi = C \exp(i\frac{\mathbf{p}.\mathbf{x}-\sqrt{m^2c^4+p^2c^2}t}{\hbar})$

To perform a WKB approximation, we seek a solution of the form:

$\psi = A(x,t)\exp(\frac{iS(x,t)}{\hbar})$

and take the leading terms in the limit $\hbar \rightarrow 0$. ($S$ is sometimes called the Hamilton-Jacobi phase function)

By subtitution, we obtain:

$((\frac {1}{c^2}\frac {\partial^2A}{\partial t^2}-\nabla^2A)+\frac{2i}{\hbar}(\frac {1}{c^2}\frac {\partial A}{\partial t} \frac {\partial S}{\partial t} -\mathbf{\nabla}A.\mathbf{\nabla}S) -\frac{A}{\hbar^2}(\frac {1}{c^2}\frac {\partial^2S}{\partial t^2}-\nabla^2S - m^2 c^2 )) = 0$

The leading term is the hamilton-Jacobi equation:

$\frac {1}{c^2}\frac {\partial^2S}{\partial t^2}-\nabla^2S - m^2 c^2 = 0$

Which can be seen to be equivalent to each equation on the main diagonal of the matrix equation written in the original answer.

Now, it is also easy to see the uniqueness of the solution. For the free particle, one can see that the non-leading terms actually vanish. i.e., the WKB approximation is exact.

The Hamilton-Jacobi phase function $S$ is just the phase of the plane wave solutions of the Klein-Gordon equation:

$ S = \mathbf{p}.\mathbf{x}-\sqrt{m^2c^4+p^2c^2}t$

On $\mathbb{R}^4$, all solutions of the free Klein-Gordon equation in Cartesian coordinates are of the form of the plane waves, which implies that the Hamilton-Jacobi phase function is linear in the Cartesian coordinates.

@lurscher 2011-07-24 15:51:03

can you elaborate on how you go from the equation you obtain after doing a trace, to saying that it is equivalent to a Hamilton-Jacobi equation?

@David Bar Moshe 2011-07-25 09:22:57

I have added an update containing the required elaboration and references. I also corrected an error: Each diagonal term of the matrix equation is equivalent to a Hamilton-Jacobi equation (no need to take the trace).

@Qmechanic 2011-07-25 20:01:09

Dear David Bar Moshe, as I'm sure you are aware, the solution isn't unique, as you seem to write (v2). We are trying to prove that it must be affine. You also seem to claim that solution necessarily must obey (additive) separation of variables. Could you elaborate why, preferably backed by a reference?

@David Bar Moshe 2011-07-26 03:35:49

@Qmechanic Of course the solution is not unique because you need boundary conditions to make it unique, but the solution form (Linear in the Cartesian coordinates) is unique. The solution of the four Hamilton-Jacobi equations on the diagonal will be with different constant coefficients which have to further satisfy the of-diagonal relations to be matrix coefficients of a Lorentz transformation. I gave a reasoning that the Hamilton-Jacobi phase function is the phase of a plane wave solution of the Klein-Gordon equation.

@David Bar Moshe 2011-07-26 03:37:22

The form uniqueness of the solution is dictated by the linearity of the solutions of the equation of motion of a free particle. Nevertheless, I'll try to find a more rigorous proof of the uniqueness of the separation of variables solution.

@Marek 2011-07-23 21:59:34

Let's first assume that the scalar product that is preserved has positive signature to show the main idea. Also, you say you don't want to assume homogeneity but this is already implicit in your equation since to form intervals differences of space-time points are used so we might as well take one of those points to be zero of a vector space (equivalently, you might be talking about the preservation of a scalar product on a tangent space to a point but this is also linear, not affine).

Let $$f : \mathbb R^2 \to \mathbb R^2, \quad (x,y) \mapsto (A(x,y), B(x,y))$$ be length-preserving and suppose $f$ is analytic with $$A(x,y) = \sum_{n,m=0}^{\infty} {a_{n,m} \over n! m!} x^n y^m, \quad B(x,y) = \sum_{n,m=0}^{\infty} {b_{n,m} \over n! m!} x^n y^m.$$

Then we have $x^2 + y^2 = A(x,y)^2 + B(x,y)^2$ for all $x,y \in \mathbb R^2$ or explicitly first $$ x^2 + y^2 = \left(\sum_{n,m=0}^{\infty} {a_{n,m} \over n! m!} x^n y^m \right)^2 + \left( \sum_{n,m=0}^{\infty} {b_{n,m} \over n! m!} x^n y^m \right)^2.$$ This immediately shows that the only non-vanishing coefficients occur when $n+m \leq 1$. We just need to investigate $n=m=0$ case but this is trivial since $f(0,0) = (a_{0,0}, b_{0,0})$.

For the $n$D case the discussion is completely analogous. For arbitrary signature some care needs to be taken since we can't use $x^2 + y^2 = 0 \rightarrow x = y =0$ anymore (perhaps one can work in $\mathbb C$ instead of $\mathbb R$ and use analytic continuation).

The last remaining ingredient of this argument is the analyticity of $f$. But this is trivial since $||f(x,y)||^2 = x^2 + y^2$ and $||\cdot||^2$ are analytic around any $(x,y) \in \mathbb R^2$.

@Ben Crowell 2011-07-24 15:55:28

The OP says s/he doesn't want to invoke the homogeneity or isotropy of space. By assuming that the space is Euclidean, you're assuming homogeneity and isotropy.

@Marek 2011-07-24 16:01:01

@Ben: I should clarify that I wasn't assuming that the space is Euclidean. All I assumed was that the scalar product with positive signature is preserved, or more generally that any scalar product is preserved (which is all OP asked for). But I'll edit the answer to make this explicit.

@Marek 2011-07-24 16:14:37

And why a downvote? This answer is certainly correct perhaps except for some technical details but if their are some minor bugs please point them out.

@Ben Crowell 2011-07-24 18:00:34

OK, I've undone the downvote. As discussed in the comments on the question, it seems to me that the OP was asking an ill-defined or self-contradictory question, but I could be wrong.

@Ben Crowell 2011-07-24 18:04:05

In your new text, you have "Also, you say you don't want to assume homogeneity but this is already implicit in your equation[...]" That's my point. The question seems ill-formed to me.

@Marek 2011-07-24 18:13:02

@Ben: after rereading the question, I agree with you. I just proved that invariance of scalar product implies linearity but I wonder whether OP wanted this or something little different.

@Qmechanic 2011-07-25 20:04:41

Dear Marek, is it possible to provide a bit more details in the Lorentzian case.

@Marek 2011-07-25 20:23:57

@Qmechanic: I'll try to get to it later. I realized the Euclidean case needs more detail as well, by the way (and then both cases can be treated simultaneously).

@Qmechanic 2011-07-23 23:51:57

Here I just want to mention that there exists a direct proof in $1+1$ dimensions using elementary arguments. Let the two coordinate patches $U_x$ and $U_y$ (which are, say, both convex sets in $\mathbb{R}^2$, containing the origin) have light-cone coordinates $x^{\pm}$ and $y^{\pm}$, respectively. The metric reads

$$ dy^{+}dy^{-} ~=~ dx^{+}dx^{-}. $$

This leads to three PDE's

$$ \frac{\partial y^{+}}{\partial x^{+}} \frac{\partial y^{-}}{\partial x^{+}} ~=~0 \qquad \qquad\Leftrightarrow\qquad \qquad\frac{\partial y^{+}}{\partial x^{+}}~=~0 \qquad\mathrm{or}\qquad \frac{\partial y^{-}}{\partial x^{+}} ~=~0 ,$$ $$ \frac{\partial y^{+}}{\partial x^{-}} \frac{\partial y^{-}}{\partial x^{-}} ~=~0\qquad \qquad\Leftrightarrow \qquad\qquad\frac{\partial y^{+}}{\partial x^{-}}~=~0 \qquad\mathrm{or}\qquad \frac{\partial y^{-}}{\partial x^{-}} ~=~0,$$ $$ \frac{\partial y^{+}}{\partial x^{+}} \frac{\partial y^{-}}{\partial x^{-}} +\frac{\partial y^{+}}{\partial x^{-}} \frac{\partial y^{-}}{\partial x^{+}} ~=~1.$$

Since $\det \frac{\partial y}{\partial x}\neq 0$, there are really only two possibilities. Either

$$\frac{\partial y^{-}}{\partial x^{+}}~=~0 ~=~\frac{\partial y^{+}}{\partial x^{-}},$$

or

$$\frac{\partial y^{+}}{\partial x^{+}}~=~0 ~=~\frac{\partial y^{-}}{\partial x^{-}}.$$

By possibly relabeling $x^{+} \leftrightarrow x^{-}$, we may assume the former. So

$$y^{+}~=~f^{+}(x^{+})\qquad \mathrm{and} \qquad y^{-}~=~f^{-}(x^{-}).$$

From the third PDE, we conclude that

$$ \frac{\partial f^{+}}{\partial x^{+}}\frac{\partial f^{-}}{\partial x^{-}} ~=~1. $$

By separation of variables, this is only possible if $\frac{\partial f^{\pm}}{\partial x^{\pm}}$ is independent of $x^{\pm}$. It follows that $y^{\pm}~=~f^{\pm}(x^{\pm})$ are affine functions. Q.E.D.

@Marek 2011-07-24 05:59:11

+1 for a general approach that can be e.g. used to derive the class of (infinitesimal) conformal transformations.

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