1
$\begingroup$

I've read this condition in a script: The Lorentz transformation must be linear because any uniform motion must be a uniform motion in any inertial frame.
All other proofs that I've read so far need some kind of homogeneity of space-time: Suppose we have two different inertial frames of reference $S$ and $S'$ and want to transform their coordinates into each other, so that we have a coordinate transformation law like: $$ {\mathrm{d}x^\mu}'= \frac{{\mathrm{d}x^\mu}'}{\mathrm{dx^\mu}}\vert_{{x^\mu}'=\tilde x}\, \mathrm{dx^\mu} $$ which is supposed to be invariant under any modification of the space-time coordinate like $\tilde x \rightarrow \tilde x'$, i.e. the Lorentz transformation looks the same at each event. Now is there a way to prove that this statement is correct or false?
When I'm not mistaken, the condition of transforming uniform linear motions into one another dictates something like, assuming motion along the x- and x'-axis, respectively, $x=vt$ and $x'=v't'$ to be true. Let us drop covariant notation as I deem it a bit overloaded for the translation along 1 axis. Also we would get $z=z'$, $y=y'$ and finally something like $x'=f(x,t)$ and $t'=g(x,t)$. What happens now? I apparently miss a condition that this enforces, since I seem to have no way to determine a general form of $f$ and $g$ at this point, which could show whether the transformation is linear or not. Could I substitute the linear motion into the transformation function, i.e. say that we must have $f=f(x-vt)$ and the same for $g$? Seems not to illegit since that is how Lorentz transformation looks like.
I'm curious about if/how this can be resolved.

$\endgroup$
  • $\begingroup$ failtrolol: "[...] uniform motion must be a uniform motion in any reference frame." -- Perhaps "any reference frame" is here meant to be understood as "any inertial reference frame". (Otherwise, "uniform motion in a non-inertial frame" seems a paradoxial notion. Though it can be said in any case that "mutual rest of participants being described" is a intrinsic, proper notion.) If so, yes: the (pairwise) motions of members of distinct inertial systems is mutually uniform, and moreover with mutually equal speed. But all that's independent of coordinate assignments, or transformations. $\endgroup$ – user12262 Nov 12 '15 at 21:06
  • 1
    $\begingroup$ Of course I meant inertial frames, I'm sorry. Clarified that in the update. Still, I'm trying to prove or disprove formally, that this condition - the transformation of motions between different inertial reference frames is a sufficient proof for linearity of Lorentz transformation or not - I think that question becomes quite clear. $\endgroup$ – failtrolol Nov 12 '15 at 21:45
3
$\begingroup$

Yes, the statement is correct. One proof I am familiar with was given by V.Fock (of Fock states, etc) in his book "Theory of Space, Time, and Gravitation". See Chap.1, Sec.8 therein (pg.20, bottom). In modern notation the idea is as follows:

In inertial frame $S$ parametrize straight lines as ($x_0 = ct$) $$ x_i = \xi_i + \beta_i s, \;\;\; i=0,1,2,3 $$ where the $\xi_i$, $\beta_i$, and $s$ are independent real parameters. Let coordinate transformations from $S$ to another inertial frame $S'$ be given by ($x'_0=ct'$) $$ x'_i = f_i(x) \equiv f_i(x_0, x_1, x_2, x_3),\;\;i=0,1,2,3 $$ Then the $f_i$-s must be such that for $x_i$ as above the transformed $x'_i$-s read $$ x'_i = \xi'_i + \beta'_i s', \;\;\; i=0,1,2,3 $$ for some real $\xi'_i$, $\beta'_i$, and $s'$.

A first step is to note that the $f_i$-s must be functions of $s$ and that from $ dx'_i/dx'_0 = \frac{dx'_i}{ds'}/\frac{dx'_0}{ds'}=$ $= \beta'_i/\beta'_0 = const$ we must also have
$$ \frac{dx'_i}{dx'_0} = \frac{\frac{df_i}{ds}}{\frac{df_0}{ds}} = \frac{\beta_j\partial^j f_i}{\beta_j\partial^j f_0} = \frac{\beta'_i}{\beta'_0} = const. $$ Taking the derivative over $s$ again, $$ \frac{d}{ds}\frac{\frac{df_i}{ds}}{\frac{df_0}{ds}}= 0 \;\; \Rightarrow \frac{\frac{d^2f_i}{ds^2}}{\frac{df_i}{ds}} = \frac{\frac{d^2f_0}{ds^2}}{\frac{df_0}{ds}} $$ gives after rearranging $$ \frac{\beta_j\beta_k\partial^j\partial^k f_i}{\beta_j\partial^j f_i} = \frac{\beta_j\beta_k\partial^j\partial^k f_0}{\beta_j\partial^j f_0} $$ The identity above has the advantage that it involves only the functions $f_i$ and the parameters defining the $x_i$-s. Since the $\beta_i$, $\xi_i$, and $s$ are independent, the $\beta_i$-s and the derivatives $\partial^jf_i$, $\partial^j\partial^kf_i$ can also be varied independently. Keep $\beta_j\neq 0$ for some $j$, put $\beta_k = 0$ for $k\neq j$, and simplify. We get $$ \frac{(\partial^j)^2f_i}{\partial^j f_i} = \frac{(\partial^j)^2f_0}{\partial^j f_0} \;\;\;\Rightarrow \;\;\; (\partial^j)^2f_i = a_j \partial^jf_i\;\; \forall i $$ for $a_j = a_j(x)$ some function of $x$. Put this back into the original identity, consider next the case where $\beta_j\neq 0$ and $\beta_k\neq 0$ for $j\neq k$, rearrange as polynomial in the $\beta$-s, and conclude that its coefficients must be null. Some algebra with the ensuing equations yields $$ 2 \partial^j\partial^k f_i = a_k \partial^j f_i + a_j \partial^k f_i,\;\;\;\forall i=0,1,2,3\;\; \text{and}\;\; j\neq k $$ Alternatively, use Fock's much more elegant argument that the derivative ratios in the original identity must be linear functions of the $\beta_i$-s to arrive at the same conclusions. The bottom line is that any transformations $f_i$ that take straight lines into straight lines must satisfy the above relations between second and first derivatives.

The 2nd step is to take into account the light speed postulate. That is, if the straight lines are light rays in $S$, then they must be light rays in $S'$. This amounts to the requirement that $$ \left(\frac{dx_1}{dx_0}\right)^2 + \left(\frac{dx_2}{dx_0}\right)^2 + \left(\frac{dx_3}{dx_0}\right)^2 = \left(\frac{dx_1/ds}{dx_0/ds}\right)^2 + \left(\frac{dx_2/ds}{dx_0/ds}\right)^2 + \left(\frac{dx_3/ds}{dx_0/ds}\right)^2 = 1 \\ \text{or} \;\;\;\beta_1^2 + \beta_2^2 + \beta_3^2 = \beta_0^2 $$ implies $$ \left(\frac{dx'_1}{dx'_0}\right)^2 + \left(\frac{dx'_2}{dx'_0}\right)^2 + \left(\frac{dx'_3}{dx'_0}\right)^2 = 1 \\ \text{or} \;\;\;\left(\frac{df_1}{ds}\right)^2 + \left(\frac{df_2}{ds}\right)^2 + \left(\frac{df_3}{ds}\right)^2 = \left(\frac{df_0}{ds}\right)^2 $$ The last condition means $$ \eta^{mn}\left(\beta_j\partial^j f_m\right)\left(\beta_k\partial^k f_n\right) = 0 $$
must be equivalent to $$ \eta^{mn}\beta_m\beta_n = 0 $$ where as usual $\eta^{jk} = 0$ for $j\neq k$, $\eta^{00} = -1$, $\eta^{jj} = 1$ otherwise. By the same argument that the $\beta_j$-s and $\partial^jf_i$-s are independent variables it gives
$$ \eta^{mn}\left(\partial^j f_m \right)\left(\partial^k f_n\right) = \eta^{jk}\lambda $$ for $\lambda = \lambda(x)$ some function of $x$. As before, taking the derivative on $x_i$-s on both sides brings in the second derivatives of the $f$-s, $$ \eta^{mn}\left[\left(\partial^i\partial^j f_m \right)\left(\partial^k f_n\right) + \left(\partial^j f_m \right)\left(\partial^i\partial^k f_n\right)\right]= \eta^{jk}\partial^i\lambda $$ Use the previously derived conditions $2 \partial^j\partial^k f_i = a_k \partial^j f_i + a_j \partial^k f_i$, as well as $\eta^{mn}\left(\partial^j f_m \right)\left(\partial^k f_n\right) = \eta^{jk}\lambda$, and obtain the following condition relating the functions $a_i$ and $\lambda$: $$ 2\eta^{jk}a_i + \eta^{ik}a_j + \eta^{ij}a_k = 2\eta^{jk}\frac{\delta^i \lambda}{\lambda} $$ Now let us check what this means. On the one hand, $$ i\neq j = k \;\; \Rightarrow \;\;a_i = \frac{\delta^i \lambda}{\lambda} $$ but then $$ i = j \neq k \;\; \text{or} \;\; i = k \neq j\;\; \Rightarrow\;\; a_k = a_j = 0 \;\; \forall j,k $$

The simple conclusion is that transformations $f_i$ that take straight lines into straight lines and are compatible with the light speed postulate must be such that $$ \partial^j\partial^k f_i = 0 \;\;\; \forall \;i,j,k = 0,1,2,3 $$ and therefore must be linear.

$\endgroup$
  • $\begingroup$ Beautiful. While it takes into account the speed of light postulate, which was not explicitly mentioned in that context, it was introduced otherwise and therefore that is a very helpful proof. Thank you. $\endgroup$ – failtrolol Nov 17 '15 at 13:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.