Landau & Lifshitz "Classical Field theory" argument for invariance of $ds^2$ In Landau & Lifshitz's "classical field theory", chapter 1, I'm getting mad with an argument used to derive the $ds^2$invariance from Einstein's postulate of the invariance of the speed of light.
Landau & Lifshitz say that if
$$ds^2= c^2dt^2-dx^2-dy^2-dz^2=0$$
then we have to have
$$ds'^2=c^2dt'^2-dx'^2-dy'^2-dz'^2=0$$
That implies, according to Landau, that $$ds^2 = A(v) ds^2.$$
I had a differential geometry exam, succeeding in it, but I can't understand what the object $ds^2$ is. If we want to interpret it as a metric, the scalar product $x^T I x$ is mapped to $x'^TI x'$
How can be it proven that there exists a function of velocity $A(v)$ such that
$$x'^T I x' = A(v) x^T I x?$$
At this moment I can't use that the transformation is linear since we don't know that for sure (if it is, please explain why).
But also if we assume linearity, it's not true that if
$x^T I x$ and $x^T M^T I M x$ have the same zeros, then $I= M^T IM$...
How can I write Landau's argument in a rigorous way? What are the precise mathematical tools and assumptions made?
 A: I would suggest you take a look at the following links (in the given order):

*

*About $g$ and $ds^2$

*General Relativity - Shorthand intuition vs Formal description

*Wikipedia: Rigorous Statement and Proof of proportionality of $ds^2$ and $ds'^2$
The first link is to a Math.SE answer which I wrote a while back, regarding the meaning of $g, ds^2$ and the expression $g_{\mu\nu}dx^{\mu}dx^{\nu}$ and how they all fit together using the usual definitions in differential geometry. The second link is to another Physics.SE answer I wrote, which is essentially about $g$ and $ds^2$ along with some other questions OP had. Finally, the last link is exactly what it says: it's a link to a Wikipedia page where the exact argument given in Landau and Lifshitz is rephrased as a proper mathematical theorem, and the proof is given (it's actually very short and elementary; no differential geometry is needed, you only need to know some basic linear algebra notions).
Lastly, I should say that most of the arguments involved are purely mathematical; i.e the existence of $A(v)$ such that $ds^2=A(v)ds'^2$. The physics only comes into play when we try to argue why (using some symmetry assumptions) $A$ is the constant function $1$, so that $ds^2=ds'^2$. For this of course, you should just refer to Landau and Lifshitz.
A: The reasoning of L&L is strictly speaking wrong. Underneath it all is a theorem about quadratic forms:
Theorem: Let $Q:\mathbb R^n\rightarrow \mathbb R$, $Q(x)=\sum_{i,j}q_{ij}x^i x^j$ be an indefinite quadratic form, and let $Z_Q=\{x\in\mathbb R^n:\ Q(x)=0\}$ be its zero locus. Let $R$ also be a quadratic form for which $Z_Q\subseteq Z_R$, i.e. each zero of $Q$ is also a zero of $R$. Then there exists a number $a\in\mathbb R$ such that $$ Q(x)=aR(x),\quad\forall x\in\mathbb R^n, $$ i.e. the two quadratic forms are proportional to one another.
The proof of this theorem is in this paper (the paper is somewhat informal and looks a bit suspicion-inducingly odd but as far as I can tell the arguments/proofs are valid). In fact two proofs are presented, one based on a diagonalization argument and one via the Nullstellensatz.
For the special case of the Minkowski metrics, a simpler proof can be given, which is presented in Schutz: A First Course In General Relativity.
First, we may suppose that the transformation from one inertial system to another is an affine transformation of the form $$ \bar x^i=A^i_{\ j}x^j+a^i, $$ where the $A$ and $a$ are constant coefficients. This follows by inertial systems being identified by the fact that free particles in them move on straight lines, i.e. their worldlines are straight, and if this property is to be preserved by the transformations between inertial systems then the transformation must be affine.
We can thus use the finite line elements $\Delta s^2$ and $\Delta \bar s^2$.These are quadratic forms $$ \Delta s^2=\eta_{ij}\Delta x^i\Delta x^j,\quad\Delta\bar s^2=\eta_{ij}\Delta\bar x^i\Delta \bar x^j=M_{ij}\Delta x^i\Delta x^j, $$where $$ M_{ij}=\eta_{kl}A^k_{\ i}A^l_{\ j} $$. Since their zeroes agree, by the above mentioned theorems we have $$ \Delta s^2=\alpha\Delta\bar s^2  $$ for some parameter $\alpha$, which is an equality between quadratic forms, i.e. it is valid for all $4$-tuples $(\Delta x^0,\Delta x^1,\Delta x^2,\Delta x^3)$.
This parameter must be independent of the coordinates, which follows from the fact that the coefficients of the quadratic forms are constants, but they can depend on parameters distinguishing the inertial frames. On physical grounds, this must be velocity, as if the frames were accelerating then the transformation between them would no longer be affine. They can't be rotation parameters or any strictly spatial parameters as otherwise homogenity and isotropy would be violated. This part of the reasoning is physical, but I see it as valid.
From this point on the argument of Landau and Lifshitz goes through without a problem.
The only really problematic argument is the proportionality of $ds^2$ and $d\bar s^2$, but that is taken care of by the theorem I mentioned.
I also note that the reasoning can be modified to work with the "infinitesimal" line element $ds^2$ instead of assuming affine transformations on physical grounds, but in my opinion the argument is clearer this way.
Edit: On affine transformations
Let $\gamma:I\rightarrow \mathbb R^4$ be a differentiable ($C^2$) parametrized curve. It is a straight line if and only if it satisfies the equation $$ \frac{d^2\gamma^i}{d\lambda^2}=k(\lambda)\frac{d\gamma^i}{d\lambda}, $$ where $k$ is any reasonable function of the parameter.
Let us now consider a transformation $$ \bar x^i=\bar x^i(x^0,x^1,x^2,x^3) $$ which is (at least locally) a $C^2$ diffeomorphism, and let $ \bar \gamma^i(\lambda)=\bar x^i(\gamma(\lambda)).$ The inverse relation is then $\gamma^i(\lambda)=x^i(\bar\gamma(\lambda))$, and we have $$ \frac{d\gamma^i}{d\lambda}=\frac{\partial x^i}{\partial \bar x^j}\frac{d\bar\gamma^j}{d\lambda}, \\ \frac{d^2\gamma^i}{d\lambda^2}=\frac{\partial^2x^i}{\partial\bar x^k\partial \bar x^i}\frac{d\bar\gamma^k}{d\lambda}\frac{d\bar\gamma^j}{d\lambda}+\frac{\partial x^i}{\partial\bar x^j}\frac{d^2\bar\gamma^j}{\partial\lambda^2}. $$ The equation in the new coordinates is thus $$ \frac{d^2\bar\gamma^i}{d\lambda^2}=k(\lambda)\frac{d\bar\gamma^i}{d\lambda}-\frac{\partial\bar x^i}{\partial x^l}\frac{\partial^2 x^l}{\partial \bar x^j\partial\bar x^k}\frac{d\bar\gamma^j}{d\lambda}\frac{d\bar\gamma^k}{d\lambda}. $$
For this to define a straight line the quadratic term must vanish and must vanish for all straight lines. However at any one parameter value $\lambda$ the derivative $d\bar\gamma^i/d\lambda$ can be anything, and thus this is only possible if $$ \frac{\partial^2 x^l}{\partial \bar x^j \partial \bar x^k}=0. $$
The general solution of this equation is $$ x^i=A^i_{\ j}\bar x^j+a^i, $$ which is an affine transformation.
A: https://en.m.wikipedia.org/wiki/Derivations_of_the_Lorentz_transformations
This article solved all my problems! But I m very curious to read your further precisations!
Linearity is not necessary to the argument...if I prove the invariance of Delta s, follows immediately, and in the most elegant way, that Lorents' transformations are nothing else But the isometries of diag(1,-1,-1,-1)...

Edit 13/07/2021
To use the "quadratic forms" theorem, i need to be convinced that a quadratic form is mapped to another quadratic form ( in the starting coordinates)
This is necessary to compare the  two monsters
$\eta_{\mu\nu} x' ^\mu x'^\nu$
And
$\eta_{\mu\nu} x^\mu x^\nu$
This is not true in general...if you take $x'^\mu = x^\mu³$,
a 2 degree polynomials is not mapped to a 2 degree polinomials....
Using differential forms requires almost the C1. Can you give me a physical or logical reason because i expect that the trasformation has to be C1?
I'm not so enrhusiastic to make hypotesis not precisely motivated! And if the nature would be ugly and bad, and the real trasformation sends (2022AD, 5,8, 0) into (65 millions ago,  Andromeda galaxy), in a not continue/differentiable manner?
