Nonlinear extension of Lorentz Group [duplicate]

Question

The Lorentz group is defined to be the set of linear transformations that leave $ds^2 = -dt^2 + |d\vec{x}|^2$ invariant. The Poincaré group contains the Lorentz group, but now we allow transformations of the form $t' = t+a$ where $a$ is some constant.

Is there a name for the group of transformations that leave $ds^2 = -dt^2 + |d\vec{x}|^2$ invariant without assuming linearity? The answers to this question here claims that any nonlinear transformation will have some "privileged" point, and hence we disregard them as nonphysical. However, I am curious nonetheless if there is a name for this group and any interest in it.

Answer 1

It's called the Lorentz group. No, that's not a joke.

• Let $A$ be the group of linear transformations that leave $$ x^2+y^2+z^2-t^2 \tag{1} $$ invariant. Nonlinear transformations that leave (1) invariant also exist, as shown in the question linked in the OP, but we've defined $A$ to include only the linear ones.

• Now consider the group of all coordinate transformations that leave $$ dx^2+dy^2+dz^2-dt^2 \tag{2} $$ invariant, and let $B$ the subgroup that leaves the origin fixed. (To clarify: the prefix $d$ means differential, not finite difference.) The definition of $B$ does not assume linearity.

The group $A$ is the Lorentz group. The group $B$ is also the Lorentz group. They have the same name because they're the same group: $A=B$. To prove this, suppose we have two different coordinate systems such that $$ dx^2+dy^2+dz^2-dt^2 = d\hat x^2+d\hat y^2+d\hat z^2-d\hat t^2, \tag{3} $$ using hats to distinguish one coordinate system from the other. The form (3) of the line element implies that in both coordinate systems, the set of geodesics through the origin is the same as the set of straight lines — lines whose coordinates are all proportional to each other. The fact that the line element has this form in both coordinate systems implies that if both coordinate systems have the same origin, then the relationship between them is linear, because all coordinate transformations preserve geodesics. (A coordinate transformation is just a smooth re-labeling of the points in spacetime, and the definition of "geodesic" doesn't depend on how we label things.) Therefore, transformations in the group $B$ are automatically linear, and from there we can easily see that it's identical to the group $A$.

Answer 2

NOTE ADDED IN PROOF The linearity for Minkowski metric preserving diffeomorphisms is treated here: Interval preserving transformations are linear in special relativity

However, one may go a step further and ask a better "foundation of SR question": what are the linear/non-linear transformations that preserve wave-fronts, and do they form a group? (it is known that in the foundations of SR light signals are treated as spherical waves propagating at speed "c").

There are two separate issues here:

1. What is the most general form of space-time transformations which leave the D'Alembert equation "conserved":

$$\Box f(x_0,x_1,x_2,x_3) =0 \Rightarrow \Box'f(x'_0,x'_1,x'_2,x'_3) =0) ~\tag{1}$$

2. Do the transformations at point 1., if non-linear, form a group?

Question 2. excludes some (if not all non-linear) solutions to point 1. because non-linearity conflicts with the group property of unique inverse transformation. Think of (as example) $\text{T:} ~ x'=\sin x^2 + a, y'=y, ~ z'=z, ~ t'= \sqrt{t^2 + 75}$. This may be a solution to problem 1., but you cannot invert it, therefore is not something the OP may seek.

If we remove from the OP the condition of being a group, then question 1. is really interesting in itself and a solution has been offered by Weyl (quoted by V.A. Fock in Appendix A of his relativity book: The theory of space, time and gravitation, Pergamon Press, 1959, +---).

Answer to 1.

The form of the most general transformation $x'_i= f_i(x_0, x_1, x_2, x_3), ~ (i=0,1,2,3)$ satisfying $(1)$ is either: $$ x'_i =\frac{x_i - \alpha_i \sum_{k=0}^{3} e_k x_k^2}{1-2 \sum_{k=0}^3 e_k \alpha_k x_k + \sum_{k=0}^{3} e_k \alpha_k^2 \sum_{l=0}^{3} e_l x_l^2 } \tag{2}$$

or $$ x'_i = a_i + \sum_{k=0}^{3} e_k a_{ik} x_k, \tag{3} $$ with $a_{ik}$ satisfying

$$ \sum_{i=0}^{3} e_i a_{ik}a_{il} = e_k \delta_{kl} \tag{4}$$

with $a, e, \alpha$ constants with respect to $x_i$.

$(3) + (4)$ are called Lorentz transformations and they do form a group, while the (2) are called Möbius (spherical) transformations and unfortunately do no form a group.