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The right spring changes length by $u_{n+1}-u_n$ and pulls atom $n$ to the right with force $C(u_{n+1}-u_n)$. The left spring contributes $-C(u_n-u_{n-1})$. Adding these restoring forces and applying Newton's second law givesUniform displacement makes both spring extensions vanish, as translational invariance requires. Search for a normal mode $u_n=u_0{e}^{i(kna-\omega t)}$. Substitution yieldsOne representative first Brillouin zone is $-\pi /a\le k\le \pi /a$. This is one positive-frequency acoustic branch: near zero wavevector, $\omega \simeq a\sqrt{C/m}|k|$, so the sound speed is $v_s=a\sqrt{C/m}$. Opposite signs of $k$ label opposite propagation directions, not a second polarization branch.

The model assumes small displacements, harmonic nearest-neighbor forces, one scalar displacement per atom and uniform equilibrium spacing. It excludes damping, anharmonic expansion and transverse polarizations. In a three-dimensional monatomic crystal there would normally be three acoustic polarizations; that counting cannot be imported into this explicitly one-dimensional scalar model.

One can derive a family containing the Lorentz transformations without assuming anything about light, but the Einstein's principle of relativity alone does not select that family member over the Galilean transformation. Extra assumptions must be stated: spatial and temporal translational symmetry, isotropy of SPACE, continuous transformations between equivalent inertial frames, reciprocity, and closure under composition. These are substantive restrictions, rather than consequences of the bare assertion that physical laws agree between inertial frames.

Here is the algebraic content. Translational symmetry makes the transformation between coincident origins a linear transformation. In one spatial direction the moving origin $x=vt$ fixes $x^{\prime }=a(v)(x-vt)$. Isotropy and the inverse transformation at $-v$ allow the standard calibrated formTo see why $K$ is universal, temporarily write the upper-right entry as $-b(v)$, leaving the lower-left entry $-v$. Equality of the two diagonal entries in a composite boost gives $b(u)v=ub(v)$, so $b(v)/v=K$ is constant for nonzero boosts. Inversion fixes $a(v{)}^2(1-K{v}^2)=1$, with the positive branch continuous at zero. This is the universality of an invariant speed under boost composition.

The resulting velocity-addition formula law and invariant quadratic expression areFor $K>0$, $C=K^{-1/2}$ is an invariant limiting speed; setting the empirically determined $C$ equal to the vacuum speed of light gives the standard Lorentz transformation. For $K=0$, $t^{\prime }=t$ and $x^{\prime }=x-vt$: an entirely consistent Galilean transformation satisfying the stated Einstein's principle of relativity. Thus the general assumptions do not establish that an invariant finite speed exists, much less identify it with light.

There is also a formal negative-$K$ branch, which has a rotation-like geometry in the time–space plane. Standard derivations exclude it using additional global assumptions, such as a preserved time orientation and unrestricted composition of admissible boosts. Without identifying those assumptions it is misleading to claim a two-way classification from the Einstein's principle of relativity alone. In three spatial dimensions isotropy requires the same constant in every direction.

The light postulate is therefore dispensable as the starting point for a conditional kinematical derivation; empirical input remains indispensable in identifying our world's value and sign of $K$. Albert Einstein's original construction uses the light postulate and Einstein synchronisation, making that empirical connection explicit.^[ref] Replacing the postulate by an independently established finite invariant speed changes the route, rather than deriving all of special relativity from relativity alone.

Assume the electrical conductor is nonmagnetic, with magnetic permeability ${\mu }_0$, and neglect end effects. Rotational symmetry and translational symmetry make the magnetic flux density azimuthal and constant along a coaxial circle. The magnetostatic Ampère law therefore givesUniform current density is $j=I/(\pi a^2)$, so $I_{\mathrm{enc}}(r)=j\pi r^2=I{r}^2/a^2$. ThusThe direction follows the right-hand rule around the electric current. For a uniform linear magnetic material, replace ${\mu }_0$ by its magnetic permeability.