Why is the Lorentz scalar product the same in all inertial frames for objects other than light?

Question

I'm having trouble understanding the consequences of Einstein's second postulate. Since the speed of light is constant in every inertial frame, it follows that: $$(c\,dt)^2-dx^2-dy^2-dz^2=(c\,dt')^2-dx'^2-dy'^2-dz'^2\quad(=0)$$ So the Lorentz scalar product is invarint, but I fail to see how that extends to objects other than light. I'm trying to understand the derivation of the Lorentz Group, so I can't just use a Lorentz Boost.

Edit: I might not have been clear enough. Maybe a better way to frame this question is: "Why do objects other than light have to transform the same way light does?"

Answer 1

The Lorentz group is defined to be the group of special linear transformations that leave the interval invariant.

I understand your issue to be this: The postulates of relativity tell us that an allowable transformation should preserve the zero set of the metric. But the Lorentz group cconsists of transformations that preserve all values of the metric. How do we reconcile this difference?

The answer is that there is no difference, because two non-degenerate quadratic forms with the same zero set must be the same form up to a multiplicative constant. Now let $g$ be the metric and $T$ a special linear transformation. If $T(g)$ and $g$ have the same zero set (as required by relativity, then they are the same form. So $T$ is automatically in the Lorentz group.

If you're concerned about the mathematical claim that preserving the zero set implies preserving the metric, it's an immediate consequence of the Hilbert Nullstellensatz, which says that two polynomials with the same zero set must have the same radical. Combine that with the fact that $g$ and $T(g)$ both have degree 2.

Answer 2

For each rotations in the planes xy, yz and zx, and for each "boost" in x, y and z directions, it is possible to examine the situation for an infinitesimal change.

For the rotations, $sin(\theta_i) \approx \theta_i$ and $cos(\theta_i) \approx 1$.

For "boosts", $sinh(\phi_i) \approx \phi_i$ and $cosh(\phi_i) \approx 1$. ($tanh(\phi_i) = v/c$)

For each case, the infinitesimal Lorentz matrix can be separated in a sum of the identity and a matrix formed only by zeros and ones (the generators), multiplied by the correspondent infinitesimal ($\theta_i$ or $\phi_i$). Below an example for a "boost" in x direction:

\begin{equation} I + \phi_1 \begin{vmatrix} 0 & -1 & 0 & 0\\ -1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{vmatrix} \end{equation}

When multiplying a 4-vector by such matrices, the distance is preserved like in the example:

$(x^0)' = x^0 -\phi_1 x^1 $
$(x^1)' = -\phi_1 x^0 + x^1$

$(x^0)'^2 - (x^1)'^2 = (x^0)^2 - (x^1)^2 - 2x^0\phi_1 x^1 + 2x^0\phi_1 x^1$ discarding the second order infinitesimal.

For a general infinitesimal transformation, the corresponding matrix is the sum of the identity and a linear combination of $\theta_i$ and $\phi_i$ multiplied by the respective generators.

That sum of matrices are the first Taylor expansion terms of an exponential, that is the Lorentz matrix for a finite general transformation.

$L = e^{\phi_1 J^1 + \phi_2 J^2 + ... + \theta_3 J^6 }$