How do we come to $c$ in the Lorentz Factor in relativity?

Question

While reading a post explaining time dilation in relativity. The author explains why we add $c$ to get to the Minkowski Metric. He explains it as

so we multiply time by the speed of light $c$ so the product $ct$ has units of metres

So that the time dimension can be considered in the same terms (units) as the spacial dimensions.

My question is how we came to use $c$ and not some other arbitrary speed. Why $c$ gets used in the Lorentz Factor. I believe it comes from Maxwells Equations for Electromagnetism but correct me if I'm wrong and point me to the right path.

the Lorentz factor: $$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$

And time dilation equation in special relativity $$t' = t\sqrt{1 - \frac{v^2}{c^2}} = \frac{t}{\gamma}$$

Edit: In the Original post $\tau$ was used instead of $t'$

Answer 1

I will answer to your question in two ways: first historical, then practical.

The first derivation of the Lorentz transformations consists in looking at some appartus made of mirrors and a beam of light. This amounts to consider the "principle of relativity" and the "principle of the constancy of the speed of light" (as noted in another post). Then you write the invariant line element $\mathrm{d}s^2$ for the light travelling in different frames and you look for the transformations that leave it invariant. This gives you the Lorentz transformations. Since you started with light you find the speed of light inside the transformations. Historically it was noted that the Maxwell equations are not invariant under Galilean transformations but under another set which corresponds to the Lorentz transformations, so it is another way to find this correspondence.

But this approach is quite problematic because it gives the light a very special role. One could make Gedanken experiments where the role of photons is replaced by gravitons for example, since the speed of "light" also appears in general relativity in places where it has nothing to do with light. Hence you could say that one sees instead the speed of the graviton. Another problem is that it would be sufficient to discover that the photon has a mass (even infinitesimally small) to break special relativity according to the first approach. So I find that it is a big epistemological problem to use the constancy of the speed of light as a postulate, and this has been regularly emphasized (for example in the works of Jean-Marc Lévy-Leblond).

A better approach is to replace the second principle by principles about the nature of spacetime: homogeneity, isotropy, causality and group structure (i.e. "good" composition laws). Using these principles you can derive the Lorentz transformations and you discover inside one parameter $c$ that characterizes the structure of spacetime (in some paper Lévy-Leblond says that it could have been named the "spacetime structure constant" if one did not have the prejudice of the speed of light). It remains to identify the value of $c$ "experimentally". According to the theory you have derived you find that a massless particle travels as the speed $c$. You know from various experiments that the photon appears massless and, to this accuracy, you can declare that $c$ is equivalent to the speed of light (but remembering that this identification would cease to be valid if the photon was found to have a mass).

With this approach one sees very clearly what is concretely our spacetimes which is not clear by just looking at mirrors and light. For example it is not clear to me that one does not need additional assumptions in the first approach to uniquely gets the Lorentz group. For example in the second approach causality is important for removing other "relativistic" groups such as the Carroll group.

Answer 2

Maxwell electromagnetic wave equation incorporates c.

From Maxwells Equations

The electromagnetic wave equation is a second-order partial differential equation that describes the propagation of electromagnetic waves through a medium or in a vacuum. The homogeneous form of the equation, written in terms of either the electric field E or the magnetic field B, takes the form:

${\displaystyle {\begin{aligned}\left(v_{ph}^{2}\nabla ^{2}-{\frac {\partial ^{2}}{\partial t^{2}}}\right)\mathbf {E} &=\mathbf {0} \\\left(v_{ph}^{2}\nabla ^{2}-{\frac {\partial ^{2}}{\partial t^{2}}}\right)\mathbf {B} &=\mathbf {0} \end{aligned}}}$

where

${\displaystyle v_{ph}={\frac {1}{\sqrt {\mu \varepsilon }}}}$

is the speed of light (i.e. phase velocity) in a medium with permeability μ, and permittivity ε, and ∇2 is the Laplace operator. In a vacuum, c = 299,792,458 meters per second, a fundamental physical constant. The electromagnetic wave equation derives from Maxwell's equations.

When Einstein developed special relativity, he knew that c would be constant for every observer, no matter how fast they were moving, even at 99.999% of c, light seems to travel at c.

He also knew that Newtons laws of motion could not deal with this, say in the addition of velocities, so he (literally) wrote an apology to Newton and accepted Maxwells idea of c being constant, which in turn meant something had to give, and that something was the idea of absolute time, which Newton, perhaps reluctantly, had accepted.

Answer 3

It was the Michelson-Morley experiment which showed the constancy of the velocity of light c, independent of the frame of the observer. Constancy of the speed of light was seemingly in contradiction with the principle of relativity of Galilei, the requirement that the equations describing the laws of physics have the same form in all admissible frames of reference.

Special relativity is precisely reflecting this apparent contradiction because it consists of nothing more than two postulates:

• the principle of relativity and
• the constancy of light, independant of the frame of the observer,

and assuming the correctness of both postulates, we can derive Lorentz transformation and the proper time formula you are citing.

Answer 4

I think it's because this way you have zero length paths for light.

For a finite interval you have

$\Delta s = c^2\Delta t ^2 -\Delta \vec{r}^2 $

And with the $c$ in there is equal to $0$ for light-like intervals.