Replacing $c$ With Infinity

Question

I am confused about replacing $c$ (the speed of light) with $\infty$. I just do not understand how this recovers the non-relativistic physics, mainly, how would this be demonstrated in the Lorentz Transformation? Would it affect simultaneity since infinity is now not defined along with time-dilation and length-contraction?

Answer 1

Take a Lorentz boost along $x$. We have: $$ \gamma = \dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}\xrightarrow{c\to \infty} 1.$$ Then $$ \begin{array} & t' &= & \gamma \left(t-\frac{v}{c^2}\;x\right)&\xrightarrow{c\to \infty} & t \\ x' &= & \gamma \left(x-v\;t\right)&\xrightarrow{c\to \infty} & x - v\; t \\ y' &=& y \\ z' &=& z, \end{array}$$ which is precisely a Galilean boost along $x$.

It is then easy to see that the rest of Lorentz transformations fall-back to Galilean boosts and rotations, thus recovering the non-relativistic transformation laws.

Answer 2

This calculation, of transformations between inertial reference frames from group postulates, ultimately finds a quantity $\kappa=-1/c^2$ that needs to be $\le0$, but what you're asking for is equivalent to setting $\kappa=0$. This special case obtains Galilean transformations; any other obtains Lorentz transformations.

Answer 3

The important factor in SR is $$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

When $c\to\infty$, $\gamma\to1$. So for what you mention in things such as time dilation ($\Delta t'=\gamma\Delta t$) or length contraction ($\gamma \Delta\ell'=\Delta\ell$), we end up with nothing different happening between inertial frames moving relative to each other. The same arguments can be made for the more general cases of Lorentz transformations, relative velocities, etc.