## Transformations in Space-Time

What are inertial systems of reference? Why are transformations between inertial systems of reference linear? A simple analysis of transformations between systems of reference.

We have two systems of reference, denoted by $\mathcal{S}$ and $\mathcal{S}'$. Each system has his own co-ordinates; where $(t, x, y, z)$ is the space-time co-ordinate for $\mathcal{S}$ and $(t', x', y', z')$ is the space-time co-ordinate for $\mathcal{S}'$. The general transformation from $\mathcal{S}$ to $\mathcal{S}'$ can be written as

(1) $(t',x',y',z') = (t'(t,x,y,z),x'(t,x,y,z),y'(t,x,y,z),z'(t,x,y,z))$

where $t'(t,x,y,z)$, $x'(t,x,y,z)$, $y'(t,x,y,z)$ and $z'(t,x,y,z)$ are unknown functions. It is convenient to use other symbols. We write $t = x_0$, $x=x_1$, $y=x_2$ and $z=x_3$. The general transformation can then be written as

(2) $x'_\mu = x'_\mu(x_\nu)$

A special path is known as $\displaystyle \frac{d^2 x_\mu}{ds^2}=0$. We call a system an inertial frame of reference, IF a free moving body will follow the path defined by $\displaystyle \frac{d^2 x_\mu}{ds^2}=0$. Now

(3) $\displaystyle \frac{d x'_\mu}{ds} = \frac{\partial x'_\mu}{\partial x_\kappa} \frac{d x_\kappa}{ds}$, thus

(4) $\displaystyle \frac{d^2 x'_\mu}{ds^2} = \frac{\partial^2 x'_\mu}{\partial x_\kappa \partial x_\lambda} \frac{d x_\kappa}{ds} \frac{d x_\lambda}{ds} + \frac{\partial x'_\mu}{\partial x_\kappa} \frac{d^2 x_\kappa}{ds^2}$

Now IF $\mathcal{S}$ and $\mathcal{S}'$ are both inertial systems of reference, then $\displaystyle \frac{d^2 x'_\mu}{ds^2} = 0$ and $\displaystyle \frac{d^2 x_\kappa}{ds^2} = 0$, therefore we obtain

(5) $\displaystyle \frac{\partial^2 x'_\mu}{\partial x_\kappa \partial x_\lambda} \frac{d x_\kappa}{ds} \frac{d x_\lambda}{ds} = 0$,

but if this is true for arbitrary values $\displaystyle \frac{d x_\kappa}{ds}$ and $\displaystyle \frac{d x_\lambda}{ds}$, then we obtain

(6) $\displaystyle \frac{\partial^2 x'_\mu}{\partial x_\kappa \partial x_\lambda} = 0$

The transformation between two inertial systems of reference satisfies $\displaystyle \frac{\partial^2 x'_\mu}{\partial x_\kappa \partial x_\lambda} = 0$. Then it is clear that transformations between two inertial systems of reference are linear transformations. As $\displaystyle \frac{\partial x'_\mu}{\partial x_\lambda} = a^\lambda_\mu$, and $\displaystyle \frac{\partial a^\lambda_\mu}{\partial x_\kappa}= 0$, thus $x'_\mu = a_\mu^\lambda x_\lambda + a_\mu$. It is convenient to set $a_\mu=0$, then we can write $x'_\mu = a_\mu^\lambda x_\lambda$. Transformations between inertial systems of reference are linear transformations.