# Deriving the Lorentz Factor

The Lorentz Factor is defined as $\gamma = 1 / \sqrt{1 - v^2 / c^2}$ with $v$ the relative velocity between two reference frames (or two objects, if you will) and $c$ the speed of light. This is the factor with which length contracts and time dilates between two reference frames. For example, if you’d sit in a rocket that was traveling at 80% the speed of light relative to your family on Earth, the Lorentz Factor would be $\gamma = 1 / \sqrt{1 - 0.8^2} = 1 \frac{2}{3}$.

What this means is that your rocket and everything in it — including you — would be shortened along the direction of motion for your family relative to the length you observe the rocket to be. In fact, its length would be contracted with factor $\gamma$; if the rocket’s length is $x$ as measured by you, your family would measure it to be $x' = x / \gamma$. In this case, they would measure the rocket to be 60% of the length your measure it to be!

Calculating the time dilation is a similar process. In your reference frame, a clock in the rocket would tick every second. If your family back on earth are able observe that clock, it would appear to tick slower. For one second to pass on your clock, they would have to wait $1 \cdot \gamma = 1 \frac{2}{3}$ seconds. In general, if a time of $t$ has passed in your frame of reference, a time of $t' = \gamma t$ has passed for your family.

In this post, I will follow Einstein’s steps in deriving this factor.

Firstly, imagine a space with clocks at certain positions, all at rest relative to each other. An observer at clock A can know the “A-time” at which events happen in the immediate proximity of clock A. An observer at clock B can know the same for clock B. However, we cannot simply assume that the A-time and B-time are the same; we will define a synchronization method for clocks.

If it takes light $t$ seconds to travel from clock A to clock B, it also takes light $t$ seconds to travel from clock B to clock A. Let a ray of light be emitted from clock A at A-time $t_A$ and let it arrive at clock B at B-time $t_B$ and be reflected back to A to arrive at A-time $t'_A$. The two clocks are synchronized if $t_B - t_A = t'_A - t_B$. Using this method, all clocks at rest relative to each other can be synchronized.

Furthermore, from observation we know that light in a vacuum travels with a constant speed $c$ regardless of the reference frame. Thus, if $AB$ is the distance between the two clocks, it holds that $\frac{2AB}{t'_A - t_A} = c$.

Now, we add a rod to this system. Call one end of the rod point A, and the other end point B. The rod travels with constant velocity $v$ relative to the stationary clocks. At both ends, clocks are placed that synchronize with the clocks in the stationary system. In other words, the indications of the clocks at points A and B are equal to the time of the stationary system at the point where they happen to be.

At both ends, we add an observer who applies the synchronization method described. A ray of light departs from point A at time $t_A$ (i.e., the stationary time), and is reflected from B back to A at time $t_B$ and reaches A again at time $t'_A$. For a stationary observer, we find that $t_B - t_A = \frac{r_{AB}}{c-v}$ and $t'_A - t_B = \frac{r_{AB}}{c + v}$ because the ray of light travels from A to B with speed $c$, but the rod moves in the same direction as the light with speed $v$, causing the speed of the light ray to be $c-v$ relative to the rod for a stationary observer. The reverse is true when the ray of light is reflected back to point A. A consequence of this is that observers moving with the rod would find the clocks to be out of sync!

Now, in stationary space let us introduce two systems of coordinates of which the X-, Y and Z-axes are parallel, and the X-axes coincide. One of the systems moves with constant velocity $v$ in direction of increasing $X$ of the other system. Call the moving system $k_m$ and the stationary system $K$. In both systems, add a measuring rod and a number of clocks. All clocks and rods are alike in all respects. The time and place of an event in this space can be defined by coordinates of $K$, $(x, y, z, t)$; and $k_m$, $(\xi, \eta, \zeta, \tau)$.

We introduce a third system, $K'$ with coordinates $(x', y', z', t')$.

\begin{aligned} x' &= x - vt \\ y' &= y \\ z' &= z \\ t' &= t \end{aligned}

A point at rest in system $k_m$ must also be at rest in this new system $K'$.

A ray of light in $K$ would travel with velocity $c$ in both direction of increasing $X$ as well as decreasing $X$. Following  from this, for an observer in $K$, the ray of light would travel with speed $c-v$ in direction of increasing $X$ relative to system $K'$, and with speed $c+v$ in the opposite direction.

At time $\tau_0$, let a ray of light be emitted from the origin of $k_m$ to a mirror at rest in $k_m$ at position $(\xi, 0, 0)$ where it arrives and is reflected at time $\tau_1$. Because the mirror is at rest in system $k_m$, it’s position can also be described by coordinates $(x', 0, 0)$ of system $K'$. The ray arrives back at the origin of $k_m$ at time $\tau_2$. Because the speed of light is constant, we have $\tau_2 - \tau_1 = \tau_1 - \tau_0$, or: $\tau_1 = \frac{1}{2} (\tau_0 + \tau_2)$.

Let’s define a function $\tau(x', y', z', t')$ which gives the time of an event in system $k_m$ given the coordinates of that event in system $K'$. We know that $\tau_1 = \frac{1}{2} (\tau_0 + \tau_2)$ and that the mirror is at position $(x', 0, 0)$ in system $K'$. The ray of light was emitted at time $\tau_0$ from $k_m$, let’s call $t'$ the time it was emitted as observed from system $K'$. Light traveling in direction of increasing $X$ has a speed of $c-v$ and a speed of $c+v$ in direction of decreasing $X$ in $K'$ because $K'$ is defined in terms of $K$. Using this, we obtain the following equations:

\begin{aligned} \tau_0 &= \tau \left(0, 0, 0, t'\right) \\ \tau_1 &= \tau \left(x', 0, 0, t' + \frac{x'}{c - v}\right) \\ \tau_2 &= \tau \left(x', 0, 0, t' + \frac{x'}{c - v} + \frac{x'}{c + v}\right) \end{aligned}

Hence, $\tau \left(x', 0, 0, t' + \frac{x'}{c - v} \right) = \frac{1}{2} \left(\tau \left(0, 0, 0, t'\right) + \tau \left(x', 0, 0, t' + \frac{x'}{c - v} + \frac{x'}{c + v}\right) \right)$.

Furthermore, all equations must be linear because of the homogeneity of the two systems. Because of this, our function can be described as the plane $\tau(x', 0, 0, t') = a \cdot x' + b \cdot t' + \tau_0$.

We have:

\begin{aligned} a &= \frac{\partial \tau}{\partial x'} \\ b &= \frac{\partial \tau}{\partial t'} \end{aligned}

Plugging in the values for $\tau_0$, $\tau_1$ and $\tau_2$ in this equation, we get:

\begin{aligned} \frac{\partial \tau}{\partial x'} x' + \frac{\partial \tau}{\partial t'} \left(t' + \frac{x'}{c-v} \right) + \tau_0 = \frac{1}{2} \bigl(&\frac{\partial \tau}{\partial t'} t' + \tau_0 + \frac{\partial \tau}{\partial x'} x' \\ & + \frac{\partial \tau}{\partial t'} \left(t' + \frac{x'}{c - v} + \frac{x'}{c + v} \right) \\ & + \tau_0 )\bigr) \end{aligned}

Cancel $\tau_0$ and $\frac{\partial \tau}{\partial t'} t'$, and set $x' = 1$:

\begin{aligned} \frac{\partial \tau}{\partial x'} + \frac{\partial \tau}{\partial t'} \left(\frac{1}{c-v} \right) &= \frac{1}{2} \left(\frac{\partial \tau}{\partial t'} \left(\frac{1}{c - v} + \frac{1}{c + v} \right) \right) \\ \frac{\partial \tau}{\partial x'} + \frac{\partial \tau}{\partial t'} \left(\frac{0.5}{c-v} - \frac{0.5}{c+v} \right) &= 0 \\ \frac{\partial \tau}{\partial x'} + \frac{\partial \tau}{\partial t'} \left(\frac{v}{c^2 - v^2} \right) &= 0 \end{aligned}

Note that we could have chosen any other emission point for the ray, so this equation is valid for all values of $(x', y', z')$.

As before we have $\tau(x', 0, 0, t') = a \cdot x' + b \cdot t' + \tau_0$. However, we now also know that $a + b \frac{v}{c^2 - v^2} = 0$.

So we have $a = -b \frac{v}{c^2 - v^2}$. Putting this into our formula for $\tau$ gives us $\tau(x', 0, 0, t') = a' \left(-\frac{v}{c^2 - v^2} x' + t'\right) + \tau_0$.

$a'$ is a function of velocity that is still unknown. For brevity, we will assume that at the origin $\tau = t' = 0$, and so $\tau(x', 0, 0, t') = a' \left(-\frac{v}{c^2 - v^2} x' + t'\right)$.

We know also that a ray of light travels with a constant speed $c$ in our system $k_m$. For a ray emitted at time $\tau = 0$ in direction of increasing $X$ we have $\xi = c \cdot \tau$.

Plugging in our equation for $\tau$, we obtain $\xi = c \cdot a' \left(-\frac{v}{c^2 - v^2} x' + t'\right)$.

In $K'$, the ray moves at speed $c-v$ relative to the X-axis of $k_m$, so we obtain $t = \frac{x'}{c - v}$.

This gives us:

\begin{aligned} \xi &= c \cdot a' \left(-\frac{v}{c^2 - v^2} x' + t'\right) \\ &= c \cdot a' \left(-\frac{v}{c^2 - v^2} x' + \frac{x'}{c - v} \right) \\ &= a' \left(-\frac{cv}{c^2 - v^2} + \frac{c}{c - v} \right) x' \\ &= a' \frac{-c^2 v + c^3}{c^3 - c^2 v - c v^2 + v^3} x' \\ &= a' \frac{(c-v) c^2}{(c-v) (c^2 - v^2)} x' \\ &= a' \frac{c^2}{c^2 - v^2} x' \end{aligned}

Analogously, if a ray were to travel along the Y-axis, we would find $\eta = c \tau = a' c \left(t' - \frac{v}{c^2 - v^2} x' \right)$. In both $k_m$ and $K$ this ray is moving at speed $c$, but in $K$ the ray has an additional velocity $v$ in direction of increasing X. Using the Pythagorean theorem we have $v^2 + v_y^2 = c^2$, or: $v_y^2 = \sqrt{c^2 - v^2}$. So in $K$, and by extension $K'$, the ray has a velocity in direction of increasing Y of $\sqrt{c^2 - v^2}$.

So in this case $t' = \frac{y}{\sqrt{c^2 - v^2}}$ and $x' = 0$. We then find $\eta = a' \frac{c}{\sqrt{c^2 - v^2}} y'$ and, repeating the same process for the Z-axis, $\zeta = a' \frac{c}{\sqrt{c^2 - v^2}} z'$.

Plugging in $x' = x - vt$, $y'=y$, $z' = z$ and $t' = t$, our equations can be written as:

\begin{aligned} \tau &= \Phi(v) \cdot \beta(v) \cdot (t - vx / c^2) \\ \xi &= \Phi(v) \cdot \beta(v) \cdot (x - vt) \\ \eta &= \Phi(v) \cdot y \\ \zeta &= \Phi(v) \cdot z \end{aligned}

With $\Phi(v) = a' \frac{c}{\sqrt{c^2 - v^2}}$ and $\beta(v) = \frac{c}{\sqrt{c^2 - v^2}}$.

In these equations, we have an as of yet unknown function $\Phi(v)$ that we have to determine. To determine the function, we introduce a fourth system of coordinates $k_s$, $(x'', y'', z'', t'')$ that has a velocity of $-v$ on the X-axis relative to system $k_m$. At time $t=0$, let the origins of $K$, $k_m$ and $k_s$ coincide and let $t'' = 0$ (i.e., $t''$ is the time in system $k_s$ which is 0 when the time $t$ in $K$ is 0).

The equations to transform from system $K$ to system $k_m$ are given above. Because the situation between systems $k_s$ and $k_m$ is simply the mirrored situation between systems $K$ and $k_m$, we can apply the same equations with opposite sign to get the equations to transform from $k_m$ coordinates to $k_s$ coordinates.

\begin{aligned} t'' &= \Phi(-v) \cdot \beta(-v) \cdot (\tau + v \xi / c^2) \\ x'' &= \Phi(-v) \cdot \beta(-v) \cdot (\xi + v \tau) \\ y'' &= \Phi(-v) \cdot \eta \\ z'' &= \Phi(-v) \cdot \zeta \end{aligned}

Plugging in the transformations for $(\xi, \eta, \zeta, \tau)$ from $K$ to $k_m$ into this set of equations will yield a transform from $K$ to $k_s$. First, note that:

\begin{aligned} \beta(-v) \beta(v) &= \frac{c}{\sqrt{c^2-(-v)^2}} \frac{c}{\sqrt{c^2-v^2}} = \frac{c}{\sqrt{c^2-v^2}} \frac{c}{\sqrt{c^2-v^2}} \\ &= \left(\frac{c}{\sqrt{c^2-v^2}}\right)^2 = \frac{c^2}{\sqrt{c^2-v^2}^2} \\ &= \frac{c^2}{c^2 - v^2} \end{aligned}

Now:

\begin{aligned} t'' &= \Phi(-v) \cdot \beta(-v) \cdot (\Phi(v) \cdot \beta(v) \cdot (t - vx / c^2) \\ & \hphantom{=} + v (\Phi(v) \cdot \beta(v) \cdot (x - vt)) / c^2) \\ &= \Phi(-v) \cdot \beta(-v) \cdot \Phi(v) \cdot \beta(v) (t - vx / c^2 + v(x - vt)/c^2) \\ &= \Phi(-v) \cdot \Phi(v) \cdot \beta(-v) \cdot \beta(v) (t - vx / c^2 + vx/c^2 - v^2t/c^2) \\ &= \Phi(-v) \cdot \Phi(v) \cdot \frac{c^2}{c^2 - v^2} (t - v^2t/c^2) \\ &= \Phi(-v) \cdot \Phi(v) \cdot \frac{c^2 (t - v^2t/c^2)}{c^2 - v^2} \\ &= \Phi(-v) \cdot \Phi(v) \cdot \frac{c^2 t (1 - v^2/c^2)}{c^2 - v^2} \\ &= \Phi(-v) \cdot \Phi(v) \cdot \frac{t (c^2 - v^2)}{c^2 - v^2} \\ &= \Phi(-v) \cdot \Phi(v) \cdot t \end{aligned}

And:

\begin{aligned} x'' &= \Phi(-v) \cdot \beta(-v) \cdot (\Phi(v) \cdot \beta(v) \cdot (x - vt) \\ & \hphantom{=} + v \Phi(v) \cdot \beta(v) \cdot (t - vx / c^2)) \\ &= \Phi(-v) \cdot \beta(-v) \cdot \Phi(v) \cdot \beta(v) ((x - vt) + vt - v^2x / c^2) \\ &= \Phi(-v) \cdot \Phi(v) \cdot \beta(-v) \cdot \beta(v) (x - vt + vt - v^2x / c^2) \\ &= \Phi(-v) \cdot \Phi(v) \cdot \frac{c^2}{c^2 - v^2} (x - v^2x / c^2) \\ &= \Phi(-v) \cdot \Phi(v) \cdot \frac{c^2 (x - v^2x / c^2)}{c^2 - v^2} \\ &= \Phi(-v) \cdot \Phi(v) \cdot \frac{c^2 x(1 - v^2 / c^2)}{c^2 - v^2} \\ &= \Phi(-v) \cdot \Phi(v) \cdot \frac{x(c^2 - v^2)}{c^2 - v^2} \\ &= \Phi(-v) \cdot \Phi(v) x \end{aligned}

And finally:

\begin{aligned} y'' &= \Phi(-v) \cdot \Phi(v) \cdot y \\ z'' &= \Phi(-v) \cdot \Phi(v) \cdot z \end{aligned}

To summarize, to transform from $K$ to $k_s$, we have the following equations:

\begin{aligned} t'' &= \Phi(-v) \cdot \Phi(v) \cdot t \\ x'' &= \Phi(-v) \cdot \Phi(v) \cdot x \\ y'' &= \Phi(-v) \cdot \Phi(v) \cdot y \\ z'' &= \Phi(-v) \cdot \Phi(v) \cdot z \end{aligned}

Note that $x''$, $y''$ and $z''$ are independent of time. Thus, $k_s$ and $K$ are at rest relative to each other. Because the origins were set to coincide at $t = 0$, this must remain the case. As such, it must hold that $(x'', y'', z'') = (x, y, z)$ and thus $\Phi(-v) \cdot \Phi(v) = 1$.

Let’s find out what $\Phi(v)$ means with regard to our systems. Say we a have stationary rod in system $k_m$ with one end on the origin and the other at $\eta = 1$. In other words, the rod is pointing upwards in system $k_m$ and has length 1. This rod, in system $K$, must also be pointing upwards and must have one end on the origin. We know that $\eta = \Phi(v) \cdot y$, thus $y = \frac{\eta}{\Phi(v)}$. In other words, in system $K$, this rod is pointing upwards and has length $\frac{\eta}{\Phi(v)}$. It is clear that because of symmetry, it does not matter whether the rod relative to system $K$ is moving in direction of increasing X or decreasing X. As such, we have $\frac{\eta}{\Phi(v)} = \frac{\eta}{\Phi(-v)}$, or: $\Phi(v) = \Phi(-v)$.

So, we have:

\begin{aligned} \Phi(-v) \Phi(v) &= 1 \\ \Phi(-v) &= \Phi(v) \\ \Phi(-v) &= \Phi(v) \cdot 1 = \Phi(v) \Phi(-v) \Phi(v) \\ 1 &= \Phi(v) \Phi(v) \\ 1 &= \Phi(v)^2 \\ \Phi(v) &= \sqrt{1} = 1 \end{aligned}

With the knowledge that $\Phi(v) = 1$, we just need a little bit more algebra to arrive at the Lorentz Factor:

\begin{aligned} \beta(v) &= \frac{c}{\sqrt{c^2 - v^2}} \\ &= \frac{1}{c^{-1} \sqrt{c^2 - v^2}} \\ &= \frac{1}{\sqrt{c^{-2} (c^2 - v^2)}} \\ &= \frac{1}{\sqrt{1 - v^2 / c^2}} = \gamma \end{aligned}