The Lorentz Factor is defined as γ=1/1−v2/c2\gamma = 1 / \sqrt{1 - v^2 / c^2} with vv the relative velocity between two reference frames (or two objects, if you will) and cc 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 γ=1/1−0.82=123\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 γ\gamma; if the rocket’s length is xx as measured by you, your family would measure it to be x′=x/γx' = 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⋅γ=1231 \cdot \gamma = 1 \frac{2}{3} seconds. In general, if a time of tt has passed in your frame of reference, a time of t′=γtt' = \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 tt seconds to travel from clock A to clock B, it also takes light tt seconds to travel from clock B to clock A. Let a ray of light be emitted from clock A at A-time tAt_A and let it arrive at clock B at B-time tBt_B and be reflected back to A to arrive at A-time tA′t'_A. The two clocks are synchronized if tB−tA=tA′−tBt_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 cc regardless of the reference frame. Thus, if ABAB is the distance between the two clocks, it holds that 2ABtA′−tA=c\frac{2AB}{t'_A - t_A} = c.

Rod with velocity vv

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 vv 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 tAt_A (i.e., the stationary time), and is reflected from B back to A at time tBt_B and reaches A again at time tA′t'_A. For a stationary observer, we find that tB−tA=rABc−vt_B - t_A = \frac{r_{AB}}{c-v} and tA′−tB=rABc+vt'_A - t_B = \frac{r_{AB}}{c + v} because the ray of light travels from A to B with speed cc, but the rod moves in the same direction as the light with speed vv, causing the speed of the light ray to be c−vc-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 vv in direction of increasing XX of the other system. Call the moving system kmk_m and the stationary system KK. 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 KK, (x,y,z,t)(x, y, z, t); and kmk_m, (ξ,η,ζ,τ)(\xi, \eta, \zeta, \tau).

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

x′=x−vty′=yz′=zt′=t \begin{aligned} x' &= x - vt \\ y' &= y \\ z' &= z \\ t' &= t \end{aligned}

A point at rest in system kmk_m must also be at rest in this new system K′K'.

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

At time τ0\tau_0, let a ray of light be emitted from the origin of kmk_m to a mirror at rest in kmk_m at position (ξ,0,0)(\xi, 0, 0) where it arrives and is reflected at time τ1\tau_1. Because the mirror is at rest in system kmk_m, it’s position can also be described by coordinates (x′,0,0)(x', 0, 0) of system K′K'. The ray arrives back at the origin of kmk_m at time τ2\tau_2. Because the speed of light is constant, we have τ2−τ1=τ1−τ0\tau_2 - \tau_1 = \tau_1 - \tau_0, or: τ1=12(τ0+τ2)\tau_1 = \frac{1}{2} (\tau_0 + \tau_2).

Let’s define a function τ(x′,y′,z′,t′)\tau(x', y', z', t') which gives the time of an event in system kmk_m given the coordinates of that event in system K′K'. We know that τ1=12(τ0+τ2)\tau_1 = \frac{1}{2} (\tau_0 + \tau_2) and that the mirror is at position (x′,0,0)(x', 0, 0) in system K′K'. The ray of light was emitted at time τ0\tau_0 from kmk_m, let’s call t′t' the time it was emitted as observed from system K′K'. Light traveling in direction of increasing XX has a speed of c−vc-v and a speed of c+vc+v in direction of decreasing XX in K′K' because K′K' is defined in terms of KK. Using this, we obtain the following equations:

τ0=τ(0,0,0,t′)τ1=τ(x′,0,0,t′+x′c−v)τ2=τ(x′,0,0,t′+x′c−v+x′c+v) \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, τ(x′,0,0,t′+x′c−v)=12(τ(0,0,0,t′)+τ(x′,0,0,t′+x′c−v+x′c+v))\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 τ(x′,0,0,t′)=a⋅x′+b⋅t′+τ0\tau(x', 0, 0, t') = a \cdot x' + b \cdot t' + \tau_0.

We have:

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

Plugging in the values for τ0\tau_0, τ1\tau_1 and τ2\tau_2 in this equation, we get:

∂τ∂x′x′+∂τ∂t′(t′+x′c−v)+τ0=12(∂τ∂t′t′+τ0+∂τ∂x′x′+∂τ∂t′(t′+x′c−v+x′c+v)+τ0)) \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 τ0\tau_0 and ∂τ∂t′t′\frac{\partial \tau}{\partial t'} t', and set x′=1x' = 1:

∂τ∂x′+∂τ∂t′(1c−v)=12(∂τ∂t′(1c−v+1c+v))∂τ∂x′+∂τ∂t′(0.5c−v−0.5c+v)=0∂τ∂x′+∂τ∂t′(vc2−v2)=0 \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′)(x', y', z').

As before we have τ(x′,0,0,t′)=a⋅x′+b⋅t′+τ0\tau(x', 0, 0, t') = a \cdot x' + b \cdot t' + \tau_0. However, we now also know that a+bvc2−v2=0a + b \frac{v}{c^2 - v^2} = 0.

So we have a=−bvc2−v2a = -b \frac{v}{c^2 - v^2}. Putting this into our formula for τ\tau gives us τ(x′,0,0,t′)=a′(−vc2−v2x′+t′)+τ0\tau(x', 0, 0, t') = a' \left(-\frac{v}{c^2 - v^2} x' + t'\right) + \tau_0.

a′a' is a function of velocity that is still unknown. For brevity, we will assume that at the origin τ=t′=0\tau = t' = 0, and so τ(x′,0,0,t′)=a′(−vc2−v2x′+t′)\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 cc in our system kmk_m. For a ray emitted at time τ=0\tau = 0 in direction of increasing XX we have ξ=c⋅τ\xi = c \cdot \tau.

Plugging in our equation for τ\tau, we obtain ξ=c⋅a′(−vc2−v2x′+t′)\xi = c \cdot a' \left(-\frac{v}{c^2 - v^2} x' + t'\right).

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

This gives us:

ξ=c⋅a′(−vc2−v2x′+t′)=c⋅a′(−vc2−v2x′+x′c−v)=a′(−cvc2−v2+cc−v)x′=a′−c2v+c3c3−c2v−cv2+v3x′=a′(c−v)c2(c−v)(c2−v2)x′=a′c2c2−v2x′ \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}

Velocity vectors in KK

Analogously, if a ray were to travel along the Y-axis, we would find η=cτ=a′c(t′−vc2−v2x′)\eta = c \tau = a' c \left(t' - \frac{v}{c^2 - v^2} x' \right). In both kmk_m and KK this ray is moving at speed cc, but in KK the ray has an additional velocity vv in direction of increasing X. Using the Pythagorean theorem we have v2+vy2=c2v^2 + v_y^2 = c^2, or: vy2=c2−v2v_y^2 = \sqrt{c^2 - v^2}. So in KK, and by extension K′K', the ray has a velocity in direction of increasing Y of c2−v2\sqrt{c^2 - v^2}.

So in this case t′=yc2−v2t' = \frac{y}{\sqrt{c^2 - v^2}} and x′=0x' = 0. We then find η=a′cc2−v2y′\eta = a' \frac{c}{\sqrt{c^2 - v^2}} y' and, repeating the same process for the Z-axis, ζ=a′cc2−v2z′\zeta = a' \frac{c}{\sqrt{c^2 - v^2}} z'.

Plugging in x′=x−vtx' = x - vt, y′=yy'=y, z′=zz' = z and t′=tt' = t, our equations can be written as:

τ=Φ(v)⋅β(v)⋅(t−vx/c2)ξ=Φ(v)⋅β(v)⋅(x−vt)η=Φ(v)⋅yζ=Φ(v)⋅z \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 Φ(v)=a′cc2−v2\Phi(v) = a' \frac{c}{\sqrt{c^2 - v^2}} and β(v)=cc2−v2\beta(v) = \frac{c}{\sqrt{c^2 - v^2}}.

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

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

t′′=Φ(−v)⋅β(−v)⋅(τ+vξ/c2)x′′=Φ(−v)⋅β(−v)⋅(ξ+vτ)y′′=Φ(−v)⋅ηz′′=Φ(−v)⋅ζ \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 (ξ,η,ζ,τ)(\xi, \eta, \zeta, \tau) from KK to kmk_m into this set of equations will yield a transform from KK to ksk_s. First, note that:

β(−v)β(v)=cc2−(−v)2cc2−v2=cc2−v2cc2−v2=(cc2−v2)2=c2c2−v22=c2c2−v2 \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:

t′′=Φ(−v)⋅β(−v)⋅(Φ(v)⋅β(v)⋅(t−vx/c2)=+v(Φ(v)⋅β(v)⋅(x−vt))/c2)=Φ(−v)⋅β(−v)⋅Φ(v)⋅β(v)(t−vx/c2+v(x−vt)/c2)=Φ(−v)⋅Φ(v)⋅β(−v)⋅β(v)(t−vx/c2+vx/c2−v2t/c2)=Φ(−v)⋅Φ(v)⋅c2c2−v2(t−v2t/c2)=Φ(−v)⋅Φ(v)⋅c2(t−v2t/c2)c2−v2=Φ(−v)⋅Φ(v)⋅c2t(1−v2/c2)c2−v2=Φ(−v)⋅Φ(v)⋅t(c2−v2)c2−v2=Φ(−v)⋅Φ(v)⋅t \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:

x′′=Φ(−v)⋅β(−v)⋅(Φ(v)⋅β(v)⋅(x−vt)=+vΦ(v)⋅β(v)⋅(t−vx/c2))=Φ(−v)⋅β(−v)⋅Φ(v)⋅β(v)((x−vt)+vt−v2x/c2)=Φ(−v)⋅Φ(v)⋅β(−v)⋅β(v)(x−vt+vt−v2x/c2)=Φ(−v)⋅Φ(v)⋅c2c2−v2(x−v2x/c2)=Φ(−v)⋅Φ(v)⋅c2(x−v2x/c2)c2−v2=Φ(−v)⋅Φ(v)⋅c2x(1−v2/c2)c2−v2=Φ(−v)⋅Φ(v)⋅x(c2−v2)c2−v2=Φ(−v)⋅Φ(v)x \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:

y′′=Φ(−v)⋅Φ(v)⋅yz′′=Φ(−v)⋅Φ(v)⋅z \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 KK to ksk_s, we have the following equations:

t′′=Φ(−v)⋅Φ(v)⋅tx′′=Φ(−v)⋅Φ(v)⋅xy′′=Φ(−v)⋅Φ(v)⋅yz′′=Φ(−v)⋅Φ(v)⋅z \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′′x'', y′′y'' and z′′z'' are independent of time. Thus, ksk_s and KK are at rest relative to each other. Because the origins were set to coincide at t=0t = 0, this must remain the case. As such, it must hold that (x′′,y′′,z′′)=(x,y,z)(x'', y'', z'') = (x, y, z) and thus Φ(−v)⋅Φ(v)=1\Phi(-v) \cdot \Phi(v) = 1.

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

So, we have:

Φ(−v)Φ(v)=1Φ(−v)=Φ(v)Φ(−v)=Φ(v)⋅1=Φ(v)Φ(−v)Φ(v)1=Φ(v)Φ(v)1=Φ(v)2Φ(v)=1=1 \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 Φ(v)=1\Phi(v) = 1, we just need a little bit more algebra to arrive at the Lorentz Factor:

β(v)=cc2−v2=1c−1c2−v2=1c−2(c2−v2)=11−v2/c2=γ \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}