Beyond a Cursory Look at the Lorentz Factor (Leo Chow)
From the archive (2022). Corrected in October 2026; see the note at the end.
This is a guest post by Leo Chow, a student at the University of Hong Kong, on special relativity. It is reproduced here with his permission.
The Lorentz factor
Let’s take a look at the Lorentz factor:
\[\gamma_v=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\]The right-hand side is a recurring structure in the theory of relativity: it appears in formulae often enough for people to give it a symbol. At first sight the structure may seem quite unmotivated, apart from a vaguely Pythagorean feel from the squares and the square root, and a hint from the minus sign that it says something about shrinking.
This post is an attempt to add more intuition to the anatomy of \(\gamma\). I suggest that an interpretation focusing on velocities instead of lengths gives a more succinct and intuitive form for the Lorentz factor:
- Assumptions: the speed of light and transverse distances are the same in all inertial frames.
- Result: the time dilation factor is \(\dfrac{1}{\sqrt{1-v^2/c^2}}\).
The usual derivation
(This revisits the standard derivation of time dilation; feel free to skip it.)
Consider a photon bouncing between two mirrors separated by \(l\). For an inertial frame moving at speed \(v\) perpendicular to \(l\), express the period \(t\) of the light clock in terms of its period \(t_0\) in the clock’s rest frame.
\[\begin{gathered} ct=\sqrt{(2l)^2+(vt)^2} \\ (ct)^2-(vt)^2=(2l)^2 \\ t^2=\frac{(2l)^2}{c^2-v^2}=\frac{(ct_0)^2}{c^2-v^2} \\ t=\frac{ct_0}{\sqrt{c^2-v^2}}=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}} \end{gathered}\]This gives the time dilation factor \(\gamma\). In the first line, the left-hand side uses the constancy of the speed of light and the right-hand side uses the constancy of transverse distances.
With this, we can already see why the Lorentz factor is the reverse projection factor \(\sec\theta\). All the proof above says is that relative velocity adds to distance but not to the speed of light, so the motion takes longer. That distance, being perpendicular to \(v\), is scaled up by the reverse projection factor \(\sec\theta\). This factor can be expressed in terms of \(c\) and \(v\) by considering a similar triangle whose sides are velocities instead of lengths, and the Lorentz factor is simply that.
Noticing the similar triangle built from velocities, you might suspect there is an alternative derivation using velocities instead of distances:
\[\text{(length, reverse projection)} \rightarrow \text{(velocity, projection)}\]A tempting but incorrect derivation
To the observer, the clock moves with speed \(v\). Part of \(c\) is “used” to provide the horizontal velocity needed to keep up with the clock, so only the vertical component of \(c\) is used to traverse the distance \(2l\). The vertical component is \(c\) reduced by \(v\), i.e. \(\sqrt{c^2-v^2}\); more conceptually, it is “the projection of \(c\) onto \(l\)”, i.e. \(c\cos\theta\). We therefore obtain \(t\) much more elegantly:
\[t=\frac{2l}{\sqrt{c^2-v^2}}=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}\]This looks like a better derivation than the first, except that it is wrong. (If it were correct, why would essentially every textbook use the clumsier version?)
We resolved the velocity vertically, resolved the path length vertically, took their ratio to get a time, and then claimed that this time is the time for the original motion. Where did it go wrong?
To answer this we need to step back and ask: why can we resolve a motion into components? We certainly can resolve a motion, but what does the ratio of resolved quantities mean? It is the duration of an imaginary situation in which a photon moves along a length \(2l\) at speed \(\sqrt{c^2-v^2}\), but photons always travel at \(c\).
So why did we get the correct result?
By resolving the velocity vertically and claiming that the journey time is unchanged, we implicitly shifted into the light clock’s rest frame in a Galilean way. A Galilean transformation is the wrong way to change frames here, but because we applied it and then implicitly undid it, we are in effect still in the observer’s frame, which is why the dilated time comes out correctly.
Put differently, the result is correct because we made a forbidden move twice: first by reducing \(c\) (the spatial part of a Galilean transformation), and second by saying the durations are equal (the temporal part of a Galilean transformation).
The problem is not resolving components. The problem is concluding that the time obtained from the ratio of resolved quantities equals the time of the unresolved motion. Concluding that requires a Galilean transformation, which is forbidden. Note the characteristic \(t'=t\) here.
We run into a similar problem if we avoid the picture above and instead consider the time for the vertical position to reach its final value. (The difference is subtle but real: one attachment is made, which, as we will see, forces a detachment elsewhere.)
We may be tempted to say that the time for the vertical position to reach its final value is, of course, the time for the position vector to reach its final value, so the method is justified as long as we don’t interpret the ratio physically as above.
That much is true by definition: the event of the position vector \(\mathbf{p}\) reaching its final value is defined by all its components \(p_i\) reaching theirs. But then we face another problem. How did we conclude that the duration \(t_y\) of the vertically resolved motion is the ratio \(L_y/v_y\), where \(L=2l\), when all we actually know is the path length over the speed,
\[t=\frac{\sqrt{L_y^2+L_x^2}}{\sqrt{v_y^2+v_x^2}}\,?\]Having detached our “vertical time” \(t_y\) from the imaginary motion along \(L\), we need a Galilean transformation back into the original frame to conclude \(t_y=L_y/v_y\), which already feels circular, if not wrong:
\[t_y=t=\frac{\sqrt{L_y^2+L_x^2}}{\sqrt{v_y^2+v_x^2}}=\frac{\sqrt{L_y^2+L_x^2-(v_xt)^2}}{\sqrt{v_y^2+v_x^2-v_x^2}}=\frac{L_y}{v_y}\]The first step uses the temporal part of the transformation, \(t'=t\); the third step uses the spatial part, \(x'=x-v_xt\).
We took a forbidden step into an imaginary realm, calculated \(t\) there, and brought it back to the real world with another forbidden step. This is reminiscent of a recurring theme in mathematics: differentiation under the integral sign, the Laplace transform, or matrix diagonalisation (transform to the eigenbasis, apply the transformation easily, return to the original basis). They all have the shape \(Q^{-1}AQ\).
In fact, even people who refuse to make the forbidden move are implicitly making it when they rearrange the expression in the first method to collect \(t\) on one side. It is a matter of whether you are willing to read the algebra as physics rather than just numbers. Compare the two methods and you can see that they play with the same triangle:
The only difference is that in the second method the triangle comes from how a change of perspective subtracts from velocities, rather than how it adds to displacement. The sides are scaled down by \(t\), so \(t\) appears in only one side length, which makes the formula for the Lorentz factor immediately obvious.
By pretending that we could project the velocity onto the transverse direction, we immediately obtain the projection factor \(\gamma^{-1}=\frac{\sqrt{c^2-v^2}}{c}\): it is \(c\) reduced by \(v\), over the original \(c\).
Side note
\[\gamma_v=\left[1-\left(\frac{v}{c}\right)^2 \right]^{-\frac{1}{2}}\]Rewritten like this, it is almost just \(\frac{c}{v}\). It is as though we started from the ratio \(\frac{c}{v}\), uncoiled it at the layer \((\cdot)^{-2}\) into \(\left(\frac{v}{c}\right)^2\), and then wedged the shell \(s(x)=1-x\) into it to stop the layers collapsing back onto each other. It can act as a wedge because \(s\) does not commute with \((\cdot)^{-2}\).
I mention this because I plan to write about how a small change in the form of a differential equation relates to the resulting change in the form of its solution, a kind of “meta-differentiation”, and this way of reconstructing expressions will show up again. As a sneak peek, consider the solution of logistic growth,
\[P=\frac{K}{\left(\frac{K-P_0}{P_0}\right)e^{-rt}+1}.\]Motivated by how similar the coefficient of the exponential looks to the larger structure outside, we flatten the expression to
\[K\left(\left(\left(K^{-1}P_0\right)^{-1}-1\right)e^{-rt}+1\right)^{-1}.\]If the shell \(s(x)=x-1\) could commute with \(e^{rt}\) and with the \((\cdot)^{-1}\) shell, the whole structure would collapse back onto exponential growth. The extra factor \(\left(1-\frac{P}{K}\right)\) in the differential equation opens up the solution \(P_0e^{rt}\) at various layers (\(KK^{-1}\), \((\cdot)^{-1}(\cdot)^{-1}\), \(-1+1\)) and then swaps the order of operations in a way that cannot be undone. The result is a knot:
\[\begin{gathered} P_0e^{rt} \\ KK^{-1}P_0e^{rt} \\ K\left[\left(K^{-1}P_0e^{rt}\right)^{-1}\right]^{-1} \\ K\left[\left(K^{-1}P_0e^{rt}\right)^{-1}-1+1\right]^{-1} \\ K\left\{\left[\left(K^{-1}P_0\right)^{-1}-1\right]e^{-rt}+1\right\}^{-1}\\ P=\frac{K}{\left(KP_0^{-1}-1\right)e^{-rt}+1} \end{gathered}\]Changes on 25 October 2026
What was wrong.
- The proper time was written \(t_o\) in some places and \(t_0\) in others.
- One line of the first derivation had \(\left(v^2/c^2\right)^2\) under the square root instead of \(v^2/c^2\).
- The common shape of the methods in the last section was written \(Q \wedge Q^{-1}\). The wedge symbol means something else; the intended form is \(Q^{-1}AQ\).
What changed. These are fixed, the typesetting is cleaned up and the wording is tightened. Leo Chow’s argument itself is unchanged. Previously titled “Leo Chow - Relativity”.