I read the light-clock example in my book which proved the time dilation formula by assuming that the speed of light is constant for all observers. But I've trouble in understanding it the other way around. Lorentz transformation is just a correction to Newtonian mechanics to account for the constant speed of light for all observers, right? I have trouble understanding how does applying this correction preserve the speed of light for all observers.

Can we start by assuming that the Lorentz transformation formulas are true and then prove that two observers $A$ and $B$ will see a light pulse moving at the same speed $c$ regardless of their relative velocity with respect to each other?

  • 4
    $\begingroup$ The relativistic velocity addition formula, derivable from the Lorentz transformation, is unambiguous; an object with speed $c$ in one inertial reference frame (IFR) has speed $c$ in all IFRs. $\endgroup$ Commented May 23, 2017 at 11:49
  • 2
    $\begingroup$ The Lorentz transformations were derived on the premise that the speed of light is a constant. You can then assume the transformations are true and show the speed of light is constant from that. It works to find constant $c$ because we assumed that in order to make the transformations. It's like I give you the equation $xy=6$ and tell you we measured $x=2$ so therefore $y=3$. If you ask "can we prove $x=2$ by first assuming $y=3$?" Yes, but that's trivial. We only have $y=3$ because we used $x=2$. If we don't trust $x=2$ then why are we starting with $y=3$? Why not $y=4$ or $y=10.568$? $\endgroup$
    – Jim
    Commented May 23, 2017 at 12:22
  • $\begingroup$ You say, "the Lorentz transformation is just a correction to Newtonian mechanics". If you understand exactly what mechanics the Lorentz transformation equations actually represent, it becomes obvious to you as to why the measuring of the speed of light by any observer is always producing the same result. $\endgroup$
    – Sean
    Commented May 23, 2017 at 17:48
  • $\begingroup$ Bloody hell I'm impressed by the amount of answers that do not mention that the Lorentz transformations assume the constant speed of light (and they are all being upvoted to oblivion). $\endgroup$
    – gented
    Commented Jun 4, 2019 at 14:39

6 Answers 6


From $ds^2=c^2dt^2-dx^2$, we see that light-speed travel is equivalent to $ds^2=0$. But $ds^2=\eta_{\mu\nu}x^\mu x^\nu$ is manifestly Lorentz-invariant, so if $ds^2=0$ holds in some reference frame it also does in others obtained by arbitrary Lorentz transformations.

  • $\begingroup$ Simplest and easiest explanation. Thank you. $\endgroup$ Commented Jun 8, 2019 at 11:57

How do you "prove" that 5-3=2? Do the "check your work" operation: final result taken with the reverse operation gets you to the starting point-- 2+3=5.$\checkmark$

The same exercise is done with the Lorentz transformation as a pedagogical tool. If the constancy of the speed of light for all observers leads to the Lorentz transformation, then the Lorentz transformation on a speed of light object should yield a constant speed. And it does. It doesn't prove that the speed of light is constant. It simply shows that the transformation is consistent with the starting axiom.

By the way, "check your work" is an important part of problem solving whether analyzing projectile motion or modeling cosmological expansion: are my solutions consistent with my starting conditions.


An enlightening (but possibly advanced) method to prove the constancy of the speed of light from the Lorentz boost transformation is to find the eigenvectors of the Lorentz boost. Two of the eigenvectors are along the light cone. The corresponding eigenvalues are equal to the doppler factor and its reciprocal. (These eigenvectors are coplanar with the 4-velocities of observers in relative motion.)

This can be compared and contrasted with the eigenvectors and eigenvalues of the Galilean transformation.

In both transformations, there are no timelike eigenvectors... that is, no preferred observers.

Now, for some details:

Given $M=\begin{pmatrix} \gamma&\beta\gamma\\ \beta\gamma&\gamma\\ \end{pmatrix}$, we set up the eigenvalue problem: $$0=\det (M-kI)=\det\begin{pmatrix} \gamma-k&\beta\gamma\\ \beta\gamma&\gamma-k\\ \end{pmatrix}=(\gamma-k)^2-(\beta\gamma)^2.$$ Solving this characteristic equation for $k$, we find $k-\gamma=\pm\beta\gamma$, which can be written as $k=\gamma(1\pm\beta)=\sqrt{\frac{1\pm\beta}{1\mp\beta}}$, which are the Doppler factors.

The eigenvector corresponding to $k=\gamma(1+\beta)$ is gotten by substitution: $$\begin{pmatrix} 0 \\ 0\end{pmatrix}= \begin{pmatrix} \gamma-(\gamma(1+\beta))&\beta\gamma\\ \beta\gamma&\gamma-(\gamma(1+\beta))\\ \end{pmatrix} \begin{pmatrix} w_t \\ w_x\end{pmatrix} = \begin{pmatrix} -\beta\gamma & \beta\gamma \\\beta\gamma & -\beta\gamma \end{pmatrix}\begin{pmatrix} w_t \\ w_x\end{pmatrix}=\begin{pmatrix} (-\beta\gamma)w_t + (\beta\gamma)w_x \\ (\beta\gamma)w_t + (-\beta\gamma)w_x\end{pmatrix}.$$

This is satisfied by vectors of the form $w_x=w_t$ –that is, the along the future-forward lightlike direction. Thus, under a Lorentz Transformation, the light-signal's velocity remains unchanged, but the future-forward component of a vector gets stretched by a factor of $k$. Similarly, the future-backward component gets reduced by a factor of $k$. (This is the basis of the Bondi k-calculus [pun intended] and methods using light-cone coordinates.)

  • $\begingroup$ Yes, this is indeed a very fun method. I'll have to take another look at Bondi calculus with this eigenvector idea in mind - it's never made much sense to me as a method for exposition (not that I've looked at it THAT hard), although I must say it's probably hard to judge a pedagogical method when one already has the concepts reasonably well sorted for oneself. $\endgroup$ Commented May 24, 2017 at 11:59

In short yes. You can try to solve it yourself. Take 2 observes A and B, moving at velocity v wrt each other. A sees a light pulse, traveling as x = ct (which means light's velocity as seen by A is dx/dt = c). Now use Lorentz transformation to find out coordinates of the pulse as seen by B. you'd see it comes to be c again.


We could use the relativistic velocity addition equation, which would show the speed of the light pulse to be independent of the relative motion between the two observers.

EDIT: Attached is a brief proof of the problem. Let an observer in frame S see an object in a reference frame moving at velocity V w.r.t. S emit a photon which travels at c. Then photon relative velocity w.r.t.S, U':


  • 2
    $\begingroup$ You could improve this answer by elaborating and actually showing the proof $\endgroup$
    – Jaywalker
    Commented May 23, 2017 at 12:29
  • $\begingroup$ Please use MathJax for mathematics and test entry for text - and you can mix the two easily. Putting a scribbled note up as an image is not considered a good way to do answer questions here. $\endgroup$ Commented May 24, 2017 at 11:15

You can use the following Lorentz Speed Transformation (derivation below) :

\begin{equation} \left|\mathbf{\dot{r'}}\right|^{2} = c^{2} - \frac{(c^{2} - |\mathbf{\dot{r}}|^2)}{\gamma^{2}(1 - v\dot{x}/c^{2})^{2}} \tag{1}\label{eq:speed-transform} \end{equation}

which applies to a light wave or a particle seen from two Inertial Frames of Reference (IFR's) in the 'standard setup' - ie an IFR $S'$ moving along the common $x/x'$ direction wrt an IFR $S$, at a velocity $v$ (where $v$ is +ve or -ve). The instantaneous velocity of the light wave/particle is $\mathbf{\dot{r}}$ in S, and $\mathbf{\dot{r'}}$ in $S'$. (Note the dot means differentiation wrt $t$ in S and differentiation wrt $t'$ in $S'$).

This formula proves the conservation of the speed $c$ for any direction of light travel, whereas with the Lorentz Velocity Transformation alone we can only prove it for directions of travel along the x, y, z axes : \begin{eqnarray} \dot{x'}& = & \frac{\dot{x} - v}{1 - v\dot{x}/c^{2}} \nonumber \\ \dot{y'} & = & \frac{\dot{y}}{\gamma(1 - v\dot{x}/c^{2})} \nonumber \\ \dot{z'} & = & \frac{\dot{z}}{\gamma(1 - v\dot{x}/c^{2})} \tag{2}\label{eq:velocity-transform} \end{eqnarray}

by putting $\mathbf{\dot{r}} = (\pm c, 0, 0), (0, \pm c, 0), (0, 0, \pm c)$ and then showing $\left|\mathbf{\dot{r'}}\right|^2 = c^{2}$.

Note the term $1 - v\dot{x}/c^{2}$ in the denominator of (\ref{eq:speed-transform}) and (\ref{eq:velocity-transform}) is always positive so long as we are dealing with the motion of a light wave or particle, since otherwise we would have : $1 \leq v\dot{x}/c^{2} \Rightarrow |v| |\dot{x}| / c^{2} \geq 1 \Rightarrow |\mathbf{\dot{r}}| \geq |\dot{x}| \geq (c/|v|) \cdot c > c$ (since $|v| < c$) - which is a contradiction, because $|\mathbf{\dot{r}}| \leq c$.

Although we initially derive the Lorentz Transformation (LT) on the basis that the speed of light is constant and rectilinear within all IFR's we should really back check that the LT formula we obtain possesses that property for ALL light transmissions - ie in all directions, since we only consider certain specific light directions in our original derivation. Moreover we should check that any speed less than $c$ maps to a speed less than $c$ under the LT.

(Various other properties of LT need to be checked also, eg that it reduces to the Galilean Transformation when $v \ll c$, and that 'causality reversal' cannot occur for 'time-like' or 'light-like' event intervals, ie the sign of $\Delta t$ cannot be flipped for such intervals - this latter property is proved by consideration of 'space-time intervals').

Plugging $|\mathbf{\dot{r}}| = c$ into the above formula (\ref{eq:speed-transform}) we obtain $|\mathbf{\dot{r'}}| = c$, and plugging in $|\mathbf{\dot{r}}| < c$ we obtain $|\mathbf{\dot{r'}}| < c$, as required.

Notice the rhs of the above formula (\ref{eq:speed-transform}) is a function of velocity $\mathbf{\dot{r}}$ (with $\dot{x} = \mathbf{\dot{r} \cdot i}$) and not speed and this correlates with the fact that the LT can map two velocities of the same speed to two velocities of different speeds - eg take velocities $(u, 0, 0)$ and $(-u, 0, 0)$ with $u \in (0, v)$ and $v > 0$, and doing the algebra we would find equal speed in $S'$ would require $u = c$, a contradiction. Thus speed in $S'$ cannot just be a function of speed in $S$. For example a uniform circular motion in the $xy$-plane in $S$ has constant speed but does not appear as a constant speed in $S'$ because on rhs of (\ref{eq:speed-transform}) $\dot{x}$ is variable.

The Lorentz Speed Transformation (\ref{eq:speed-transform}) can be derived from the Lorentz Velocity Transformation (\ref{eq:velocity-transform}) as follows. From (\ref{eq:velocity-transform}) we have :

\begin{eqnarray*} \left|\mathbf{\dot{r'}}\right|^{2}(1 - v\dot{x}/c^{2})^2 & = & (\dot{x} - v)^{2} + (\dot{y}/\gamma)^2 + (\dot{z}/\gamma)^2 \\ & = & |\mathbf{\dot{r}}|^2 + \left(\frac{1}{\gamma^{2}} - 1\right)\dot{y}^{2} + \left(\frac{1}{\gamma^{2}} - 1\right)\dot{z}^{2} - 2v\dot{x} + v^{2} \\ & = & |\mathbf{\dot{r}}|^2 - \frac{v^{2}}{c^{2}}(\dot{y}^{2} + \dot{z}^{2}) - 2v\dot{x} + v^{2}, \\ & & \mbox{(since} \hspace{0.5em} \frac{1}{\gamma^{2}} - 1 = -\frac{ v^{2} }{ c^{2} } \mbox{)} \\ & = & |\mathbf{\dot{r}}|^2 - \frac{v^{2}}{c^{2}}(|\mathbf{\dot{r}}|^2 - \dot{x}^{2}) - 2v\dot{x} + v^{2} \\ & = & \frac{1}{\gamma^{2}}|\mathbf{\dot{r}}|^2 + \frac{v^{2}}{c^{2}}\dot{x}^{2} - 2v\dot{x} + v^{2} \\ & = & \frac{1}{\gamma^{2}}(|\mathbf{\dot{r}}|^2 - c^{2}) + \frac{v^{2}}{c^{2}}\dot{x}^{2} - 2v\dot{x} + \left(v^{2} + \frac{c^{2}}{\gamma^{2}}\right) \\ & = & \frac{1}{\gamma^{2}}(|\mathbf{\dot{r}}|^2 - c^{2}) + \frac{v^{2}}{c^{2}}\dot{x}^{2} - 2v\dot{x} + c^{2}, \\ & & \mbox{(since} \hspace{0.5em} v^{2} + \frac{c^{2}}{\gamma^{2}} = v^{2} + c^{2}\left(1 - \frac{v^{2}}{c^{2}}\right) = c^{2} \mbox{)} \\ & = & \frac{1}{\gamma^{2}}(|\mathbf{\dot{r}}|^2 - c^{2}) + \left(\frac{v\dot{x}}{c} - c\right)^{2} \\ & = & \frac{|\mathbf{\dot{r}}|^2 - c^{2}}{\gamma^{2}} + c^{2}\left(1 - \frac{v\dot{x}}{c^{2}}\right)^{2} \\ \end{eqnarray*}

from which (\ref{eq:speed-transform}) now follows.


Not the answer you're looking for? Browse other questions tagged or ask your own question.