The velocity of light in vacuum is the same for all inertial observers. This means $$\frac{dx}{dt}=\frac{dx'}{dt'}=c\\ \Rightarrow {(c\,dt)}^2-dx^2=0={(c\,dt')}^2-{dx'}^2.$$ I think that without further work, it is not obvious that for arbitrary nonzero values of ${(c\,dt)}^2-dx^2$, the equality $${(c\,dt)}^2-dx^2={(c\,dt')}^2-{dx'}^2$$ will hold true.

Given the postulate of the constancy of the speed of light, if we were to find how $(t,x)$ must transform, we can make use of the first equality only - a weaker condition than the second equality. But it is usually derived using the stronger condition (second equality) which assumes the equality for arbitrary values of ${(c\,dt)}^2-dx^2$.

If we strictly have to follow the postulate, IMO, we must derive the Lorentz transformation equations in two steps as follows.

Step 1. First, we assume $$c\,t'=A\,c\,t+B\,x,\quad x'=K\,c\,t+D\,x.$$

Step 2 Then make use of two conditions ${(c\,dt')}^2-{dx'}^2=0$ and ${(c\,dt)}^2-dx^2=0$.

We start with $${(c\,dt')}^2-{dx'}^2=0\\ \Rightarrow (A^2-K^2)\,c^2dt^2 + (B^2-D^2)\,dx^2 + 2(AB-KD)\,c\,dt\,dx=0$$ Now the only condition that we can use is $c\,dt=\pm dx$, which is insufficient to find all the four unknown constants.

Does this mean that one cannot derive the Lorentz transformation equations from the constancy of the speed of light only?

  • $\begingroup$ With $c>0$ and $dt>0$ you would have $cdt>0$, so your condition can only be $cdt=dx$ . A variable can not be positive and negative at the same time. $\endgroup$
    – Thomas
    May 8, 2021 at 15:05
  • $\begingroup$ $dx/dt=\pm c$ implies light moving along $\pm x$ axis. Think about the light cone diagram. In fact, the first postulate implies $dx/dt=\pm c=dx'/dt'$. $\endgroup$ May 8, 2021 at 15:10
  • $\begingroup$ You would then have to write $x_1=ct$ and $x_2=-ct$ . You can not assume $x=ct$ and $x=-ct$ at the same time. You might as well be saying that $1=-1$ $\endgroup$
    – Thomas
    May 8, 2021 at 15:27
  • $\begingroup$ I am not assuming that. I will use $dx=cdt$ and $dx=-cdt$, as two separate conditions. That's obvious. In any case, it doesn't solve the problem at hand. $\endgroup$ May 8, 2021 at 15:35

2 Answers 2


Yes, the constancy of the speed of light is not sufficient. You need additional assumptions. These notes by Victor Yakovenko provide a derivation of the general coordinate transformation

$$\pmatrix{x'\\t'}=\frac{1}{\sqrt{1+v^2/a}} \pmatrix{1 & -v\\v/a & 1}\pmatrix{x\\t}$$

where $a$ is some parameter with dimensions of velocity squared. This derivation makes the following assumptions:

  1. The coordinate transformation should be linear (you already assumed this in step 1)
  2. Space is isotropic, so e.g. the length of a moving ruler is the same if it's moving to the left as if it were moving to the right
  3. The composition of two transformations is another transformation
  4. Transformations depend only on the relative velocity between frames

This yields 3 viable possibilities. Either $a>0$, $a<0$, or $a\rightarrow \infty$ (the latter case results in the Galilean transformations). However, if we additionally demand that there exists an invariant speed $c$ such that objects moving with speed $c$ in one frame are moving with the same speed in every other frame, then the only possibility is that $a = -c^2 < 0$.

There are many routes to the Lorentz transformations which make different assumptions, but the point of my answer is that the assumption of the constancy of the speed of light is not sufficient all by itself. There must be other (reasonable) physically-motivated assumptions about the structure and symmetries of spacetime to go along with it.

  • 1
    $\begingroup$ $a$ should be $-c^2$ not $-1/c^2$ $\endgroup$
    – Thomas
    May 8, 2021 at 16:18
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    $\begingroup$ It's fun to see this since Yakovenko taught my c. mech course in grad school and so I saw him give precisely this derivation. $\endgroup$
    – zeldredge
    May 8, 2021 at 16:53
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    $\begingroup$ @Thomas It's not clear what you mean by "there isn't really any way to bring in the speed of light in a natural way". Constancy of the speed of light is an independent assumption. The purpose of the derivation starting from the points 1. 2. 3. 4 is to demonstrate that it is possible to defer the assumption of the constancy of the speed of light to the very last step. That demonstrates the independency. It appears you qualify the choice between 3 possibilities as 'bringing in the speed of light in an unnatural way', but it's not clear how that should constitute something "unnatural". $\endgroup$
    – Cleonis
    May 8, 2021 at 17:27
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    $\begingroup$ @Thomas I am aware that massive objects move at different speeds in different frames. What I said was that if you want there to be an invariant speed, then that fixes $a$ to be minus the square of that speed. Empirically, we have observed that the universe we occupy does have such a speed - the speed of light - which is the final piece required in the derivation of the Lorentz transformation formula. $\endgroup$
    – J. Murray
    May 8, 2021 at 18:00
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    $\begingroup$ @Thomas My answer takes for granted that you know what an inertial reference frame is and what the corresponding coordinates specify, since that is the context in which the question was asked. Your questions are not unimportant, but the answers to them are prerequisites for this one. $\endgroup$
    – J. Murray
    May 8, 2021 at 18:28

I have seen the Lorentz transformation derived in a very similar way you approached it:

Let me first rewrite your final equation without the differentials (there is little point in using these if the speed of light is assumed as constant)

$$(1)\:\:\:(A^2-K^2)c^2t^2 +(B^2-D^2)x^2 +2(AB-KD)ctx =0$$

If you subtract furthermore the equation $c^2t^2-x^2=0$ from this you get

$$(2)\:\:\:(A^2-K^2-1)c^2t^2 +(B^2-D^2+1)x^2 +2(AB-KD)ctx =0$$

If that is supposed to be true for all x at given t, each of the coefficients to the different powers of x have to be zero separately, so

$$(3)\:\:\:A^2-K^2-1=0$$ $$(4)\:\:\:B^2-D^2+1=0$$ $$(5)\:\:\:AB-KD=0$$

Now additionally you obviously have to use the constraint that the primed and unprimed frames are moving relatively to each other (after all, that is why we are doing this in the first place). So the additional conditions we have are $$(6)\:\:\:x'=0 => x=vt$$ $$(7)\:\:\:x=0 => x'=-vt$$

Inserting this into your transformation $$(8)\:\:\:x'=Dx+Kct$$ $$(9)\:\:\:ct'=Act+Bx$$

yields then $$(10)\:\:\:K=-\frac{v}{c}D$$ $$(11)\:\:\:A=D$$ $$(12)\:\:\:B=K$$

Inserting this into (3) and (5) yields

$$(13)\:\:\:D^2(1-\frac{v^2}{c^2})-1=0 => D=\sqrt{\frac{1}{1-\frac{v^2}{c^2}}}$$ $$(14)\:\:\:B=-\frac{v}{c}\sqrt{\frac{1}{1-\frac{v^2}{c^2}}}$$.

However, there is in my opinion a problem with this derivation:

The quadratic equations that lead to Eq.(1) $$(15)\:\:\: x^2=c^2t^2 ;\:\:\: x'^2=c^2t'^2$$ have the solutions $$(16)\:\:\: x_1=ct ;\:\:\: x_1'=ct'$$ $$(17)\:\:\: x_2=-ct ;\:\:\: x_2'=-ct'$$


$$(18)\:\:\: x_2= -x_1$$ $$(19)\:\:\: x_2'= -x_1'$$

We can rewrite the transformation (8) for the two solutions as

$$(20)\:\:\:x_1'=Dx_1+Kct$$ $$(21)\:\:\:x_2'=Dx_2+Kct$$

However, by inserting (18),(19) into (21) we get

$$(22)\:\:\:-x_1'=-Dx_1+Kct$$ i.e. $$(23)\:\:\:x_1'=Dx_1-Kct$$

which is inconsistent with (20) unless $K=0$ i.e. unless $v=0$, which doesn't make any sense.

  • $\begingroup$ The original equations which led to (1) include $ct' = Act + Bx$ and $x' = Kct + Dx$, which are not invariant under $(x,x')\rightarrow(-x,-x')$. $\endgroup$
    – J. Murray
    May 10, 2021 at 19:18
  • $\begingroup$ @J.Murray You are trying to find a linear transformation that is consistent with $x^2=c^2t^2$, $x'^2=c^2t'^2$ The latter equations are invariant under $(x,x′)→(−x,−x′)$ $\endgroup$
    – Thomas
    May 10, 2021 at 19:59
  • $\begingroup$ Right ... but why would you expect your linear transformation to be? $x^2=4$ is invariant under $x\rightarrow -x$, but $x=2$ obviously isn't, nor is $x=-2$. $\endgroup$
    – J. Murray
    May 10, 2021 at 20:06
  • $\begingroup$ @J.Murray You missed out the crucial part: from $x'^2=4$ you have then also the two solutions $x_1'=2$ and $x_2'=-2$, that is the primed variable changes sign when the unprimed changes sign $\endgroup$
    – Thomas
    May 10, 2021 at 21:41
  • $\begingroup$ I think perhaps you are missing the crucial part. $x' = \gamma(x-vt)$ and $-x' = \gamma (-x -vt) \iff x' = \gamma(x+vt)$ are both perfectly valid boosts. They're not the same boost, just as $x=2$ and $x=-2$ are not the same solutions to $x^2=4$. $\endgroup$
    – J. Murray
    May 10, 2021 at 21:46

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