This approach is seeming intuitive to me as I can visualize what's going on at each step and there's not much complex math. But I'm not sure if I'm on the right track or if I'm making some mistakes. Here it is:

$A$ has set up a space-time co-ordinate system with some arbitrary event along his world-line as the origin. He assigns $(t,x)$ as the co-ordinates of the events around him. $A$ observes $B$ to be traveling at velocity $+v$. $B$ passes $A$ at the origin in $A's$ co-ordinates.

We need to find $(t',x')$ co-ordinates from $B's$ point of view, assuming he also sets up the same event as the origin as $A$ does (the event lies on both their worldlines)

Since all inertial frames are equivalent, $B$ must observe $A$ as moving with $-v$ speed.

If $A'$ worldline has co-ordinates $(t, 0)$ in $A's$ view, then the same worldline should be $(t,-vt)$ in $B's$ view, assuming absolute time (we keep the $t$ co-ordinate unchanged)

If we drop absolute time as a requirement, then $(t,0)$ from $A's$ frame can transform to $(\gamma t, -\gamma v t)$ in $B's$ frame. This is so the speed of $-v$ is still preserved. $\gamma$ is the stretching/squeezing factor and it should only depend on $v$ (because time is homogenous, so the stretching should be by a constant factor. The stretching factor can't depend on $t$ because the absolute $t$ values depend on the arbitrarily chosen origin).

Now we know that $(t,0)$ from $A's$ view transforms to $(\gamma t, -\gamma vt)$ in $B's$ view. By symmetry, $(t,0)$ from $B's$ view transforms to $(\gamma t, \gamma vt)$ in $A's$ view.

So the transformation from $A's$ frame to $B's$ frame transforms points of the form $(\gamma t, \gamma vt)$ to $(t,0)$, and points of the form $(t,0)$ to $(\gamma t, -\gamma vt)$.

Now we look at an object $C$ at rest relative to $A$. Its worldline is a verticle line in $A's$ frame parallel to the $t$ axis. Assuming the distance between $A$ and $C$ is $d$ in $A's$ frame, $B$ passes $C$ at the co-ordinate $(\frac{d}{v}, d)$ in $A's$ frame. Since this point lies on $B's$ worldline, it transforms to $(\frac{d}{\gamma v}, 0)$ in $B's$ frame.

This is where the intersection point of $B's$ and $C's$ worldlines gets transformed to. About the rest of the points on $C'$ worldline, if we shift the origin of $B$ to be the intersection point $(\frac{d}{v}, d$), the situation of transforming $C's$ worldline to $B's$ frame is identical to the one where where we transformed $A's$ worldline to $B'$ frame (as $C$ is also moving at $-v$ wrt $B$. It's just that the intersection point of the worldlines of $C$ and $B$ is different is different from that of $A$ and $B$)

So after transformation, the intersection point transforms to $(\frac{d}{\gamma v},0)$, and the rest of the points on $C's$ worldline transform to the line having the slope $-v$, containing the point $(\frac{d}{\gamma v},0)$, and having the same stretching of $\gamma$ from that point (as $C$ is also moving at $-v$ wrt $B$, it should have the same stretching as $A$)

So now we have the method to transform all the vertical lines (and hence every point) in $A's$ frame to $B's$ frame. Depending on the $\gamma (v)$ function, the transformation should be unique.

I don't know how to get the $\gamma (v)$ function, but is the rest of my above thinking correct or are there any holes in this? I think $\gamma(v)<1$ should correspond to rotations, $=1$ to Gallilean transformation and $>1$ to Lorentz transformation.

Update - I just tried to calculate using my method in the post with $\gamma (v)=\frac{1}{\sqrt{1-v^2}}$. It does give the same values as Lorentz transformation

I tried transforming $(x,t)=(5,2)$, a random point, with $v=0.5$.

First, we calculate where the worldline $(0.5t,t)$ meets the line $(5,t)$. This intersection point is $(5,10)$ and it will map to $(0,\frac{10}{\gamma})$ after the transformation.

For now, I will shift this point to the origin (and shift the line $(5,t)$ to the $t$ axis) to perform stretching of $\gamma$ on the shifted version of the line.

The point $(5,2)$ lands at the point $(5-5,2-10)=(0,-8)$ after the shift. Now we change the velocity (slope) of this shifted line to $-0.5$ to change to other frame's perspective. Now the point $(0,-8)$ lands at the point $(4,-8)$.

Now we stretch this line by $\gamma$. The point $(4,-8)$ now lands at $(4.62,-9.23)$

Now we finally shift the origin to $(0,\frac{10}{\gamma}=(0,8.66)$. The point $(4.62,-9.23)$ lands at $(4.62,-0.57)$

If we use the Lorentz transform formula on the point $(2,5)$, using $c=1$, $v=0.5$, we also get $(4.62,-0.57)$

I think deriving $\gamma (v)=\frac{1}{\sqrt{1-v^2}}$ should just be a matter of adding the requirement $T_{-v} (T_{v} (x,y))=(x,y)$, to the method in the post, right? This is equivalent to saying that we recover back the original point after we switch back to the original frame.

The end result of this derivation is the same as Lorentz transformations. Does anyone think there are unjustified steps in the derivation?

Update- I tried to derive the expression for $\gamma (v)$, but it involves the use of a faster than light inertial frame.

Using the method in the post, we first derive the transformation of points of the form $(x,0)$ (points on the horizontal axis of a frame) in terms of $\gamma (v)$.

It is : $x'=\gamma x$, $t'=\frac{x}{v}(\frac{1}{\gamma}-\gamma)$

It' s slope is equal to $\frac{x'}{t'}=\frac{\gamma ^2 v}{1-\gamma ^2}$

The worldline of $B$ (the frame we're transforming to) has a slope $v$ in $A's$ frame. If we consider a frame $C$ whose time axis is the same as $A's$ space axis and whose space axis is the same as $A's$ time axis, then $B'$ worldline has slope $\frac{1}{v}$ in $C's$ frame ($C$ is the faster than light inertial frame here).

Now principle of relativity implies that $C's$ worldline (which is the space axis of $A$) seen in $B's$ frame also has slope of absolute value $\frac{1}{v}$. The sign of the slope can change as sign only refers to direction.

So we have $\frac{\gamma ^2 v}{1-\gamma ^2}= \frac{1}{v}$ or $=\frac{-1}{v}$

This gives $\gamma=\frac{1}{\sqrt{1+v^2}}$ or $\gamma=\frac{1}{\sqrt{1-v^2}}$. We accept the latter formula based on experimental evidence.

Does the above work? It involves use of faster than light inertial frames, but I don't think special relativity rules out the existence of those frames as I've read that they're still speculated. Special relativity only rules out slower than light objects getting accelerated to light speed.

  • 1
    $\begingroup$ The Lorentz transformation derived by Einstein in the frame of Special Relativity is based on two principles : (1) The principle of relativity and (2) the principle of the invariant speed of light. I think that in your remarkable effort you use (1) but I don't see any use of (2). May be this is the reason you could not determine the $\gamma$-factor. $\endgroup$
    – Frobenius
    Commented Jul 10, 2020 at 13:37
  • 1
    $\begingroup$ I would strongly recommend this article: iopscience.iop.org/article/10.1088/1367-2630/ab76f7. It goes through a derivation very parallel to your own near the beginning. (Although the bulk of the paper is arguing that the root(1+v^2) formula should not in fact be thrown out). $\endgroup$
    – Dast
    Commented Jul 14, 2020 at 11:42
  • $\begingroup$ @Dast Thanks. I'd go through it $\endgroup$
    – Ryder Rude
    Commented Jul 14, 2020 at 11:47

3 Answers 3


Your derivation, up to before the point where you mention $C$, is the standard derivation. Once you have assumed the linearity of the transform, i.e. that $B$ has position ($\gamma t, \pm\gamma v t$), the standard derivation continues with the observation that the "length" (it's not positive semi-definite in Minkowski space) of a four-vector $x^\mu \leftrightarrow (t, x)$, defined (setting $c=1$) as $$s^2 = t^2 - x^2,$$ is invariant under Lorentz transformation.

The proof that $\gamma^{-2} = 1 - v^2$ follows by calculating $s^2$ in the frame of $A$ and $B$ and equating these expressions. (The "other" sign, $\tilde\gamma^{-2} = 1+v^2$ does not preserve the four-length above.)

Invariance of this length is equivalent to the statement that $c=1$ for all observers. (An excellent, concise discussion is given throughout Ch. 2 of Rindler's Essential Relativity; see p.31ff, Sec. 2.6 of the Revised Second Edition, in particular.)

As commented by Frobenius, you have not used this fact anywhere in your derivation.

Your mention of 'faster than light inertial frames' is outside of the purview of Einstein's theoretical framework. In particular, you will violate the order of events for those at causal separation. Causality is a nice thing to maintain.

  • 1
    $\begingroup$ Actually, the requirement that the speed of light be the same for all observers may be satisfied even if the interval $s^2=t^2-x^2$ is not conserved. The only requirement is that $s^2=0$ be an invariant statement. There is a larger class of transformations that satisfy this requirement. $\endgroup$ Commented Jul 14, 2020 at 16:49
  • $\begingroup$ Nothing in the standard derivation that we're discussing requires the conformal group, to which you refer. We might stipulate, to remove scale transformations from consideration, that we're dealing with a set of theories that include massive particles. $\endgroup$
    – MarkWayne
    Commented Jul 14, 2020 at 17:16
  • 2
    $\begingroup$ I was just concerned because you say that the invariance of length is equivalent to the statement that $c=1$ for all observers. As I understand it (and I may be wrong!), the statement that $c=1$ for all observers only becomes equivalent to the invariance of $s^2$ when we add the additional requirement that constant velocities remain constant in all reference frames. $\endgroup$ Commented Jul 14, 2020 at 17:21

Your reasoning is correct, but it seems to me that introducing the changing of $B's$ origin was kind or arbitrary. Usually, when we talk about Lorentz transformations, we do not talk about looking at individual lines from different frames in reference. Instead, a much easier way would be to actually make a change in the whole co-ordinate system itself.

When you are talking about normal situations, when we consider only two co-moving people, the normal Lorentz transformations hold, without actually needing spacetime diagrams. These are $$t' = \frac{t - vx}{\sqrt{1- v^2}}$$ $$ x' = \frac{x - vt}{\sqrt{1 - v^2}}$$ assuming we choose units with $c = 1$. These are the Lorentz transformations that you reasoned with only $A$ nd $b$ in account.

Now, when we rather talk about more objects, like in the situation where one frame of reference has both $A$ and $C$, we prefer to use spacetime diagrams. But, then it is becomes less convenient to do the Lorentz transformations on each and every worldline and measure values. So, we do a trick (which I will briefly describe here, and leave the nitty-gritties to you):

  1. Take any one object, for example $A$, and look how its worldline transforms in $B's$ frame. In your case, that was the first part of your derivation.
  2. Using the results you obtain create a matrix. By now, its seems apparent where this is going.
  3. Take the matrix you obtained and then apply it to the entire $A$ frame of reference (the co-ordinate system where $A$ and $C$ are located). This is standard linear algebra.

The new co-ordinate system that you get will contain every worldline ($A$ and $C$ in this case), from the reference frame of $B$. Yes, your reasoning was correct and works, but is not very convenient for frames with many objects.

Update: I have been asked in the comments to touch upon the second update in the question. It is a way of deriving the $\gamma(v)$ function. But here is why that mathematically yields results but is not that correct.

  1. Anything involving faster than light travel is generally avoided due to the fact that special relativity tells us that nothing can travel faster than light. So, even in derivations and mathematics, it is preferable to avoid faster than light situation or reference frames.
  2. Time is not interchangeable. While spacetime diagrams are a very powerful way to model relativistic situations, they seem to imply that time and space can switch places by rotations. Physically, this is not possible: sure, you can make changes to the time and space axes, but you cannot switch them. (In very high gravity situations, like inside black holes, time and space can switch but then we need to take into account general relativity, which in this context is out of scope.)

So, though you get results, that is not actually the correct way to do it. Instead here are some clues to what you can do.

  1. Leverage symmetry: symmetry is a very important concept in physics, and in deriving the $\gamma(v)$ there are two main symmetries which come into play. I won't just straightforwardly reveal them, instead for now, I will just say that they have to do with directions of velocities and the fact that all observers consider themselves at rest.
  2. Substitute: When I learned the derivation of $\gamma(v)$, it used the fact that Lorentz transformations work both ways: $A$ to $B$ or $B$ to $A$. So you substitute the values of one into another and work your way through it.
  3. Work your way through it: You inevitable have to do some very length but easy algebra. So instead of trying to find shortcuts, just go through it the ugly way, and the results you get will be simple and elegant.

If these sounded vague, it is because I don't want to give away the derivation steps completely. Rather, it is meant as an exercise to the reader to try and do this task.

  • $\begingroup$ Though the reasoning can also be converted into a formula in terms of $\gamma (v)$. The reasoning is only for derivation, it's not for practical calculations. Though what about the last update which derives $\gamma (v)$, but by using the existence of faster than light inertial frames? $\endgroup$
    – Ryder Rude
    Commented Jul 15, 2020 at 9:13
  • $\begingroup$ The matrix is also just the same formula in a different notation. $\endgroup$
    – Ryder Rude
    Commented Jul 15, 2020 at 9:19
  • $\begingroup$ @Ryder Rude, yes you are correct, but the matrix formulation is more convenient, and sometimes convenience always helps :) For the second update, see my updated answer. $\endgroup$
    – PNS
    Commented Jul 15, 2020 at 10:32

Deriving the proper Lorentz transformation is a little complicated. It involves assumptions about the allowed form of the transformation law and the requirement that the speed of light be constant.

One assumption is that the transformation may only be a function of the relative velocity of the two frames (and not, for example, which point in space is being considered). Another assumption is that there is no length contraction along directions perpendicular to the direction of the boost. With these assumptions, the Lorentz transformation (for a boost along the $x$-axis) must be of the form $$t'=\Lambda_{tt}(v)t+\Lambda_{tx}(v)x$$ $$x'=\Lambda_{xt}(v)t+\Lambda_{xx}(v)x$$ $$y'=y$$ $$z'=z,$$ where the $\Lambda$s are, for now, arbitrary functions of $v$. For simplicity, I will work only in two spacial dimensions ($x$ and $y$). I'm not sure whether it's possible to show that the Lorentz transformation is the only one that works in only one spacial direction.

Now we impose the requirement that the speed of light be constant. Suppose that there is a particle travelling at the speed of light in one reference frame. For simplicity, let's have it pass through the origin. Then for any point $(t,x,y)$ along the particle's world line, we must have $(\sqrt{x^2+y^2})/t=c$, or equivalently $$x^2+y^2-c^2t^2=0.$$ In the boosted reference frame, we must also have $$0=x'^2+y'^2-c^2t'^2=(\Lambda_{xt}t+\Lambda_{xx}x)^2+y^2-c^2(\Lambda_{tt}t+\Lambda_{tx}x)^2$$ $$=(\Lambda_{xt}^2-c^2\Lambda_{tt})t^2+(\Lambda_{xx}^2-c^2\Lambda_{tx}^2)x^2+y^2+2(\Lambda_{xt}\Lambda_{xx}-c^2\Lambda_{tt}\Lambda_{tx})tx.$$ Now we can take advantage of the fact that this must be true for any light-like path. For example, this must be true for $tx>0$ (a particle travelling forward) and for $tx<0$ (a particle travelling backward). Therefore, the cross-terms must independently cancel: $$\text{1) }\Lambda_{xt}\Lambda_{xx}-c^2\Lambda_{tt}\Lambda_{tx}=0.$$ Now if we throw out the cross-terms and use $y^2=c^2t^2-x^2$ to eliminate $y^2$, we end up with $$0=(\Lambda_{xt}^2-c^2\Lambda_{tt}+c^2)t^2+(\Lambda_{xx}^2-c^2\Lambda_{tx}^2-1)x^2.$$ This must be true for any $t^2$ and $x^2$ (for any values of these two variables, we can find $y^2$ such that $x^2+y^2-c^2t^2=0$ is satisfied). Therefore, we must have $$\text{2) }\Lambda_{xt}^2-c^2\Lambda_{tt}^2+c^2=0,\text{ and}$$ $$\text{3) }\Lambda_{xx}^2-c^2\Lambda_{tx}^2-1=0.$$ As a side note, you can easily check that these requirements are sufficient to ensure that the invariant interval will be preserved: $$x^2+y^2-c^2t^2=x'^2+y'^2-c^2t'^2.$$ This came from both the requirement that the speed of light is constant and the assumption we made in writing down the general transformation law.

Now we have three equations for four unknowns. We can get an additional equation by considering the point $(t,0,0)$. In the boosted frame, this point should transform to $(\Lambda_{tt}(v)t,\Lambda_{xt}(v)t,0)$. This point should be moving at speed $-v$ after a boost by $v$. This might be considered the definition of a boost of velocity $v$. This gives, $$\text{4) }\frac{\Lambda_{xt}(v)}{\Lambda_{tt}(v)}=-v.$$

Now it's just a matter of solving the four equations. Let's define $\gamma=\Lambda_{tt}(v)$. Then from equation 4), we get $\Lambda_{xt}(v)=-\gamma v$. We can combine this with equation 2) to get $\gamma$: $$\gamma^2v^2-c^2\gamma^2+c^2=0$$ $$\implies \gamma=\frac{c}{\sqrt{c^2-v^2}}=\frac{1}{\sqrt{1-v^2/c^2}}.$$ Meanwhile, from equation 1), we get $$-\gamma v\Lambda_{xx}-c^2\gamma\Lambda_{tx}=0$$ $$\implies \Lambda_{tx}=-v\Lambda_{xx}/c^2.$$ We can combine this with equation 3) to get $$\Lambda_{xx}^2-v^2\Lambda_{xx}^2/c^2=1\implies \Lambda_{xx}=\frac{1}{\sqrt{1-v^2/c^2}}=\gamma.$$ Then $\Lambda_{tx}=-v\Lambda_{xx}/c^2=-\gamma v/c^2$. All together, this gives the expected form of the Lorentz transformation: $$t'=\gamma\left(t-\frac{vx}{c^2}\right)$$ $$x'=\gamma(x-vt)$$ $$y'=y$$ $$z'=z$$

  • $\begingroup$ $B$ is moving at $v$ wrt $A$. $A's$ worldline $(0,t)$ (in $A's$ frame) maps to $(-\gamma vt, \gamma t)$ in $B's$ frame. By symmetry, $B's$ worldline $(0,t)$ is $(\gamma vt, \gamma t)$ in $A's$ frame. So transformation from $A's$ frame to $B's$ frame takes points of the form $(\gamma vt, \gamma t)$ to $(0,t)$. Hence it takes $(d, \frac{d}{v})$ to $(0,\frac{d}{\gamma v})$ $\endgroup$
    – Ryder Rude
    Commented Jul 15, 2020 at 9:36
  • $\begingroup$ Oh, I see. Thanks for the clarification. Looking at it more closely, I see that $\gamma(\frac{d}{v}-\frac{dv}{c^2})=\frac{1}{\sqrt{1-v^2/c^2}}\frac{d-dv^2/c^2}{v}=\frac{d}{\gamma v}$, so your result is equivalent and correct. I'll update my post. $\endgroup$ Commented Jul 15, 2020 at 14:41

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