I am trying to better understand the Lorentz Transformation on a fundamental level and gain some intuition of it. In the Lorentz Transformation, the derivative of x' with respect to x must be a nonzero constant. We know that it is nonzero because spatial coordinates are correlated between reference frames (that is to say, the Lorentz Transformation must be reversible, and all of space in one reference frame cannot all be in one place or have one coordinate through all of space as that makes no physical sense and is not reversible). We also know that it must be a constant because space is homogeneous (which is just an axiom of our universe) and because if it were dependent on x, it would depend on position and thus be different in different positions, so space would not be homogeneous. What other derivatives can we find just from fundamental features of our universe (i.e. before deriving the Lorentz Transformation through algebraic manipulation and the like. I want to understand the transformation more fundamentally and intuitively)? For example, can we find intuitively or from external fundamental features of our universe dx/dt in terms of coordinates? From these, what can else we tell about the Lorentz Transformation? What other derivatives, values, and relationships can we find from them? More specifically, I want to find as much information about the Lorentz Transformations as I can (primarily that dt'/dx /=0) by using only some fundamental, intuitive properties of our universe (such as the homogeneity of space, dx'/dx /= 0). Can we find some properties of the Lorentz Transformation as described earlier from just a few properties and values that are intuitive and simple in our universe? To be clear, I am talking about the relationships of coordinates between different frames of reference.

  • $\begingroup$ This nice video may help. $\endgroup$ Sep 28, 2020 at 14:23
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    $\begingroup$ @relayman357 Yeah, that was a good video. I already watched all of the videos in that series. I guess I am trying to gain a deeper understanding of the lorentz transformation than shown there (specifically in its derivation and such), and I am trying to do this through its derivatives. I also want to learn just how the derivatives of the Lorentz Transformation work. Thanks for linking it though, I appreciate it and maybe it will help others who see this question. $\endgroup$ Sep 28, 2020 at 17:44
  • $\begingroup$ It is very enlightening to study the Lorentz transformations from a more mathematical perspective. Learning about the Lorentz group, Minkowski space and its isometries is going to give you a much more concrete understanding. It is hard to appreciate the value of the Lorentz transformations if you are just working purely with coordinates and vector components. $\endgroup$
    – Charlie
    Oct 1, 2020 at 20:09
  • $\begingroup$ @Charlie, Alright, perhaps that will help me even more. Could you please give me some tips or ideas on what I should study to gain a better fundamental, intuitive sense of (and understanding of) the Lorentz transformations? It would also be appreciated if you could explain some of this to me (preferably in a more simple and intuitive way). $\endgroup$ Oct 5, 2020 at 14:33

2 Answers 2


To say that there is an $x$-dependent shift in $t$, under a coordinate transform, is just a back-door way of saying that infinite speed is relative, not absolute. This is in contrast to the situation in non-relativistic physics, where infinite speed is absolute.

In Relativity, however, the absolute speed is finite and non-zero. There's only room for one speed to be absolute (lest they all be absolute), so infinity must give way. Therefore, under a change to a moving frame, infinite speed may transform to a fast but finite speed.

To best understand this, you should step back for a broader perspective.

Space-Time Geometries, Classified By Which Speed Is The Absolute Speed
Consider the different ways of combining spatial geometry with time into a single chrono-geometry for "space-time" that leaves the following invariant: $$β dt^2 - α dx^2, \hspace 1em dt \frac{∂}{∂t} + dx \frac{∂}{∂x}, \hspace 1em β \left(\frac{∂}{∂x}\right)^2 - α \left(\frac{∂}{∂t}\right)^2,$$ for different settings of the parameters $α$ and $β$. We'll confine our attention to combining the time dimension $t$ with just one spatial dimension $x$, because that's all we'll need in most of what follows.

In the cases where $αβ ≥ 0$, this gives us a geometry with an absolute speed. We can write it as a quantity in units of length per unit time $ᴄ = \sqrt{β/α}$, or as a quantity rendered in marathoner's units of time per unit length $ᴐ = \sqrt{α/β}$. Included in this are the cases $ᴄ = ∞$ (or $ᴐ = 0$), when $α = 0$ and $β ≠ 0$, where the absolute speed is infinite; and $ᴄ = 0$ (or $ᴐ = ∞$), when $α ≠ 0$ and $β = 0$ where the absolute speed is zero. Otherwise, $αβ > 0$ and the absolute speed is both finite and non-zero and is what is otherwise known as "(in vacuum) light speed".

The cases where $αβ < 0$ are 4-dimensional Euclidean geometry, where $t$ is not a time dimension at all, but a 4th spatial dimension, and where our chrono-geometry, or "space-time", is actually just a time-less space. There is no concept of speed, at all, in this geometry and it is replaced by the concept of slope.

In the case where $α = 0$ and $β = 0$, all speeds are absolute; and in place of the above geometric invariants, you would have these (putting back in the other two spatial dimensions as $y$ and $z$) as your invariants: $$dt^2, \hspace 1em dx^2 + dy^2 + dz^2, \hspace 1em dx \frac{∂}{∂x} + dy \frac{∂}{∂y} + dz \frac{∂}{∂z}, \hspace 1em \left(\frac{∂}{∂t}\right)^2, \hspace 1em \left(\frac{∂}{∂x}\right)^2 + \left(\frac{∂}{∂y}\right)^2 + \left(\frac{∂}{∂z}\right)^2.$$

"Simultaneity" Is A Circumlocution For "Infinite Speed"
The infinite speed is the speed at which simultaneous happens. So, talking about simultaneity is just a way of talking, through the back door, about infinite speed. Thus, when you say that simultaneity is absolute, you're saying that the infinite speed is absolute, i.e. $α = 0$.

Otherwise, the infinite speed is relative, and simultaneous transforms - in a moving frame of reference - to a fast, but finite, speed. In the 4D Euclidean case it could even transform to 0 (i.e. a 90 degree rotation), because in that case, speeds are not speeds at all, but slopes.

So, in the cases where $α ≠ 0$, coordinate transforms involving $t$ will have an offset involving $x$ - as this is what it means to have "simultaneous" transforming to a fast, but finite, speed.

The Geometries' Coordinate Transforms
A transform that preserves the geometric invariants has the following infinitesimal form: $$δx = -βυt, \hspace 1em δt = -αυx,$$ which represents a change to a moving frame of reference - or a "boost". The parameter $υ$ then represents the infinitesimal boost. In finite form, this becomes $$x' = γ(x - βwt), \hspace 1em t' = γ(t - αwx).$$ As for what the scale factor $γ$ ought to be, this can be determined by imposing a principle of "reciprocity" of requiring the scale factor to be independent of the direction of boost. That means that the opposite boost - the inverse of this one - should have the form: $$x = γ(x' + βwt'), \hspace 1em t = γ(t' + αwx'),$$ which implies $$γ^2\left(1 - αβw^2\right) = 1 \hspace 1em⇒\hspace 1em γ = ±\frac{1}{\sqrt{1 - αβw^2}}.$$ So, there are actually up to two types of transforms that come out of this: one involving a flip of the coordinates $$x' = -x, \hspace 1em t' = -t,$$ and the other involving transforms that preserve the signs $$x' = \frac{x - βwt}{\sqrt{1 - αβw^2}}, \hspace 1em t' = \frac{t - αwx}{\sqrt{1 - αβw^2}}.$$

Alternatively, if you don't use any reciprocity principle, then you could just treat $γ$ as an independent parameter, factoring out a scale factor $Λ ≠ 0$ and writing $γ = Λ/\sqrt{1 - αβw^2}$. This is equivalent to adding rescaling $(x',t') = (Λx,Λt)$ as a separate transform, where $Λ ≠ 0$. So, then you are talking about two sets of transforms: the ones just laid out, along with the rescaling transforms, that includes the sign-flip as a special case.

In all cases, the boost parameter $w$ is restricted by $αβw^2 < 1$; but this restriction is only non-tautological when $αβ > 0$.

We can look at the transform as a boost involving a slowness parameter $n = αw$ and rewrite it as $$x' = \frac{x - nt/ᴐ^2}{\sqrt{1 - (n/ᴐ)^2}}, \hspace 1em t' = \frac{t - nx}{\sqrt{1 - (n/ᴐ)^2}}.$$ This connects with the case of the geometry $ᴐ = ∞$ and $ᴄ = 0$ where the absolute speed is zero: $$x' = x, \hspace 1em t' = t - nx.$$

Or, we can look at it as a boost involving a parameter $v = βw$ for speed in the usual, more familiar, sense of the term and rewrite it as: $$x' = \frac{x - vt}{\sqrt{1 - (v/ᴄ)^2}}, \hspace 1em t' = \frac{t - vx/ᴄ^2}{\sqrt{1 - (v/ᴄ)^2}}.$$ Included in this is the case $ᴄ = ∞$ and $ᴐ = 0$ of the geometry that has an infinite absolute speed and "absolute simultaneity": $$x' = x - vt, \hspace 1em t' = t.$$

In the 4D Euclidean case, we could just write $$γ = \cos{θ}, \hspace 1em \sqrt{-αβ} w = \tan{θ},$$ and write the transform as an ordinary Euclidean rotation: $$x' = x \cos{θ} - \text{sgn}(β) \sqrt{-\frac{β}{α}} t \sin{θ}, \hspace 1em t' = t \cos{θ} - \text{sgn}(α) \sqrt{-\frac{α}{β}} x \sin{θ}.$$ Unlike the other cases, this includes the sign-flip $(x',t') = (-x,-t)$ with $θ = 180°$.

As for dimensional analysis, if you call $S$ the dimension of speed, with $[υ] = S = [w]$, treating it independently of the dimensions of length $L = [x]$ and time duration $T = [t]$, then we can assign the following dimensions, $[α] = T/(LS)$, $[β] = L/(ST)$, we well as $[ᴄ] = L/T = [v]$ and $[ᴐ] = T/L = [n]$. This, of course, leads naturally to the question of whether we should also have another coefficient $κ$ with the dimension $[κ] = S/(TL)$, which would yield $[κα] = 1/L^2$ and $[βκ] = 1/T^2$, just as $αβ$ yields $[αβ] = 1/S^2$, but that's another discussion for another time.

Only One Speed Is Absolute, Or Else They All Are
So, $t$ must have an offset involving $x$, in any transform predicated on there being a finite, non-zero absolute "speed of light".

As to your original query - seen from this perspective - if you try to force $t' = t$ on a Lorentz transform, which, itself, was entirely predicated on the absolute speed being both finite and non-zero, then you'll be slipping in the infinity as a second absolute speed. But, the only way to get two separate speeds to be absolute is for them all to be. This can be seen as follows.

Under boosts, a speed $V = x/t$ (for uniform motion to/from the origin $(x,t) = (0,0)$) transforms as: $$V' = \frac{x'}{t'} = \frac{γ(x - βwt)}{γ(t - αwx)} = \frac{V - βw}{1 - αwV}.$$ Alternatively, we could consider slowness $Λ = t/x$ (again, as uniform motion to/from the origin $(x,t) = (0,0)$) to transform as: $$Λ' = \frac{t'}{x'} = \frac{γ(t - αwx)}{γ(x - βwt)} = \frac{Λ - αw}{1 - βwΛ}.$$

The only way to get two separate speeds $0 ≤ V_0 < V_1 ≤ ∞$ to be absolute with any boost $w ≠ 0$ is with: $$V_0 = \frac{V_0 - βw}{1 - αwV_0}, \hspace 1em V_1 = \frac{V_1 - βw}{1 - αwV_1},$$ or $$αw\left(V_0\right)^2 = βw = αw\left(V_1\right)^2,$$ or, since $w ≠ 0$: $$α\left(V_0\right)^2 = β = α\left(V_1\right)^2.$$ Thus $α\left(\left(V_1\right)^2 - \left(V_0\right)^2\right) = 0$, or since $\left(V_1\right)^2 > \left(V_0\right)^2$, it follows that $α = 0$, and by virtue of the above relations, that $β = 0$. That's the case of the geometry, where all speeds are absolute.

So, there is only room for one absolute speed, barring the case of them all being so. Thus, if you make it finite and non-zero, then infinity has to give way, and infinite speeds become relative. In particular, its transform under a boost, expressing it as zero slowness $Λ = 0$ is: $$Λ' = \frac{Λ - αw}{1 - βwΛ} = -αw = -n.$$ Expressing it as infinite speed $V = ∞$, the transform is: $$V' = \frac{V - βw}{1 - αwV} = -\frac{1}{αw} = -\frac{β}{αv} = -\frac{ᴄ^2}{v}.$$

Therefore, simultaneous transforms to faster-than-light speed, and - conversely - for every faster-than-light speed there is a frame of reference in which it transforms to infinity: the speed of being simultaneous.

Probing The Geometry With The Constitutive Relations In Electromagnetism
Now, the one way you can probe this geometry is to look at the relations that connect the electric displacement $𝐃$ and magnetic induction $𝐁$ to the electric force $𝐄$ and magnetic field $𝐇$. Since these fields transform, under infinitesimal boosts $𝛖$, as: $$δ𝐃 = +α𝛖×𝐇, \hspace 1em δ𝐄 = +β𝛖×𝐁, \hspace 1em δ𝐇 = -β𝛖×𝐃, \hspace 1em δ𝐁 = -α𝛖×𝐄,$$ then a relation that is isotropic in one frame: $$𝐃 = ε𝐄, \hspace 1em 𝐁 = μ𝐇,$$ where $ε$ and $μ$ are characteristic of the background medium, will transform in another frame in finite form to: $$𝐃 + α𝐆×𝐇 = ε(𝐄 + β𝐆×𝐁), \hspace 1em 𝐁 - α𝐆×𝐄 = μ(𝐇 - β𝐆×𝐃).$$ The velocity $𝐆$, which has dimension $[𝐆] = S$, then indicates the speed of the medium, itself. In the case where $βεμ = α$, if $εμ|𝐆|^2 < 1$, it is provably equivalent to the $𝐆 = 𝟎$ case, and $𝐆$ drops out. Otherwise, it remains.

If the relative signs of the additions to $𝐁$ and $𝐇$ disagree, and those of the additions to $𝐃$ and $𝐄$, too, then you're in a 4D Euclidean space and there is no time dimension at all.

If the $β$ additions are absent, while the $α$ ones remain, then you are in a space-time where the absolute speed is zero and where even things with non-zero momentum remain in place without moving anywhere. That's the Alice In Wonderland Universe, and is named after the story's author, Lewis Carroll.

If the $α$ additions are absent, while the $β$ ones remain, then you are in a space-time where the absolute speed is infinite and simultaneity is absolute. In that case, the equations are equivalent to what Thomson wrote, after correcting for Maxwell's omission of the $𝐆×𝐃$ term. They are also equivalent to what Hertz and Lorentz wrote. As for Heaviside ... I haven't yet deciphered his writings, but likely so.

That's the non-relativistic case. And, yes: that also includes Lorentz.

If both additions are present and their signs agree, then you are in the world governed (ironically) by the "Lorentz transforms", where the absolute speed is finite and non-zero. In that case, the equations are equivalent to what Einstein and Laub wrote, as well as Minkowski, in 1908. That's the Maxwell-Minkowski relations.

Finally, if all the additions are absent, so that the relations are equivalent to their "stationary" $𝐆 = 𝟎$ form, independently of what $ε$ and $μ$ are, then all speeds are absolute, and you're basically back in the world prescribed by the ancient ones of the Hellenistic Era, as far as we can determine ... but you never know about the Stoic Logicians. They might have known better. They were Vulcans.

Classification Of Geometries Based On Deformations Of The Static Group
The (mathematically) perceptive reader will recognize that I've surveyed the issue from the standpoint of "deformation theory" - which is actually the correct way to comprehensively handle this matter - considering a subclass of all the deformations of the "static group" (i.e. the kinematic group associated with the case all speeds are absolute). There were 5 groups alluded to, a sub-family of the 14 laid out by Bacry and Lévy-Leblond (BLL) and these, in turn, are just a subclass of all that there is. The family I laid out here lies resides in a 2-parameter-space, while the BLL family resides in a 3-parameter-space, while the full class is non-uniform and has multiple overlapping branches in parameter space.

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    $\begingroup$ I've never heard of the analogy of simultaneity being an 'infinite speed' being used to show that simultaneity must be relative if there is some other absolute speed, and I like the intuition behind it--especially because it leaves so little (if any) room for alternative conclusions given our axioms. I also appreciate the geometric proofs you used, and how general they were. The quick derivation of the Lorentz factor also helped. Admittedly, I did find the transformations involving the 'slowness factor' difficult to wrap my head around, but overall this is a great explanation. $\endgroup$ yesterday
  • $\begingroup$ Slowness? You're closer to there than you realize. Those space-time diagrams. Do you notice they always put time on the vertical axis? Doesn't that seem a little strange to you? The stock market graphs don't do that. They go horizontal, as do other trend-lines. But anyway ... if time is on the vertical axis, then what's the slope of the worldlines on it? In what units? $\endgroup$
    – NinjaDarth
    1 hour ago

So like if I were asking this question to the Mathematics Stack Exchange I would say something like “I know that $\operatorname{SO}(3, 1)$ and $\operatorname{SO}(4)$ have $\operatorname{SO}(3)$ as a subgroup, and probably bigger ones like $\operatorname{SL}(4, \mathbb R),$ are those the only 4x4 options? Or are there others too?” Here $\operatorname{SO}(3)$ is the group of rotations preserving the Euclidean norm, and we're asking what 4x4 matrix groups have that as a subgroup.

In terms of the physics, we attach less importance to what is possible and more importance to what is actual. So these groups are indeed studied by several of my peers but generally in the contexts of quantum field theory where they describe something about the sorts of particles which exist. I’m from a condensed-matter background so I have less ability to answer those things than a particle physicist.

The relativity of simultaneity

So if I reduce the Lorentz transform to its simplest presentation, the idea is that we want to make sure that people agree on the speed that light is going. You imagine that Alice is moving past Bob at some speed $v \ll c$, and maybe Alice presses a button which turns on the lights—we call this an event. Now the light that communicates that this has happened, expands outward from Alice as a thin sphere. Let’s say that it’s a sphere for Bob.

In Bob’s understanding, we would say that Alice is off-center from the sphere, the sphere is centered on some origin $0$ and located off at $x^2 + y^2 + z^2 = c^2 t^2$ whereas Alice is off-center at $(x, y, z) = (v t, 0, 0).$ So one edge is closer to Alice at a distance $t/(c + v)$ and the other is further at a distance $t/(c - v).$ But, if we want to make the speed of light constant for everyone, then when we switch to Alice’s frame of reference we need to put Alice at the center of her sphere. All of the other directions $y,z$ they both agree on, the sphere is tangent to Alice's motion: it is purely this $x$-axis where something needs to change.

The core claim of special relativity is a phenomenon called relativity of simultaneity. Everything else will be derived as a second-order consequence of this claim and we can ignore length contraction and time dilation to first order. The claim is that Alice disagrees with Bob about what is simultaneous. Alice agrees that the light passed this point at distance $t/(c + v)$—but thinks that happened about $t(1 - v/c)$ ago. Like, Bob's clock there said $t$ at the time, but it has always been out-of-sync with the clock Alice carries by this amount. And Alice agrees that the light will pass this point at distance $t/(c-v)$, but thinks that will happen at $t(1 + v/c)$ or so from now.

In other words when we are setting both their clocks to the same zero and calling some instant a time $t$ later as “right now,” to first order Bob sees that “at time 0” some clock at $x = ct$ showed time 0 and sees that “right now” that clock is showing time $t$, and the light is just hitting that clock right now. But Alice thinks that “at time zero” it showed a time $vx/c^2$ and “right now” it is showing a time $t + vx/c^2.$ She agrees that the light passed it when it showed time $t$ but she disputes that this time is “right now,” because the clock was not properly synchronized for her to start with. Instead she thinks that this event happened at time $t' = t - vx/c^2$ for her. In relativity two people who are at the same point agree on what “right now” means at that point. But they disagree on what time “right now” means at far-off locations. If I am on Voyager 2, traveling at about 3.3 AU/year towards a distant star 500 light-years away, and you think based on your complicated equations of stellar evolution that this star is going supernova today, then I think that it went supernova nine and a half days ago. Of course neither of us will get to see the result until about 500 years from now but we will both end up being right when we see our respective results, it is just that what “right now” meant to us at that distance was fundamentally different.

Deriving the Lorentz transform

So if you imagine that Bob built a line of clocks that he thought were all in-sync and all showed $0$ at time $t=0$, Alice thinks that the clock at coordinate $x$ is behind where it should be by a factor $x v/c^2$ (or ahead of where it should be, if $x$ is negative and therefore the factor is negative). This also needs to be understood as a fundamental property of acceleration which we did not appreciate before now because the speed of light is so fast. It is just a property of our universe that if you accelerate with acceleration $\alpha$ in the $x$-direction you see an effect that is not explainable as a Doppler shift or anything else, where clocks ahead of you by a coordinate $x$ appear to tick faster a rate of $(1 + \alpha x/c^2)$ seconds per second (or tick slower if $x$ is negative, you get the picture). Indeed there must be a surface at $x = -c^2/\alpha$ where clocks appear to stand still, this is what we call an “event horizon,” signals of light from before a certain distance cannot reach a constantly-accelerating observer in relativity.

In other words before relativity we connected Alice to Bob with the Galilean transformation, which I will write here with $w = ct$ and $\beta = v/c$ as $$\begin{bmatrix}w'\\x'\\y'\\z'\end{bmatrix} = \begin{bmatrix}1&0&0&0\\-\beta&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix} \begin{bmatrix}w\\x\\y\\z\end{bmatrix},$$ but now we have a theory that to first-order must be instead $$\begin{bmatrix}w'\\x'\\y'\\z'\end{bmatrix} = \begin{bmatrix}1&-\beta&0&0\\-\beta&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix} \begin{bmatrix}w\\x\\y\\z\end{bmatrix}.$$ Call this matrix $\bar L(\beta).$ There are now three ways to proceed. One is to construct $\bar L(\beta) \bar L(-\beta)$ on the idea that if Bob sees Alice moving forward at velocity $v_x = +\beta c,$ then Alice should see Bob moving backwards with velocity $v_x = -\beta c$ and so transforming forwards and then backwards should take us back to where we started, so you get the identity matrix $\operatorname{diag}(1, 1, 1, 1)$. But it doesn’t here, it takes us to $\operatorname{diag}(1-\beta^2, 1-\beta^2, 1, 1).$ And the idea is to say “well this was a first-order theory, I can fudge the matrix by dividing its first two rows by $1/\sqrt{1 - \beta^2}$ and that will propagate entirely through the argument to give me $\operatorname{diag}(1, 1, 1, 1)$.” And this works, but maybe it’s not the most stable foundation possible. Another approach is to consider light beams that travel in various “train experiments” and work out these factors much more directly as coming from the Pythagorean theorem, $ct$ being a hypotenuse of a right triangle with base $v t$ and some fixed height $h$.

But my favorite is to force the first-order theory to give you the answer. We try to accelerate by some parameter $\phi$, in $N$ steps of size $\phi/N$, and we therefore form $$L(\phi) = \lim_{N\to\infty} [\bar L(\phi/N)]^N.$$Matrix exponentiation requires an eigenbasis, but an eigenbasis is very easy to come by: $[1, 1, 0, 0]$ and $[-1, 1, 0, 0]$ are clear eigenvectors to join $[0, 0, 1, 0]$ and $[0, 0, 0, 1].$ So one can work out that in fact,$$ L(\phi) = \begin{bmatrix}\cosh\phi&-\sinh\phi&0&0\\-\sinh\phi&\cosh\phi&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix},$$and now $L(\phi)L(-\phi) = I$ via the rules for hyperbolic sines and hyperbolic cosines that $\cosh^2\phi - \sinh^2\phi = 1.$ This connects to the other two approaches by identifying that actually $\beta = \tanh\phi = \sinh \phi/\cosh\phi,$ at which point one can work out that $\cosh^{-2}\phi = 1 - \tanh^2\phi$ and therefore $\cosh\phi = 1/\sqrt{1 - \beta^2}.$ So the same parameter reappears but in a much more logically rigorous way that assures us that indeed, every other effect in relativity comes from the relativity of simultaneity compounded with the Galilean transformation.

To get the full Lorentz group one composes this boost operation with the rotations and one thereby gets a group of all linear transforms that preserve the Lorentz-norm $w^2 - x^2 - y^2 - z^2.$

I think that’s pretty elegant but the fundamental question of “which other groups of these 4x4 matrices have SO(3) as a subgroup?” I think is also a valid question which you might ask a mathematician.

  • $\begingroup$ Alright, so I already know about the relativity of simultaneity. Actually, that is what incited this question; I want to better fundamentally understand why it must occur and why it has the value it does (-vx/c^2). Could you please elaborate on just why this must be the value and why it changes with respect to x (the basis of the relativity of simultaneity) rather than some other variable (such as v) or not at all. I was trying to use basic derivatives in our universe such as dx'/dx /= 0 to show that, fundamentally, dt'/dx /= 0, and thus time does shift over space (simultaneity is relative). $\endgroup$ Oct 5, 2020 at 14:30
  • $\begingroup$ I mean, you can't make that argument because the Galilean transformation has $(\partial t'/\partial x)_{t,y,z} = 0$ (you have to be a bit precise about what you are holding constant here, in this case it is $t,x,y$) and there is nothing mathematically inconsistent about pre-Einstein relativity of this form; it just does not square with our idea that everybody agrees every light pulse is moving at the exact same speed $c$. The form $-vx/c^2$ is the only form which is linear and to first-order guarantees the latter property. $\endgroup$
    – CR Drost
    Oct 6, 2020 at 3:59
  • $\begingroup$ Oh, alright. I was trying to see if there was something mathematically inconsistent about Galilean relativity that required special relativity. I know why -vx/c^2 is linear: there is no x in the derivative with respect to x, so the derivative with respect to x is constant, which is a requirement for space to be homogeneous. However, why is it the only term that allows for C to be invariant? Also, can we still find some basic, fundamental properties of the Lorentz Transformations with some simple derivatives (still, dt'/dx does not equal 0) if we include that dx/dt=dx'/dt' for c? $\endgroup$ Oct 12, 2020 at 15:29
  • $\begingroup$ So just work it out to first order. We clearly want the first order theory to contain the Galilean $x'=x-vt$. Considering the trajectory $x=ct$ we can see that this must map to $x'=(c-v)t$ and to have this be $x'=ct'$ we must state that $t'=\frac{c-v}c t,$ no? And then you have the first-order transform $$\begin{bmatrix}1&-\beta\\-\beta&1\end{bmatrix}$$. Einstein was somewhat luckier in that Lorentz had already derived this for him: all of Einstein’s 1905 “miracle year” papers essentially take someone else's idea and tell physicists to take it more seriously because it has real consequences. $\endgroup$
    – CR Drost
    Oct 12, 2020 at 20:55
  • $\begingroup$ That does make sense. I looked over your answer a few times until it made sense to me. However, this does not seem to show that t' is a function of x at all. Am I missing something here? Also, how did you get that matrix from x'=x-vt and t'=((c-v)/c)t? It doesn't seem to follow as, again, you didn't show that t' is a function of x.... What am I missing? $\endgroup$ Oct 15, 2020 at 13:54

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