# B Trying to understand Lorentz Transformations

#### Ibix

Science Advisor
Yes, that is exactly what I am asking.

Dale: It's a difficult concept for me. I understand what the general form of an equation is. I just don't see how people originally came up with the idea that the general equation took that form.
Originally, Einstein didn't come up with it that way - he used his train thought experiment. But then people showed that the transforms are simply an expression of symmetries, and this is a more abstract but much more powerful way of thinking, that underlies all fundamental physics post-Einstein.

The transforms simply relate coordinates on spacetime, so they can only depend on the coordinates and the relative velocity of the frames. There are no other physical concepts that are relevant. The transforms have to be linear because if they aren't there's either a special place or a special time or both. We can arbitrarily choose an origin that the coordinates share, and we can arbitrarily choose our x axis to lie parallel to the relative velocity of the frames.

With those assertions, the most general transformation we can possibly write is$$\begin{eqnarray*}t'&=&At+Bx+Cy+Dz\\ x'&=&Et+Fx+Gy+Hz\\ y'&=&It+Jx+Ky+Lz\\ z'&=&Mt+Nx+Oy+Pz\end{eqnarray*}$$where the capital letters are constant. We are not making any assumptions about these constants. Many of them may be zero - but we'll prove that, we won't assume it.

All the off-diagonal elements including $y$ or $z$ must be zero from symmetry. Simply rotate your coordinate axes 180° about the x axis and $y$ becomes $-y$ - but this shouldn't have any effect on $x'$ or $t'$, and the only way for it to have no effect if those cross terms are zero. We cannot make this argument about off diagonal terms with $x$ because there is a difference if we flip the x axis (the velocity changes sign) and we can't make this argument about $t$ because flipping the direction of time also flips the direction of velocity.

That tells us that our four equations are just
$$\begin{eqnarray*}t'&=&At+Bx\\ x'&=&Et+Fx\\ y'&=&Ky\\ z'&=&Pz\end{eqnarray*}$$Adding that the inverse transforms must look the same as the forward transforms except for the sign of the velocity (i.e., the principle of relativity) means that we want $y'=Ky$ and $y=y'/K$ to look the same which means $K=1$, and similarly $P=1$.

This is where you started. Sweetsprings already posted the derivation from here.

• Dale

#### FactChecker

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The Lorentz equation for time, though, doesn't jump out at me in the same way as being intuitive.
People may have worked on this for a long time. You should not assume that it was intuitively obvious to the casual observer.

• nitsuj and Ibix

#### Mister T

Science Advisor
Gold Member
In the equation for t, why is x being added into the equation?
Note that if the coefficient B has a value of zero then the term containing x will be zero. So the possibility is there, but it's only a possibility.

#### Dale

Mentor
In the equation y = mx + b, the first time I read it and the description for it it made sense to me right away. It makes sense because I visualize that what it's describing is a straight light which isn't curved, and I understand why we use each of the terms. The Lorentz equation for time, though, doesn't jump out at me in the same way as being intuitive.
If you do not include time then you cannot distinguish between a particle that travels in a straight line at a constant velocity and a particle which travels in a straight line at an accelerating velocity. So you lose the critical distinction between inertial and non-inertial coordinates. Once you include time as a coordinate, then a general linear transformation includes time. Leaving it out is making an assumption.

It may not be intuitive, but sometimes that is why you have to use the math, to make logical leaps that are not intuitive. For you, this is one of those times. You specifically need to use the math a little more until you develop some new intuition.

#### pervect

Staff Emeritus
Science Advisor
One way of formulating special relativity is to postulate that the speed of light equal to "c" for all observers. This goal can't be accomplished if one limits oneself to the transform t'=t.

A longish quote from Einstein's 1905 paper might clarify the assumptions made:

Examples of this sort, together with the unsuccessful attempts to discoverany motion of the earth relatively to the “light medium,” suggest that thephenomena of electrodynamics as well as of mechanics possess no properties corresponding to the idea of absolute rest. They suggest rather that, as hasalready been shown to the first order of small quantities, the same laws of electrodynamics and optics will be valid for all frames of reference for which theequations of mechanics hold good.1We will raise this conjecture (the purport of which will hereafter be called the “Principle of Relativity”) to the statusof a postulate, and also introduce another postulate, which is only apparently irreconcilable with the former, namely, that light is always propagated in emptyspace with a definite velocitycwhich is independent of the state of motion of the emitting body. These two postulates suffice for the attainment of a simple and consistent theory of the electrodynamics of moving bodies based on Maxwell’s theory for stationary bodies. The introduction of a “luminiferous ether” will prove to be superfluous inasmuch as the view here to be developed will not require an “absolutely stationary space” provided with special properties....
In the end though, it's a matter of whether one wants to learn how the theory works, or not. Ideally, one would let experiment decide whether relativity or classical Newtonian mechanics best matched experimental evidence. In practice, perhaps, if one is a student, one doesn't bother to learn a new theory unless one is pretty sure it's going to be a better theory.

A bit of motivation as to why classical Newtonian mechanics doesn't work quite right might be helpful in motivating one to learn special relativity. I am rather fond of "The Ultimate Speed"

There is a peer-reviewed paper associated with this video that goes into more detail. The video is less dry than the paper, one reason I link to it, though in general one needs to be cautious about accepting any video one may encounter on the internet.

Assuming that one has already decided that special relativity is worth learning, it becomes simple. One can't assume that t'=t and still learn special relativity. The question of why one should learn special relativity is more complex, but it boils down to the fact that it works very well. One can't really understand this point, about how well it works, until one learns the theory in the first place, however.

There are other routes to learning and formulating the theory than EInstein's original approach, but they'll all lead to similar results, where t' is not equal to t.

• gmax137

#### Ibix

Science Advisor
People may have worked on this for a long time. You should not assume that it was intuitively obvious to the casual observer.
Indeed. The mathematical tools to derive the Lorentz transforms from symmetry arguments existed at least as far back as Newton. Nobody did it (edit: or as PeroK points out below, nobody published it) for around 250 years.

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#### PeroK

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Hmmm. Are you saying that as long as the equation is giving the right results, I should just learn it and use it and not get too bogged down in how it originated?
In many ways, SR can be the opposite of that. Everything ought to make sense and, in the end, you wonder how no one thought of it before Einstein.

The problem here is perhaps one of mathematical experience and technique. For example, you see posts on here where some typical mathematical operation (squaring both sides of an equation, say) is seen by the inexperienced as a great leap of imagination.

At the next level, you are viewing a set of linear equations as something almost mystical that a mathematician dreamed up out of nowhere. But, once you have a bit of mathematical experience, it's the most obvious thing to do.

At the most basic level you are asking: why on Earth would anyone ever come up with the idea of analysing a possible linear relationship between time and space coordinates?

And our answer is, well no one did for several hundred years, but why not try? Maybe someone in 1850 did try but was too appalled by the consequences and didn't even publish!?

There is an element of hindsight, of course, that the great symmetries of time, space, energy and momentum etc. have been studied in depth and we know that SR could have been built on these.

I know some texts on SR simply produce the Lorentz Transformation as a postulate. That would be too much like pulling a rabbit out of a hat for me. But, developing it from symmetry and homogeneity principles seems very natural to me, even if it relies on a fairly modern view of physics.

It's good that you want to understand why this approach is valid. But, you must also try to develop the scientific capabilities of your brain. And sometimes that means going with the flow and seeing where it leads. It feels to me sometimes that there's a part of your brain that wants to reject this stuff and, obviously, you do want to learn it. So, perhaps you yourself have to start beating up on the reactionary part of your brain - instead of asking us to do it • Dale and Ibix

#### stevendaryl

Staff Emeritus
Science Advisor
In the equation y = mx + b, the first time I read it and the description for it it made sense to me right away. It makes sense because I visualize that what it's describing is a straight light which isn't curved, and I understand why we use each of the terms. The Lorentz equation for time, though, doesn't jump out at me in the same way as being intuitive.
I would say that if you have two different inertial coordinate systems systems in one spatial and one time dimension, $(x,t)$, and $(x',t')$ we can describe their "disagreement" using four factors:
1. They may disagree about what objects are at rest.
2. They may disagree about the length of objects.
3. They may disagree about the time between two events.
4. They may disagree about which events are simultaneous (take place at the same time).
(There are actually a couple of other sources of disagreement that I'm going to ignore: 5. They can disagree about the direction in which $x$ increases. 6. They can disagree about the origin of the $(x,t)$ coordinate system.)

Effect 1 leads to the transformation for $x'$ involving both $x$ and $t$.
Effect 4 leads to the transformation for $t'$ involving both $x$ and $t$.
Effects 2 and 3 are matters of scaling, in a sense.

Galilean transformations only involve effect 1.

Why would different inertial coordinate systems disagree about which events are simultaneous? Well, you have to go through, as Einstein did, the thought process of asking how do you establish that distant clocks are synchronized. One possibility is that you bring two clocks together, synchronize them, and then take them to different locations. But how do you know that moving clocks around doesn't affect them? Another possibility is that you use some signal with a known speed (such as light) to synchronize distant clocks, and take into account the transit time. But if two different coordinate systems disagree about whether objects are at rest, then they will disagree about transit time, as well (it takes longer to catch up to a moving object, if it's moving away, and it takes less time to catch up to a moving object, if it's moving toward you).

So if you're willing to buy that clock synchronization might be relative to an inertial coordinate system, then effects 2 and 3 follow. The way that you would measure the length of a moving object is by noting that one end is at location $x$ at the same time that the other end is at location $x+L$. You subtract those two coordinates to get a length $L$. But if different coordinate systems disagree about whether clocks are synchronized, then they will disagree about whether the two ends were measured at the same time. So they will disagree about the length of the moving object. There is a similar reason that different inertial coordinate systems may disagree about the time between events.

• Dale

#### robphy

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(Hoping not to divert the thread... but replying to a side comment by PeroK...)

At the most basic level you are asking: why on Earth would anyone ever come up with the idea of analysing a possible linear relationship between time and space coordinates?

And our answer is, well no one did for several hundred years, but why not try? Maybe someone in 1850 did try but was too appalled by the consequences and didn't even publish!?
If anyone was able to do that, it would have been Felix Klein.
https://de.wikisource.org/wiki/Felix_Klein

In 1872, he developed his Erlangen Program
https://en.wikipedia.org/wiki/Erlangen_program
which essentially laid the foundations for Minkowski spacetime.
(Klein essentially unified elliptic, hyperbolic, and Euclidean geometry via projective geometry.
The scheme was extended to the "Cayley-Klein Geometries" (9 in 2-dimensions),
of which Minkowski spacetime, Galilean spacetime, and the DeSitter spacetimes are included (but apparently not interpreted physically).

Here's a curious comment from Toretti's Philosophy of Geometry from Riemann to Poincaré, p.129
https://books.google.com/books?id=EcLrCAAAQBAJ&printsec=frontcover&dq=torretti&hl=en&sa=X&ved=0ahUKEwio6bP235bhAhUkUt8KHZydCAIQ6AEIKDAA#v=onepage&q=finite sequence of rotations&f=false The passage in Klein that Toretti refers to is
on p.189 of Klein's Vorlesungen uber Nicht-Euklidische Geometrie (1926)
(which [if i decoded the preface correctly] is partly based on lectures from 1892 and 1893
[possibly these handwritten lectures https://archive.org/details/nichteuklidische01klei ],
plus additional material the editors put together). In that passage, (update: here are the figures Klein refers to... looks like worldlines in Minkowski and Galilean spacetimes. Note how the lines bunch up near the [light-cone] edge in 114.)  I transcribed the section
Felix Klein (1926 posthumous) said:
Die Erfahrung zeigt uns, daß wir stets durch eine endliche Anzahl derartiger Drehungen in die Nähe der Ausgangslage zurückkommen, ja sogar über sie hinausgelangen können. Diese Eigenschaft ist aber bei der hyperbolischen und parabolischen Messung von Winkeln nicht vorhanden, da wir dort durch die entsprechende Abtragung untereinander kongruen ter Winkel nie über bestimmte Grenzlagen herauskommen können (vgl. die Abb. 114 und 120). In einer Maßbestimmung, die in der Außenwelt anwendbar ist, muß also sowohl die räumliche, wie auch die ebene Winkelmessung elliptisch sein.
and fed it into Google translate (bolding mine)
Google Translate of Klein said:
Experience shows us that we can always come back by a finite number of such rotations in the vicinity of the starting position, and even get beyond them. However, this property is not present in the hyperbolic and parabolic measurement of angles, since we can never get beyond certain boundary layers there by the corresponding removal of congruent angles (see Figures 114 and 120). In a measure that is applicable in the outside world, so both the spatial, as well as the level angle measurement must be elliptical.
In hindsight, the hyperbolic and parabolic measurement of angles [between radial lines... thought of as inertial worldlines] refer to rapidity in Minkowski spacetime [so velocity $v=c\tanh\theta$ ] and in Galilean spacetime [here, where, it turns out, Galilean-rapidity coincides with velocity [up to a constant carrying units] ].
So, the point is:
the foundations were there.. but maybe not yet realized in the physical world.

(As I said, these lectures were published posthumously in 1926 and drawn on material from 1892 and 1893 onward...
but it's not clear how far onward. Was this passage before 1905 (when Einstein published) or 1908 (when Minkowski formulated "spacetime geometry")?)

From the Wikisource link, it might be interesting to read the transcription of
https://de.wikisource.org/wiki/Über_die_geometrischen_Grundlagen_der_Lorentzgruppe (1910)
(which Google Translate translates to "About the geometric foundations of the Lorentz group")
which can be read via Google Translate [in the Chrome browser].

My \$0.02.
(Sorry for intruding.)

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• PeroK and Ibix

#### Histspec

With those assertions, the most general transformation we can possibly write is$$\begin{eqnarray*}t'&=&At+Bx+Cy+Dz\\ x'&=&Et+Fx+Gy+Hz\\ y'&=&It+Jx+Ky+Lz\\ z'&=&Mt+Nx+Oy+Pz\end{eqnarray*}$$where the capital letters are constant. We are not making any assumptions about these constants. Many of them may be zero - but we'll prove that, we won't assume it.

All the off-diagonal elements including $y$ or $z$ must be zero from symmetry.
The above transformation is already a complete Lorentz transformation (in matrix form $\mathbf{x=g}\mathbf{x}'$ or $\mathbf{x'}=\mathbf{g}^{-1}\mathbf{x}$) leaving invariant $-x_{0}^{2}+x_{1}^{2}+x_{2}^{2}+x_{3}^{2}$ by using the conditions

$$\begin{matrix}\mathbf{g}^{{\rm T}}\mathbf{A}\mathbf{g}=\mathbf{A}\\ \mathbf{A}=\mathrm{diag}(-1,1,1,1) \end{matrix}$$

One can see, that the Lorentz transformation and its group can be understood as a branch of linear algebra.

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• Ibix

#### Histspec

If anyone was able to do that, it would have been Felix Klein.
https://de.wikisource.org/wiki/Felix_Klein

In 1872, he developed his Erlangen Program
https://en.wikipedia.org/wiki/Erlangen_program
which essentially laid the foundations for Minkowski spacetime.
Indeed. Like many other mathematicians in the 19th century, Klein analyzed the transformations leaving invariant the interval $-x_{0}^{2}+x_{1}+x_{2}$ or $-x_{0}^{2}+x_{1}+x_{2}+x_{3}$ which he related to hyperbolic geometry (or planes/spaces of constant negative curvature).
https://en.wikipedia.org/wiki/History_of_Lorentz_transformations#Klein_(1871–1897)

For instance, in his 1896-lecture (pp. 13–14) on the theory of the top, he specifically identified one of these coordinates with time (even though later in the book, he assured the readers that he didn't imply any "metaphysical" interpretation of his formulas on non-Euclidean geometry):

$$\begin{matrix}x^{2}+y^{2}+z^{2}-t^{2}=0\\ =(x+iy)(x-iy)+(z+t)(z-t)=0\\ x+iy:x-iy:z+t:t-z=\zeta_{1}\zeta_{2}^{\prime}:\zeta_{2}\zeta_{1}^{\prime}:\zeta_{1}\zeta_{1}^{\prime}:\zeta_{2}\zeta_{2}^{\prime}\\ \frac{\zeta_{1}}{\zeta_{2}}=\zeta\rightarrow\zeta=\frac{x+iy}{t-z}=\frac{t+z}{x-iy}\\ X^{2}+Y^{2}+Z^{2}-T^{2}=0=\ \text{etc.}\\ \zeta=\frac{\alpha Z+\beta}{\gamma Z+\delta}\rightarrow\begin{matrix}\zeta_{1}=\alpha Z_{1}+\beta Z_{2}, & \zeta_{1}^{\prime}=\bar{\alpha}Z_{1}^{\prime}+\bar{\beta}Z_{2}^{\prime}\\ \zeta_{2}=\gamma Z_{1}+\delta Z_{2}, & \zeta_{2}^{\prime}=\bar{\gamma}Z_{1}^{\prime}+\bar{\delta}Z_{2}^{\prime} \end{matrix}\\ (\alpha\delta-\beta\gamma=1) \end{matrix}\quad\begin{array}{c|c|c|c|c} & X+iY & X-iY & T+Z & T-Z\\ \hline x+iy & \alpha\bar{\delta} & \beta\bar{\gamma} & \alpha\bar{\gamma} & \beta\bar{\delta}\\ \hline x-iy & \gamma\bar{\beta} & \delta\bar{\alpha} & \gamma\bar{\alpha} & \delta\bar{\beta}\\ \hline t+z & \alpha\bar{\beta} & \beta\bar{\alpha} & \alpha\bar{\alpha} & \beta\bar{\beta}\\ \hline t-z & \gamma\bar{\delta} & \delta\bar{\gamma} & \gamma\bar{\gamma} & \delta\bar{\delta} \end{array}$$

This type of Lorentz transformation in terms of linear fractional (Möbius) transformations and spin transformations became important in relativistic quantum theory beginning in the 1920ies.

#### Nugatory

Mentor
Hmmm. Are you saying that as long as the equation is giving the right results, I should just learn it and use it and not get too bogged down in how it originated?
It depends on what you're trying to accomplish. "How it originated" is a question for the history of science, and that history is full of wrong turns and unnecessary detours.

Thus, if your goal is to understand the physics you'll want to learn from a good modern treatment that starts with the equations that best capture what we've learned with the benefit of decades or centuries of hindsight.

If your goal is to understand how humanity built this awe-inspiring intellectual structure from the raw materials of math and experiment (and for the non-specialist this question may be more important) then you'll want to study how it originated. But even then you're best off starting with a solid grasp of the modern treatment. It's a lot easier to follow the writings of Marco Polo or Lewis and Clark if you have access to a modern map; and if you don't understand the geometric formulation of special relativity you will be unable to see that it is hidden and struggling to emerge from the earliest papers.

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• FactChecker and Dale

#### NoahsArk

Gold Member
Thank you for the responses.

Based on the feedback seems like I need to work on solving problems involving the transformations in order to try and build my intuition more.

#### Mister T

Science Advisor
Gold Member
Making up your own problems to solve can be even more helpful.

#### sweet springs

The SR relation, square of invariant distance,

c2t2−x2=c2(t1)2−(x1)2​
I would say this relation assures that Lorentz transformation is linear for x,y,z and t. For example if the transformation includes $x^2$, term $x^4$ should appear in the above formula. That's a mess.

#### arydberg

You seem to have difficulty grasping the fact that SR is totally outside your conception of reality. You are trying to make sense out of it. You cannot. x and t have values in a lab frame. When viewed from a moving frame we get x' and t'. They do not agree with x and t

#### NoahsArk

Gold Member
I'm now stuck on how to derive the LTs. This is the part I get stuck on which is shown in this video at 11min 15 sec:

When the Taylor and Wheeler book gets to that part of the derivation I also get stuck.

Basically, what he is saying is that if you have an equation of this form: aT2 + bX2 = cT2 - dX2, then you have to conclude that the coeficients "a" and "c" must be equal, because you have a T2 on both sides of the equation, and you also have to conclude, since there are X2s on both sides, that "b" and "d" must also be equal. This is not true- you can have many values of "a" which are different from "c", and the equation will still be equal so long as "b" and "d" are also different by the right amounts.

Please let me know what I am missing.

#### FactChecker

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If the coefficients, a, b, c, and d, are constants rather than functions of T and/or X, then you must have a=c and b=d. I think the initial assumption was that they are constants.

#### sweet springs

This is not true- you can have many values of "a" which are different from "c", and the equation will still be equal so long as "b" and "d" are also different by the right amounts.
Yes, if you carefully choose special values of X and T, i,e. tag of special event happening in special place and special time for given a, b, c and d.
But the teacher is so ambitious that tag of any place any time events, e.g. my writing this article here now, your reading this article there later, my death in some place and in sometime in near future, your birth, etc., follow the same formula with same a,b,c and d, to have correspond tag of different number in another IFR. It means that for any values of X and T the equation must hold.

The formula in the lecture something like
$$FT^2+GXT+HX^2=c^2T^2-X^2$$ so
$$(F-c^2)T^2+GXT+(H+1)X^2=0$$
Here all the coefficients of X^2, XT,T^2 should vanish so that this equation holds even if X and T may take any value tag for any events.

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#### Ibix

Science Advisor
Basically, what he is saying is that if you have an equation of this form: aT2 + bX2 = cT2 - dX2, then you have to conclude that the coeficients "a" and "c" must be equal, because you have a T2 on both sides of the equation, and you also have to conclude, since there are X2s on both sides, that "b" and "d" must also be equal. This is not true- you can have many values of "a" which are different from "c", and the equation will still be equal so long as "b" and "d" are also different by the right amounts.
$$\begin{eqnarray*}aT^2+bX^2&=&cT^2-dX^2\\ (b+d)X^2&=&(c-a)T^2\end{eqnarray*}$$
You are correct that, for any chosen $X$ and $T$ this can be made true by picking carefully chosen values of $(b+d)$ and $(c-a)$. However it will not be true for (almost) all other values of $X$ and $T$. So such transforms would not be globally applicable - and the whole point is to find a relationship that works for any $X$ and $T$.

The only solution that applies for any $X$ and $T$ is where $(b+d)=0$ and $(c-a)=0$.

#### NoahsArk

Gold Member
Thank you for the responses. It's starting to make more sense but I'm still a bit stuck:

If the coefficients, a, b, c, and d, are constants rather than functions of T and/or X, then you must have a=c and b=d.
Even if they are constants, why can't they have different values?

Here all the coefficients of X^2, XT,T^2 should vanish so that this equation holds even if X and T may take any value tag for any events.
The only solution that applies for any X and T is where (b+d)=0 and (c−a)=0.
Say b = 2 and d = 4 and c = 9 and a = 3. Now the equation holds true for all T and X values, even though b + d and c - a do not = 0.

#### Ibix

Science Advisor
Even if they are constants, why can't they have different values?
Because then they are not constants. They change with position and/or time.
Say b = 2 and d = 4 and c = 9 and a = 3. Now the equation holds true for all T and X values, even though b + d and c - a do not = 0.
No it doesn't. That gives you $6X^2=6T^2$ which only holds true for $X=\pm T$. If you put in $X=6$ and $T=22$ (for example) your values do not produce an equality.

#### stevendaryl

Staff Emeritus
Science Advisor
Say b = 2 and d = 4 and c = 9 and a = 3. Now the equation holds true for all T and X values, even though b + d and c - a do not = 0.
The original equation was:

$aT^2 + b X^2 = c T^2 - dX^2$

This has to hold for all $X$ and $T$. So it has to hold in the case $T=0$ and $X=1$. In that case, we have:

$a \cdot 0 + b \cdot 1 = c \cdot 0 - d \cdot 1$

So we conclude: $b=-d$

It also has to hold when $X=0$ and $T=1$. So:

$a \cdot 1+ b \cdot 0 = c \cdot 1 - d \cdot 0$

So we conclude: $a = c$.

• Dale

#### Dale

Mentor
Please let me know what I am missing.
The equation needs to hold for all values of X and all values of T.

#### FactChecker

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Suppose $a,b,c,d$ are constants and $(b+d)X^2=(c-a)T^2$ for all values of $X , T$. Pick any set of values, $X_1$ and $T_1$. We have $(b+d)X_1^2=(c-a)T_1^2$.

Now leave $X$ equal to $X_1$ and let $T_2 = T_1+1$.

Then $(c-a)T_1^2=(b+d)X_1^2=(c-a)T_2^2=(c-a)(T_1+1)^2=(c-a)(T_1^2 + 2T_1+1)$.

This gives $0 = 2(c-a)(T_1+1)$. So either $c=a$ or $T_1=-1$.

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