# Easiest possible way to derive the Lorentz transformation

• I
I put the level for this thread as I, but anything from B to A is acceptable here.

I'm hoping this isn't too imprecise, but what are the easiest or simplest (or fastest) ways to derive the Lorentz transformation equations you know? I am not after blatant corner cutting here, by the way. Just either low level math requirements OR simplest, most direct methods.

This is open to your interpretation of "easiest," but if anyone has any ideas, here are the two main definitions of easiest I mean:

(1) Easiest mathematically: what is the simplest way to get the Lorentz transformation equations in terms of lowest math level required.

(2) Easiest in terms of expediency/eloquence, etc.: regardless of mathematical complexity, which brings out the transformation fastest, or simplest? (as an example of the kind of thing I'm looking for in this definition of "simple," you can derive equations of motion of a system using vectors and Newton's laws, and worrying about various constraint forces, or you can use Lagrangian mechanics and not have to worry about much of that. In this case I would consider Lagrangian mechanics "simpler.")

Again, I hope what I'm asking is comprehensible, but if you understand what I'm saying, please post your method of choice that you feel is simplest/easiest (whichever definition of simple or easy you prefer).

atyy

As a response to my own thread to give an idea of what I'm after:

For category 1: The easiest I've seen is to start with the generic transformation forms of:

x = ax' + bct'
t = fct' + gx'

And then utilize the two postulates, and then solve for the four constants. For example, since the speed of light is c in both frames:

$$\frac{x}{t} = \frac{a\frac{x'}{t'} + bc}{cf + g\frac{x'}{t'}}$$
which in the case of light is:
$$c = \frac{ac + bc}{cf + gc}$$

So you'd use that information and solve or the constants, with a being the Lorentz factor when you've finished.

For category 2, the most general yet simplest way I know is get a general transformation equation for flat spacetime which has both the Galileo transform and Lorentz, and simply choose a finite speed limit to get the Lorentz. This way is a lot more rigorous than the previous way, but it gives the cool result of the Galileo transformations and Lorentz transformations having the same form, but merely separated by your choice of what k equals in the following:

$$x' = \frac{x - vt}{\sqrt{1 - kv^2}}$$
and
$$t' = \frac{t - kvx}{\sqrt{1 - kv^2}}$$

and the difference between the two will simply be whether k equals zero or 1 (with v/c factor popping out with appropriate choice of units).

I am under the impression there is something far more simple and elogant than the above though, something I'm guessing that has something to do with the Poincaré group.

atyy
Orodruin
Staff Emeritus
Homework Helper
Gold Member
For a light signal from the origin ##t^2 - x^2 = 0## in any inertial frame. Therefore search for a (linear - homogeneous space etc etc) transformation that keeps ##s^2 = t^2 - x^2## invariant for all ##x## and ##t##. The general linear transformation is on the form
$$t' = At + Bx, \quad Ct + Dx$$
and therefore
$$t'^2 - x'^2 = (A^2 - C^2)t^2 - (D^2 - B^2)x^2 + 2(AB - CD)xt = t^2 - x^2.$$
This gives ##A^2 - C^2 = D^2 - B^2 = 1##. These are hyperbolae that can be parametrised ##A = \cosh(\theta)##, ##C = -\sinh(\theta)##, ##D = \cosh(\phi)##, ##B = -\sinh(\phi)##. Now ##AB = CD## directly leads to ##\sinh(\theta-\phi) = 0## and therefore ##\theta = \phi##. The Lorentz transformation is therefore
$$t' = \cosh(\theta) t - \sinh(\theta) x, \quad x' = \cosh(\theta) x - \sinh(\theta) t.$$
Done.

You can of course introduce ##v## as the speed with which ##x' = 0## moves in the original frame. This will give you ##v = x/t = \tanh(\theta)## and therefore ##\cosh(\theta) = \gamma(v)## and ##\sinh(\theta) = v\gamma(v)##.

vela, atyy, FactChecker and 6 others
That is a very sneaky way. I like it. Especially because of how easy it is to see that ##A^2 - C^2 = D^2 - B^2 = 1##. Oh, and how it utilizes the hyperbolic trigonometric identities (and the easiest ones to remember at that).

Orodruin
Staff Emeritus
Homework Helper
Gold Member
That is a very sneaky way. I like it. Especially because of how easy it is to see that ##A^2 - C^2 = D^2 - B^2 = 1##. Oh, and how it utilizes the hyperbolic trigonometric identities (and the easiest ones to remember at that).
It becomes even better if you start by showing the class the exact same way of deriving 2D rotations. The only difference is that you get ##\phi=-\theta## and trigonometric ones instead of hyperbolic.

atyy
It becomes even better if you start by showing the class the exact same way of deriving 2D rotations. The only difference is that you get ##\phi=-\theta## and trigonometric ones instead of hyperbolic.
I've had trouble with hyperbolic identities, but I've pretty well trained my eye that if I see anything like a2 + b2, or ##\sqrt{a^2 - b^2}## to instantly think trig identities. Reading through your derivation twice, I was surprised that I easily recognized ##cosh^2 (ϕ) - sinh^2(ϕ) = 1##. But yeah, I like this method, and it's going to make me want to reconsider running away from this form of parameterization that results in special relativity being described via hyperbolic functions.

Orodruin
Staff Emeritus
Homework Helper
Gold Member
I've had trouble with hyperbolic identities, but I've pretty well trained my eye that if I see anything like a2 + b2, or ##\sqrt{a^2 - b^2}## to instantly think trig identities. Reading through your derivation twice, I was surprised that I easily recognized ##cosh^2 (ϕ) - sinh^2(ϕ) = 1##. But yeah, I like this method, and it's going to make me want to reconsider running away from this form of parameterization that results in special relativity being described via hyperbolic functions.
Honestly, anybody interested in working with SR should start by reviewing the hyperbolic identities. It just makes the rest of the work much less cumbersome.

Klystron, vela and SiennaTheGr8
For a light signal from the origin ##t^2 - x^2 = 0## in any inertial frame. Therefore search for a (linear - homogeneous space etc etc) transformation that keeps ##s^2 = t^2 - x^2## invariant for all ##x## and ##t##. The general linear transformation is on the form
$$t' = At + Bx, \quad Ct + Dx$$
Missing an ##x'## in the second equation I believe. Also should be ##AB - CD = 0## directly leads to . . . (I reckon).

Orodruin
Staff Emeritus
Homework Helper
Gold Member
Missing an x′x′x' in the second equation I believe.
Yes.

Also should be ##AB - CD = 0## directly leads to . . . (I reckon).
That is just a rewriting of AB = CD ...

Yes.

That is just a rewriting of AB = CD ...
Agreed, hence It is merely a reckon ;) I just thought it fitted the visual flow better as I read it.

robphy
Homework Helper
Gold Member
While arguments about linear transformations are nice (and found in Einstein's paper)
and
arguments that introduce $t^2-x^2$ early on and hint at hyperbolic trig functions are nice,
I think they are abstract in the sense that they are somewhat distant from measurements that one would make.
So, if your target audience are math students, that might be okay.

I suggested Bondi's method, which uses radar measurements and the principle of relativity and simple algebra.
The signature of the square interval arises as a result. (Secretly, it uses the eigenbasis of the Lorentz transformation, where the Doppler factors are the eigenvalues and, of course, the eigenvectors are lightlike.)

Concerning various attempts to derive the Lorentz transformations...
Here's an ancient thread from 2006
https://www.physicsforums.com/threa...vations-of-the-lorentz-transformation.123103/
and another from 2005
https://www.physicsforums.com/threa...challenge-for-experts-only.83373/#post-694535

atyy
pervect
Staff Emeritus
I suggested Bondi's method, which uses radar measurements and the principle of relativity and simple algebra.
I second Bondi's approach as one of the most accessible ways of deriving the Lorentz transformation, using very physical arguments and high-school algebra.

Also, if one can get the reader to actually draw and understand space-time diagrams, Robphy's approach using rotated graph paper (see for instance https://arxiv.org/abs/1111.7254 or his physics forums insight article) is quite nice and very basic, without the focus on the Lorentz transformation. The key part is understanding how to draw the space-time diagram of a light clock for an observer at rest and for a moving observer. Getting to this point seems the hardest, if one gets the reader past this point the mathematical part is basically trivial.

My experience is that it seems hard to get readers to understand space-time diagrams. They resist drawing them themselves, and they also resist looking at diagrams others have drawn.

atyy and Sorcerer
robphy
Homework Helper
Gold Member
My experience is that it seems hard to get readers to understand space-time diagrams. They resist drawing them themselves, and they also resist looking at diagrams others have drawn.
It seems to me the tools that make help make relativity easier (e.g. spacetime diagrams, geometrical interpretations, rapidity and hyperbolic trig functions, 4-vectors) are not widely used because they aren't in introductory physics textbooks, and they aren't in there because Einstein didn't use them... and initially had some resistance to them. Many introductory textbooks seem to focus too much on the historical / pseudohistorical development of relativity and consider arriving at the Lorentz Transformations as the pinnacle of the presentation, regardless of whether students actually understand what is going on.

Often I hear that these tools are "too advanced" or "too mathematical".

(end of rant)

P.S. @pervect , thanks for the shout-out.

Dale
Orodruin
Staff Emeritus
Homework Helper
Gold Member
While arguments about linear transformations are nice (and found in Einstein's paper)
and
arguments that introduce $t^2-x^2$ early on and hint at hyperbolic trig functions are nice,
I think they are abstract in the sense that they are somewhat distant from measurements that one would make.
So, if your target audience are math students, that might be okay.

I suggested Bondi's method, which uses radar measurements and the principle of relativity and simple algebra.
The signature of the square interval arises as a result. (Secretly, it uses the eigenbasis of the Lorentz transformation, where the Doppler factors are the eigenvalues and, of course, the eigenvectors are lightlike.)
I do not have anything against Bondi's approach, but I do not think it is in the spirit of this thread. It may satisfy (1), but I do not think it satisfies (2).

I do not have anything against Bondi's approach, but I do not think it is in the spirit of this thread. It may satisfy (1), but I do not think it satisfies (2).
I’m open to both or either. I do think your hyperbolic function approach is a very nice middle ground. The math is not horribly difficult and its done in just a few lines.

robphy
Homework Helper
Gold Member
I’m open to both or either. I do think your hyperbolic function approach is a very nice middle ground. The math is not horribly difficult and its done in just a few lines.
In my opinion, trigonometry is so much more intuitive compared to "gamma factors" (=$\cosh\theta$).
In fact, when formulated trigonometrically, most textbook relativity problems
reduce to a mathematical problem
of either finding an angle (rapidity) [when finding a velocity]
or finding a side of a right triangle [when finding a length or a time or an energy or a momentum].
Of course, one needs to interpret physically...
.. but the whole business of "is it primed or unprimed?" or "should I use the time-dilation equation? or the length-contraction equation? or what?"
can be minimized.

If only students could learn to draw a spacetime diagram to realize that it's often really a trig problem...
...oh that's right... rapidity is too advanced.

m4r35n357 and Sorcerer
From
\begin{align}
t^2 - x^2 = (t - x)(t + x) = e^{-\phi}(t - x)e^{\phi}(t + x) = (t'-x')(t'+x') = t'^2 - x'^2
\end{align}
we have
\begin{align}
t' &= \frac{1}{2}[(t' + x') + (t' - x')] \\
&= \frac{1}{2} [e^{\phi}(t + x) + e^{-\phi}(t - x)] \\
&= \frac{1}{2} [\cosh \phi (t + x) + \sinh \phi (t + x) + \cosh \phi(t - x) - \sinh \phi (t - x)] \\
&= \frac{1}{2} [2 \cosh \phi t + 2 \sinh \phi x] \\
&= t \cosh \phi + x \sinh \phi , \\
x' &= \frac{1}{2}[(t' + x') - (t' - x')] \\
&= \frac{1}{2} [e^{\phi}(t + x) - e^{-\phi}(t - x)] \\
&= t \sinh \phi + x \cosh \phi
\end{align}
and on choosing a frame such that ##x' = 0## we find ##x = - u t = - \frac{\sinh \phi}{\cosh \phi} t = - (\tanh \phi ) t## which gives, on using ##\cosh^2 \phi - \sinh^2 \phi = 1## that ##1 - u^2 = \frac{1}{\cosh^2 \phi}## or ##\cosh \phi = \frac{1}{\sqrt{1-u^2}}## and ##\sinh \phi = \frac{u}{\sqrt{1-u^2}}## i.e.
\begin{align}
t' &= t \cosh \phi + x \sinh \phi \\
&= \frac{1}{\sqrt{1-u^2}} (t + u x) \\
x' &= t \sinh \phi + x \cosh \phi \\
&= \frac{1}{\sqrt{1-u^2}}(x + u t)
\end{align}
(Zee, Einstein's Gravity in a Nutshell', p 170)

atyy
In my opinion, trigonometry is so much more intuitive compared to "gamma factors" (=$\cosh\theta$).
In fact, when formulated trigonometrically, most textbook relativity problems
reduce to a mathematical problem
of either finding an angle (rapidity) [when finding a velocity]
or finding a side of a right triangle [when finding a length or a time or an energy or a momentum].
Of course, one needs to interpret physically...
.. but the whole business of "is it primed or unprimed?" or "should I use the time-dilation equation? or the length-contraction equation? or what?"
can be minimized.

If only students could learn to draw a spacetime diagram to realize that it's often really a trig problem...
...oh that's right... rapidity is too advanced.
As a student who has always preferred the algebraic approach over the geometric one, here’s some insight, I suppose.

Things like time and space coordinates are conceptually easy to comprehend. So, my guess is students feel more comfortable with the more complicated math of the usual way SR is introduced in text books (e.g., x = γ(x’ + vt’) ) because the things that make up the math are things that are more intuitively familiar (time, distance and speed).

Of course, the downside is that concepts like the relativity of simultaneity are NOT intuitively familiar, and are hard to see in the algebra for students at that level.

So it seems in the end the spacetime diagram approach would be better for student understanding, but GETTING there is a big problem because, as you point out somewhat facetiously, notions like rapidity aren’t something students are used to, while simple ideas like time and distance have been thrown at them since Physics 1 or earlier.

PeroK
robphy
Homework Helper
Gold Member
Things like time and space coordinates are conceptually easy to comprehend. So, my guess is students feel more comfortable with the more complicated math of the usual way SR is introduced in text books (e.g., x = γ(x’ + vt’) ) because the things that make up the math are things that are more intuitively familiar (time, distance and speed).
[snip]
So it seems in the end the spacetime diagram approach would be better for student understanding, but GETTING there is a big problem because, as you point out somewhat facetiously, notions like rapidity aren’t something students are used to, while simple ideas like time and distance have been thrown at them since Physics 1 or earlier.
I have no problem with time and space coordinates... they correspond to the sides of triangles.
My problem is with velocity (=slope) instead of rapidity (=angle).

When finding components of a vector
with magnitude and angle $\theta$, or magnitude and slope $m$,
do you use angles (as in $\cos\theta$ and $\sin\theta$) ?
or slopes (as in $\frac{1}{\sqrt{1+m^2}}$ and $\frac{m}{\sqrt{1+m^2}}$ ) ?

If a triangle is not in some standard position and you want the components with respect to some other tilted axes,
do you rotate the triangle into the standard position, then determine the components?
[This is what is done when someone says... let's transform into the rest frame... to avoid having to do "slope composition".]

I think using the invariant interval as a starting point is an interesting exercise, but sort of cheating. In particular it skips the helpful (IMO) step of comparing those "k" (=1/c^2) values which shows the distinction between Galilean and Minkowski spacetimes.

Mathematically the difference seems to come down to using the determinant of the LT rather than the proper time to generate the relationship between A, B, C, D but I'm not sure what the implications of that are.

I suppose I just prefer the spacetime interval to emerge as a consequence of the LT, not the other way round.

robphy
Homework Helper
Gold Member
From
\begin{align}
t^2 - x^2 = (t - x)(t + x) = e^{-\phi}(t - x)e^{\phi}(t + x) = (t'-x')(t'+x') = t'^2 - x'^2
\end{align}
[snip]
(Zee, Einstein's Gravity in a Nutshell', p 170)
This is essentially Bondi's approach... $e^{\phi}$ is Bondi's $k$, the Doppler Factor, which is the eigenvalue of the Lorentz boost.
$(t+x)$ and $(t-x)$ are essentially the radar-measurements, which are secretly light-cone coordinates.
However, Bondi doesn't start with the invariant... then magically write "1" in a fancy way, without motivation.... that is, you have to know the answer ahead of time.

atyy
I think using the invariant interval as a starting point is an interesting exercise, but sort of cheating. In particular it skips the helpful (IMO) step of comparing those "k" (=1/c^2) values which shows the distinction between Galilean and Minkowski spacetimes.

Mathematically the difference seems to come down to using the determinant of the LT rather than the proper time to generate the relationship between A, B, C, D but I'm not sure what the implications of that are.

I suppose I just prefer the spacetime interval to emerge as a consequence of the LT, not the other way round.
What I like about it, though, is that it essentially starts very thing that makes special relativity what it is. Instead of going through the "relative" stuff first, it starts with the invariant stuff. Obviously that's all preference. And I suppose you lose the historical motivation for getting to special relativity, which I guess is a big part of the conceptual argument for it given to students (and that's also probably why the old method of teaching it didn't put as much emphasis on spacetime diagrams as early education in special relativity seems to be moving toward today).

atyy
If only students could learn to draw a spacetime diagram to realize that it's often really a trig problem...
...oh that's right... rapidity is too advanced.
Can rapidity be used in GR?

robphy
Homework Helper
Gold Member
Can rapidity be used in GR?
In the tangent space at an event, yes.

Dale
Mentor
2020 Award
Many introductory textbooks seem to focus too much on the historical / pseudohistorical development
Yes, and this presentation leaves it as a disconnected set of weird facts and concepts. For me I didn’t get it for 7 years until I stumbled on spacetime diagrams and four vectors. A week later and I “got” Relativity.

I would have greatly benefited from someone introducing those tools far earlier.

vela, m4r35n357 and robphy