Easiest possible way to derive the Lorentz transformation

  • I
  • Thread starter Sorcerer
  • Start date
  • #1
281
52
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).
 

Answers and Replies

  • #2
281
52
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.
 
  • #3
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
18,485
8,388
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)##.
 
  • Like
Likes vela, atyy, FactChecker and 6 others
  • #4
281
52
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).
 
  • #5
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
18,485
8,388
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.
 
  • #6
281
52
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.
 
  • #7
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
18,485
8,388
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.
 
  • Like
Likes Klystron, vela and SiennaTheGr8
  • #8
654
148
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).
 
  • #9
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
18,485
8,388
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 ...
 
  • #10
654
148
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.
 
  • #11
robphy
Science Advisor
Homework Helper
Insights Author
Gold Member
6,288
1,613
While arguments about linear transformations are nice (and found in Einstein's paper)
and
arguments that introduce [itex]t^2-x^2[/itex] 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.

In an earlier thread,
[ https://www.physicsforums.com/threads/deriving-lorentz-transformation.957302/#post-6070325 ]
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
 
  • #12
pervect
Staff Emeritus
Science Advisor
Insights Author
10,147
1,298
In an earlier thread,
[ https://www.physicsforums.com/threads/deriving-lorentz-transformation.957302/#post-6070325 ]
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.
 
  • Like
Likes atyy and Sorcerer
  • #13
robphy
Science Advisor
Homework Helper
Insights Author
Gold Member
6,288
1,613
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.
 
  • #14
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
18,485
8,388
While arguments about linear transformations are nice (and found in Einstein's paper)
and
arguments that introduce [itex]t^2-x^2[/itex] 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.

In an earlier thread,
[ https://www.physicsforums.com/threads/deriving-lorentz-transformation.957302/#post-6070325 ]
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).
 
  • #15
281
52
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.
 
  • #16
robphy
Science Advisor
Homework Helper
Insights Author
Gold Member
6,288
1,613
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" (=[itex]\cosh\theta[/itex]).
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.
 
  • Like
Likes m4r35n357 and Sorcerer
  • #17
309
41
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)
 
  • #18
281
52
In my opinion, trigonometry is so much more intuitive compared to "gamma factors" (=[itex]\cosh\theta[/itex]).
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.
 
  • #19
robphy
Science Advisor
Homework Helper
Insights Author
Gold Member
6,288
1,613
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 [itex]\theta[/itex], or magnitude and slope [itex]m[/itex],
do you use angles (as in [itex]\cos\theta [/itex] and [itex]\sin\theta [/itex]) ?
or slopes (as in [itex]\frac{1}{\sqrt{1+m^2}} [/itex] and [itex]\frac{m}{\sqrt{1+m^2}} [/itex] ) ?

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".]
 
  • #20
654
148
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.
 
  • #21
robphy
Science Advisor
Homework Helper
Insights Author
Gold Member
6,288
1,613
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... [itex] e^{\phi}[/itex] is Bondi's [itex]k[/itex], the Doppler Factor, which is the eigenvalue of the Lorentz boost.
[itex](t+x) [/itex] and [itex](t-x) [/itex] 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.
 
  • #22
281
52
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).
 
  • #23
atyy
Science Advisor
14,751
3,257
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?
 
  • #25
Dale
Mentor
Insights Author
2021 Award
32,757
9,857
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.
 
  • Like
Likes vela, m4r35n357 and robphy
  • #26
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
21,494
12,807
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.

First, I'm not convinced that if relativity is taught one way or another it makes much difference to the key concepts that need to be understood by the student. Of course it can be taught badly or it can be taught well. But, assuming a well-presented set of material, how much difference does it make? I believe it would need an proper academic study of a large number of students .

Second, from a personal point of view, when I was learning SR the only thing I gave up on was the spacetime diagrams that tried to show two reference frames at once. I just didn't get it. Not the concept, but how to read things off the diagram - especially when the axes were not at right angles. I didn't like that at all!

Third, although I use diagrams and try to visualise things as much as possible, algebra has a huge advantage in that as soon as you draw a diagram it must have specifics. E.g. algebraically I can have a general quadratic: ##ax^2 + bx + c##. But, I cannot draw a general quadratic. To draw it, I have to take some specific values for ##a, b, c##. But, with algebra I can plough on generally for any and all values of ##a, b, c##.

Fourth, when I encountered the hyperbolic trig functions that replaced ##v/c## etc. I thought it was a neat idea and could make the algebra easier. But, if I then have to visualise ##\theta## as some (abstract) angle between four-vectors, then that is a big effort for me. Especially if I'm going to use it geometrically to help solve a problem. For me, ##\cosh## and ##\sinh## are functions with certain properties and, unlike ##\cos## and ##\sin##, I can't say I have a geometric intuition for them.

In summary, I could imagine being taught SR initially with a high dependence on the geometric aspects and having a rough time. Then, I could imagine finding a textbook that presented it based on the familiar concepts of space and time and an algebraic derivation of the Lorentz Transformation and thinking why didn't we do it this way from the start?!
 
  • #27
atyy
Science Advisor
14,751
3,257
In the tangent space at an event, yes.

Would that also be true for imaginary numbers (ict)?
 
  • #28
robphy
Science Advisor
Homework Helper
Insights Author
Gold Member
6,288
1,613
First, I'm not convinced that if relativity is taught one way or another it makes much difference to the key concepts that need to be understood by the student. Of course it can be taught badly or it can be taught well. But, assuming a well-presented set of material, how much difference does it make? I believe it would need an proper academic study of a large number of students .

Yes, a study is needed.
The only study I am aware of is one by Rachel Scherr @U.Wash
http://www.physics.umd.edu/perg/papers/scherr/dissertation/index.html
There must have been more since then.

PeroK said:
Second, from a personal point of view, when I was learning SR the only thing I gave up on was the spacetime diagrams that tried to show two reference frames at once. I just didn't get it. Not the concept, but how to read things off the diagram - especially when the axes were not at right angles. I didn't like that at all!

Agreed... that's one of the reasons why I developed my "rotated graph paper" and my "spacetime trigonometry" methods.

If relativity is about the "geometry of spacetime", I used to wonder where the "geometry" is.
In the most abstract treatments, it's in the invariant structures (e.g. spacetime intervals, curvature tensors, etc..).

One of the advocates of the spacetime diagram was JL Synge, whose textbooks in the 1950s-1960s were among the first to really use them.
Just a few quotes...
  • JL Synge - Relativity: the General Theory 1960 preface said:
    for a simple space-time diagram will often bring out the inner meaning of a mass of calculations. Surely one of the reasons why the general theory of relativity remains a mystery to so many physicists is that they do not realize how easy it is to form a qualitative geometrical image of what is going on. It is in fact easier to deal with space-time diagrams, which remain fixed, than with the kinematical pictures of Newtonian mechanics.

    [snip]
    It is to support Minkowski's way of looking at relativity that I find myself pursuing the hard path of the missionary. When, in a relativistic discussion, I try to make things clearer by a space-time diagram, the other participants look at it with polite detachment and, after a pause of embarrassment as if some childish indecency had been exhibited, resume the debate in their own terms.
  • JL Synge - Relativity: the Special Theory 1956 p.63 said:
    We are not embarking on a programme of "graphical. relativity". Our space-time diagrams are to be used as a mathematician or physicist uses rough sketches, rather than as an architect or engineer uses blueprints. The diagrams are to serve as guides for the mind. Anyone who studies relativity without understanding how to use simple space-time diagrams is as much inhibited as a student of functions of a complex variable who does not understand the Argand diagram.
However, I think one can get more quantitative information from spacetime diagrams using "rotated graph paper" and "spacetime trigonometry" methods and make them more accessible to the general student. So, I've been using them in my explanations here and elsewhere in an attempt to show how they give meaning (physically and mathematically) to the various equations one might use in the problem.



PeroK said:
Third, although I use diagrams and try to visualise things as much as possible, algebra has a huge advantage in that as soon as you draw a diagram it must have specifics. E.g. algebraically I can have a general quadratic: ##ax^2 + bx + c##. But, I cannot draw a general quadratic. To draw it, I have to take some specific values for ##a, b, c##. But, with algebra I can plough on generally for any and all values of ##a, b, c##.

Sure... both geometry and algebra have advantages and disadvantages.
I'm suggesting to use both!
It's fair to say that most introductory treatments (and most discussions here and elsewhere) don't use the spacetime diagram much.
So, we have algebra accompanied by long discussions of the meaning of terms and equations.
"A spacetime diagram is worth a thousand words."


PeroK said:
Fourth, when I encountered the hyperbolic trig functions that replaced ##v/c## etc. I thought it was a neat idea and could make the algebra easier. But, if I then have to visualise ##\theta## as some (abstract) angle between four-vectors, then that is a big effort for me. Especially if I'm going to use it geometrically to help solve a problem. For me, ##\cosh## and ##\sinh## are functions with certain properties and, unlike ##\cos## and ##\sin##, I can't say I have a geometric intuition for them.

Actually, what is important
is not the value of the angle (which can be visualized as the arc-length along the unit hyperbola)
but the idea
that there is an "angle" [the rapidity] between two timelike directions and that one can form "right triangles"
and that "[itex]\tanh(\rm angle)[/itex]" [the [itex](v/c)[/itex]-factor] is "opposite/adjacent = slope"
and that "[itex]\cosh(\rm angle)[/itex]" [the [itex]\gamma[/itex]-factor] is "adjacent/hypotenuse = projection onto the adjacent leg".
Assuming students are familiar with "[itex]\cos\theta[/itex]" from ordinary trigonometry,
I would think (with sufficient motivation) the "angle"-based "[itex]\cosh(\rm rapidity)[/itex]" might be easier
than the "slope"-based [itex]\gamma=\frac{1}{\sqrt{1-(v/c)^2}} [/itex].
(Does one use [itex]\frac{1}{\sqrt{1+m^2}} [/itex] much in ordinary trigonometry?)



PeroK said:
In summary, I could imagine being taught SR initially with a high dependence on the geometric aspects and having a rough time. Then, I could imagine finding a textbook that presented it based on the familiar concepts of space and time and an algebraic derivation of the Lorentz Transformation and thinking why didn't we do it this way from the start?!

If not taught well, I agree... and, as you say, this could apply to whatever method.

I don't have to imagine...
I see... that the typical textbook "based on the familiar concepts of space and time and an algebraic derivation of the Lorentz Transformation" leaves most students confused (is it primed or unprimed? Should I use time-dilation or length-contraction?). Appealing to geometry (suitably presented, and connected to what they already know from Euclidean geometry and trigonometry) gives students another crutch with which to explore Special Relativity.

Again, I don't see why one has to choose one or the other.
Spacetime diagrams and algebra can work together.... especially if one can learn to read the algebra from the diagrams.



  • Let me mention that there is an introductory textbook that tries to develop the spacetime diagram:Tom Moore's Six Ideas that Shaped Physics - Unit R
    https://www.amazon.com/dp/0077600959/?tag=pfamazon01-20
  • One last comment:
    the ordinary position-vs-time graph is a diagram that has an underlying non-euclidean geometry, where "elapsed wristwatch time" plays the role of a "length".. so that two objects tracked for one second of elapsed time trace out on a position-vs-time diagram different curves that have the same "length"... the usual Pythagorean theorem doesn't apply to these curves [even if straight]. However, we've learned to read such diagrams reasonably well.
 
Last edited:
  • Like
Likes Dale and SiennaTheGr8
  • #30
469
174
As a few people have mentioned already, the rapidity ##\phi## and hyperbolic functions are worth using even if you eschew spacetime diagrams. The rapidity simplifies much of the algebra, but it also provides a geometric understanding of what you're doing.

Rapidity is nice in the full 3+1D picture, too. Just vectorize it: ##\vec \phi = \phi \hat \phi##, with ##\hat \phi = \hat v##. Here are some four-vectors in units where ##c=1##, with only rapidity, hyperbolic functions, and invariants (overdot means proper-time derivative):

Four-velocity (future-pointing unit vector): ##\vec V = (\cosh{\phi}, \, \hat \phi \sinh{\phi})##

Infinitesimal four-displacement: ##d \vec R = \vec V \, d \tau = d \tau \, ( \cosh{\phi}, \, \hat \phi \sinh{\phi} ) ##

Four-momentum: ##\vec P = m \vec V = m \, ( \cosh{\phi}, \, \hat \phi \sinh{\phi} ) ##

Four-acceleration: ##\vec A = \dot{\vec V} = (\dot{\phi} \, \sinh{\phi}, \, \hat \phi \dot{\phi} \cosh{\phi} + \dot{\hat \phi} \sinh{\phi}) ##

Four-force (for constant ##m##): ##\vec F = \dot{\vec P} = m \vec A = m \, (\dot{\phi} \, \sinh{\phi}, \, \hat \phi \dot{\phi} \cosh{\phi} + \dot{\hat \phi} \sinh{\phi})##

Look how easy it is now to obtain, say, the negative squared magnitude of the four-acceleration ##\vec A## (i.e., the squared proper acceleration):

##\begin{align} - \vec A \cdot \vec A &= (\hat \phi \dot{\phi} \cosh{\phi} + \dot{\hat \phi} \sinh{\phi})^2 - (\dot{\phi} \, \sinh{\phi})^2 \nonumber \\ &= \dot{\phi}^2 + (\dot{\hat \phi} \sinh{\phi})^2 \nonumber \end{align}##

(because ##\hat \phi \cdot \dot{\hat \phi} = 0##, ##\hat \phi ^2 = 1##, and ##\cosh^2{\phi} - \sinh^2{\phi} = 1##). In the special case of rectilinear motion, ##\dot{\hat \phi} = \vec 0##, and the proper acceleration is simply the invariant ##\dot{\phi}## (proper-time derivative of the rapidity).
 

Related Threads on Easiest possible way to derive the Lorentz transformation

Replies
17
Views
803
Replies
13
Views
3K
Replies
2
Views
1K
  • Last Post
Replies
27
Views
4K
Replies
1
Views
7K
H
Replies
1
Views
834
Replies
18
Views
739
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
22
Views
2K
Top