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What amount of acceleration involved is too high for SR to apply?

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  What amount of acceleration involved is too high for SR to apply? amirjf nin <amirjfnin@aim.com> - 2026-03-31 17:12 -0400
    Re: What amount of acceleration involved is too high for SR to apply? Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2026-04-01 02:14 +0200
    Re: What amount of acceleration involved is too high for SR to apply? ram@zedat.fu-berlin.de (Stefan Ram) - 2026-04-02 03:07 +0000
      Re: What amount of acceleration involved is too high for SR to apply? Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2026-04-02 14:05 +0200
      Re: What amount of acceleration involved is too high for SR to apply? ram@zedat.fu-berlin.de (Stefan Ram) - 2026-04-03 00:21 +0000
        Re: What amount of acceleration involved is too high for SR to apply? Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2026-04-03 04:46 +0200
          Re: What amount of acceleration involved is too high for SR to apply? Ross Finlayson <ross.a.finlayson@gmail.com> - 2026-04-03 11:50 -0700
            Re: What amount of acceleration involved is too high for SR to apply? Ross Finlayson <ross.a.finlayson@gmail.com> - 2026-04-03 15:39 -0700
    Re: What amount of acceleration involved is too high for SR to apply? ram@zedat.fu-berlin.de (Stefan Ram) - 2026-04-02 23:03 +0000
      Re: What amount of acceleration involved is too high for SR to apply? Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2026-04-03 02:25 +0200

#670578 — What amount of acceleration involved is too high for SR to apply?

What amount of acceleration involved is too high for SR to apply? 
Numbers please.

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#670580

amirjf nin wrote:
> What amount of acceleration involved is too high for SR to apply? 

There is no (3-)acceleration whose Euclidean norm (which is what you
probably mean by "amount") is too large for SR to apply.

How did you get that idea in the first place?

-- 
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.

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#670604

amirjf nin <amirjfnin@aim.com> wrote or quoted:
>What amount of acceleration involved is too high for SR to apply? 
>Numbers please.

  This is a question of how you define "special relativity":

  - When accelerated motion is being described, do you deem
    this to be SR?

  - When the observer himself is being accelerated, is this 
    still SR?

  In his early 1905 paper, Einstein focused on inertial frames.

  But in 1973, the book "Gravitation" (Misner et al.) said,

|Accelerated motion and accelerated observers can be analyzed
|using special relativity.
(page 163).

  So, it seems, today, both kinds of acceleration are deemed to
  still be part of SR.

  Only when spacetime is /curved/ it's deemed to be GR. (See 
  chapters 6 and 7 of that book.)

  (The metric for an accelerated observer in SR [using Rindler
  coordinates] looks like a homogeneous gravitational field,
  but even this is still SR!)

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#670609

Stefan Ram wrote:
> amirjf nin <amirjfnin@aim.com> wrote or quoted:
>> What amount of acceleration involved is too high for SR to apply? 
>> Numbers please.
> 
>   This is a question of how you define "special relativity":

No, it is not.

>   - When accelerated motion is being described, do you deem
>     this to be SR?

Yes.  Contrary to common misconception, acceleration is well-defined in SR.
There are in fact several concepts of acceleration in SR.

>   - When the observer himself is being accelerated, is this 
>     still SR?

Yes.

>   In his early 1905 paper, Einstein focused on inertial frames.

In that paper, he already discussed a "(slowly accelerated) electron":

<https://onlinelibrary.wiley.com/doi/epdf/10.1002/andp.19053221004>

page 27

>   But in 1973, the book "Gravitation" (Misner et al.) said,
> 
> |Accelerated motion and accelerated observers can be analyzed
> |using special relativity.
> (page 163).
> 
>   So, it seems, today, both kinds of acceleration are deemed to
>   still be part of SR.

A way to handle acceleration in SR is by considering momentarily inertial
reference frames:

<https://en.wikipedia.org/w/index.php?title=Acceleration_(special_relativity)&oldid=1338748125>

>   Only when spacetime is /curved/ it's deemed to be GR. (See 
>   chapters 6 and 7 of that book.)

GR applies *particularly* when the spacetime is curved; but NOT *only*, as
can be shown by deriving the equation for the special-relativistic elapsed
proper time from the Schwarzschild metric (instead of from the Lorentz
transformation):

       ds^2 = -(1 - 2m/r) dt^2 + (1 - 2m/r)^-1 dr^2 + r^2 (dΩ)^2,
              (dΩ)^2 = (dθ)^2 + sin^2(θ) (dφ)^2
            = -dt^2 [(1 - 2m/r) - (1 - 2m/r)^-1 (dr/dt)^2 + r^2 (dΩ/dt)^2]
            = -dt^2 [(1 - 2m/r) - (1 - 2m/r)^-1 (dr/dt)^2], (dΩ)^2 = 0
  <==>   dτ = dt √[(1 - 2m/r) - (1 - 2m/r)^-1 (dr/dt)^2]
  <==>   dτ = dt √(1 - v^2), m = 0 or r → ∞
  <==>   dτ = dt/γ(v)
  <==>  ∆τ := ∫_W dτ = ∆t/γ(v).  QED.

  [You can derive the Minkowski metric for a flat spacetime the same way:
   as m = 0 or r → ∞, the result is the Minkowski metric in spherical
   coordinates.]

This should not be surprising, though; any good new theory contains an
experimentally well established theory as a special case:

StarTalk: Why Science Doesn’t Make Laws Anymore
<https://youtu.be/EVJdwD7coQ4?si=4CymnorwlbEMvsit>

-- 
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.

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#670623

ram@zedat.fu-berlin.de (Stefan Ram) wrote or quoted:
>This is a question of how you define "special relativity":

  In his 1914 work "The formal foundation of the general theory
  of relativity" [1] Einstein did not yet use "special theory
  of relativity" [2], but he wrote,

|This view recalls that of the original (more special) theory
|of relativity, in which an electric charge moving in a
|magnetic field . . . 
[3]

  , using "more special" [4].

  But in his 1915 work "On the General Theory of Relativity." [5]
  he then writes,

|Just as the special theory of relativity is founded on the
|postulate that its equations should be covariant with respect
|to linear, orthogonal transformations, so the theory
|presented here rests on the postulate of the covariance of
|all systems of equations with respect to transformations of
|determinant 1.
[6]

  . This might be the first occurence of the term "special theory
  of relativity" [2]. So, it seems that at this time accelerated
  observers were not yet part of the special theory!

  Original German wordings:

  [1]

|Die formale Grundlage der allgemeinen Relativitätstheorie

  [2]

|spezielle Relativitätstheorie

  [3]

|Es erinnert diese Auffassung an diejenige der ursprünglichen
|(spezielleren) Relativitätstheorie, daß man die auf eine in
|einem Magnetfelde bewegte elektrische Masse . . .

  [4]

|Zur allgemeinen Relativitätstheorie.

  [5]

|Wie die spezielle Relativitätstheorie auf das Postulat
|gegründet ist, daß ihre Gleichungen bezüglich linearer,
|orthogonaler Transformationen kovariant sein sollen, so ruht
|die hier darzulegende Theorie auf dem Postulat der Kovarianz
|aller Gleichungssysteme bezüglich Transformationen von der
|Substitutionsdeterminante 1.

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#670625

Stefan Ram wrote:
> ram@zedat.fu-berlin.de (Stefan Ram) wrote or quoted:
>> This is a question of how you define "special relativity":
> 
>   In his 1914 work "The formal foundation of the general theory
>   of relativity" [1] Einstein did not yet use "special theory
>   of relativity" [2], but he wrote,
> 
> |This view recalls that of the original (more special) theory
> |of relativity, in which an electric charge moving in a
> |magnetic field . . . 
> [3]
> 
>   , using "more special" [4].
> 
>   But in his 1915 work "On the General Theory of Relativity." [5]
>   he then writes,
> 
> |Just as the special theory of relativity is founded on the
> |postulate that its equations should be covariant with respect
> |to linear, orthogonal transformations, so the theory
> |presented here rests on the postulate of the covariance of
> |all systems of equations with respect to transformations of
> |determinant 1.
> [6]
> 
>   . This might be the first occurence of the term "special theory
>   of relativity" [2]. So, it seems that at this time accelerated
>   observers were not yet part of the special theory!

/Non sequitur./  "Linear, orthogonal transformation" has nothing inherently
to do with "uniform motion along a straight line", i.e. inertial motion.

Instead, a transformation T is *linear* if

  T(a X + b Y) = a T(X) + b T(Y),

where a and b are scalars, and X and Y are vectors in the transformation's
domain (in the degenerate case those can also be scalars, but let's not
confuse ourselves with ambiguities).

Such a transformation can be represented as matrix--vector multiplication:

  X' = T(X) = T X,

where T may be represented as a matrix of row dimension m and colum
dimension n for an X of dimension n and an X' of dimension m, because the
vector X can be represented as a linear combination of basis vectors,

  X = x_1 B_1 + x_2 B_2 + ...,

where the x_i are scalars, the components of X, and the B_i are the basis
vectors.  Then if the transformation is linear,

  T(X) = T(x_1 B_1 + x_2 B_2 + ...) = x_1 T(B_1) + x_2 T(B_2) + ...

and it suffices to transform the basis vectors; then to obtain X' in the
basis of X, find the corresponding components of X' -- x_1', x_2', etc. --
in the basis of X.  For example, consider the transformation matrix

  T = [1 0].
      [2 1]

When it is applied to a vector, it produces, for example

  [1 0] [2] = [2].
  [2 1] [1]   [5]

But the input vector there can also be written

        [2] = 2 [1] + 1 [0]
        [1]     [0]     [1]

and it actually suffices to transform the basis vectors (1, 0)^T and (0, 1)^T:

  [1 0] [1] = [1],  [1 0] [0] = [0].
  [2 1] [0]   [2]   [2 1] [1]   [1]

It is not a coincidence that the transformed basis vectors look exactly like
the first and second column of the transformation matrix.  (See the link
below.)  Anyhow, if we put this together:

  [1 0] [2] = [1 0] (2 [1] + 1 [0])
  [2 1] [1]   [2 1] (  [0]     [1])

            = 2 [1 0] [1] + [1 0] [0]
                [2 1] [0]   [2 1] [1]

            = 2 [1] + 1 [0]
                [2]     [1]

            = [2] + [0]
              [4]   [1]

            = [2]
              [5]

as required.

The Lorentz transformation is such a (linear) transformation.  For motion of
the primed frame (t', x', y', z') in the x-direction of the unprimed frame
(t, x, y, z), relative to the latter frame at the speed v:

    t' = γ [t - (v/c^2) x]
    x' = γ (x - v t)
    y' = y
    z' = z.

That it is indeed linear is not trivial to show, let alone see, unless you
change the time coordinate in a clever way, by multiplying the time
transformation equation by c:

  c t' = γ [c t - (v/c) x  ]
    x' = γ [x   - (v/c) c t]
    y' = y
    z' = z.

Now the coordinates can be relabeled, and for reasons that will become
apparent later, it is done this way:

  (x^0, x^1, x^2, x^3)^T := (c t, x, y, z)^T.

Then the transformation reads

  (x')^0 = γ [x^0 - (v/c) x^1]
  (x')^1 = γ [x^1 - (v/c) x^0]
  (x')^2 = x^2
  (x')^3 = x^3.

  [Notice the symmetry between the first equation, the time transformation,
   and the second equation in which the first spatial coordinate is
   transformed.  Again, that is not a coincidence.]

It is further useful to define β := v/c.

  (x')^0 = γ [x^0 - β x^1]
  (x')^1 = γ [x^1 - β x^0]
  (x')^2 = x^2
  (x')^3 = x^3.

And the transformation *can* be written as matrix--vector multiplication,
where one introduces the *4-vectors* X' and X, and the Lorentz
transformation matrix L:

  [(x')^0]    [ γ     -γ β   0   0] [x^0]
  [(x')^1] =  [-γ β    γ     0   0] [x^1]
  [(x')^2]    [ 0      0     1   0] [x^2]
  [(x')^3]    [ 0      0     0   1] [x^3]
  `---.--'    `---------.---------' `-.-'
      X'                L             X

as required.

Notice that, for example, multiplying the first row of the Lorentz
transformation matrix with the position 4-vector X gives

  γ x^0 - γ β x^1 + 0 x^2 + 0 x^3 = γ [x^0 - β x^1]

which is indeed what (x')^0 should be.

This is essentially what Minkowski did in 1908: introduce a 4-dimensional
manifold that he called "space-time" (in the original German: "Raum-Zeit";
now overwhelmingly "spacetime" and "Raumzeit" instead, respectively).

That the determinant of this matrix/transformation is indeed 1 is again not
trivial to show (although it is just an algebraic exercise).  However,
noticing that

  γ(v) = 1/√(1 - v^2/c^2) = 1/√(1 - β²),

one can write

  γ² (1 - β²) = γ² - γ² β² = 1.

There is a known mathematical identity: [1]

  cosh^2(x) - sinh^2(x) = 1.

So one can identify

  cosh(w) = γ,
  sinh(w) = γ β,

and find

  tanh(w) = sinh(w)/cosh(w) = β = v/c.

w = artanh(β) = artanh(v/c) is thus called *rapidity*.  [It is very useful
in discussing constant accelerations, as I will show another time.]

Then the Lorentz transformation matrix can be written

      [ cosh(w)  -sinh(w)  0  0]
  L = [-sinh(w)   cosh(w)  0  0].
      [       0         0  1  0]
      [       0         0  0  1]

And now it is much easier to show, using the aforementioned identity, that
the determinant of this matrix (and thus of the transformation) is indeed 1.
 By Laplace expansion along the fourth row or column, and again along the
third row or column, one finds

        | cosh(w)  -sinh(w)  0  0|   | cosh(w)  -sinh(w)  0|
  |L| = |-sinh(w)   cosh(w)  0  0| = |-sinh(w)   cosh(w)  0|
        |       0         0  1  0|   |       0         0  1|
        |       0         0  0  1|

      = | cosh(w)  -sinh(w)| = cosh^2(w) - sinh^2(w) = 1.
        |-sinh(w)   cosh(w)|

   [This is one example of many where a Physics student's required
    first- and second-semester Mathematics course "Linear Algebra"
    turns out to be very useful.  So hang on, it's worth it :)  BTDT.]

It is important that the determinant is equal to 1 because that means that
scale is preserved: there is no stretching or squishing in any direction,
and no inversion of direction, of any 4-vector.  [As you can find
beautifully explained graphically for Euclidean space here:
<https://www.youtube.com/watch?v=Ip3X9LOh2dk&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab&index=6>]

In fact, a Lorentz transformation matrix and in general the Lorentz
transformation is -- as one can see now -- a *rotation* (matrix) in a
hyperbolic geometry [the matrix for a counter-clockwise rotation by the
angle α in 3-dimensional Euclidean space has cos(α) and sin(α), and
different signs instead]; and like with all rotations, the determinant
MUST be equal to 1 as a rotation preserves lengths (and thus angles in
3-dimension Euclidean space, too).

<https://en.wikipedia.org/w/index.php?title=Rapidity&oldid=1327771719>

In group theory, the Lorentz transformations form a group, the Lorentz
group, that is isomorph to the special orthogonal group SO(3, 1) [rotations
in 3-dimensional Euclidean space are isomorph to SO(3)], the group whose
members can be represented by matrices M that satisfy M^T M = I, i.e. M^T =
M^-1 (*orthogonal* matrices, where I means the identity matrix) and |M| = 1
(therefore "special").

<https://en.wikipedia.org/w/index.php?title=Lorentz_group&oldid=1345728644>

___
[1] This identity follows from the definition of the hyperbolic functions,
    where cosh is defined as the even part of the exponential function,
    and sinh as its odd part.  Any function can be split into a part that
    is even [g(-x) = g(x)] and a part that is odd [h(-x) = -h(x)]:

       f(x) = [f(x) + f(-x)]/2 + [f(x) - f(-x)]/2 =: g(x) + h(x).
      g(-x) = [f(-x) + f(x)]/2 =  [f(x) + f(-x)]/2 =  g(x),
      h(-x) = [f(-x) - f(x)]/2 = -[f(x) - f(-x)]/2 = -h(x).

    For the exponential function:

        e^x = [e^x + e^{-x}]/2 + [e^x - e^{-x}]/2 =: cosh(x) + sinh(x).

    Then

      cosh^2(x) = [e^x + e^{-x}]^2/2^2
                = [e^{2x} + 2 e^x e^{-x} + e^{-2x}]/4; [binomial theorem]
                = [e^{2x} + 2 + e^{-2x}]/4,

      sinh^2(x) = [e^x - e^{-x}]^2/2^2
                = [e^{2x} - 2 + e^{-2x}]/4,

    and finally

      cosh^2(x) - sinh^2(x)
        = [e^{2x} + 2 + e^{-2x} - e^{2x} + 2 - e^{-2x}]/4,
        = 4/4
        = 1.
-- 
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.

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#670643

On 04/02/2026 07:46 PM, Thomas 'PointedEars' Lahn wrote:
> Stefan Ram wrote:
>> ram@zedat.fu-berlin.de (Stefan Ram) wrote or quoted:
>>> This is a question of how you define "special relativity":
>>
>>    In his 1914 work "The formal foundation of the general theory
>>    of relativity" [1] Einstein did not yet use "special theory
>>    of relativity" [2], but he wrote,
>>
>> |This view recalls that of the original (more special) theory
>> |of relativity, in which an electric charge moving in a
>> |magnetic field . . .
>> [3]
>>
>>    , using "more special" [4].
>>
>>    But in his 1915 work "On the General Theory of Relativity." [5]
>>    he then writes,
>>
>> |Just as the special theory of relativity is founded on the
>> |postulate that its equations should be covariant with respect
>> |to linear, orthogonal transformations, so the theory
>> |presented here rests on the postulate of the covariance of
>> |all systems of equations with respect to transformations of
>> |determinant 1.
>> [6]
>>
>>    . This might be the first occurence of the term "special theory
>>    of relativity" [2]. So, it seems that at this time accelerated
>>    observers were not yet part of the special theory!
>
> /Non sequitur./  "Linear, orthogonal transformation" has nothing inherently
> to do with "uniform motion along a straight line", i.e. inertial motion.
>
> Instead, a transformation T is *linear* if
>
>    T(a X + b Y) = a T(X) + b T(Y),
>
> where a and b are scalars, and X and Y are vectors in the transformation's
> domain (in the degenerate case those can also be scalars, but let's not
> confuse ourselves with ambiguities).
>
> Such a transformation can be represented as matrix--vector multiplication:
>
>    X' = T(X) = T X,
>
> where T may be represented as a matrix of row dimension m and colum
> dimension n for an X of dimension n and an X' of dimension m, because the
> vector X can be represented as a linear combination of basis vectors,
>
>    X = x_1 B_1 + x_2 B_2 + ...,
>
> where the x_i are scalars, the components of X, and the B_i are the basis
> vectors.  Then if the transformation is linear,
>
>    T(X) = T(x_1 B_1 + x_2 B_2 + ...) = x_1 T(B_1) + x_2 T(B_2) + ...
>
> and it suffices to transform the basis vectors; then to obtain X' in the
> basis of X, find the corresponding components of X' -- x_1', x_2', etc. --
> in the basis of X.  For example, consider the transformation matrix
>
>    T = [1 0].
>        [2 1]
>
> When it is applied to a vector, it produces, for example
>
>    [1 0] [2] = [2].
>    [2 1] [1]   [5]
>
> But the input vector there can also be written
>
>          [2] = 2 [1] + 1 [0]
>          [1]     [0]     [1]
>
> and it actually suffices to transform the basis vectors (1, 0)^T and (0, 1)^T:
>
>    [1 0] [1] = [1],  [1 0] [0] = [0].
>    [2 1] [0]   [2]   [2 1] [1]   [1]
>
> It is not a coincidence that the transformed basis vectors look exactly like
> the first and second column of the transformation matrix.  (See the link
> below.)  Anyhow, if we put this together:
>
>    [1 0] [2] = [1 0] (2 [1] + 1 [0])
>    [2 1] [1]   [2 1] (  [0]     [1])
>
>              = 2 [1 0] [1] + [1 0] [0]
>                  [2 1] [0]   [2 1] [1]
>
>              = 2 [1] + 1 [0]
>                  [2]     [1]
>
>              = [2] + [0]
>                [4]   [1]
>
>              = [2]
>                [5]
>
> as required.
>
> The Lorentz transformation is such a (linear) transformation.  For motion of
> the primed frame (t', x', y', z') in the x-direction of the unprimed frame
> (t, x, y, z), relative to the latter frame at the speed v:
>
>      t' = γ [t - (v/c^2) x]
>      x' = γ (x - v t)
>      y' = y
>      z' = z.
>
> That it is indeed linear is not trivial to show, let alone see, unless you
> change the time coordinate in a clever way, by multiplying the time
> transformation equation by c:
>
>    c t' = γ [c t - (v/c) x  ]
>      x' = γ [x   - (v/c) c t]
>      y' = y
>      z' = z.
>
> Now the coordinates can be relabeled, and for reasons that will become
> apparent later, it is done this way:
>
>    (x^0, x^1, x^2, x^3)^T := (c t, x, y, z)^T.
>
> Then the transformation reads
>
>    (x')^0 = γ [x^0 - (v/c) x^1]
>    (x')^1 = γ [x^1 - (v/c) x^0]
>    (x')^2 = x^2
>    (x')^3 = x^3.
>
>    [Notice the symmetry between the first equation, the time transformation,
>     and the second equation in which the first spatial coordinate is
>     transformed.  Again, that is not a coincidence.]
>
> It is further useful to define β := v/c.
>
>    (x')^0 = γ [x^0 - β x^1]
>    (x')^1 = γ [x^1 - β x^0]
>    (x')^2 = x^2
>    (x')^3 = x^3.
>
> And the transformation *can* be written as matrix--vector multiplication,
> where one introduces the *4-vectors* X' and X, and the Lorentz
> transformation matrix L:
>
>    [(x')^0]    [ γ     -γ β   0   0] [x^0]
>    [(x')^1] =  [-γ β    γ     0   0] [x^1]
>    [(x')^2]    [ 0      0     1   0] [x^2]
>    [(x')^3]    [ 0      0     0   1] [x^3]
>    `---.--'    `---------.---------' `-.-'
>        X'                L             X
>
> as required.
>
> Notice that, for example, multiplying the first row of the Lorentz
> transformation matrix with the position 4-vector X gives
>
>    γ x^0 - γ β x^1 + 0 x^2 + 0 x^3 = γ [x^0 - β x^1]
>
> which is indeed what (x')^0 should be.
>
> This is essentially what Minkowski did in 1908: introduce a 4-dimensional
> manifold that he called "space-time" (in the original German: "Raum-Zeit";
> now overwhelmingly "spacetime" and "Raumzeit" instead, respectively).
>
> That the determinant of this matrix/transformation is indeed 1 is again not
> trivial to show (although it is just an algebraic exercise).  However,
> noticing that
>
>    γ(v) = 1/√(1 - v^2/c^2) = 1/√(1 - β²),
>
> one can write
>
>    γ² (1 - β²) = γ² - γ² β² = 1.
>
> There is a known mathematical identity: [1]
>
>    cosh^2(x) - sinh^2(x) = 1.
>
> So one can identify
>
>    cosh(w) = γ,
>    sinh(w) = γ β,
>
> and find
>
>    tanh(w) = sinh(w)/cosh(w) = β = v/c.
>
> w = artanh(β) = artanh(v/c) is thus called *rapidity*.  [It is very useful
> in discussing constant accelerations, as I will show another time.]
>
> Then the Lorentz transformation matrix can be written
>
>        [ cosh(w)  -sinh(w)  0  0]
>    L = [-sinh(w)   cosh(w)  0  0].
>        [       0         0  1  0]
>        [       0         0  0  1]
>
> And now it is much easier to show, using the aforementioned identity, that
> the determinant of this matrix (and thus of the transformation) is indeed 1.
>   By Laplace expansion along the fourth row or column, and again along the
> third row or column, one finds
>
>          | cosh(w)  -sinh(w)  0  0|   | cosh(w)  -sinh(w)  0|
>    |L| = |-sinh(w)   cosh(w)  0  0| = |-sinh(w)   cosh(w)  0|
>          |       0         0  1  0|   |       0         0  1|
>          |       0         0  0  1|
>
>        = | cosh(w)  -sinh(w)| = cosh^2(w) - sinh^2(w) = 1.
>          |-sinh(w)   cosh(w)|
>
>     [This is one example of many where a Physics student's required
>      first- and second-semester Mathematics course "Linear Algebra"
>      turns out to be very useful.  So hang on, it's worth it :)  BTDT.]
>
> It is important that the determinant is equal to 1 because that means that
> scale is preserved: there is no stretching or squishing in any direction,
> and no inversion of direction, of any 4-vector.  [As you can find
> beautifully explained graphically for Euclidean space here:
> <https://www.youtube.com/watch?v=Ip3X9LOh2dk&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab&index=6>]
>
> In fact, a Lorentz transformation matrix and in general the Lorentz
> transformation is -- as one can see now -- a *rotation* (matrix) in a
> hyperbolic geometry [the matrix for a counter-clockwise rotation by the
> angle α in 3-dimensional Euclidean space has cos(α) and sin(α), and
> different signs instead]; and like with all rotations, the determinant
> MUST be equal to 1 as a rotation preserves lengths (and thus angles in
> 3-dimension Euclidean space, too).
>
> <https://en.wikipedia.org/w/index.php?title=Rapidity&oldid=1327771719>
>
> In group theory, the Lorentz transformations form a group, the Lorentz
> group, that is isomorph to the special orthogonal group SO(3, 1) [rotations
> in 3-dimensional Euclidean space are isomorph to SO(3)], the group whose
> members can be represented by matrices M that satisfy M^T M = I, i.e. M^T =
> M^-1 (*orthogonal* matrices, where I means the identity matrix) and |M| = 1
> (therefore "special").
>
> <https://en.wikipedia.org/w/index.php?title=Lorentz_group&oldid=1345728644>
>
> ___
> [1] This identity follows from the definition of the hyperbolic functions,
>      where cosh is defined as the even part of the exponential function,
>      and sinh as its odd part.  Any function can be split into a part that
>      is even [g(-x) = g(x)] and a part that is odd [h(-x) = -h(x)]:
>
>         f(x) = [f(x) + f(-x)]/2 + [f(x) - f(-x)]/2 =: g(x) + h(x).
>        g(-x) = [f(-x) + f(x)]/2 =  [f(x) + f(-x)]/2 =  g(x),
>        h(-x) = [f(-x) - f(x)]/2 = -[f(x) - f(-x)]/2 = -h(x).
>
>      For the exponential function:
>
>          e^x = [e^x + e^{-x}]/2 + [e^x - e^{-x}]/2 =: cosh(x) + sinh(x).
>
>      Then
>
>        cosh^2(x) = [e^x + e^{-x}]^2/2^2
>                  = [e^{2x} + 2 e^x e^{-x} + e^{-2x}]/4; [binomial theorem]
>                  = [e^{2x} + 2 + e^{-2x}]/4,
>
>        sinh^2(x) = [e^x - e^{-x}]^2/2^2
>                  = [e^{2x} - 2 + e^{-2x}]/4,
>
>      and finally
>
>        cosh^2(x) - sinh^2(x)
>          = [e^{2x} + 2 + e^{-2x} - e^{2x} + 2 - e^{-2x}]/4,
>          = 4/4
>          = 1.
>


Thanks for writing. Derivations are nice.

Here the account of the "hyperbolic-trigonometric" and making
for ideas for the "determinantal analysis" as it is, has that
these are _techniques_ of analysis, as for what _concepts_
they're intended to entail.

For example, when Einstein wrote "this is the way we do", that's
not the same necessarily as "this is how and what we do", about,
"this is why we do".

So, for reading from Weyl's "Levels of Infinity", about Hilbert's
account of "invariant theory" and the "Nullstellensatz", which
is an account of the "invariant theory" making for determinantal
analysis, there's that the Desarguesian (or Arguesian) about which
is to be making an account for "projective geometry" for "invariant"
theory, explains that that technique is secondary to the analysis,
the results.

Then invariant theory of course is quite de rigeur these days
in the mathematical physics, it's directly related to the accounts
of Noether's theorem that symmetries make conservation laws,
it "is" the "invariant theory" or theory of invariants. So,
Hilbert's account of the _Nullstellensatz_, "empty star", about
the inner product (the usual scalar product the determinant in
the determinantal analysis) making a boundary about which is
symmetries making for an account of the invariant, and thusly
about the account of "not symmetry-breaking", where "symmetry-breaking"
in accounts like these of "accelerated co-moving frames" is the
usual setting in relativity theory looking for physics, has that
this sort of technique is due "invariant theory" since Hilbert.


So, about Desargues theorem and the resulting _identities_ that
result from the _constructions_, geometrically, that then being
a sort of second level after triangle/Cauchy-Schwarz inequality,
the deconstructive account is about what "symmetry-flex" instead
of "symmetry-breaking" is, why there's an account of "continuity
law" more thorough than "conservation law".


Here for example there's a paper of Maddux "Identities Generalizing
the Theorems of Pappus and Desargues", these being fundamental in
the projective geometry, as describes some ideas about that besides
the determinantal analysis and including things like the "cumulants"
and "orthogonants" and for things like Schwartz functions about
the broader setting of linear algebra and including out into the
generalized matrix products and inverses and the matroid setting,
about why the technique described isn't so much a part of the
mathematical model the physical model its interpretation the theory,
instead just another "linearisation".

https://www.mdpi.com/2073-8994/13/8/1382


So, these being the things, here for example in this video essay
"Reading Foundations: replete anti-reductionism", after the
recent "continuous quanta", "double relativity", and "rational
rational algebras", is about then the accounts of completions
of rational algebras, here about the wider setting of the
"determinantal analysis", as that's usually termed "Hodge dual",
about "voiding the Nullstellensatz".

https://en.wikipedia.org/wiki/Hilbert%27s_Nullstellensatz


Rational? It's a continuum mechanics, ....

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#670649

On 04/03/2026 11:50 AM, Ross Finlayson wrote:
> On 04/02/2026 07:46 PM, Thomas 'PointedEars' Lahn wrote:
>> Stefan Ram wrote:
>>> ram@zedat.fu-berlin.de (Stefan Ram) wrote or quoted:
>>>> This is a question of how you define "special relativity":
>>>
>>>    In his 1914 work "The formal foundation of the general theory
>>>    of relativity" [1] Einstein did not yet use "special theory
>>>    of relativity" [2], but he wrote,
>>>
>>> |This view recalls that of the original (more special) theory
>>> |of relativity, in which an electric charge moving in a
>>> |magnetic field . . .
>>> [3]
>>>
>>>    , using "more special" [4].
>>>
>>>    But in his 1915 work "On the General Theory of Relativity." [5]
>>>    he then writes,
>>>
>>> |Just as the special theory of relativity is founded on the
>>> |postulate that its equations should be covariant with respect
>>> |to linear, orthogonal transformations, so the theory
>>> |presented here rests on the postulate of the covariance of
>>> |all systems of equations with respect to transformations of
>>> |determinant 1.
>>> [6]
>>>
>>>    . This might be the first occurence of the term "special theory
>>>    of relativity" [2]. So, it seems that at this time accelerated
>>>    observers were not yet part of the special theory!
>>
>> /Non sequitur./  "Linear, orthogonal transformation" has nothing
>> inherently
>> to do with "uniform motion along a straight line", i.e. inertial motion.
>>
>> Instead, a transformation T is *linear* if
>>
>>    T(a X + b Y) = a T(X) + b T(Y),
>>
>> where a and b are scalars, and X and Y are vectors in the
>> transformation's
>> domain (in the degenerate case those can also be scalars, but let's not
>> confuse ourselves with ambiguities).
>>
>> Such a transformation can be represented as matrix--vector
>> multiplication:
>>
>>    X' = T(X) = T X,
>>
>> where T may be represented as a matrix of row dimension m and colum
>> dimension n for an X of dimension n and an X' of dimension m, because the
>> vector X can be represented as a linear combination of basis vectors,
>>
>>    X = x_1 B_1 + x_2 B_2 + ...,
>>
>> where the x_i are scalars, the components of X, and the B_i are the basis
>> vectors.  Then if the transformation is linear,
>>
>>    T(X) = T(x_1 B_1 + x_2 B_2 + ...) = x_1 T(B_1) + x_2 T(B_2) + ...
>>
>> and it suffices to transform the basis vectors; then to obtain X' in the
>> basis of X, find the corresponding components of X' -- x_1', x_2',
>> etc. --
>> in the basis of X.  For example, consider the transformation matrix
>>
>>    T = [1 0].
>>        [2 1]
>>
>> When it is applied to a vector, it produces, for example
>>
>>    [1 0] [2] = [2].
>>    [2 1] [1]   [5]
>>
>> But the input vector there can also be written
>>
>>          [2] = 2 [1] + 1 [0]
>>          [1]     [0]     [1]
>>
>> and it actually suffices to transform the basis vectors (1, 0)^T and
>> (0, 1)^T:
>>
>>    [1 0] [1] = [1],  [1 0] [0] = [0].
>>    [2 1] [0]   [2]   [2 1] [1]   [1]
>>
>> It is not a coincidence that the transformed basis vectors look
>> exactly like
>> the first and second column of the transformation matrix.  (See the link
>> below.)  Anyhow, if we put this together:
>>
>>    [1 0] [2] = [1 0] (2 [1] + 1 [0])
>>    [2 1] [1]   [2 1] (  [0]     [1])
>>
>>              = 2 [1 0] [1] + [1 0] [0]
>>                  [2 1] [0]   [2 1] [1]
>>
>>              = 2 [1] + 1 [0]
>>                  [2]     [1]
>>
>>              = [2] + [0]
>>                [4]   [1]
>>
>>              = [2]
>>                [5]
>>
>> as required.
>>
>> The Lorentz transformation is such a (linear) transformation.  For
>> motion of
>> the primed frame (t', x', y', z') in the x-direction of the unprimed
>> frame
>> (t, x, y, z), relative to the latter frame at the speed v:
>>
>>      t' = γ [t - (v/c^2) x]
>>      x' = γ (x - v t)
>>      y' = y
>>      z' = z.
>>
>> That it is indeed linear is not trivial to show, let alone see, unless
>> you
>> change the time coordinate in a clever way, by multiplying the time
>> transformation equation by c:
>>
>>    c t' = γ [c t - (v/c) x  ]
>>      x' = γ [x   - (v/c) c t]
>>      y' = y
>>      z' = z.
>>
>> Now the coordinates can be relabeled, and for reasons that will become
>> apparent later, it is done this way:
>>
>>    (x^0, x^1, x^2, x^3)^T := (c t, x, y, z)^T.
>>
>> Then the transformation reads
>>
>>    (x')^0 = γ [x^0 - (v/c) x^1]
>>    (x')^1 = γ [x^1 - (v/c) x^0]
>>    (x')^2 = x^2
>>    (x')^3 = x^3.
>>
>>    [Notice the symmetry between the first equation, the time
>> transformation,
>>     and the second equation in which the first spatial coordinate is
>>     transformed.  Again, that is not a coincidence.]
>>
>> It is further useful to define β := v/c.
>>
>>    (x')^0 = γ [x^0 - β x^1]
>>    (x')^1 = γ [x^1 - β x^0]
>>    (x')^2 = x^2
>>    (x')^3 = x^3.
>>
>> And the transformation *can* be written as matrix--vector multiplication,
>> where one introduces the *4-vectors* X' and X, and the Lorentz
>> transformation matrix L:
>>
>>    [(x')^0]    [ γ     -γ β   0   0] [x^0]
>>    [(x')^1] =  [-γ β    γ     0   0] [x^1]
>>    [(x')^2]    [ 0      0     1   0] [x^2]
>>    [(x')^3]    [ 0      0     0   1] [x^3]
>>    `---.--'    `---------.---------' `-.-'
>>        X'                L             X
>>
>> as required.
>>
>> Notice that, for example, multiplying the first row of the Lorentz
>> transformation matrix with the position 4-vector X gives
>>
>>    γ x^0 - γ β x^1 + 0 x^2 + 0 x^3 = γ [x^0 - β x^1]
>>
>> which is indeed what (x')^0 should be.
>>
>> This is essentially what Minkowski did in 1908: introduce a 4-dimensional
>> manifold that he called "space-time" (in the original German:
>> "Raum-Zeit";
>> now overwhelmingly "spacetime" and "Raumzeit" instead, respectively).
>>
>> That the determinant of this matrix/transformation is indeed 1 is
>> again not
>> trivial to show (although it is just an algebraic exercise).  However,
>> noticing that
>>
>>    γ(v) = 1/√(1 - v^2/c^2) = 1/√(1 - β²),
>>
>> one can write
>>
>>    γ² (1 - β²) = γ² - γ² β² = 1.
>>
>> There is a known mathematical identity: [1]
>>
>>    cosh^2(x) - sinh^2(x) = 1.
>>
>> So one can identify
>>
>>    cosh(w) = γ,
>>    sinh(w) = γ β,
>>
>> and find
>>
>>    tanh(w) = sinh(w)/cosh(w) = β = v/c.
>>
>> w = artanh(β) = artanh(v/c) is thus called *rapidity*.  [It is very
>> useful
>> in discussing constant accelerations, as I will show another time.]
>>
>> Then the Lorentz transformation matrix can be written
>>
>>        [ cosh(w)  -sinh(w)  0  0]
>>    L = [-sinh(w)   cosh(w)  0  0].
>>        [       0         0  1  0]
>>        [       0         0  0  1]
>>
>> And now it is much easier to show, using the aforementioned identity,
>> that
>> the determinant of this matrix (and thus of the transformation) is
>> indeed 1.
>>   By Laplace expansion along the fourth row or column, and again along
>> the
>> third row or column, one finds
>>
>>          | cosh(w)  -sinh(w)  0  0|   | cosh(w)  -sinh(w)  0|
>>    |L| = |-sinh(w)   cosh(w)  0  0| = |-sinh(w)   cosh(w)  0|
>>          |       0         0  1  0|   |       0         0  1|
>>          |       0         0  0  1|
>>
>>        = | cosh(w)  -sinh(w)| = cosh^2(w) - sinh^2(w) = 1.
>>          |-sinh(w)   cosh(w)|
>>
>>     [This is one example of many where a Physics student's required
>>      first- and second-semester Mathematics course "Linear Algebra"
>>      turns out to be very useful.  So hang on, it's worth it :)  BTDT.]
>>
>> It is important that the determinant is equal to 1 because that means
>> that
>> scale is preserved: there is no stretching or squishing in any direction,
>> and no inversion of direction, of any 4-vector.  [As you can find
>> beautifully explained graphically for Euclidean space here:
>> <https://www.youtube.com/watch?v=Ip3X9LOh2dk&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab&index=6>]
>>
>>
>> In fact, a Lorentz transformation matrix and in general the Lorentz
>> transformation is -- as one can see now -- a *rotation* (matrix) in a
>> hyperbolic geometry [the matrix for a counter-clockwise rotation by the
>> angle α in 3-dimensional Euclidean space has cos(α) and sin(α), and
>> different signs instead]; and like with all rotations, the determinant
>> MUST be equal to 1 as a rotation preserves lengths (and thus angles in
>> 3-dimension Euclidean space, too).
>>
>> <https://en.wikipedia.org/w/index.php?title=Rapidity&oldid=1327771719>
>>
>> In group theory, the Lorentz transformations form a group, the Lorentz
>> group, that is isomorph to the special orthogonal group SO(3, 1)
>> [rotations
>> in 3-dimensional Euclidean space are isomorph to SO(3)], the group whose
>> members can be represented by matrices M that satisfy M^T M = I, i.e.
>> M^T =
>> M^-1 (*orthogonal* matrices, where I means the identity matrix) and
>> |M| = 1
>> (therefore "special").
>>
>> <https://en.wikipedia.org/w/index.php?title=Lorentz_group&oldid=1345728644>
>>
>>
>> ___
>> [1] This identity follows from the definition of the hyperbolic
>> functions,
>>      where cosh is defined as the even part of the exponential function,
>>      and sinh as its odd part.  Any function can be split into a part
>> that
>>      is even [g(-x) = g(x)] and a part that is odd [h(-x) = -h(x)]:
>>
>>         f(x) = [f(x) + f(-x)]/2 + [f(x) - f(-x)]/2 =: g(x) + h(x).
>>        g(-x) = [f(-x) + f(x)]/2 =  [f(x) + f(-x)]/2 =  g(x),
>>        h(-x) = [f(-x) - f(x)]/2 = -[f(x) - f(-x)]/2 = -h(x).
>>
>>      For the exponential function:
>>
>>          e^x = [e^x + e^{-x}]/2 + [e^x - e^{-x}]/2 =: cosh(x) + sinh(x).
>>
>>      Then
>>
>>        cosh^2(x) = [e^x + e^{-x}]^2/2^2
>>                  = [e^{2x} + 2 e^x e^{-x} + e^{-2x}]/4; [binomial
>> theorem]
>>                  = [e^{2x} + 2 + e^{-2x}]/4,
>>
>>        sinh^2(x) = [e^x - e^{-x}]^2/2^2
>>                  = [e^{2x} - 2 + e^{-2x}]/4,
>>
>>      and finally
>>
>>        cosh^2(x) - sinh^2(x)
>>          = [e^{2x} + 2 + e^{-2x} - e^{2x} + 2 - e^{-2x}]/4,
>>          = 4/4
>>          = 1.
>>
>
>
> Thanks for writing. Derivations are nice.
>
> Here the account of the "hyperbolic-trigonometric" and making
> for ideas for the "determinantal analysis" as it is, has that
> these are _techniques_ of analysis, as for what _concepts_
> they're intended to entail.
>
> For example, when Einstein wrote "this is the way we do", that's
> not the same necessarily as "this is how and what we do", about,
> "this is why we do".
>
> So, for reading from Weyl's "Levels of Infinity", about Hilbert's
> account of "invariant theory" and the "Nullstellensatz", which
> is an account of the "invariant theory" making for determinantal
> analysis, there's that the Desarguesian (or Arguesian) about which
> is to be making an account for "projective geometry" for "invariant"
> theory, explains that that technique is secondary to the analysis,
> the results.
>
> Then invariant theory of course is quite de rigeur these days
> in the mathematical physics, it's directly related to the accounts
> of Noether's theorem that symmetries make conservation laws,
> it "is" the "invariant theory" or theory of invariants. So,
> Hilbert's account of the _Nullstellensatz_, "empty star", about
> the inner product (the usual scalar product the determinant in
> the determinantal analysis) making a boundary about which is
> symmetries making for an account of the invariant, and thusly
> about the account of "not symmetry-breaking", where "symmetry-breaking"
> in accounts like these of "accelerated co-moving frames" is the
> usual setting in relativity theory looking for physics, has that
> this sort of technique is due "invariant theory" since Hilbert.
>
>
> So, about Desargues theorem and the resulting _identities_ that
> result from the _constructions_, geometrically, that then being
> a sort of second level after triangle/Cauchy-Schwarz inequality,
> the deconstructive account is about what "symmetry-flex" instead
> of "symmetry-breaking" is, why there's an account of "continuity
> law" more thorough than "conservation law".
>
>
> Here for example there's a paper of Maddux "Identities Generalizing
> the Theorems of Pappus and Desargues", these being fundamental in
> the projective geometry, as describes some ideas about that besides
> the determinantal analysis and including things like the "cumulants"
> and "orthogonants" and for things like Schwartz functions about
> the broader setting of linear algebra and including out into the
> generalized matrix products and inverses and the matroid setting,
> about why the technique described isn't so much a part of the
> mathematical model the physical model its interpretation the theory,
> instead just another "linearisation".
>
> https://www.mdpi.com/2073-8994/13/8/1382
>
>
> So, these being the things, here for example in this video essay
> "Reading Foundations: replete anti-reductionism", after the
> recent "continuous quanta", "double relativity", and "rational
> rational algebras", is about then the accounts of completions
> of rational algebras, here about the wider setting of the
> "determinantal analysis", as that's usually termed "Hodge dual",
> about "voiding the Nullstellensatz".
>
> https://en.wikipedia.org/wiki/Hilbert%27s_Nullstellensatz
>
>
> Rational? It's a continuum mechanics, ....
>
>

Hilbert's Nullstellensatz - theorem of zeros
(Planck Institute)
29:30



https://www.youtube.com/watch?v=MnRN8-M0sTs&t=1770



Rational radical? Radical rational? Seems opposites, ....

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#670622

amirjf nin <amirjfnin@aim.com> wrote or quoted:
>What amount of acceleration involved is too high for SR to apply? 
>Numbers please.

  I would not apply to a program, when it says, "Our astronauts are 
  required to stand up to a." when a > 3g.

  However, this actually is only a rough estimate. The details of my
  g tolerance depend on the duration and direction of the acceleration!

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#670624

Stefan Ram wrote:
> amirjf nin <amirjfnin@aim.com> wrote or quoted:
>> What amount of acceleration involved is too high for SR to apply? 
>> Numbers please.
> 
>   I would not apply to a program, when it says, "Our astronauts are 
>   required to stand up to a." when a > 3g.

That would be a premature decision, because the human body is able to
withstand up to 10 g -- but it depends on how long the acceleration
takes (the larger the acceleration, the shorter the duration before
damage occurs), and it depends on the orientation of the body:

The human body can withstand large accelerations perpendicular to it
better than large accelerations parallel to it.  One of the reasons
for that is that the blood flow to the brain and thus its oxygenation is
being disturbed or even prevented when the body is accelerated upwards
(in the direction of the head because of the inertia of the blood; this
leads to a blackout).  Also acceleration in the direction of the face
("eyeballs in") is easier to take than acceleration in the opposite
direction ("eyeballs out") for the same reason (just with e.g. the eyeballs
instead).

>   However, this actually is only a rough estimate. The details of my
>   g tolerance depend on the duration and direction of the acceleration!

Exactly.

<https://en.wikipedia.org/wiki/G-force#Human_tolerance>

And none of this has to do with the question.  It was posted too late
for an April Fools joke, too. *shrug*

-- 
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.

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