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.
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.)
What amount of acceleration involved is too high for SR to apply?
Numbers please.
This is a question of how you define "special relativity":
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!
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!
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' = +| [t - (v/c^2) x]
x' = +| (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' = +| [c t - (v/c) x ]
x' = +| [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 = +| [x^0 - (v/c) x^1]
(x')^1 = +| [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 +# := v/c.
(x')^0 = +| [x^0 - +# x^1]
(x')^1 = +| [x^1 - +# 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] [ +| -+| +# 0 0] [x^0]
[(x')^1] = [-+| +# +| 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
+| x^0 - +| +# x^1 + 0 x^2 + 0 x^3 = +| [x^0 - +# 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
+|(v) = 1/reU(1 - v^2/c^2) = 1/reU(1 - +#-#),
one can write
+|-# (1 - +#-#) = +|-# - +|-# +#-# = 1.
There is a known mathematical identity: [1]
cosh^2(x) - sinh^2(x) = 1.
So one can identify
cosh(w) = +|,
sinh(w) = +| +#,
and find
tanh(w) = sinh(w)/cosh(w) = +# = v/c.
w = artanh(+#) = 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 +# in 3-dimensional Euclidean space has cos(+#) and sin(+#), 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.
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' = +| [t - (v/c^2) x]
x' = +| (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' = +| [c t - (v/c) x ]
x' = +| [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 = +| [x^0 - (v/c) x^1]
(x')^1 = +| [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 +# := v/c.
(x')^0 = +| [x^0 - +# x^1]
(x')^1 = +| [x^1 - +# 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] [ +| -+| +# 0 0] [x^0]
[(x')^1] = [-+| +# +| 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
+| x^0 - +| +# x^1 + 0 x^2 + 0 x^3 = +| [x^0 - +# 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
+|(v) = 1/reU(1 - v^2/c^2) = 1/reU(1 - +#-#),
one can write
+|-# (1 - +#-#) = +|-# - +|-# +#-# = 1.
There is a known mathematical identity: [1]
cosh^2(x) - sinh^2(x) = 1.
So one can identify
cosh(w) = +|,
sinh(w) = +| +#,
and find
tanh(w) = sinh(w)/cosh(w) = +# = v/c.
w = artanh(+#) = 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 +# in 3-dimensional Euclidean space has cos(+#) and sin(+#), 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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