Linearity of Lorentz transformations
Bill Hobba
linearity of inertial system transformations was called into question. I
believe that the symmetry properties of an inertial reference frame implies
it must be linear. By symmetry properties I mean homogeneity and isotropy
in space and homogeneity in time.
I have seen a number of proofs this must be the case an outline of one I
will give below.
let x' = f(x,t); delta x'/delta x is the ratio of a rod of length delta x to
the same rod in the moving system. Homogeneity of space and time implies
this must be independent of x or t. In taking the limit we have dx'/dx = c1
where c1 is not dependant on either x or t. Integrating we have x' = c1x +
g(t). Taking the derivative wrt to t and noting again from homogeny in
space and time it must be independent of x or t we have x' = c1x + c2t + c3.
A similar proof follows for the time coordinate.
Does anyone see anything wrong with the proof, in particular does anyone see
how is it possible to have an extra term v/c f(x) added to the Lorentz
transform?
Thanks
Bill
Mark
news:3f2ef...@news.iprimus.com.au...
The Lorentz transformations relate to actual space and time measurements.
Special Relativity abolishes the idea of absolute space and absolute time...
prefering to use inertial frames of reference.
Read Mach for further understanding, a good primar for Einsteint... as
Einstein tried to incorporate his ideas into General Relativity.
You're on the right track!
;)
Bill Hobba
Mark replied:
>
> The Lorentz transformations relate to actual space and time measurements.
>
> Special Relativity abolishes the idea of absolute space and absolute
time...
> prefering to use inertial frames of reference.
>
> Read Mach for further understanding, a good primar for Einsteint... as
> Einstein tried to incorporate his ideas into General Relativity.
>
> You're on the right track!
My proof related to inertial reference frames only so GR is not required; SR
is all that is needed. The lorentz transformations are an integral part of
SR and in no way incorporate the idea of absolute space and time - they
describe how space time measurements in one inertial coordinate system
relate to space time measurements in another coordinate system. Note that
Landau in Mechanics defines inertial reference frames as those that posses
the symmetry properties I stated and is equivalent to the usual definition
via the POR.
BTW I am not a supporter of Mach's principle as I have never seen an exact
statement of it that is experimentally verifiable or refutable. If you know
of one I would like to hear it. However I have seen quite a few and non
really look that good to me.
To me the fundamental idea of GR is that the metric must be a dynamical
variable and have its own lagrangian - once you assume that the EFE more or
less follow.
Thanks
Bill
Perfectly Innocent
> I believe that the symmetry properties of an inertial reference frame
> implies it must be linear.
Homogeneity and isotropy is properly understood as applying to space,
not coordinates. Furthermore, explicit examples exist proving that
non-linear functions are allowed to be the change of coordinate
transformations between physically equivalent inertial frames of
reference in a relativistic, isotropic and homogeneous space.
See http://www.everythingimportant.org/relativity/generalized.htm
I take the principle of relativity to mean the physical equivalence
of all inertial frames of reference.
Eugene Shubert
Perfectly Innocent
> I believe that the symmetry properties of an inertial reference frame implies
> it must be linear.
>
> I have seen a number of proofs this must be the case an outline of one I
> will give below.
>
> let x' = f(x,t); delta x'/delta x is the ratio of a rod of length delta x to
> the same rod in the moving system. Homogeneity of space and time implies
> this must be independent of x or t.
> Thanks
> Bill
If delta x'/delta x varied like Exp(kt), I'd call that homogeneous.
Eugene Shubert
http://www.everythingimportant.org/relativity/generalized.htm
Ken S. Tucker
I quite agree, although with a caveat below....
>Mark replied:
>>
>> The Lorentz transformations relate to actual space and time
measurements.
>>
>> Special Relativity abolishes the idea of absolute space and
absolute time...
>> prefering to use inertial frames of reference.
>>
>> Read Mach for further understanding, a good primar for Einsteint...
as
>> Einstein tried to incorporate his ideas into General Relativity.
>>
>> You're on the right track!
>
>My proof related to inertial reference frames only so GR is not
required;
>SR is all that is needed.
Bill, you're using finite relations in the transformation, (quote from
above),
x' = c1x + c2t + c3, (Hobba1)
something that is difficult to do when relating
any CS at a point, (General Covariance). The quoted equation relates
CS's separated in location. It becomes unreasonably difficult to
compare
distance origins without the benefit of GR even if no g-fields are
present,
You go on to say,
"A similar proof follows for the time coordinate."
presuming a solution like,
t = (x' - c1x - c3)/c2, which is getting complicated.
The reason for the complication is due to non-coincident origins.
But a transformation of location will first simplify the relation of
CS's K and K' to have a coincident origin. When this is done,
your question,
" Does anyone see anything wrong with the proof, in particular does
anyone see
how is it possible to have an extra term v/c f(x) added to the
Lorentz
transform?"
That's right, you can then generally substitute
g_i4 = - g_ij dx^j/dx^4
(which is your v/c f(x) term ).
> The lorentz transformations are an integral part of
>SR and in no way incorporate the idea of absolute space and time -
they
>describe how space time measurements in one inertial coordinate
system
>relate to space time measurements in another coordinate system. Note
that
>Landau in Mechanics defines inertial reference frames as those that
posses
>the symmetry properties I stated and is equivalent to the usual
definition
>via the POR.
>
>BTW I am not a supporter of Mach's principle
Me too, an inertial force is sensed by a free-falling object
if it deviates from a geodesic solution, and this solution is
the solution of Newton's 1st law of motion.
>as I have never seen an exact
>statement of it that is experimentally verifiable or refutable. If
you know
>of one I would like to hear it. However I have seen quite a few and
non
>really look that good to me.
>
>To me the fundamental idea of GR is that the metric must be a
dynamical
>variable
Well Bill, the metric defined by,
g_i4 = - g_ij dx^j/dx^4
is termed the *dynamic non-orthogonal spacetime metric*
>and have its own lagrangian - once you assume that the EFE more or
>less follow.
>
>Thanks
>Bill
Your welcome, and thanks for the good question,
Ken S. Tucker
Bill Hobba
> Bill, you're using finite relations in the transformation, (quote from
> above),
I have no idea what your trying to say.
Ken S. Tucker wrote;
>
> x' = c1x + c2t + c3, (Hobba1)
>
> something that is difficult to do when relating
> any CS at a point, (General Covariance).
The principle of general covariance (really the principle of general
invariance but I will not be that picky) is a GR concept - I am dealing with
SR here so it is a concept that is not applicable.
Ken S. Tucker wrote;
'The quoted equation relates CS's separated in location. It becomes
unreasonably difficult to compare distance origins without the benefit of GR
even if no g-fields are present,'
Again this has nothing to do with GR - it is an inertial frame so gravity is
precluded.
Ken S. Tucker wrote;
> You go on to say,
> "A similar proof follows for the time coordinate."
>
> presuming a solution like,
>
> t = (x' - c1x - c3)/c2, which is getting complicated.
No it is the time coordinate in the ' system ie
t' = c4x + c5t + c6
Ken S. Tucker wrote;
> The reason for the complication is due to non-coincident origins.
Coincidence of origins will remove the constant at the end but is linear in
either case ie if the origins are coincident then the transformation is:
x' = c1x + c2t and t' = c3x + c4t.
Ken S. Tucker wrote;
> But a transformation of location will first simplify the relation of
> CS's K and K' to have a coincident origin. When this is done,
> your question,
>
> " Does anyone see anything wrong with the proof, in particular does
> anyone see
> how is it possible to have an extra term v/c f(x) added to the
> Lorentz
> transform?"
>
> That's right, you can then generally substitute
>
> g_i4 = - g_ij dx^j/dx^4
>
> (which is your v/c f(x) term ).
Again I do not follow. If a term v/cf(x) was added then the transformation
would be:
x' = c1x + c2t + c3 + v/cf(x). Taking the ratio of a rod of length delta x
in the x system to the rod measured in the x' system we have delta x'/delta
x = c1deltax + v/cf(xstart) - v/cf(xend)/deltax which is not independent of
position as the homogeneity property of an inertial reference frame implied
it should be. Remember we are dealing with SR here and inertial reference
frames.
Thanks
Bill
Bill Hobba
> > I believe that the symmetry properties of an inertial reference frame
implies
> > it must be linear.
> >
> > I have seen a number of proofs this must be the case an outline of one I
> > will give below.
> >
> > let x' = f(x,t); delta x'/delta x is the ratio of a rod of length delta
x to
> > the same rod in the moving system. Homogeneity of space and time
implies
> > this must be independent of x or t.
Perfectly Innocent wrote:
> If delta x'/delta x varied like Exp(kt), I'd call that homogeneous.
Remember in this sense homogeneous in time means that the same experiment
done at a different time will give the same results. It is a physical
requirement - not a mathematical one. Thus if you added your term to the
equation you would have the ratio of a rod of length delta x stationary in
the non ' system to the same rod measured in the ' system as exp(kt) which
is dependant on when the experiment is carried out. The homogeneity
property is broken - you can tell when the experiment was carried out and is
not allowed by the properties of an inertial reference frame.
Thanks
Bill
Perfectly Innocent
mathematically permissible for a moving rod to experience expansion or
contraction during the time it accelerates out of its stationary frame
of reference. What's stopping a moving rod from returning to its point
of origin smaller or larger? The expanding or shrinking effect could
go like exp(kt) for as long as the rod maintains a constant velocity
v. Conceivably, this might be made to work in 3 spatial dimensions.
Every observer could say that every other frame is shrinking uniformly
in time and, akin to the twin paradox, all returning twins could end
up YOUNGER and smaller.
Can you prove that homogeneity and isotropy alone disallows this
possibility?
Eugene Shubert
http://www.everythingimportant.org/relativity/generalized.htm
Bilge
>linearity of inertial system transformations was called into question. I
>believe that the symmetry properties of an inertial reference frame implies
>it must be linear. By symmetry properties I mean homogeneity and isotropy
>in space and homogeneity in time.
that the length of the rod is unchanged under an infinitessimal
displacement. Since the infinitessimal displacement depends only
on the first derivatives, and a finite displacement is built
from a series of infinitessimal ones, the transform is linear.
Ken S. Tucker
Ok, Bill, pardon the top post, I may be confused but
I'm not doing GR when I use,
ds^2 = g_uv dx^u dx^v (kst1)
since this is an abbreviated way of writing
ds^2 = c^t^2 - dx^2 -dy^2 -dz^2 (kst2)
when g11...g33 =-1, g00=1.
So (kst1) is SR when the metrics are constants, nothing
precludes the use of tensor analysis merely because one
is doing SR.
Would you agree that Eq. (kst1) essentially contains
everything necessary about the Lorentz transform?
For example in SR, it is possible to integrate (kst1) to find,
(after specifyng metrics that are constant in integration)
s^2 = g_uv x^u x^v + constant of integration. (kst3)
Obviously Eqs (kst1,3) are expressed in the prime system,
ds^2 = g'_uv dx'^u dx'^v (kst1')
s^2 = g'_uv x'^u x'^v + constant' of integration. (kst3')
Any SR or Lorentz transform must comply with kst1,3
and kst'1,3 and resolve kst2. So if you begin with
these equations and derive what you want, that should
be ok.
That said, you can substitute the following metrics,
g00.....g33 =1 and g14 = -v/c in Eq. (kst1) to get
(I'll just use t and x),
ds^2 = g00 dt^2 + 2 g01 dt dx + g11 dx^2
= dt^2 -2 v/c dt dx + dx^2
v = dx/dt
ds^2 = dt^2 - 2 (dx/dt) dt dx + dx^2 = dt^2 - dx^2
and is Eq. (kst2).
Regards
Ken S. Tucker
Bill Hobba
Bill Hobba wrote:
> >In another thread and in some private correspondence the issue of the
> >linearity of inertial system transformations was called into question.
I
> >believe that the symmetry properties of an inertial reference frame
implies
> >it must be linear. By symmetry properties I mean homogeneity and
isotropy
> >in space and homogeneity in time.
>
Bilge replied:
>That alone does imply the linearity.
Of course I agee.
Bilge said
'Your "proof" just states that the length of the rod is unchanged under an
infinitessimal displacement. Since the infinitessimal displacement depends
only on the first derivatives, and a finite displacement is built from a
series of infinitessimal ones, the transform is linear.'
I am a little confused here. I said delta x'/delta x is independent of x or
t where delta x' and delta x are finite. I then take the limit to give the
derivative and integrate back up to obtain the original function f(x,t). I
probably am missing something. Could you clarify?
Thanks
Bill
Bill Hobba
Bilge said
> 'Your "proof" just states that the length of the rod is unchanged under an
> infinitessimal displacement. Since the infinitessimal displacement depends
> only on the first derivatives, and a finite displacement is built from a
> series of infinitessimal ones, the transform is linear.'
>
Dill Hobba replied (misspelling intended)
> I am a little confused here. I said delta x'/delta x is independent of x
or
> t where delta x' and delta x are finite. I then take the limit to give
the
> derivative and integrate back up to obtain the original function f(x,t).
I
> probably am missing something. Could you clarify?
God I am stupid. I see Bilge's point. The fact that detla x'/delta x = c
where c is impendent of x or t implies more or less by the definition of
linearity it must be linear (same increase in x always leads to same
increase in x') without the calculus. That comes from having a math
background where you try to prove things without understanding what they
mean. Thus taking the origin as the start point in both systems you have
from the fact above ax = c1x for a fixed t say t = 0. Same for the origin
(ie ax=0) delta x/delta t = c. Thus the origin transforms as ax = c2t. To
obtain the full function you add the distance the rod moved from the origin
to the length of the rod to get x' = c1x + c2t.
Thanks
Bill
Bill Hobba
> Bilge said
> > 'Your "proof" just states that the length of the rod is unchanged under
an
> > infinitessimal displacement. Since the infinitessimal displacement
depends
> > only on the first derivatives, and a finite displacement is built from a
> > series of infinitessimal ones, the transform is linear.'
> >
>
> Dill Hobba replied (misspelling intended)
> > I am a little confused here. I said delta x'/delta x is independent of
x
> or
> > t where delta x' and delta x are finite. I then take the limit to give
> the
> > derivative and integrate back up to obtain the original function f(x,t).
> I
> > probably am missing something. Could you clarify?
>
Bill Hobba said
> God I am stupid. I see Bilge's point. The fact that detla x'/delta x = c
> where c is impendent of x or t implies more or less by the definition of
> linearity it must be linear (same increase in x always leads to same
> increase in x') without the calculus. That comes from having a math
> background where you try to prove things without understanding what they
> mean. Thus taking the origin as the start point in both systems you have
> from the fact above ax = c1x for a fixed t say t = 0. Same for the origin
> (ie ax=0) delta x/delta t = c. Thus the origin transforms as ax = c2t.
To
> obtain the full function you add the distance the rod moved from the
origin
> to the length of the rod to get x' = c1x + c2t.
After reading what I wrote a few typo's and the way I said things may not
convey exactly what I was trying to say. As Bilge said finite displacements
are built from infinitesimal ones so the ratio of an infinitesimal rod in
one frame to the rod viewed in the other is independent of where the rod
is - this leads to the ratio of finite rods being the same ie linear.
However me being the way I am I kind of like the calculus argument ie dx'/dx
= c1 then integrating up to get x' = c1x + c(t) etc. Of course since there
are more than one variable I really should have used partial derivatives.
Thanks
Bill
Bill Hobba
> > I believe that the symmetry properties of an inertial reference frame
> > implies it must be linear.
>
Eugine Shubert replied.
> Homogeneity and isotropy is properly understood as applying to space,
> not coordinates. Furthermore, explicit examples exist proving that
> non-linear functions are allowed to be the change of coordinate
> transformations between physically equivalent inertial frames of
> reference in a relativistic, isotropic and homogeneous space.
>
> See http://www.everythingimportant.org/relativity/generalized.htm
>
> I take the principle of relativity to mean the physical equivalence
> of all inertial frames of reference.
It applies to the points in a reasonable coordinate such as a Cartesian one
(basically a stationary one) As points have no objective existence apart
from a coordinate system your statement does not even make much sense. I
have seen your counter example and does not obey the space homogeneity
property - the v/cf(x) term forbids it - you can tell the difference between
points.
If you believe in the POR then you believe that an inertial system is
homogeneous and isotropic. Simply shift all the clocks a certain amount and
mover and /or rotate you coordinate system and you again have an inertial
coordinate system. Indeed that's is how Landau in his book Mechanics
defines an inertial coordinate system.
After thinking about what Bilge wrote I have now come up with the most
elegant way I know to prove it.
Consider the transformation as a vector relation L(X) where X = (x1, x2, x3,
t). For small delta x from the calculus we know that L(X + delta x) = L(X)
+ A(delta x) where A is a 4x4 matrix. Now from homgeinity in space and time
A must not depend on X - if it did you would be able to tell the difference
between differnt points and times by the value of A. Thus L(X) = L(0 +
delta x1 + delta x2 +++ delta xn) where each delta xi is small and adds up
to X. Hence L(X) = L(0) + A(delta x1) + A(delta x2) ++++ A(delta xn) = L(0)
+ A(delta x1 + delta x2 ++++ delta xn) = L(0) + A(X) = C + A(X) where C is
the vector L(0). Thus the transformation is linear.
For you to still allow your v/cf(x) term you must show the fault in my
reasoning.
Thanks
Bill
Tom Roberts
> I believe that the symmetry properties of an inertial reference frame implies
Rather than pounding on "linear or not?", let me suggest that the
definition of "inertial coordinates" is:
Inertial coordinates are such that _ANY_ object that moves
inertially traces a uniform straight line wrt the coordinates.
Linearity of transforms between inertial coordinates follows
immediately. Remarkably, one does not need to define what "moving
inertially" actually means....
This trivially dismisses Eugene Shubert's objection.
Tom Roberts tjro...@lucent.com
David McAnally
>In another thread and in some private correspondence the issue of the
>linearity of inertial system transformations was called into question. I
>believe that the symmetry properties of an inertial reference frame implies
>it must be linear. By symmetry properties I mean homogeneity and isotropy
>in space and homogeneity in time.
>I have seen a number of proofs this must be the case an outline of one I
>will give below.
>let x' = f(x,t); delta x'/delta x is the ratio of a rod of length delta x to
>the same rod in the moving system.
I hate to be picky, but if you are going to hold t constant, then
delta x'/delta x is actually the ratio of a rod of length x' which
is stationary in the primed frame with respect to the length of
the same rod in the unprimed frame. This matches the fact that
the coefficient of x in the formula for x' is greater than or
equal to 1, since that corresponds to length contraction (delta x
is less than delta x').
> Homogeneity of space and time implies
>this must be independent of x or t. In taking the limit we have dx'/dx = c1
>where c1 is not dependant on either x or t. Integrating we have x' = c1x +
>g(t). Taking the derivative wrt to t and noting again from homogeny in
>space and time it must be independent of x or t we have x' = c1x + c2t + c3.
>A similar proof follows for the time coordinate.
>Does anyone see anything wrong with the proof, in particular does anyone see
>how is it possible to have an extra term v/c f(x) added to the Lorentz
>transform?
No, I don't see anything wrong with the proof, and if you have the
extra term, then the metric after the transformation is not the
Minkowski metric, so I don't see how anybody could claim that there
is such an extra term.
David McAnally
--------------
Perfectly Innocent
>
> Rather than pounding on "linear or not?", let me suggest that the
> definition of "inertial coordinates" is:
>
> Inertial coordinates are such that _ANY_ object that moves
> inertially traces a uniform straight line wrt the coordinates.
The problem with your definition is that it's entirely a definition.
It's completely empty of physical content. Be aware that my NON-LINEAR
TRANSFORMATION in exercise 1 and 2 of
http://www.everythingimportant.org/relativity/generalized.htm has the
property that every "inertial" observer traces out equal distances in
equal *PROPER* time intervals.
> Linearity of transforms between inertial coordinates follows
> immediately. Remarkably, one does not need to define what "moving
> inertially" actually means....
Unremarkably, if you don't want to grapple with physics, then you've
left in the ignorance of not understanding physics.
> This trivially dismisses Eugene Shubert's objection.
Unremarkably, trivial definitions often dismiss substantive ideas.
For example:
Inertial frames in VSL (variable speed of light) special relativity
don't move linearly but you're quite content with throwing out this
interesting generalization of SR with a definition. Why? For the sole
reason of winning an empty argument or, possibly, to keep the debate
out of territory unfamiliar to you but accessible to me?
Eugene Shubert
http://www.everythingimportant.org
Perfectly Innocent
> "Bill Hobba" <bho...@iprimus.com.au> writes:
>
> >Does anyone see anything wrong with the proof, in particular does anyone see
> >how is it possible to have an extra term v/c f(x) added to the Lorentz
> >transform?
>
> No, I don't see anything wrong with the proof, and if you have the
> extra term, then the metric after the transformation is not the
> Minkowski metric, so I don't see how anybody could claim that there
> is such an extra term.
>
> David McAnally
>
> --------------
David,
The point about the extra term is a simple test in SR understanding.
See exercise 1 and 2 of
http://www.everythingimportant.org/relativity/generalized.htm
David McAnally
>David,
Your questions at the end of the webpage completely ignore the second
postulate. You have transformation equations
x' = x_0 + ((x-x_0) - v (t-t_0) + v/c \zeta(x))/sqrt(1-v^2/c^2),
t' = t_0 + ((t-t_0) - v/c^2 (x-x_0) - 1/c \zeta(x))/sqrt(1-v^2/c^2)
+ 1/c \zeta(x').
In order to be consistent with the second postulate, ct'-x' must
be functionally dependent on ct-x and v only, and ct'+x' must
be functionally dependent on ct+x and v only. Now,
ct'+x' = ct_0 + x_0 + {(c-v) (t-t_0) + (1-v/c) (x-x_0)
- (1-v/c) \zeta(x)}/sqrt(1-v^2/c^2) + \zeta(x')
= ct_0 + x_0 + sqrt((c-v)/(c+v)) [c (t-t_0) + (x-x_0)]
- sqrt((c-v)/(c+v)) \zeta(x) + \zeta(x'),
so, since ct'+x' and sqrt((c-v)/(c+v)) [c (t-t_0) + (x-x_0)]
are functionally dependent on ct+x and v only, it is required
that -sqrt((c-v)/(c+v)) \zeta(x) + \zeta(x') be functionally
dependent on ct+x and v only. Also,
ct'-x' = ct_0 - x_0 + {(c+v) (t-t_0) - (1+v/c) (x-x_0)
- (1+v/c) \zeta(x)}/sqrt(1-v^2/c^2) + \zeta(x')
= ct_0 + x_0 + sqrt((c+v)/(c-v)) [c (t-t_0) - (x-x_0)]
- sqrt((c+v)/(c-v)) \zeta(x) + \zeta(x'),
so, since ct'-x' and sqrt((c+v)/(c-v)) [c (t-t_0) - (x-x_0)]
are functionally dependent on ct-x and v only, it is required
that -sqrt((c+v)/(c-v)) \zeta(x) + \zeta(x') be functionally
dependent on ct-x and v only.
Since -sqrt((c-v)/(c+v)) \zeta(x) + \zeta(x') must be
functionally dependent on ct+x and v only, then its
derivative with respect to v, evaluated at v=0, must
be functionally dependent on ct+x only, and so
\zeta(x)/c - (t-t_0) \zeta'(x) + \zeta(x) \zeta'(x)/c
must be functionally dependent on ct+x only.
Since -sqrt((c+v)/(c-v)) \zeta(x) + \zeta(x') must be
functionally dependent on ct-x and v only, then its
derivative with respect to v, evaluated at v=0, must
be functionally dependent on ct-x only, and so
- \zeta(x)/c - (t-t_0) \zeta'(x) + \zeta(x) \zeta'(x)/c
must be functionally dependent on ct-x only.
So
\zeta(x)/c - (t-t_0) \zeta'(x) + \zeta(x) \zeta'(x)/c = F(ct+x),
- \zeta(x)/c - (t-t_0) \zeta'(x) + \zeta(x) \zeta'(x)/c = G(ct-x),
(*)
for some functions F and G. Upon subtraction, this yields that
2 \zeta(x)/c = F(ct+x) - G(ct-x).
Differentiation with respect to t yields 0 = c F'(ct+x) - c G'(ct-x),
so that F'(ct+x) = G'(ct-x) = K for some constant K, and so
F(ct+x) = K (ct+x) + L, G(ct-x) = K (ct-x) + M for some constants
L and M. So 2 \zeta(x)/c = 2 K x + L - M, so that \zeta(x) = K c x + B
for B constant. Substituting into (*), then
K x + B/c - K c (t-t_0) + K (K c x + B) = K (ct+x) + L,
- K x - B/c - K c (t-t_0) + K (K c x + B) = K (ct-x) + M.
Equating the coefficients of t on both sides of either equation,
we get -Kc = Kc, and so K = 0. It follows that the only function
\zeta(x) consistent with the second postulate is constant, and
so the transformation is an element of the Poincare group.
On final note, the transformations from the earlier part of the
page form a one-dimensional subgroup of the conformal group
for 1+1 dimensional space. The conformal group is well-known,
and is infinite-dimensional. The generators for the Lie algebra
of the conformal group are (ct+x)^(n+1) d/d(ct+x) for all integers
n and (ct-x)^(n+1) d/d(ct-x) for all integers n, where d/d(ct+x)
is partial differentiation with respect to ct+x where ct-x is held
constant, and d/d(ct-x) is partial differentiation with respect
to ct-x where ct+x is held constant. The Lie algebra of the
conformal group is isomorphic to the direct sum of a pair of
centerless real Virasoro algebras.
David McAnally
---------------
Perfectly Innocent
>
> >David,
>
> >The point about the extra term is a simple test in SR understanding.
> >See exercise 1 and 2 of
> >http://www.everythingimportant.org/relativity/generalized.htm
>
> Your questions at the end of the webpage completely ignore the second
> postulate.
Dear David,
I'd be very happy to entertain your interpretation of Einstein's
second postulate. Is it a law of physics?
My clock is set to Central Standard Time. Clocks on the West Coast
are set to Pacific Standard Time. Is this convention a violation of
the laws of physics?
Do you know how to use general transformation equations?
True or false: If you use my transformation equations and have two
synchronized clocks and slowly transport one of them to any convenient
distance D and then measure the speed of light, i.e., D/(t2-t1), then
the measured answer will be c. (t1 is the time on the stationary clock
when the light pulse is sent. t2 is the time when the light arrives at
the slowly transported clock. Take the limit of ultraslow transport
for a perfect value of c).
If you believe that Einstein's clock synchronization is sacrosanct,
does that mean you are willing to abandon the second postulate when it
comes to generalizing SR to a circle? Does that mean you believe in an
absolute time order on SxR? Same question for (S^2)xR and (S^3)xR? How
do you explain the paradox of the great illumination?
http://www.everythingimportant.org/relativity/simultaneity.htm
Eugene Shubert
David McAnally
>Does anyone see anything wrong with the proof, in particular does anyone see
>how is it possible to have an extra term v/c f(x) added to the Lorentz
>transform?
By ignoring the second postulate. If the coordinates are to reflect
the fact that the speed of light is c in both frames of reference,
then the transformations as on the relevant webpage (as supplied by
Perfectly Innocent) require that f(x) be constant.
David McAnally
--------------
Tom Roberts
> Tom Roberts <tjro...@lucent.com> wrote in message news:<bh47sm$e...@netnews.proxy.lucent.com>...
>>Rather than pounding on "linear or not?", let me suggest that the
>>definition of "inertial coordinates" is:
>> Inertial coordinates are such that _ANY_ object that moves
>> inertially traces a uniform straight line wrt the coordinates.
> The problem with your definition is that it's entirely a definition.
> It's completely empty of physical content.
Coordinates are geometrical, not physical. So it makes good sense to
describe them in geometrical terms, as I did.
One can use physically-based coordinates (e.g. laid out with
a meter stick), and that can give them physical content.
But that is not essential.... Traditionally physicists
always did this (including Einstein's 1905 paper), but
since the advent of GR it is known to be too restrictive.
> Be aware that my NON-LINEAR
> TRANSFORMATION in exercise 1 and 2 of
> http://www.everythingimportant.org/relativity/generalized.htm has the
> property that every "inertial" observer traces out equal distances in
> equal *PROPER* time intervals.
OF COURSE an inertially-moving object traces out equal distances in
equal proper time intervals -- that's a direct consequence of timelike
geodesics in a flat manifold. But that's not the issue, the issue is
whether it traces a uniform straight line WRT THE COORDINATES (i.e. the
coordinates are linear functions of proper time or some other affine
parameter). After all, the goal is to define inertial coordinates. Your
coordinates do not do that except for some special choices of zeta(.).
Don't forget to include y and z, and consider paths not
along the axes....
>>Linearity of transforms between inertial coordinates follows
>>immediately. Remarkably, one does not need to define what "moving
>>inertially" actually means....
> Unremarkably, if you don't want to grapple with physics, then you've
> left in the ignorance of not understanding physics.
But this is "grappling" with geometry, not physics.
> Inertial frames in VSL (variable speed of light) special relativity
> don't move linearly but you're quite content with throwing out this
> interesting generalization of SR with a definition. Why?
Because if inertially-moving objects don't move in uniform straight
lines wrt the coordinates, there's no point in calling the coordinates
"inertial" -- that is far too enormous a break with Newtonian usage.
Tom Roberts tjro...@lucent.com
David McAnally
to perfectl...@as-if.com (Perfectly Innocent):
>On final note, the transformations from the earlier part of the
>page form a one-dimensional subgroup of the conformal group
>for 1+1 dimensional space. The conformal group is well-known,
>and is infinite-dimensional. The generators for the Lie algebra
>of the conformal group are (ct+x)^(n+1) d/d(ct+x) for all integers
>n and (ct-x)^(n+1) d/d(ct-x) for all integers n, where d/d(ct+x)
>is partial differentiation with respect to ct+x where ct-x is held
>constant, and d/d(ct-x) is partial differentiation with respect
>to ct-x where ct+x is held constant. The Lie algebra of the
>conformal group is isomorphic to the direct sum of a pair of
>centerless real Virasoro algebras.
At n = -1, the differential operators above are
d/d(ct+x) = 1/(2c) d/dt + 1/2 d/dx,
d/d(ct-x) = 1/(2c) d/dt - 1/2 d/dx,
and generate the group of translations (t' = t + a, x' = x + b).
At n = 0, the operators are
(ct-x) d/d(ct-x) = (1/2) (t d/dt + x d/dx) - 1/2 (1/c x d/dt + c t d/dx),
(ct+x) d/d(ct+x) = (1/2) (t d/dt + x d/dx) + 1/2 (1/c x d/dt + c t d/dx),
so that
(ct+x) d/d(ct+x) + (ct-x) d/d(ct-x) = t d/dt + x d/dx,
(ct+x) d/d(ct+x) - (ct-x) d/d(ct-x) = 1/c x d/dt + c t d/dt.
The first of these operators generates the group of dilations
(t' = at, x' = ax), and the second of these operators generates
the group of Lorentz boosts (t' = t cosh a + (x/c) sinh a,
x' = x cosh a + c t sinh a).
At n = 1,
(ct+x)^2 d/d(ct+x) + (ct-x)^2 d/d(ct-x)
= (c t^2 + x^2/c) d/dt + 2 c t x d/dx,
(ct+x)^2 d/d(ct+x) - (ct-x)^2 d/d(ct-x)
= (c^2 t^2 + x^2) d/dx + 2 t x d/dt.
These two operators generate the group of special conformal
transformations:
t' = [t - a (c t^2 - x^2/c)]/V, x' = [x - b (c t^2 - x^2/c)]/V,
where V = 1 - 2 (c a t - b x/c) + (c a^2 - b^2/c) (c t^2 - x^2/c).
These six operators form a Lie subalgebra of the Lie algebra of
the conformal group. The Lie subalgebra is isomorphic to so(2,2)
(so that they generate a group locally isomorphic to SO(2,2)).
This corresponds to the result that if p+q > 2, then the conformal
group for a pseudo-Euclidean space of dimension p+q is locally
isomorphic to SO(p+1,q+1) (the group generated by translations,
dilations, rotations, Lorentz boosts and special conformal
transformations).
David McAnally
--------------
David McAnally
<snip>
>At n = 0, the operators are
>(ct-x) d/d(ct-x) = (1/2) (t d/dt + x d/dx) - 1/2 (1/c x d/dt + c t d/dx),
>(ct+x) d/d(ct+x) = (1/2) (t d/dt + x d/dx) + 1/2 (1/c x d/dt + c t d/dx),
>so that
>(ct+x) d/d(ct+x) + (ct-x) d/d(ct-x) = t d/dt + x d/dx,
>(ct+x) d/d(ct+x) - (ct-x) d/d(ct-x) = 1/c x d/dt + c t d/dt.
That should be:
(ct+x) d/d(ct+x) - (ct-x) d/d(ct-x) = 1/c x d/dt + c t d/dx.
>The first of these operators generates the group of dilations
>(t' = at, x' = ax), and the second of these operators generates
>the group of Lorentz boosts (t' = t cosh a + (x/c) sinh a,
>x' = x cosh a + c t sinh a).
David McAnally
--------------
Perfectly Innocent
> Coordinates are geometrical, not physical. So it makes good sense to
> describe them in geometrical terms, as I did.
Why are my coordinates any less geometrical than yours?
> > Be aware that my NON-LINEAR
> > TRANSFORMATION in exercise 1 and 2 of
> > http://www.everythingimportant.org/relativity/generalized.htm has the
> > property that every "inertial" observer traces out equal distances in
> > equal *PROPER* time intervals.
>
> OF COURSE an inertially-moving object traces out equal distances in
> equal proper time intervals. ... After all, the goal is to define
> inertial coordinates.
I just did.
Eugene Shubert
http://www.everythingimportant.org/relativity/
Perfectly Innocent
David,
I am fairly confident that our disagreement is based on different
presuppositions. The speed of light is c is my equations. We simply
define speed differently. Furthermore, I suspect that you're using
simultaneity and imposing this obsolete and delusive prejudice onto
your equations whereas I am certain that I interpret my equations
without this ancient and subtle prejudice.
Eugene Shubert
http://www.everythingimportant.org/relativity
http://www.everythingimportant.org/relativity/simultaneity.htm
David McAnally
>D.McAnally@i'm_a_gnu.uq.net.au (David McAnally) wrote in message news:<bh98uj$9o1$1...@bunyip.cc.uq.edu.au>...
>> "Bill Hobba" <bho...@iprimus.com.au> writes:
>>
>> >Does anyone see anything wrong with the proof, in particular does anyone see
>> >how is it possible to have an extra term v/c f(x) added to the Lorentz
>> >transform?
>>
>> By ignoring the second postulate. If the coordinates are to reflect
>> the fact that the speed of light is c in both frames of reference,
>> then the transformations as on the relevant webpage (as supplied by
>> Perfectly Innocent) require that f(x) be constant.
>>
>> David McAnally
>>
>> --------------
>David,
>I am fairly confident that our disagreement is based on different
>presuppositions. The speed of light is c is my equations.
Not if you define the velocity in the primed frame of reference
by the traditional definition of dx'/dt'. If you define velocity
by the traditional definition of dx'/dt', then that requires that
f(x) is a constant.
>We simply
>define speed differently.
Yes. I define the velocity as the derivate of the spatial
coordiantes with respect to the temporal coordiante.
>Furthermore, I suspect that you're using
>simultaneity
You are welcome to suspect what you want, even if you are completely
wrong.
>and imposing this obsolete and delusive prejudice onto
>your equations whereas I am certain that I interpret my equations
>without this ancient and subtle prejudice.
No. You are using a completely unfamiliar and unknown definition
for velocity.
I asked before why the first half of the webpage was so completely
devoted to determining a one-dimensional subgroup of the conformal
equation. Your complaints about the fact that I define the
velocity in the primed frame to equal dx'/dt' make it all the
more astounding that you even bothered trying to come up with
the subgroup.
David McAnally
--------------
Perfectly Innocent
> perfectl...@as-if.com (Perfectly Innocent) writes:
>
> >David,
>
> >I am fairly confident that our disagreement is based on different
> >presuppositions. The speed of light is c is my equations.
>
> Not if you define the velocity in the primed frame of reference
> by the traditional definition of dx'/dt'. If you define velocity
> by the traditional definition of dx'/dt', then that requires that
> f(x) is a constant.
The traditional definition of velocity being dx'/dt' does indeed evoke
a notion of simultaneity, which I refuse to depend upon.
> >We simply
> >define speed differently.
>
> Yes. I define the velocity as the derivate of the spatial
> coordiantes with respect to the temporal coordiante.
The traditional definition of velocity, as the time rate of change of
position, is based on the spooky notion of tracking distant motion
according to an absolute, stationary clock. And that's a source of
trouble that you don't even realize.
> >Furthermore, I suspect that you're using
> >simultaneity
>
> You are welcome to suspect what you want, even if you are completely
> wrong.
If you're certain, then let's discuss concrete counterexamples to your
beliefs. Where's the error in my derivation of the Lorentz
transformation based as it is on a Galilean definition of time?
http://www.everythingimportant.org/relativity/
Where's the physical content of velocity as dx'/dt' if we were to
invoke a change of coordinates from the Einsteinian synchronization to
a Galilean synchronization?
http://www.everythingimportant.org/relativity/
> >and imposing this obsolete and delusive prejudice onto
> >your equations whereas I am certain that I interpret my equations
> >without this ancient and subtle prejudice.
>
> No. You are using a completely unfamiliar and unknown definition
> for velocity.
In my studies of generalized SR, I use proper velocity to understand
the physics. Proper velocity is very well known. Because there is no
frame that moves at light speed, for that special case, I used a
reasonable, operational definition to define the speed of light:
> I asked before why the first half of the webpage was so completely
> devoted to determining a one-dimensional subgroup of the conformal
> equation. Your complaints about the fact that I define the
> velocity in the primed frame to equal dx'/dt' make it all the
> more astounding that you even bothered trying to come up with
> the subgroup.
Regarding: http://www.everythingimportant.org/relativity/generalized.htm
I have an interest in generalizations of SR that preserve the
equivalence of all frames of reference (the principle of relativity).
To that end it occurred to me that examining groups that generalize
the Lorentz transformation might be reasonable objects to investigate.
Why is a group with one space dimension, one time dimension and one
parameter describing a continuum of different states of motion
one-dimensional?
Eugene Shubert
http://www.everythingimportant.org/relativity/simultaneity.htm
Bill Hobba
> Bill Hobba wrote:
> > I believe that the symmetry properties of an inertial reference frame
implies
> > it must be linear. [...]
>
Tom Roberts replied:
> Rather than pounding on "linear or not?", let me suggest that the
> definition of "inertial coordinates" is:
>
> Inertial coordinates are such that _ANY_ object that moves
> inertially traces a uniform straight line wrt the coordinates.
>
> Linearity of transforms between inertial coordinates follows
> immediately. Remarkably, one does not need to define what "moving
> inertially" actually means....
>
> This trivially dismisses Eugene Shubert's objection.
Of course you can define it that way - in fact some books I know define it
similarly and it is probably the most elegant formulation. The reason I
choose the symmetry is my background (I initially leant SR and more advanced
mechanics from Landau) where Landau defines it by way of symmetry
properties. The only concern I have is it is not immediately obvious to a
dumb nut like me why you seek to define it that way. But my slowness in no
way diminishes its advantages.
Just out of interest page 11 of Rindler - Introduction to Special Relativity
gives a proof of why it must be linear by considering the ticks on an
inertial clock and a bit of calculus. But as I said I can be bit slow and
someone like yourself or Bilge may give me a hint to something simple I am
missing.
Thanks
Bill
Bilge
>I have an interest in generalizations of SR that preserve the
>equivalence of all frames of reference (the principle of relativity).
There exists such a theory: general relativity.
Perfectly Innocent
But do you understand the meaning of the principle of relativity?
Do you think that homogeneity and isotropy does or does not imply
the Lorentz transformation?
Eugene Shubert
http://www.everythingimportant.org
Bilge
>news:<slrnbkaj56....@radioactivex.lebesque-al.net>...
>> Perfectly Innocent:
>>
>> >I have an interest in generalizations of SR that preserve the
>> >equivalence of all frames of reference (the principle of relativity).
>>
>> There exists such a theory: general relativity.
>
>But do you understand the meaning of the principle of relativity?
>Do you think that homogeneity and isotropy does or does not imply
>the Lorentz transformation?
What's you're point?