A Nonlinear Group of Lorentz-Equivalent Transformations
Shubee
transformations given in exercise 1 of http://www.everythingimportant.org/relativity/generalized.htm
form a mathematical group. It looks right to me but I'd like it
confirmed by unbiased physicists. Here is his solution that I copied
from his post at http://groups.google.com/group/sci.physics.relativity/msg/67b70dea713f28de
------------------------------------
Let c=1 (for simplicity)
Define gamma_v = 1/sqrt(1-v^2)
>From the exercise:
T_v(x,t)=(x',t') where
x'= x0 + gamma_v((x-x0)-v(t-t0)+vf(x))
t'= t0 + gamma_v((t-t0)-v(x-x0)-f(x)) + f(x')
That means we have:
T_w(T_v(x,t)) = T_w(x',t') = (x'',t'')
where x',t' are as above, so
x''= x0 + gamma_w([x'-x0]-w[t'-t0]+wf(x'))
t''= t0 + gamma_w([t'-t0]-w[x'-x0]-f(x')) + f(x'')
Replace x',t' with expansion from T_v above..
x''= x0 + gamma_w(
[x0 + gamma_v((x-x0)-v(t-t0)+vf(x))-x0]
-w[t0 + gamma_v((t-t0)-v(x-x0)-f(x)) +
f(x0+gamma_v((x-x0)-v(t-t0)+vf(x)))-t0]
+wf(x0 + gamma_v((x-x0)-v(t-t0)+vf(x)))
)
t''= t0 + gamma_w(
[t0 + gamma_v((t-t0)-v(x-x0)-f(x)) +
f(x0+gamma_v((x-x0)-v(t-t0)+vf(x)))-t0]
-w[x0 + gamma_v((x-x0)-v(t-t0)+vf(x))-x0]
-f(x0 + gamma_v((x-x0)-v(t-t0)+vf(x)))
) + f(x'')
Do some simple canceling and expanding..
x''= x0 + gamma_w(
gamma_v((x-x0)-v(t-t0)+vf(x))
-w.gamma_v((t-t0)-v(x-x0)-f(x))
-w.f(x0+gamma_v((x-x0)-v(t-t0)+vf(x)))
+w.f(x0+gamma_v((x-x0)-v(t-t0)+vf(x)))
)
t''= t0 + gamma_w(
gamma_v((t-t0)-v(x-x0)-f(x))
+ f(x0+gamma_v((x-x0)-v(t-t0)+vf(x)))
-w.gamma_v((x-x0)-v(t-t0)+vf(x))
- f(x0+gamma_v((x-x0)-v(t-t0)+vf(x)))
) + f(x'')
Do some more canceling and factoring...
x''= x0 + gamma_w.gamma_v(
(x-x0)-v(t-t0)+vf(x)-w.(t-t0)+wv(x-x0)+wf(x)
)
t''= t0 + gamma_w.gamma_v(
(t-t0)-v(x-x0)-f(x)-w(x-x0)+wv(t-t0)-wvf(x)
) + f(x'')
Do some factoring...
x''= x0 + gamma_w.gamma_v(
(1+vw)(x-x0) -(v+w)(t-t0) +(v+w)f(x)
)
t''= t0 + gamma_w.gamma_v(
(1+vw)(t-t0) -(v+w)(x-x0) -(1+vw)f(x)
) + f(x'')
and some more factoring...
x''= x0 + (1+vw).gamma_w.gamma_v(
(x-x0) -(v+w)/(1+vw)(t-t0) +(v+w)/(1+vw)f(x)
)
t''= t0 + (1+vw).gamma_w.gamma_v(
(t-t0) -(v+w)/(1+vw)(x-x0) -f(x)
) + f(x'')
In the above we have:
gamma_w.gamma_v = (1/sqrt(1-v^2))(1/sqrt(1-w^2))
gamma_w.gamma_v = 1/sqrt((1-v^2)(1-w^2))
gamma_w.gamma_v = 1/sqrt(1-v^2-w^2+v^2w^2))
So...
(1+vw).gamma_w.gamma_v = (1+vw)/sqrt(1-v^2-w^2+v^2w^2))
Also in the above we have (v+w)/(1+vw)
let U = (v+w)/(1+vw) .. we can write that as (v(+)w).
then...
gamma_U = 1/sqrt(1-((v+w)/(1+vw))^2)
= 1/sqrt(1-(v^2+2vw+w^2)/(1+vw)^2)
= 1/(sqrt((1+vw)^2-v^2-2vw-w^2)/sqrt((1+vw)^2)
= 1/(sqrt(1+2vw+v^2w^2-v^2-2vw-w^2)/(1+vw))
= (1+vw)/sqrt(1-v^2-w^2+v^2w^2)
= (1+vw).gamma_w.gamma_v
so we can replace appropriate terms with U and gamma_U...
x''= x0 + gamma_U((x-x0) -U(t-t0) +U.f(x))
t''= t0 + gamma_U((t-t0) -U(x-x0) -f(x)) + f(x'')
so...
T_[U](x,t) = T_[v(+)w](x,t) = T_w(T_v(x,y)) = (x'',t'')
Please let me know if there are any errors in the above. Its tricky
when typing in moderately complex formulas like that as plain text.
------------------------------------
Any questions?
Eric Gisse
[...]
>
> Any questions?
Yea.
Where is the proof that it obeys the properties required of it in
order to form a group?
Oh, and when are you going to post your treatment of energy and
momentum in your 'derivation' of special relativity?
Dono
> I'd like to congratulate Jeckyl for proving that the nonlinear
> form a mathematical group. It looks right to me but I'd like it
> confirmed by unbiased physicists. Here is his solution that I copied
Just one, and not a question but a statement: the transforms are
aphysical. The presence of the arbitray function f(x) in the right
hand side makes it that way (the presence of f(x') in the expression
for t' is even worse) . To prove that this is the case, look at the
following:
The Lorentz transforms (derived By Einstein using a physical process)
look like this:
x'=gamma(v).(x-vt)
t'=gamma(v).(t-vx)
Now, if you have an object in frame S moving with speed u-dx/dt along
the x-axis and you want to find its speed u' in frame S' you only need
to calculate:
u'=dx'/dt'=dx'/dt.dt/dt'=(dx/dt-v)/(1-v.dx/dt)=(u-v)/(1-uv)
Now, try obtaining u'=dx'/dt' with the simplified Shitbert transforms:
x'=gamma(v).(x-vt+vf(x))
t'=gamma(v).(t-vx-f(x))+f(x')
Foe extra points, try calculating the acceleration in S' when you know
the acceleration in S :-)
Dono
> Any questions?
Yes, one more : can you apply the Shitbert tranforms to the wave
equation and show that it is "Shitbert invariant"?
You know, try (nabla^2)E-d^2E/dt^2=0 and see what you get (if you ever
get past the derivatives). You know, Shitbert, this is why
mathematicians with a BS only don't make physicists.
Shubee
> On Jul 30, 5:24 pm, Shubee <e.Shu...@gmail.com> wrote:
>
> > I'd like to congratulate Jeckyl for proving that the nonlinear
> > transformations given in exercise 1 of
>> http://www.everythingimportant.org/relativity/generalized.htm
> > form a mathematical group. It looks right to me but I'd like it
> > confirmed by unbiased physicists.
>
They are just as physical as Lorentz transformations. You just don't
know how to interpret them.
Deep in the google archives one can find my account of finding that
nonlinear solution to the Shubertian clock model. I said that I had
computed time dilation with it and discovered that for a traveling
twin that takes a trip at speed v (defined by a Shubertian invariant
at http://www.everythingimportant.org/relativity/special.pdf ) and
then turns around with zero turnaround time and returns, will find
that his clock time differs with his stay at home twin by the usual
gamma. I will be updating http://www.everythingimportant.org/relativity/special.pdf
will that calculation.
> [The remainder of the crap snipped]
How many times have I told you that you can only compute physically
meaningful quantities by using invariants?
http://www.everythingimportant.org/relativity/generalized.htm
Shubee
Dono
Then you will not have any trouble in computing dx'/dt'.
[rest of Shitbertian physics snipped as BS]
Eric Gisse
Ok.
>From what physical principle are they derived from?
What experimental evidence agrees with your arbitrary functions?
[...]
schoenf...@gmail.com
What is the foundational principle you used to derive the Lorentz
transformation? What's the difference between that and the usual
derivations?
Shubee
There is no law of physics that demands that clocks be synchronized.
Traditional SR presupposes that Einstein's synchronization procedure
is sacrosanct.
http://www.everythingimportant.org/relativity/synchronization.htm
Shubee
http://www.everythingimportant.org/relativity/special.pdf
YBM
> On Jul 31, 3:42 pm, schoenfeld....@gmail.com wrote:
>> On Jul 31, 10:24 am, Shubee <e.Shu...@gmail.com> wrote:
>> [...]
>>
>>> Any questions?
>> What is the foundational principle you used to derive the Lorentz
>> transformation? What's the difference between that and the usual
>> derivations?
>
> There is no law of physics that demands that clocks be synchronized.
> Traditional SR presupposes that Einstein's synchronization procedure
> is sacrosanct.
Of course not (but you've never understand what synchronization is about,
you confuse it with clock tweaking), Einstein's convention is only the
more practical (being compatible with Newton's second law, it can even
be equivalently realized with something else than light as a signal).
If you were a mathematician you would have heard about Roger Penrose
and the kind of alternative sychronization procedures he studied.
Sam Wormley
> Traditional SR presupposes that Einstein's synchronization procedure
> is sacrosanct.
>
Why that's crazy... How can yo misinterpret that? Cite please!
Shubee
I did cite one important reference to prove that:
http://www.everythingimportant.org/relativity/synchronization.htm
Here is another:
http://arxiv.org/abs/gr-qc/0409105
Shubee
schoenf...@gmail.com
> Traditional SR presupposes that Einstein's synchronization procedure
> is sacrosanct.
So you can't answer a simple question. There is some mathematical step
somewhere, assuming the derivation is valid, which contains the
essence of the transformation equation (which is really just a
hyperbolic rotation). Either you can cite it or you cannot.
[...]
Shubee
> So you can't answer a simple question. There is some mathematical step
> somewhere, assuming the derivation is valid, which contains the
> essence of the transformation equation (which is really just a
> hyperbolic rotation). Either you can cite it or you cannot.
You asked for a foundational principle. I'm happy to be more precise.
The usual approach to the Lorentz transformation is to presuppose no
absolute frame of reference. I believe that it's more instructive to
derive a more general theory.
The physicist J. H. Field also sees the importance of beginning with
less dogmatic assumptions. Consider his commentary:
"A much weaker statement of the Relativity Principle than Einstein's
first postulate is sufficient to derive the Lorentz Transformation."
"It was recognised at an early date by Ignatowsky and Frank and Rothe
that Einstein's second postulate was not necessary to derive the
Lorentz Transformation. The questions then arise: what are the weakest
postulates which are sufficient to derive it and what is their minimum
number?"
In fulfillment of Dr. Field's objective, I believe that I have
discovered the leanest derivation and the weakest postulate. It's easy
to see that all the empirical results from Einsteinian SR can be
obtained from a simple tautology. There are laws of physics that are
the same in all frames of reference and there may be laws of physics
that aren't.
I believe that Einstein assumed too much. He thought that all physical
laws are the same in all inertial frames of reference. My weaker axiom
set only requires that my definition of time be similarly defined in
all inertial frames of reference. Therefore, my simple tautology is
inviolable and consistent. There are laws of physics that are Lorentz
invariant and there may be laws of physics that aren't.
Ultimately, I believe that Einstein's theory is overly simplistic. My
generalization is obviously more complex. Every mathematician should
see the beauty of it. Einstein's theory could be false and my theory
could be true. But if my theory is false, then so is Einstein's
theory.
Shubee
http://www.everythingimportant.org/relativity/special.pdf
Dono
> I believe that Einstein assumed too much. He thought that all physical
> laws are the same in all inertial frames of reference.
It dates back way before Einstein , Shitbert.
Physicists (real ones, not pretenders like you) make very heavy use of
this "thought".
> There are laws of physics that are Lorentz
> invariant and there may be laws of physics that aren't.
>
...and there is Shitbertian physics that is full of shit!
> Ultimately, I believe that Einstein's theory is overly simplistic. My
> generalization is obviously more complex. Every mathematician should
> see the beauty of it.
...and every physicist sees its imbecility. You can't even calculate
dx'/dt' with your transforms. You can't even prove that Maxwell's
equations are frame invariant.
> Einstein's theory could be false and my theory
> could be true.
Not if it is unable to predict a simple thing like dx'/dt'.......
> But if my theory is false, then so is Einstein's
> theory.
>
Shitbert, theories can only be falsified by experiment.
Now,the case of Shitbertian physics is different, since yours is not
even a theory.
Eric Gisse
[...]
So shooby....what experimental evidence contradicts modern relativity?
What evidence supports your idiotic re-invention of it?
Shubee
It's simply a more general theory, in keeping with Hilbert's
philosophy.
http://www.everythingimportant.org/relativity/special.pdf
(Section 2) http://arxiv.org/abs/physics/9811050
(Sections 1.1 & 1.2)
Shubee
Dono
> On Jul 31, 9:29 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > On Jul 31, 8:02 pm, Shubee <e.Shu...@gmail.com> wrote:
> > [...]
>
> > So shooby....what experimental evidence contradicts modern relativity?
> > What evidence supports your idiotic re-invention of it?
>
> It's simply a more general theory, in keeping with Hilbert's
> philosophy.http://www.everythingimportant.org/relativity/special.pdf
> (Sections 1.1 & 1.2)
>
> Shubee
In other words...no answer. Or a non-sequitur.
Shitbert is only looking to plug his theory, Shitbertian physics :-)
Eric Gisse
> On Jul 31, 9:29 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > On Jul 31, 8:02 pm, Shubee <e.Shu...@gmail.com> wrote:
> > [...]
>
> > So shooby....what experimental evidence contradicts modern relativity?
> > What evidence supports your idiotic re-invention of it?
>
> It's simply a more general theory, in keeping with Hilbert's
> philosophy.http://www.everythingimportant.org/relativity/special.pdf
> (Sections 1.1 & 1.2)
>
> Shubee
So let me get this straight...
Despite there being no observed experimental contradiction with
special relativity, and despite that the theory is _already_ axiomatic
as per Hilbert's desire, you still think that physicists got it WRONG?
You have got to be kidding me.
schoenf...@gmail.com
I tried to understand what you were doing and I think I have a
picture. It looks like you started out with the Lorentz transform and
worked back and tried to find a general relationship between time
elapsed for observers.
But you present your paper from bottom-up when, in my opinion, it
would've been easier to start with the Lorentz transform and work down
to the "new idea" (which I presume is the sliding ruler/
synchronization concept).
Anyway, I suspect your paper contains an error. In your 3rd universe,
you require that the i'th ruler move opposite the j'th ruler and you
express this mathematically as u_ij = -u_ji.
But, if ruler 1 moves opposite ruler 2 and ruler 2 moves opposite
ruler 3, then ruler 1 does NOT move opposite ruler 3 as there are only
2 directions to move in (the number of signs you have). Hence u_ij !=
u_ji for i=1, j=3. You only have 2 sliding rulers as ruler 1 and ruler
3 don't slide w.r.t to each other.
You can fix that by adding a degree of translational freedom, but this
would make your sliding rulers to be sliding planes in that case. In
the general case, N sliding rulers are really N sliding (N-1) planes
in an N-space.
That might mean you can no longer rely on the argument you make in the
middle of page 4 which allows you to negate one time equation to get
the other. You must ASSERT this to be the case for the general case -
i.e. u_ij = -u_ji would be an axiom.
Anyway, there may be something interesting there but I suspect in
between equation (6) and (7) you insert, via assumption, the essence
of the Lorentz transform.
Shubee
>
> I tried to understand what you were doing and I think I have a
> picture. It looks like you started out with the Lorentz transform and
> worked back and tried to find a general relationship between time
> elapsed for observers.
>
> But you present your paper from bottom-up when, in my opinion, it
> would've been easier to start with the Lorentz transform and work down
> to the "new idea" (which I presume is the sliding ruler/
> synchronization concept).
>
> Anyway, I suspect your paper contains an error. In your 3rd universe,
> you require that the i'th ruler move opposite the j'th ruler and you
> express this mathematically as u_ij = -u_ji.
>
> But, if ruler 1 moves opposite ruler 2 and ruler 2 moves opposite
> ruler 3, then ruler 1 does NOT move opposite ruler 3 as there are only
> 2 directions to move in (the number of signs you have). Hence u_ij !=
> u_ji for i=1, j=3. You only have 2 sliding rulers as ruler 1 and ruler
> 3 don't slide w.r.t to each other.
Thanks for your careful reading of my paper. Of course, to solve my
system of functional equations I had to assume that u_ij != 0 for all
i,j. So there is no error.
> You can fix that by adding a degree of translational freedom, but this
> would make your sliding rulers to be sliding planes in that case. In
> the general case, N sliding rulers are really N sliding (N-1) planes
> in an N-space.
>
> That might mean you can no longer rely on the argument you make in the
> middle of page 4 which allows you to negate one time equation to get
> the other. You must ASSERT this to be the case for the general case -
> i.e. u_ij = -u_ji would be an axiom.
You are right in pointing out that u_ij = -u_ji is an axiom. It's a
very important axiom in my system. I intent to update my paper with a
separate section just to discuss my axioms and those of Einstein.
> Anyway, there may be something interesting there but I suspect in
> between equation (6) and (7) you insert, via assumption, the essence
> of the Lorentz transform.
I think that you've already identified my most important insight,
which is the axiom u_ij = -u_ji. The importance of my general approach
is that it generates nonlinear solutions very quickly. I will be
demonstrating that in my update.
Also, I freely admit what the essence of the Lorentz transform is. "My
weaker axiom set only requires that my definition of time be similarly
defined in all inertial frames of reference."
Note that there are linear and nonlinear solutions to this constraint.
The foundation of it all is the axiom u_ij = -u_ji as you've surmised.
Thanks for your kind words.
Shubee
http://www.everythingimportant.org/relativity/special.pdf
YBM
> You are right in pointing out that u_ij = -u_ji is an axiom. It's a
> very important axiom in my system. I intent to update my paper with a
> separate section just to discuss my axioms and those of Einstein.
Five years without noticing a so simple hidden axiom in your own stupid
paper... and you haven't even identified it yourself... Hilbert could
be proud of you.
You are a joke...
Shubee
Isaac Newton didn't write up his greatest accomplishments till the end
of his life. It doesn't mean that he hadn't figured it out many years
before.
The belief that the author of the very intelligent and original paper
http://www.everythingimportant.org/relativity/special.pdf
isn't aware of his own assumptions, is ludicrous.
It's not like the world is begging me to write out the solutions to
all my claims. And it's not like I have an academic sponsor to
recommend my paper for posting at the arXiv server. But this thread
did open as a congratulatory thank you to Jeckyl for confirming my
claim that the nonlinear transformations given in exercise 1 of
http://www.everythingimportant.org/relativity/generalized.htm form a
mathematical group.
Would you care to confirm that the claim of exercise 2 is true also?
Shubee
http://groups.google.com/group/sci.physics.relativity/msg/eb5ffd21cfa1ba1a
Sam Wormley
That Hilbert and his students supplied significant portions of the
mathematical infrastructure required for quantum mechanics and general
relativity, there is no doubt.
http://plato.stanford.edu/search/searcher.py?query=Hilbert
Hilbert's Program and Gödel's incompleteness theorems
http://plato.stanford.edu/entries/hilbert-program/#4
Shubee
I have many more sections to write up for the case of 1 spatial
dimension before I entertain mathematical generalizations up to the
furthest limit where time is defined by infinitely many Euclidean N-
spaces passing through each other.
> > That might mean you can no longer rely on the argument you make in the
> > middle of page 4 which allows you to negate one time equation to get
> > the other. You must ASSERT this to be the case for the general case -
> > i.e. u_ij = -u_ji would be an axiom.
>
> You are right in pointing out that u_ij = -u_ji is an axiom. It's a
> very important axiom in my system. I intent to update my paper with a
> separate section just to discuss my axioms and those of Einstein.
>
> > Anyway, there may be something interesting there but I suspect in
> > between equation (6) and (7) you insert, via assumption, the essence
> > of the Lorentz transform.
>
> I think that you've already identified my most important insight,
> which is the axiom u_ij = -u_ji.
I wish to apologize for the carelessness of this remark. I certainly
don't believe that my most important insight is the axiom u_ij = -
u_ji. Although equal and opposite proper velocities in my system
certainly is an axiom, I believe that the key insight of my paper is
encapsulated in my two fundamental equations for Xi_2, which are
equations (4) and (5):
t_1 = -x_2/u_12 + g_1(x_1)
t_2 = -x_1/u_21 + g_2(x_2)
These nonlinear equations are the complete solution for clock time in
any two inertial frames of reference. Granted, these equations are
derived from the axiom u_ij = -u_ji but the real cleverness is in the
definition of clock time and in recognizing the infinite array of very
natural coordinate clocks.
The current approach to SR is to begin with the equations
x' = f(x,t,v)
t' = g(x,t,v)
and then to presuppose a group structure.
It seems to me that anyone who reads my paper carefully will agree
that the old way is obsolete and unenlightening and will eventually be
discarded for my more physically illuminating approach because a
concept of time is built right into it.
Shubee
http://www.everythingimportant.org/relativity/special.pdf
Dono
> On Aug 1, 6:18 am, YBM <ybm...@nooos.fr> wrote:
>
> > Shubee wrote :
>
> > > You are right in pointing out that u_ij = -u_ji is an axiom. It's a
> > > very important axiom in my system. I intent to update my paper with a
> > > separate section just to discuss my axioms and those of Einstein.
>
> > Five years without noticing a so simple hidden axiom in your own stupid
> > paper... and you haven't even identified it yourself... Hilbert could
> > be proud of you.
>
> > You are a joke...
>
> Isaac Newton didn't write up his greatest accomplishments till the end
> of his life. It doesn't mean that he hadn't figured it out many years
> before.
>
> isn't aware of his own assumptions, is ludicrous.
>
> It's not like the world is begging me to write out the solutions to
> all my claims. And it's not like I have an academic sponsor to
> recommend my paper for posting at the arXiv server. But this thread
> did open as a congratulatory thank you to Jeckyl for confirming my
> mathematical group.
> Would you care to confirm that the claim of exercise 2 is true also?
>
You still can't do anything with the Shitbert transforms....Try
calculating dx'/dt' from dx/dt.
Try showing the invariance of Maxwell's equations.....
t'Hoft will ask you that as soon as you try to publish again. Actually
all the serious editors will ask for this.
Shubee
> Anyway, I suspect your paper contains an error. In your 3rd universe,
> you require that the i'th ruler move opposite the j'th ruler and you
> express this mathematically as u_ij = -u_ji.
>
> But, if ruler 1 moves opposite ruler 2 and ruler 2 moves opposite
> ruler 3, then ruler 1 does NOT move opposite ruler 3 as there are only
> 2 directions to move in (the number of signs you have). Hence u_ij !=
> u_ji for i=1, j=3. You only have 2 sliding rulers as ruler 1 and ruler
> 3 don't slide w.r.t to each other.
You are right in saying that "if ruler 1 moves opposite ruler 2 and
ruler 2 moves opposite [to] ruler 3, then ruler 1 does NOT
[necessarily] move opposite [to] ruler 3." That is correct. I should
add that u_ij != 0 for all i,j. I will add that correction to
equations (6) and (7) next time I update my paper.
But it seems that you're saying more than that. So perhaps you're
misinterpreting what I mean by u_ij. I wrote the following on page 5
of http://www.everythingimportant.org/relativity/special.pdf
"Because all points of the line Gamma_2 have equal proper velocities,
and likewise for the line Gamma_1, we shall say that the proper
velocity of Gamma_2 with respect to Gamma_1 is u_12 and that the
proper velocity of Gamma_1 with respect to Gamma_2 is u_21."
u_ij = - u_ji simply means that "proper velocities" are equal and
opposite. There can be infinitely many lines in 1 spatial dimension
that are sliding against each other. Any two of them will certainly
obey u_ij = - u_ji. Please see the exact definition of u_ij on page
5.
Shubee
http://www.everythingimportant.org/relativity/special.pdf
Eric Gisse
[...]
> The current approach to SR is to begin with the equations
>
> x' = f(x,t,v)
> t' = g(x,t,v)
>
> and then to presuppose a group structure.
No, shooby, that is wrong. Try again.
[...]
Shubee
Granted, there are many different ways to derive the Lorentz
transformation. But certainly that route is popular. See
http://arxiv.org/PS_cache/physics/pdf/0302/0302045v1.pdf
http://o.castera.free.fr/pdf/onemorederivation.pdf
http://www.courses.fas.harvard.edu/%7Ephys16/Textbook/ch10.pdf
http://ocw.mit.edu/NR/rdonlyres/Physics/8-033Fall-2006/C70EE5E1-F712-4664-85FC-C305E60E310F/0/lecture4_kinem1.pdf
http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.cmp/1103858408
Shubee
http://www.everythingimportant.org/relativity/special.pdf
Ben Rudiak-Gould
> I'd like to congratulate Jeckyl for proving that the nonlinear
> transformations given in exercise 1 of http://www.everythingimportant.org/relativity/generalized.htm
> confirmed by unbiased physicists.
Well, if I understand correctly what you're doing, then it's obviously a
group. There's no need to do all that work. If G is a group, and H is a
subgroup of G, and g is some element of G, then { gxg^-1 | x in H } is a
subgroup of G. So, let G be some large group that includes your
transformations (such as the diffeomorphism group) and let H be the Lorentz
group, and let g be some position-dependent reparameterization of time (or
any other element of G for that matter), and you're done.
-- Ben
Eric Gisse
[...]
Group structure comes later, not first.
Read their derivations of the Lorentz transformation, then read yours.
Notice the difference in content.
Shubee
Obviously. But you couldn't believe all the professional physicists
that I had stumped by that easy exercise. Do you care to comment on
exercise number 2 from http://www.everythingimportant.org/relativity/generalized.htm
?
Shubee
http://www.everythingimportant.org/relativity/special.pdf
Eric Gisse
Dono
So, Shitbert
It took roughly 6 months for us to extract from you the invertibility
and the transitivity properties.
How long will it take to get you to admit that you can't calculate dx'/
dt' from your "transforms"? This one looks like a "forever"
YBM
>> Obviously. But you couldn't believe all the professional physicists
>> that I had stumped by that easy exercise. Do you care to comment on
>> exercise number 2 fromhttp://www.everythingimportant.org/relativity/generalized.htm
>> ?
>
> "Do my work for me, please"
>
>> Shubeehttp://www.everythingimportant.org/relativity/special.pdf
even worse : "do my elementary work on the consequences of my nonsense"
BTW, let's see : it will be interesting when this people will arrive on the
contradictions we pointed out years ago...
We could even write Eugene Shubert's responses in advance, couldn't we ?
Eric Gisse
Dude, Jeckyl had to do it for him. He didn't do it himself.
Dono
I know :-),
But this time he's stuck for good :-) He's not going to come back from
the last list I gave him. He's as good as dead. :-)
Eric Gisse
He doesn't care.
He acknowledges there is no experimental error in special relativity.
He presses forward.
He acknowledges he hasn't actually posted how his framework treats
energy-and momentum, despite how important it is to physicists. He
presses forward.
He lies about physicists not caring about invariants in a theory as an
excuse to not provide his own. He presses forward.
A snake eating itself makes more progress than he does, though the
motion is similar.
Dono
Good analysis :-)
schoenf...@gmail.com
> On Aug 1, 3:06 am, schoenfeld....@gmail.com wrote:
>
> > Anyway, I suspect your paper contains an error. In your 3rd universe,
> > you require that the i'th ruler move opposite the j'th ruler and you
> > express this mathematically as u_ij = -u_ji.
>
> > But, if ruler 1 moves opposite ruler 2 and ruler 2 moves opposite
> > ruler 3, then ruler 1 does NOT move opposite ruler 3 as there are only
> > 2 directions to move in (the number of signs you have). Hence u_ij !=
> > u_ji for i=1, j=3. You only have 2 sliding rulers as ruler 1 and ruler
> > 3 don't slide w.r.t to each other.
>
> You are right in saying that "if ruler 1 moves opposite ruler 2 and
> ruler 2 moves opposite [to] ruler 3, then ruler 1 does NOT
> [necessarily] move opposite [to] ruler 3." That is correct. I should
> add that u_ij != 0 for all i,j. I will add that correction to
> equations (6) and (7) next time I update my paper.
>
> But it seems that you're saying more than that. So perhaps you're
> misinterpreting what I mean by u_ij. I wrote the following on page 5
>
> "Because all points of the line Gamma_2 have equal proper velocities,
> and likewise for the line Gamma_1, we shall say that the proper
> velocity of Gamma_2 with respect to Gamma_1 is u_12 and that the
> proper velocity of Gamma_1 with respect to Gamma_2 is u_21."
>
> u_ij = - u_ji simply means that "proper velocities" are equal and
> opposite. There can be infinitely many lines in 1 spatial dimension
> that are sliding against each other. Any two of them will certainly
> obey u_ij = - u_ji. Please see the exact definition of u_ij on page
> 5.
Okay. The statement u_ij=-u_ji is equivalent to the PoR. I was
thinking about this, but I'll need some more time to look at it. I'll
get back to you. It's an interesting derivation (assuming it's
correct). I might add, ANY derivation of the Lorentz transform is
worthy of at the very least a cursory analysis. It's odd how other
posters seem to dismiss anything that does not come straight out of
some perceived authority on the subject.
> Shubeehttp://www.everythingimportant.org/relativity/special.pdf
Shubee
Chimpanzee societies are all about recognizing who the dominant alpha
males are and submitting to their authority.
Shubee
http://www.everythingimportant.org/relativity/special.pdf
The_Man
Why evidence do you have that you any more about biology than you do
about math or physics?
>
> Shubeehttp://www.everythingimportant.org/relativity/special.pdf- Hide quoted text -
>
> - Show quoted text -
Shubee
Ben,
The importance of this nonlinear Lorentz-equivalent transformation
group isn't that it's so obviously isomorphic to the Lorentz group.
Its importance is that it is a concrete nonlinear solution to the
spacetime clock model developed in http://www.everythingimportant.org/relativity/special.pdf
and it refutes the popular myth that spacetime transformations between
inertial frames of reference must be linear.
Have you never heard of that misconception?
Shubee
http://www.everythingimportant.org/relativity/special.pdf
Dono
> and it refutes the popular myth that spacetime transformations between
> inertial frames of reference must be linear.
>
...only in your deluded mind. You can't do jack shit with your so-
called "transforms". Show how you calculate dx'/dt' , Shitbert......
Eric Gisse
> On Aug 1, 4:44 pm, Ben Rudiak-Gould <br276delet...@cam.ac.uk> wrote:
>
>
>
> > Shubee wrote:
> > > I'd like to congratulate Jeckyl for proving that the nonlinear
> > > transformations given in exercise 1 of
> >>http://www.everythingimportant.org/relativity/generalized.htm
> > > form a mathematical group. It looks right to me but I'd like it
> > > confirmed by unbiased physicists.
>
> > Well, if I understand correctly what you're doing, then it's obviously a
> > group. There's no need to do all that work. If G is a group, and H is a
> > subgroup of G, and g is some element of G, then { gxg^-1 | x in H } is a
> > subgroup of G. So, let G be some large group that includes your
> > transformations (such as the diffeomorphism group) and let H be the Lorentz
> > group, and let g be some position-dependent reparameterization of time (or
> > any other element of G for that matter), and you're done.
>
> > -- Ben
>
> Ben,
>
> The importance of this nonlinear Lorentz-equivalent transformation
> group isn't that it's so obviously isomorphic to the Lorentz group.
The Lorentz group is linear, dipshit. They are not isomorphic.
If you want to shut me up, write down the isomorphism that relates
them or do something simpler: show that the quantity x^2 - c^2t^2 is
preserved under a coordinate transformation.
[...]
Shubee
There is nothing that I could do to shut you up permanently. You just
move from one delusion to the next.
Shubee
http://www.everythingimportant.org/relativity/special.pdf
Eric Gisse
Then stop posting to sci.physics, because you will know no peace as
long as you continue spewing stupidities.
>
> Shubeehttp://www.everythingimportant.org/relativity/special.pdf
Shubee
You call trivialities utter stupidities. And you judge utter
stupidities to be amazingly profound.
Shubee
http://www.everythingimportant.org/relativity/special.pdf
Eric Gisse
You can't even answer my simple questions, and you expect
to...well...I have no idea what you expect to do. What do you expect
to do, anyway?
Special relativity contains no observational error. Your best scenario
is the production of something utterly equivalent - a reinvention of
the wheel. You will be ignored if only because special relativity is
derivable through other much simpler methods.
>
> Shubeehttp://www.everythingimportant.org/relativity/special.pdf
T.M. Sommers
>
> The importance of this nonlinear Lorentz-equivalent transformation
> group isn't that it's so obviously isomorphic to the Lorentz group.
Let's see the proof, then.
--
Thomas M. Sommers -- t...@nj.net -- AB2SB
T.M. Sommers
In other words, you can't do it. There is only one way to prove
that you can do it, and that is to do it. Coming up with more
excuses not to do it only proves that you can't.
In other words, put up or shut up.
schoenf...@gmail.com
Figure 2 and subsequent remarks say that \Gamma_2 moves to the right
of \Gamma_1 and vice-versa. But, w.r.t to which coordinate system?
Your universe contained only \Gamma_1 and 2. It would make more sense
to say that \Gamma_2 moves to the right w.r.t \Gamma_1 and \Gamma_1 to
the left w.r.t \Gamma_2. An introduction of a third ruler who might
observe \Gamma_1 and \Gamma_2 moving to the right w.r.t to itself
still preserves u_ij=-u_ji. In fact, that is what u_ij=-u_ji tells you
- there is no preferred coordinate system leading to the principle of
relativity.
Okay, here is a serious error in your paper.
In page 4 you have
[1] t_1 = -x_2
[2] t_2 = x_1
In page 5 you introduce a "time scaling factors" for both rulers
called a and b. You say for CONSISTENCY IN UNITS a=b and assign u to
that value. Thus you make replacements
[3] t_1 <- t_1/u
[4] t_2 <- t_2/u
That's fine.
But, then you say
[5] t_1 = -x_2/u
[6] t_2 = x_1/u
which does not follow from substitution of [3] and [4] into [1] and
[2]. Equations [5] and [6] break from your sliding ruler idea and move
into, what looks like, some "rotating something" arrangement
(recognized by equation form only). You begin to treat 'u' as a
function not as a constant, which does not follow from a=b.. Later on
when you make g_i(u_ij, x_i)=g(u_ij)x_i you pretty much make
transformations rotations. It looks like you discard the circular
solutions later on and get the hyperbolic solution (which is the
Lorentz transform).
Shubee
I believe that those directions are independent of all coordinate
systems. I believe that they are true for all observers.
> Your universe contained only \Gamma_1 and 2. It would make more sense
> to say that \Gamma_2 moves to the right w.r.t \Gamma_1 and \Gamma_1 to
> the left w.r.t \Gamma_2.
I see your point. I'll see how to clarify (or simplify) the language.
> An introduction of a third ruler who might
> observe \Gamma_1 and \Gamma_2 moving to the right w.r.t to itself
> still preserves u_ij=-u_ji. In fact, that is what u_ij=-u_ji tells you
> - there is no preferred coordinate system leading to the principle of
> relativity.
It seems to me that we can have u_ij=-u_ji and an absolute frame of
reference at the same time. But you'll have to wait for my update to
the paper to understand why. That's going to be a new section.
> Okay, here is a serious error in your paper.
>
> In page 4 you have
> [1] t_1 = -x_2
> [2] t_2 = x_1
>
> In page 5 you introduce a "time scaling factors" for both rulers
> called a and b. You say for CONSISTENCY IN UNITS a=b and assign u to
> that value. Thus you make replacements
> [3] t_1 <- t_1/u
> [4] t_2 <- t_2/u
> That's fine.
Didn't you draw those arrows backwards?
> But, then you say
>
> [5] t_1 = -x_2/u
> [6] t_2 = x_1/u
>
> which does not follow from substitution of [3] and [4] into [1] and
> [2].
Doesn't it follow if you draw the arrows as I do in the paper? Let a =
b = u. At the top of page 5, I wrote
t_1 -> t_1/a
t_2 -> t_2/b
If a = b = u, then this change of units can be written
t_1 -> t_1/u
t_2 -> t_2/u
Note that the change indicated by the arrow means divide by u.
You acknowledge that the initial definition of t_1 and t_2 are
[1] t_1 = -x_2
[2] t_2 = x_1
Then by following the arrow means that we divide by u. So
[5] t_1 = -x_2/u
[6] t_2 = x_1/u
is correct.
> Equations [5] and [6] break from your sliding ruler idea and move
> into, what looks like, some "rotating something" arrangement
> (recognized by equation form only).
Given that these rulers are sliding, then every point of Gamma_1 and
Gamma_2 is essentially a clock. Pick a point a_1 on Gamma_1. The time
at the point a_1 will depend on whatever point of Gamma_2 is flying by
at that instant. The original equation for clock time for the point
a_1 on Gamma_1 will be t_1 = -x_2. Suppose you are infinitesimally
small and positioned at a_1 of the line Gamma_1. At one instant, the
point x_2 =0 of Gamma_2 flies by. So your clock time at that instant
is t_1 = -x_2 = -0 = 0. At a later instant you notice that x_2 =-1 of
Gamma_2 flies by. Computing the time for that event gives you t_1 = -
x_2 = -(-1) = 1. You wait some more and then see x_2 =-2 of Gamma_2
flying by. Computing the time for that instant gives you t_1 = -x_2 = -
(-2) = 2. Thus computing the time when you see the points 0, -1, -2
translates into the clock reading time 0, 1 and then 2 respectively.
Suppose we change the units we measure time by. Let u=1/2. Then our
mathematical clock will read time 0, 2, 4 when the points 0, -1, -2
fly by. That's not a rotation. It's exactly what I said it was. It's a
mere change of units.
I thank you for trying to read and make sense out of my paper.
Shubee
http://www.everythingimportant.org/relativity/special.pdf
schoenf...@gmail.com
Sorry I misunderstood the arrows. Forget that. I still think though
this step where you do g_i(u_ij, x_i)=g(u_ij)x_i establishes the
necessary structure to give some sort of geometry. I didnt get much
time to take a thorough look though. I'll refrain from commenting
until I look at it properly ;-).
Shubee
> Shubee wrote:
>
> > There is nothing that I could do to shut you up permanently.
> > You just move from one delusion to the next.
>
> In other words, you can't do it.
Any child mathematician can do it.
> There is only one way to prove
> that you can do it, and that is to do it.
And what do I get out of it?
> Coming up with more excuses not
> to do it only proves that you can't.
And why don't you prove that the two groups are not isomorphic, if
that is what you believe? I'm always in the mood to chuckle or laugh
hysterically.
> In other words, put up or shut up.
Sam Wormley, Pmb, YBM, Dono and Eric Gisse are all predictable
machines. Perhaps they are sufficiently well-conditioned for
your entertainment needs to jump through your hoops for you.
Shubee
http://www.everythingimportant.org/relativity/special.pdf
The_Man
Hi,
Am I predictable, too? I know that I am only a mere chemist.:-)
But I noticed something about your transformations. As v-> 0, it seems
that they don't converge to the Galilean transformation. Since it
doesn't, how can your transforms be isomorphic to the Lorentz
transforms, which do?
>
> Shubeehttp://www.everythingimportant.org/relativity/special.pdf
Shubee
> Hi,
>
> Am I predictable, too? I know that I am only a mere chemist.:-)
>
> But I noticed something about your transformations. As v-> 0, it seems
> that they don't converge to the Galilean transformation. Since it
> doesn't, how can your transforms be isomorphic to the Lorentz
> transforms, which do?
When I substitute v=0 into the transformation given in exercise 1 of
http://www.everythingimportant.org/relativity/generalized.htm
I get x' = x and t' = t. What do you get?
Likewise, substituting v=0 into the Galilean transformation
x' = x-vt and t' = t
also produces x' = x and t' = t.
Good thinking though.
Shubee
http://www.everythingimportant.org/relativity/special.pdf
T.M. Sommers
> On Aug 3, 10:30 pm, "T.M. Sommers" <t...@nj.net> wrote:
>>Shubee wrote:
>>
>>>There is nothing that I could do to shut you up permanently.
>>>You just move from one delusion to the next.
>>
>>In other words, you can't do it.
>
> Any child mathematician can do it.
Since you obviously can't do it, then you aren't even a child
mathematician.
>>There is only one way to prove
>>that you can do it, and that is to do it.
>
> And what do I get out of it?
You'll prove that you can do it.
>>Coming up with more excuses not
>>to do it only proves that you can't.
>
> And why don't you prove that the two groups are not isomorphic, if
> that is what you believe?
What I believe does not matter. You are the one making the
claim, and it is up to you to prove it.
> I'm always in the mood to chuckle or laugh
> hysterically.
So is everyone else, and as long as you keep making claims you
can't prove everyone else will be laughing at you.
>>In other words, put up or shut up.
>
> Sam Wormley, Pmb, YBM, Dono and Eric Gisse are all predictable
> machines. Perhaps they are sufficiently well-conditioned for
> your entertainment needs to jump through your hoops for you.
In other words, you can't prove your claim.
T.M. Sommers
> On Aug 4, 7:40 am, The_Man <me_so_hornee...@yahoo.com> wrote:
>
>>Am I predictable, too? I know that I am only a mere chemist.:-)
>>
>>But I noticed something about your transformations. As v-> 0, it seems
>>that they don't converge to the Galilean transformation. Since it
>>doesn't, how can your transforms be isomorphic to the Lorentz
>>transforms, which do?
>
> When I substitute v=0 into the transformation given in exercise 1 of
> http://www.everythingimportant.org/relativity/generalized.htm
> I get x' = x and t' = t. What do you get?
Doesn't that depend on what zeta(x) is?
At any rate, the important question is not what happens when v =
0, but what happens when v is small but non-zero. Small enough
to ignore terms in v^2 but not terms in v.
Dono
>
> - Show quoted text -
Shitbert,
You've been had again. Eric just sunk you with another torpedo.
If you start with the Shitbert so-called "transforms:
x'=gamma(v).(x-vt+vf(x))
t'=gamma(v).(t-vx-f(x))+f(x')
and you TRY to calculate the Lorentz INVARIANT
x'^2-t'^2
it is CHILD PLAY to show you that you are UNABLE to get rid of the
terms in f(x'). So, the Shitbert group is NOT isomorphic with the
Lorentz group.
Even you can do this calculation. You have been had, Shitbert, thanks
for the laughs :-)
Dono
> I'd like to congratulate Jeckyl for proving that the nonlinear
> form a mathematical group. It looks right to me but I'd like it
> from his post athttp://groups.google.com/group/sci.physics.relativity/msg/67b70dea713...
>
> ------------------------------------
>
> Let c=1 (for simplicity)
>
> Define gamma_v = 1/sqrt(1-v^2)
>
> >From the exercise:
>
> T_v(x,t)=(x',t') where
> x'= x0 + gamma_v((x-x0)-v(t-t0)+vf(x))
> t'= t0 + gamma_v((t-t0)-v(x-x0)-f(x)) + f(x')
>
> That means we have:
>
> T_w(T_v(x,t)) = T_w(x',t') = (x'',t'')
> where x',t' are as above, so
>
> x''= x0 + gamma_w([x'-x0]-w[t'-t0]+wf(x'))
> t''= t0 + gamma_w([t'-t0]-w[x'-x0]-f(x')) + f(x'')
>
> Replace x',t' with expansion from T_v above..
>
> x''= x0 + gamma_w(
> [x0 + gamma_v((x-x0)-v(t-t0)+vf(x))-x0]
> -w[t0 + gamma_v((t-t0)-v(x-x0)-f(x)) +
> f(x0+gamma_v((x-x0)-v(t-t0)+vf(x)))-t0]
> +wf(x0 + gamma_v((x-x0)-v(t-t0)+vf(x)))
> )
> t''= t0 + gamma_w(
> [t0 + gamma_v((t-t0)-v(x-x0)-f(x)) +
> f(x0+gamma_v((x-x0)-v(t-t0)+vf(x)))-t0]
> -w[x0 + gamma_v((x-x0)-v(t-t0)+vf(x))-x0]
> -f(x0 + gamma_v((x-x0)-v(t-t0)+vf(x)))
> ) + f(x'')
>
> Do some simple canceling and expanding..
>
> x''= x0 + gamma_w(
> gamma_v((x-x0)-v(t-t0)+vf(x))
> -w.gamma_v((t-t0)-v(x-x0)-f(x))
> -w.f(x0+gamma_v((x-x0)-v(t-t0)+vf(x)))
> +w.f(x0+gamma_v((x-x0)-v(t-t0)+vf(x)))
> )
> t''= t0 + gamma_w(
> gamma_v((t-t0)-v(x-x0)-f(x))
> + f(x0+gamma_v((x-x0)-v(t-t0)+vf(x)))
> -w.gamma_v((x-x0)-v(t-t0)+vf(x))
> - f(x0+gamma_v((x-x0)-v(t-t0)+vf(x)))
> ) + f(x'')
>
> Do some more canceling and factoring...
>
> x''= x0 + gamma_w.gamma_v(
> (x-x0)-v(t-t0)+vf(x)-w.(t-t0)+wv(x-x0)+wf(x)
> )
> t''= t0 + gamma_w.gamma_v(
> (t-t0)-v(x-x0)-f(x)-w(x-x0)+wv(t-t0)-wvf(x)
> ) + f(x'')
>
> Do some factoring...
> x''= x0 + gamma_w.gamma_v(
> (1+vw)(x-x0) -(v+w)(t-t0) +(v+w)f(x)
> )
> t''= t0 + gamma_w.gamma_v(
> (1+vw)(t-t0) -(v+w)(x-x0) -(1+vw)f(x)
> ) + f(x'')
>
> and some more factoring...
>
> x''= x0 + (1+vw).gamma_w.gamma_v(
> (x-x0) -(v+w)/(1+vw)(t-t0) +(v+w)/(1+vw)f(x)
> )
> t''= t0 + (1+vw).gamma_w.gamma_v(
> (t-t0) -(v+w)/(1+vw)(x-x0) -f(x)
> ) + f(x'')
>
> In the above we have:
>
> gamma_w.gamma_v = (1/sqrt(1-v^2))(1/sqrt(1-w^2))
> gamma_w.gamma_v = 1/sqrt((1-v^2)(1-w^2))
> gamma_w.gamma_v = 1/sqrt(1-v^2-w^2+v^2w^2))
>
> So...
>
> (1+vw).gamma_w.gamma_v = (1+vw)/sqrt(1-v^2-w^2+v^2w^2))
>
> Also in the above we have (v+w)/(1+vw)
>
> let U = (v+w)/(1+vw) .. we can write that as (v(+)w).
>
> then...
>
> gamma_U = 1/sqrt(1-((v+w)/(1+vw))^2)
> = 1/sqrt(1-(v^2+2vw+w^2)/(1+vw)^2)
> = 1/(sqrt((1+vw)^2-v^2-2vw-w^2)/sqrt((1+vw)^2)
> = 1/(sqrt(1+2vw+v^2w^2-v^2-2vw-w^2)/(1+vw))
> = (1+vw)/sqrt(1-v^2-w^2+v^2w^2)
> = (1+vw).gamma_w.gamma_v
>
> so we can replace appropriate terms with U and gamma_U...
>
> x''= x0 + gamma_U((x-x0) -U(t-t0) +U.f(x))
> t''= t0 + gamma_U((t-t0) -U(x-x0) -f(x)) + f(x'')
>
> so...
>
> T_[U](x,t) = T_[v(+)w](x,t) = T_w(T_v(x,y)) = (x'',t'')
>
> Please let me know if there are any errors in the above. Its tricky
> when typing in moderately complex formulas like that as plain text.
>
> ------------------------------------
>
> Any questions?
Shitbert,
You've been had again. Eric just sunk you with another torpedo.
If you start with the Shitbert so-called "transforms:
x'=gamma(v).(x-vt+vf(x))
t'=gamma(v).(t-vx-f(x))+f(x')
and you TRY to calculate the Lorentz INVARIANT
x'^2-t'^2
it is CHILD PLAY to show you that you are UNABLE to get rid of the
terms in f(x'). So, the Shitbert group is NOT isomorphic with the
Lorentz group.
Even YOU can do this calculation.Not even your crank friend Jerkyl can
get you out of this. You have been had, Shitbert, thanks
for the laughs :-)
The_Man
> On Aug 4, 7:40 am, The_Man <me_so_hornee...@yahoo.com> wrote:
>
> > Hi,
>
> > Am I predictable, too? I know that I am only a mere chemist.:-)
>
> > But I noticed something about your transformations. As v-> 0, it seems
> > that they don't converge to the Galilean transformation. Since it
> > doesn't, how can your transforms be isomorphic to the Lorentz
> > transforms, which do?
>
> I get x' = x and t' = t. What do you get?
>
> Likewise, substituting v=0 into the Galilean transformation
> x' = x-vt and t' = t
> also produces x' = x and t' = t.
>
> Good thinking though.
I apologize.. I screwed up just a little. I shoud have written,
instead of "as v -> 0", rather "v/c -> 0"
If you take the Lorentz transform as v/c -> 0, it becomes the galilean
transformation.
The galilean tranformation isn't
x' = x, and t = t'.
It's x'=x-vt, and t'=t.
We are not trying to find the transforms AT v=0; that would be
trivial. We want the transform for v small relative (pun intended) to
c..
The Lorentz transforms in 2 D are :
x' = gamma (x -vt)
t' = gamma ( t - vx/c^2)
Right? With v very small compared to c, v/c ->0. Therefore, we can
replace v/c by zero, but not v itself
Then, the Lorentz transformation becomes the galilean transformation:
x'= x -vt
t' =t.
Now let's take the Shubertian transformations
x' = gamma (x - vt) + vf(x)
t' = gamma [t - v -f (x)] + f(x')
where I took the liberty of synchronizing the coordinate at t0.
Now, as v/c -> gamma -> 0. No problem. That gives for Shubertian
transform
x' = x-vt + vf(x)
t' = t - v - f(x) + f(x')
which ISN'T the Galilean transformation - not by a long shot.
The only way it works, is if f(x) and f(x') are IDENTICALLY CONSTANTS
(Which is a trivial modification of Lorentz, equivalent to a shift of
axes).
Maybe the physicists could help me out here, and let me know if I
mucked it up :-(
>
> Shubeehttp://www.everythingimportant.org/relativity/special.pdf
Shubee
> Shubee wrote:
> > On Aug 3, 10:30 pm, "T.M. Sommers" <t...@nj.net> wrote:
> >>Shubee wrote:
>
> >>>There is nothing that I could do to shut you up permanently.
> >>>You just move from one delusion to the next.
>
> >>In other words, you can't do it.
>
> > Any child mathematician can do it.
>
> Since you obviously can't do it, then you aren't even a child
> mathematician.
You have no proof that I can't do it.
> >>There is only one way to prove
> >>that you can do it, and that is to do it.
>
> > And what do I get out of it?
>
> You'll prove that you can do it.
I already know I can do it and that the exercise is trivial.
> >>Coming up with more excuses not
> >>to do it only proves that you can't.
Can you do the exercise?
> > And why don't you prove that the two groups are not isomorphic, if
> > that is what you believe?
>
> What I believe does not matter. You are the one making the
> claim, and it is up to you to prove it.
Who is going to benefit from my proof? What evidence is there what you
would recognize an explicitly stated isomorphism? Do you understand my
derivation of the Lorentz transformation at
http://www.everythingimportant.org/relativity/special.pdf ?
> > I'm always in the mood to chuckle or laugh
> > hysterically.
>
> So is everyone else, and as long as you keep making claims you
> can't prove everyone else will be laughing at you.
There's no doubt that Ben Rudiak-Gould knows that I'm right. The two
groups are clearly isomorphic. What evidence is there that you or any
one else on this newsgroup understands elementary group theory and
will be persuaded that I'm right?
Shubee
http://www.everythingimportant.org/relativity/special.pdf
Shubee
> Shubee wrote:
> > On Aug 4, 7:40 am, The_Man <me_so_hornee...@yahoo.com> wrote:
>
> >>Am I predictable, too? I know that I am only a mere chemist.:-)
>
> >>But I noticed something about your transformations. As v-> 0, it seems
> >>that they don't converge to the Galilean transformation. Since it
> >>doesn't, how can your transforms be isomorphic to the Lorentz
> >>transforms, which do?
>
> > When I substitute v=0 into the transformation given in exercise 1 of
> >http://www.everythingimportant.org/relativity/generalized.htm
> > I get x' = x and t' = t. What do you get?
>
> Doesn't that depend on what zeta(x) is?
Not in this instance.
Shubee
http://www.everythingimportant.org/relativity/special.pdf
T.M. Sommers
> On Aug 4, 7:55 am, "T.M. Sommers" <t...@nj.net> wrote:
>>Shubee wrote:
>>>On Aug 3, 10:30 pm, "T.M. Sommers" <t...@nj.net> wrote:
>>>>Shubee wrote:
>>>
>>>>>There is nothing that I could do to shut you up permanently.
>>>>>You just move from one delusion to the next.
>>>>
>>>>In other words, you can't do it.
>>>
>>>Any child mathematician can do it.
>>
>>Since you obviously can't do it, then you aren't even a child
>>mathematician.
>
> You have no proof that I can't do it.
The fact that you consistently refuse to do the proof is all the
proof I need.
>>>>There is only one way to prove
>>>>that you can do it, and that is to do it.
>>>
>>>And what do I get out of it?
>>
>>You'll prove that you can do it.
>
> I already know I can do it and that the exercise is trivial.
Everyone else already knows that you can't do it.
>>>>Coming up with more excuses not
>>>>to do it only proves that you can't.
>
> Can you do the exercise?
Why should I when you can't? It's not my claim that needs
proving, it's yours.
>>>And why don't you prove that the two groups are not isomorphic, if
>>>that is what you believe?
>>
>>What I believe does not matter. You are the one making the
>>claim, and it is up to you to prove it.
>
> Who is going to benefit from my proof? What evidence is there what you
> would recognize an explicitly stated isomorphism? Do you understand my
> derivation of the Lorentz transformation at
> http://www.everythingimportant.org/relativity/special.pdf ?
I am not the issue. I am making no claims that contradict
accepted physics. You are, and it is up to your to prove your claim.
>>>I'm always in the mood to chuckle or laugh
>>>hysterically.
>>
>>So is everyone else, and as long as you keep making claims you
>>can't prove everyone else will be laughing at you.
>
> There's no doubt that Ben Rudiak-Gould knows that I'm right. The two
> groups are clearly isomorphic. What evidence is there that you or any
> one else on this newsgroup understands elementary group theory and
> will be persuaded that I'm right?
What evidence is there that you can do the proof?
As I said before, put up or shut up.
T.M. Sommers
> On Aug 4, 8:07 am, "T.M. Sommers" <t...@nj.net> wrote:
>>Shubee wrote:
>>>On Aug 4, 7:40 am, The_Man <me_so_hornee...@yahoo.com> wrote:
>>
>>>>Am I predictable, too? I know that I am only a mere chemist.:-)
>>>
>>>>But I noticed something about your transformations. As v-> 0, it seems
>>>>that they don't converge to the Galilean transformation. Since it
>>>>doesn't, how can your transforms be isomorphic to the Lorentz
>>>>transforms, which do?
>>>
>>>When I substitute v=0 into the transformation given in exercise 1 of
>>>http://www.everythingimportant.org/relativity/generalized.htm
>>>I get x' = x and t' = t. What do you get?
>>
>>Doesn't that depend on what zeta(x) is?
>
> Not in this instance.
Why not?
>>At any rate, the important question is not what happens when v =
>>0, but what happens when v is small but non-zero. Small enough
>>to ignore terms in v^2 but not terms in v.
Did you understand this?
The_Man
No, he doesn't. I have shown above in a second post that his
Shubertian transform doesn't reduce to the Galilean transform at low
v. Which means that not only is it NOT isomorphic to Lorentz (as Eric
has shown already), but is COMPLETELY unphysical.
>
> --
> Thomas M. Sommers -- t...@nj.net -- AB2SB- Hide quoted text -
Shubee
> On Aug 4, 10:51 am, Shubee <e.Shu...@gmail.com> wrote:
>
>
>
> > On Aug 4, 7:40 am, The_Man <me_so_hornee...@yahoo.com> wrote:
>
> > > Hi,
>
> > > Am I predictable, too? I know that I am only a mere chemist.:-)
>
> > > But I noticed something about your transformations. As v-> 0, it seems
> > > that they don't converge to the Galilean transformation. Since it
> > > doesn't, how can your transforms be isomorphic to the Lorentz
> > > transforms, which do?
>
> > When I substitute v=0 into the transformation given in exercise 1
That equation is wrong. Take another look at exercise 1 at
http://www.everythingimportant.org/relativity/generalized.htm
If you hold v constant and let c -> infinity, then the transformation
clearly becomes the Galilean transformation.
Shubee
http://www.everythingimportant.org/relativity/special.pdf
Shubee
> On Jul 30, 5:24 pm, Shubee <e.Shu...@gmail.com> wrote:
>
>
>
> > I'd like to congratulate Jeckyl for proving that the nonlinear
> > transformations given in exercise 1 of
>> http://www.everythingimportant.org/relativity/generalized.htm
> > form a mathematical group. It looks right to me but I'd like it
> > confirmed by unbiased physicists. Here is his solution that I copied
Unfortunately, like Eric Gisse, you don't even know the definition of
a group isomorphism. It follows logically then that wouldn't recognize
a valid proof of my nonlinear group being isomorphic to the Lorentz
group.
Shubee
http://www.everythingimportant.org/relativity/special.pdf
Dono
>
>
>
> > Even YOU can do this calculation.Not even your crank friend Jerkyl can
> > get you out of this. You have been had, Shitbert, thanks
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
Shitbert,
The Lorentz transforms have x^2-t^2 as an invariant, the so-called
Shitbert "transforms" don't. You wouldn't know but this invariant is
essential in deriving a lot of useful physical properties (you are
just a Bull_Shitter in math, no physics knowledge) Gisse cooked your
goose.
hahahahahaha , thanks for the laughs , Shitbert
Shubee
Unfortunately, like Eric Gisse, you don't even know the definition
of a group isomorphism. It follows logically then that you wouldn't
recognize a valid proof of my nonlinear group being isomorphic to
the Lorentz group.
Shubee
http://www.everythingimportant.org/relativity/special.pdf
> Even YOU can do this calculation.Not even your crank friend Jerkyl can
Dono
>
>
>
> > Even YOU can do this calculation.Not even your crank friend Jerkyl can
> > get you out of this. You have been had, Shitbert, thanks
The_Man
> On Aug 4, 8:16 am, The_Man <me_so_hornee...@yahoo.com> wrote:
>
>
>
>
>
> > On Aug 4, 10:51 am, Shubee <e.Shu...@gmail.com> wrote:
>
> > > On Aug 4, 7:40 am, The_Man <me_so_hornee...@yahoo.com> wrote:
>
> > > > Hi,
>
> > > > Am I predictable, too? I know that I am only a mere chemist.:-)
>
> > > > But I noticed something about your transformations. As v-> 0, it seems
> > > > that they don't converge to the Galilean transformation. Since it
> > > > doesn't, how can your transforms be isomorphic to the Lorentz
> > > > transforms, which do?
>
> > > When I substitute v=0 into the transformation given in exercise 1
> If you hold v constant and let c -> infinity, then the transformation
> clearly becomes the Galilean transformation.
I used the version that Jeckyl used to prove some of the group
properties. But, be that as it may, you can't let c-> infinity,
because light isn't infinitely fast, and this restriction isn't
necessary to reduce the Lorentz transformation to the Galilean. BTW,
f(x) / g(x) doesn't necessarily go to zero as g(x) goes to infinity,
since f(x) might grow to infinity too, and inf / inf isn't zero.
Nevertheless, Exercise 2 fails. It is impossible to show that the
Shubertian transform is "physically" identical to Lorentz, since you
need an UNPHYSICAL assumption to reduce your transform to the
galilean, whereas Lorentz needed none such.
>
> Shubeehttp://www.everythingimportant.org/relativity/special.pdf
>
>
>
> > where I took the liberty of synchronizing the coordinate at t0.
>
> > Now, as v/c -> gamma -> 0. No problem. That gives for Shubertian
> > transform
>
> > x' = x-vt + vf(x)
> > t' = t - v - f(x) + f(x')
>
> > which ISN'T the Galilean transformation - not by a long shot.
> > The only way it works, is if f(x) and f(x') are IDENTICALLY CONSTANTS
> > (Which is a trivial modification of Lorentz, equivalent to a shift of
> > axes).
>
> > Maybe the physicists could help me out here, and let me know if I
> > mucked it up :-(- Hide quoted text -
Shubee
Jeckyl selected units of measurement, such a light years per year, so
that c=1. That's acceptable. I used units of spacetime measurement so
that c -> infinity. Consider the speed of light in units of inches per
second, then inches per minute, hour, month, year, decade, century,
millennia, etc. Numerically, we can make c -> infinity in this way. It
has the same effect as keeping c constant and letting v -> 0.
Shubee
http://www.everythingimportant.org/relativity/special.pdf
T.M. Sommers
>
> Jeckyl selected units of measurement, such a light years per year, so
> that c=1. That's acceptable. I used units of spacetime measurement so
> that c -> infinity. Consider the speed of light in units of inches per
> second, then inches per minute, hour, month, year, decade, century,
> millennia, etc. Numerically, we can make c -> infinity in this way. It
> has the same effect as keeping c constant and letting v -> 0.
You're joking, aren't you? If you change the units for c, you
have to change the units for everything else, too.
Shubee
But the equation stays the same, doesn't it?
Shubee
http://www.everythingimportant.org/relativity/special.pdf
Shubee
>
> > Shubee wrote:
>
> > > Jeckyl selected units of measurement, such a light years per year, so
> > > that c=1. That's acceptable. I used units of spacetime measurement so
> > > that c -> infinity. Consider the speed of light in units of inches per
> > > second, then inches per minute, hour, month, year, decade, century,
> > > millennia, etc. Numerically, we can make c -> infinity in this way. It
> > > has the same effect as keeping c constant and letting v -> 0.
>
> > You're joking, aren't you? If you change the units for c, you
> > have to change the units for everything else, too.
I kept the units of distance in inches. That works, doesn't it?
And the equations stay the same, don't they?
Shubee
T.M. Sommers
But you are changing the unit of time, so you must change it
everywhere.
You can't change the physics by fiddling with units.
T.M. Sommers
The equation itself stays the same, but all the values of all the
variables change along with c so as to keep c constant,
relatively speaking. That is, as the numerical value of c
increases with the change of units, the numerical value of v also
changes, so v/c stays the same.
Shubee
> Shubee wrote:
> > On Aug 4, 9:27 am, Shubee <e.Shu...@gmail.com> wrote:
> >>On Aug 4, 9:24 am, "T.M. Sommers" <t...@nj.net> wrote:
> >>>Shubee wrote:
>
> >>>>Jeckyl selected units of measurement, such a light years per year, so
> >>>>that c=1. That's acceptable. I used units of spacetime measurement so
> >>>>that c -> infinity. Consider the speed of light in units of inches per
> >>>>second, then inches per minute, hour, month, year, decade, century,
> >>>>millennia, etc. Numerically, we can make c -> infinity in this way. It
> >>>>has the same effect as keeping c constant and letting v -> 0.
>
> >>>You're joking, aren't you? If you change the units for c, you
> >>>have to change the units for everything else, too.
>
> > I kept the units of distance in inches. That works, doesn't it?
>
> But you are changing the unit of time, so you must change it
> everywhere.
Everyone understands that when you change the units of time, it
applies everywhere. How does that triviality contradict my argument?
> You can't change the physics by fiddling with units.
I thought that we were discussing taking a mathematical limit.
Shubee
http://www.everythingimportant.org/relativity/special.pdf
Shubee
Were you not trying to take the limit as v -> 0? How is that different
than letting c -> infinity?
Shubee
http://www.everythingimportant.org/relativity/special.pdf
Ben Rudiak-Gould
> Do you care to comment on exercise number 2 from
> http://www.everythingimportant.org/relativity/generalized.htm
> ?
I tend to stop responding in threads that have grown out of control like
this one has. But since people seem to trust my pronouncements, I'm going to
make a few.
1. These transformations do form a group.
2. It's group-isomorphic to the Lorentz group.
3. It's not "indistinguishable" from the Lorentz group, to the
extent that "indistinguishable" has a mathematical meaning,
because along with the group structure these elements have
an interpretation as actions on (x,y,z,t), and they can be
distinguished that way.
4. You can define physical theories that have this as their
symmetry group. There's a natural isomorphism between the
set of such theories and the set of Lorentz-invariant
theories, so one might say in this sense that the group is
(physically) indistinguishable from the Lorentz group.
All of these things should be obvious if you know a bit of group theory.
Math aside, all I really want to know is what point you (Shubee) are trying
to make here. My best guess is that you're trying to make people understand
that special relativity isn't about coordinate systems and transformations
between them. If so, great; that's the point I've been trying to make for
years. But it's evidently being lost on everyone here (including me!), so
however much you might like your presentation, you may have to look for a
new one.
In another post you say
> it refutes the popular myth that spacetime transformations between
> inertial frames of reference must be linear.
But your reference frames aren't inertial. I don't think you can define a
theory with this symmetry group in which Newton's first law holds with
respect to (x,y,z,t), unless f is linear.
-- Ben
Shubee
I understand that my second theorem (exercise 2) doesn't say anything
about the correspondence principle. Let f(x) be the arbitrary function
in exercise 1 and 2 of http://www.everythingimportant.org/relativity/generalized.htm
Write f(x) = cg(x)
x0 = a
t0 = b
Then theorem 2 only says that in the limit as c-> infinity, the
resulting nonlinear transformation is physically indistinguishable
from the Galilean transformation.
In other words, the transformation equations
x' = x -v(t-b) +vg(x)
t' = t -g(x) +g[ x -v(t-b) + vg(x) ]
form a group isomorphic to the Galilean group.
Shubee
http://www.everythingimportant.org/relativity/special.pdf
The_Man
> On Aug 4, 8:29 am, "T.M. Sommers" <t...@nj.net> wrote:
>
>
>
>
>
> > Shubee wrote:
> > > On Aug 4, 8:07 am, "T.M. Sommers" <t...@nj.net> wrote:
> > >>Shubee wrote:
> > >>>On Aug 4, 7:40 am, The_Man <me_so_hornee...@yahoo.com> wrote:
>
> > >>>>Am I predictable, too? I know that I am only a mere chemist.:-)
>
> > >>>>But I noticed something about your transformations. As v-> 0, it seems
> > >>>>that they don't converge to the Galilean transformation. Since it
> > >>>>doesn't, how can your transforms be isomorphic to the Lorentz
> > >>>>transforms, which do?
>
> > >>>When I substitute v=0 into the transformation given in exercise 1 of
> > >>>http://www.everythingimportant.org/relativity/generalized.htm
> > >>>I get x' = x and t' = t. What do you get?
>
> > >>Doesn't that depend on what zeta(x) is?
>
> > > Not in this instance.
>
> > Why not?
>
> > >>At any rate, the important question is not what happens when v =
> > >>0, but what happens when v is small but non-zero. Small enough
> > >>to ignore terms in v^2 but not terms in v.
>
> > Did you understand this?
>
> I understand that my second theorem (exercise 2) doesn't say anything
> about the correspondence principle. Let f(x) be the arbitrary function
> Write f(x) = cg(x)
> x0 = a
> t0 = b
>
> Then theorem 2 only says that in the limit as c-> infinity, the
> resulting nonlinear transformation is physically indistinguishable
> from the Galilean transformation.
>
> In other words, the transformation equations
>
> x' = x -v(t-b) +vg(x)
> t' = t -g(x) +g[ x -v(t-b) + vg(x) ]
>
> form a group isomorphic to the Galilean group.
But "c" isn't a variable; it's a constant :-)
Allowing v/c -> 0 works out OK for x', but your time t' is screwed up.
The 1/c factor multiplies zeta(x'). 1/c is small, but it isn;t zero,
and it doesnt approach zero at loew velocities. The product can only
become zero with a certain selection of zeta(x') - probably only a
trivial one at that.
Though I am only a chemist.
>
> Shubeehttp://www.everythingimportant.org/relativity/special.pdf- Hide quoted text -
Androcles
"Ben Rudiak-Gould" <br276d...@cam.ac.uk> wrote in message
news:f92e3f$p9o$1...@gemini.csx.cam.ac.uk...
: Shubee wrote:
: > Do you care to comment on exercise number 2 from
: > http://www.everythingimportant.org/relativity/generalized.htm
: > ?
:
: I tend to stop responding in threads that have grown out of control like
: this one has. But since people seem to trust my pronouncements, I'm going
to
: make a few.
:
: 1. These transformations do form a group.
:
: 2. It's group-isomorphic to the Lorentz group.
NOTHING is isomorphic to time because time is not a vector, it has no
additive inverse.
End.
Shubee
> Shubee wrote:
> > Do you care to comment on exercise number 2 from
> >http://www.everythingimportant.org/relativity/generalized.htm
> > ?
>
> I tend to stop responding in threads that have grown out of control like
> this one has. But since people seem to trust my pronouncements, I'm going to
> make a few.
Thank you Ben. I greatly appreciate your help.
> 1. These transformations do form a group.
>
> 2. It's group-isomorphic to the Lorentz group.
>
> 3. It's not "indistinguishable" from the Lorentz group, to the
> extent that "indistinguishable" has a mathematical meaning,
> because along with the group structure these elements have
> an interpretation as actions on (x,y,z,t), and they can be
> distinguished that way.
I make the precise point of saying "physically indistinguishable" in
exercise 2.
> 4. You can define physical theories that have this as their
> symmetry group. There's a natural isomorphism between the
> set of such theories and the set of Lorentz-invariant
> theories, so one might say in this sense that the group is
> (physically) indistinguishable from the Lorentz group.
The two theories are "physically indistinguishable" in the sense that
the laws of physics do not change when we adopt a different clock
synchronization procedure.
> All of these things should be obvious if you know a bit of group theory.
Very true.
> Math aside, all I really want to know is what point you (Shubee) are trying
> to make here. My best guess is that you're trying to make people understand
> that special relativity isn't about coordinate systems and transformations
> between them. If so, great; that's the point I've been trying to make for
> years. But it's evidently being lost on everyone here (including me!), so
> however much you might like your presentation, you may have to look for a
> new one.
>
> In another post you say
>
> > it refutes the popular myth that spacetime transformations between
> > inertial frames of reference must be linear.
>
> But your reference frames aren't inertial.
My reference frames are most certainly inertial according to a
coordinate independent definition.
> I don't think you can define a theory with this symmetry group
> in which Newton's first law holds with respect to (x,y,z,t),
> unless f is linear.
>
> -- Ben
Ah, but if we are going to discuss the laws of physics in a way that's
independent of a preferred clock synchronization scheme, then we need
to define an inertial frame of reference without demanding that clocks
be synchronized. That's the whole point of my very general formulation
of SR. In section 4 of http://www.everythingimportant.org/relativity/special.pdf
I derive the most general solution for clock time in any two inertial
frames of reference.
The fundamental equations representing inertial frames of reference
are nonlinear. For the universe Xi_2, consisting of two inertial
frames of reference in one spatial dimension, it's easy to see that
t_1 = -x_2/u_12 + g_1(x_1)
t_2 = -x_1/u_21 + g_2(x_2)
u_ij = -u_ji = u
The g_i(x_i) are arbitrary synchronization functions.
Shubee
http://www.everythingimportant.org/relativity/special.pdf
T.M. Sommers
> On Aug 4, 8:29 am, "T.M. Sommers" <t...@nj.net> wrote:
>>Shubee wrote:
>>>On Aug 4, 8:07 am, "T.M. Sommers" <t...@nj.net> wrote:
>>>
>>>>At any rate, the important question is not what happens when v =
>>>>0, but what happens when v is small but non-zero. Small enough
>>>>to ignore terms in v^2 but not terms in v.
>>>
>>Did you understand this?
>
> I understand that my second theorem (exercise 2) doesn't say anything
> about the correspondence principle. Let f(x) be the arbitrary function
> in exercise 1 and 2 of http://www.everythingimportant.org/relativity/generalized.htm
> Write f(x) = cg(x)
> x0 = a
> t0 = b
Utterly irrelevant to what I was saying.
> Then theorem 2 only says that in the limit as c-> infinity, the
> resulting nonlinear transformation is physically indistinguishable
> from the Galilean transformation.
You can't take the limit as c->infinity without changing the
physics, because c is physically a constant. Furthermore, you
explained that you are letting c->infinity by simply changing
units, which I explained elsewhere is a no go.
T.M. Sommers
> On Aug 4, 9:51 am, "T.M. Sommers" <t...@nj.net> wrote:
>>Shubee wrote:
>>>On Aug 4, 9:27 am, Shubee <e.Shu...@gmail.com> wrote:
>>>>On Aug 4, 9:24 am, "T.M. Sommers" <t...@nj.net> wrote:
>>>>>Shubee wrote:
>>>>
>>>>>>Jeckyl selected units of measurement, such a light years per year, so
>>>>>>that c=1. That's acceptable. I used units of spacetime measurement so
>>>>>>that c -> infinity. Consider the speed of light in units of inches per
>>>>>>second, then inches per minute, hour, month, year, decade, century,
>>>>>>millennia, etc. Numerically, we can make c -> infinity in this way. It
>>>>>>has the same effect as keeping c constant and letting v -> 0.
>>>>>
>>>>>You're joking, aren't you? If you change the units for c, you
>>>>>have to change the units for everything else, too.
>>>>
>>>I kept the units of distance in inches. That works, doesn't it?
>>
>>But you are changing the unit of time, so you must change it
>>everywhere.
>
> Everyone understands that when you change the units of time, it
> applies everywhere. How does that triviality contradict my argument?
By counteracting the change in numerical value of c.
>>You can't change the physics by fiddling with units.
>
> I thought that we were discussing taking a mathematical limit.
Of an allegedly physical equation.
T.M. Sommers
> On Aug 4, 9:53 am, "T.M. Sommers" <t...@nj.net> wrote:
>>Shubee wrote:
>>>On Aug 4, 9:24 am, "T.M. Sommers" <t...@nj.net> wrote:
>>>>Shubee wrote:
>>>
>>>>>Jeckyl selected units of measurement, such a light years per year, so
>>>>>that c=1. That's acceptable. I used units of spacetime measurement so
>>>>>that c -> infinity. Consider the speed of light in units of inches per
>>>>>second, then inches per minute, hour, month, year, decade, century,
>>>>>millennia, etc. Numerically, we can make c -> infinity in this way. It
>>>>>has the same effect as keeping c constant and letting v -> 0.
>>>>
>>>>You're joking, aren't you? If you change the units for c, you
>>>>have to change the units for everything else, too.
>>>
>>>But the equation stays the same, doesn't it?
>>
>>The equation itself stays the same, but all the values of all the
>>variables change along with c so as to keep c constant,
>>relatively speaking. That is, as the numerical value of c
>>increases with the change of units, the numerical value of v also
>>changes, so v/c stays the same.
>
> Were you not trying to take the limit as v -> 0?
Not me originally; someone else. But that doesn't matter.
> How is that different
> than letting c -> infinity?
Read what I wrote above. Do you really not understand? Let me
put it another way. By just changing units, you are not really
letting c approach infinity. The numerical value of c might
grow, but it is physically the same finite speed. That is far
different from letting a speed actually increase or decrease.
Shubee
> > Do you care to comment on exercise number 2 from
> >http://www.everythingimportant.org/relativity/generalized.htm
> > ?
>
> I tend to stop responding in threads that have grown out of control like
> this one has. But since people seem to trust my pronouncements, I'm going to
> make a few.
Ben,
Your expertise carries a lot of weight with me. And you are right
about the lack of control in this thread. If you'd like to discuss
coordinate independent formulations of SR with me in a new thread, we
can begin anew at sci.math. I would like that also. You are also very
welcome to discuss this with me at http://www.everythingimportant.org/viewforum.php?f=14
where I am authorized to delete any unwanted noise.
Shubee
T.M. Sommers
Let me put it more mathematically explicit. You are letting the
numerical value of c grow by changing the units used for time.
In other words, every time a variable with dimensions of time
appears it is multiplied by some conversion factor k. So every t
in the equation is multiplied by k and every velocity is divided
by k.
Let's look at the numerator of the second term of the first equation:
(x - x0) - v (t - t0) + zeta(x) v/c.
This becomes, after the unit change:
(x - x0) - v/k (tk - t0 k) + zeta(x) v/k / c/k.
It should be immediately obvious that all the ks cancel, so you
are left, in effect, with the original values. That is, c has
not grown to infinity, or to anything else, either.
Eric Gisse
[...]
> Maybe the physicists could help me out here, and let me know if I
> mucked it up :-(
Close enough. Chemists don't have to worry about this shit, so it's
ok.
When it is said the Lorentz transformations reduce to the Galilean
transformations, this is called the "correspondence principle". It is
the same thing as demanding that quantum mechanics reduces to
classical mechanics, and that general relativity reduces to Newtonian
gravitation.
To prove correspondence, you Taylor expand the parameter in question
and toss out the higher order terms. This is the mathematically
correct way of proving things, which may surprise Shooby to no fucking
end.
>
>
>
> > Shubeehttp://www.everythingimportant.org/relativity/special.pdf
Eric Gisse
[...]
>
> Were you not trying to take the limit as v -> 0? How is that different
> than letting c -> infinity?
...because you have terms that aren't just powers of v/c. Plus we want
the _low_ velocity limit, not the zero velocity limit.
This is cute. You pretend your group is isomorphic to the Lorentz
group but you can't show that any of the invariants exist in your
group, or that it actually reduces to the Galilean transformations for
the low velocity limit.
Shooby, did anyone ever teach you about the power series expansion?
>
> Shubeehttp://www.everythingimportant.org/relativity/special.pdf
Dono
> To prove correspondence, you Taylor expand the parameter in question
> and toss out the higher order terms. This is the mathematically
> correct way of proving things, which may surprise Shooby to no fucking
> end.
It is called the Schlomlich-Roche remainder and needs to converge to
0. In Shitbert's case , he's stuck with the frigging ARBITRARY
function f(x') that DOESN'T satisfy any of the conditions. You are
cooking his gise one after the other :-)
Dono
>
> - Show quoted text -
You are trying to convince an autonatic answering machine that he's
wrong . Actually your odds of convincing the answering machine are
higher :-)
Shubee
> This is cute. You pretend your group is isomorphic to the Lorentz
> group
It's a trivial conclusion for Ben Rudiak-Gould and myself:
http://groups.google.com/group/sci.physics.relativity/msg/9f9a1aa34b49ff2e
> but you can't show that any of the invariants exist in your
> group, or that it actually reduces to the Galilean transformations for
> the low velocity limit.
I already computed the Newtonian limit:
http://groups.google.com/group/sci.physics/msg/4d0ebff048c2e21e
Shubee
Eric Gisse
> On Aug 4, 3:15 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > This is cute. You pretend your group is isomorphic to the Lorentz
> > group
>
> It's a trivial conclusion for Ben Rudiak-Gould and myself:
...and yourself? You have done nothing but assert assert assert. I
have not seen ONE proof from *you* as of yet.
Is this how you did math homework in school? Did they let you get away
with asserting that the result is true without proof?
>
> http://groups.google.com/group/sci.physics.relativity/msg/9f9a1aa34b4...
#1 wasn't proved by you - Jeckyl did it. All that can be stated is
that they form _a_ group. Not that they are related to the Lorentz
group or any other group for that matter.
#2 has not been proven.
http://groups.google.com/group/sci.physics.relativity/msg/311837abf78fd3d1?dmode=source
What he says is true but is not directly relevant because you have
_NOT_ shown that your group is a subgroup of anything.
#3 is what we have been saying all along.
#4 depends on a big set of assumptions which have not been shown to
hold for your transformation group. If you want to assert you have a
Lorentz invariant theory, a great start would be to SHOW THAT IT IS
LORENTZ INVARIANT. Why do you think I was asking you if the quantity
x^2 - c^2t^2 was preserved under one of your transformations?
>
> > but you can't show that any of the invariants exist in your
> > group, or that it actually reduces to the Galilean transformations for
> > the low velocity limit.
>
> I already computed the Newtonian limit:
>
> http://groups.google.com/group/sci.physics/msg/4d0ebff048c2e21e
IT IS WRONG.
Galilean transformations:
x' = x - vt
t' = t
>
> Shubee
Shubee
> On Aug 4, 2:53 pm, Shubee <e.Shu...@gmail.com> wrote:
>
> > On Aug 4, 3:15 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > > This is cute. You pretend your group is isomorphic to the Lorentz
> > > group
>
> > It's a trivial conclusion for Ben Rudiak-Gould and myself:
>
> that they form _a_ group. Not that they are related to the Lorentz
> group or any other group for that matter.
I thought it was too obvious for me to prove and too impossible for
you to understand.
> #2 has not been proven.
When Ben Rudiak-Gould addressed point 1,2,3,4, he said, "All of these
things should be obvious if you know a bit of group theory."
> #3 is what we have been saying all along.
I also agree with Ben Rudiak-Gould's point #3 in the sense that there
is a mathematical distinction between differing clocks synchronization
schemes.
http://groups.google.com/group/sci.physics/msg/ba2b82c6cf480db6
> #4 depends on a big set of assumptions which have not been shown to
> hold for your transformation group. If you want to assert you have a
> Lorentz invariant theory, a great start would be to SHOW THAT IT IS
> LORENTZ INVARIANT. Why do you think I was asking you if the quantity
> x^2 - c^2t^2 was preserved under one of your transformations?
Because you don't understand group theory?
> > > but you can't show that any of the invariants exist in your
> > > group, or that it actually reduces to the Galilean transformations for
> > > the low velocity limit.
>
> > I already computed the Newtonian limit:
>
> >http://groups.google.com/group/sci.physics/msg/4d0ebff048c2e21e
>
> IT IS WRONG.
>
> Galilean transformations:
>
> x' = x - vt
> t' = t
Just as the nonlinear transformation group in exercise 1 of
http://www.everythingimportant.org/relativity/generalized.htm is
"physically indistinguishable" from the Lorentz group, the
transformations
x' = x -v(t-b) +vg(x)
t' = t -g(x) +g[ x -v(t-b) + vg(x) ]
form a group isomorphic to and is "physically indistinguishable" from
the Galilean group.
http://groups.google.com/group/sci.physics.relativity/msg/ba2b82c6cf480db6
Shubee
http://www.everythingimportant.org/relativity/special.pdf