Re: Deriving space-time interval
Sue...
> Hello,
>
> How was the following formula derived? Is it related to the Lorenz
> factor?
>
> s^2 = r^2 - (ct)^2
>
> Thanks,
> Paul W
More than ya want to know:
http://farside.ph.utexas.edu/teaching/em/lectures/node113.html
Sue...
PD
> Hello,
>
> How was the following formula derived? Is it related to the Lorenz
> factor?
>
> s^2 = r^2 - (ct)^2
>
> Thanks,
> Paul W
It can be "derived" a number of ways from the postulate that the speed
of light is constant, independent of reference frame. You can find
Einstein's original 1905 paper, or you can read Taylor and Wheeler's
Spacetime Physics for two different approaches.
It's better to say, however, that it was *guessed* and the results of
that guess were compared to experiment, which went exceptionally well.
PD
Koobee Wublee
> How was the following formula derived? Is it related to the Lorenz
> factor?
>
> s^2 = r^2 - (ct)^2
Do you mean the following?
** ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2
If so, start with the Lorentz transformation.
** dt' = (dt - B dx / c) / sqrt(1 - B^2)
** dx' = (dx - B c dt) / sqrt(1 - B^2)
** dy' = dy
** dz' = dz
Where
** B = v / c
You will find the following is true through simple algebra.
c^2 dt'^2 - dx'^2 - dy'^2 - dz'^2 =
c^2 dt^2 - dx^2 - dy^2 - dz^2 == ds^2
Koobee Wublee
> On Mar 1, 12:21 pm, "Paul Whymark" <1337...@gmail.com> wrote:
> > s^2 = r^2 - (ct)^2
>
> It's better to say, however, that it was *guessed* and the results of
> that guess were compared to experiment, which went exceptionally well.
I am sure you stand by the Lorentz transformation. From the Lorentz
transformation, all you need is the following equation.
ds^2 = c^2 dt'^2 - dx'^2 = c^2 dt^2 - dx^2
The above equation would not be able to predict the results of any
experiments.
To *Guess* the spacetime as (ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2), it
takes a lot of fudging and self-hypnotism. To accept (ds^2 = c^2 dt^2
- dx^2 - dy^2 - dz^2), you must abandon the Lorentz transformation.
On the other hand, to accept the Lorentz transformation, you must
embrace (ds^2 = c^2 dt^2 - dx^2) and not (ds^2 = c^2 dt^2 - dx^2 -
dy^2 - dz^2). There is no other way regardless any *guessing* game
you have.
Sue...
> On Mar 1, 10:59 am, "PD" <TheDraperFam...@gmail.com> wrote:
>
> > On Mar 1, 12:21 pm, "Paul Whymark" <1337...@gmail.com> wrote:
> > > s^2 = r^2 - (ct)^2
>
> > It's better to say, however, that it was *guessed* and the results of
> > that guess were compared to experiment, which went exceptionally well.
>
<<I am sure you stand by the Lorentz transformation. From the Lorentz
transformation, all you need is the following equation.
ds^2 = c^2 dt'^2 - dx'^2 = c^2 dt^2 - dx^2 }
>>
<< Note: if you know about complex numbers you will
notice that the space part enters as if it were imaginary
R2 = (ct)2 + (ix)2 + (iy)2 + (iz)2 = (ct)2 + (ir)2
>>
http://www.aoc.nrao.edu/~smyers/courses/astro12/speedoflight.html
Dirk Van de moortel
"Koobee Wublee"
aka Australopithecus Afarensis
aka Scholarly Fungi
aka Time Traveler
aka Lordly Amoeba
aka Ibn Battuta
aka Marco Polo
<koobee...@gmail.com> wrote in message news:1172906654.0...@64g2000cwx.googlegroups.com...
> On Mar 1, 10:59 am, "PD" <TheDraperFam...@gmail.com> wrote:
>> On Mar 1, 12:21 pm, "Paul Whymark" <1337...@gmail.com> wrote:
>
>> > s^2 = r^2 - (ct)^2
>>
>> It's better to say, however, that it was *guessed* and the results of
>> that guess were compared to experiment, which went exceptionally well.
>
> I am sure you stand by the Lorentz transformation.
We are sure you don't understand that transformation, troll:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/LorentzTale.html
You don't even understand the meanings of the variables.
Dirk Vdm
Eric Gisse
Yippie, another post in which KW pontificates about a subject he has
no knowledge in.
First off, the Lorentz transformation is easily derived from the full
line element.
As far as deriving the metric is concerned, you can start off with
SR's postulates and derive the distance relation, or you can simply
postulate the metric and be done with it. Group theory is also another
approach. Demand there be a finite maximal speed and some other minor
things like invertibility, and you get SO(3,1) - the Lorentz group.
Why do you even post to this newsgroup? You don't even have a
rudimentary understanding of either of the relativity theories, or
even multivariable calculus.
PD
> On Mar 1, 10:59 am, "PD" <TheDraperFam...@gmail.com> wrote:
>
> > On Mar 1, 12:21 pm, "Paul Whymark" <1337...@gmail.com> wrote:
> > > s^2 = r^2 - (ct)^2
>
> > It's better to say, however, that it was *guessed* and the results of
> > that guess were compared to experiment, which went exceptionally well.
>
> I am sure you stand by the Lorentz transformation. From the Lorentz
> transformation, all you need is the following equation.
>
> ds^2 = c^2 dt'^2 - dx'^2 = c^2 dt^2 - dx^2
>
> The above equation would not be able to predict the results of any
> experiments.
Sure it is. You have to play with it a little, but there is quite a
bit that can be gleaned from the above. As an example, Einstein
derived from this the rate of slow down of a clock in a circular
motion around an inertial clock, right there in his 1905 paper.
There's a *ton* that can be predicted from this simple equation.
That's the beauty of it. If you'd like some examples, I suggest you
work some problems, many of which have been compared to real
measurements in real experiments, at the end of chapters in Spacetime
Physics, by Taylor and Wheeler.
I don't know where you got the idea that the above equation would not
be able to predict the results of any experiments.
>
> To *Guess* the spacetime as (ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2), it
> takes a lot of fudging and self-hypnotism. To accept (ds^2 = c^2 dt^2
> - dx^2 - dy^2 - dz^2), you must abandon the Lorentz transformation.
> On the other hand, to accept the Lorentz transformation, you must
> embrace (ds^2 = c^2 dt^2 - dx^2) and not (ds^2 = c^2 dt^2 - dx^2 -
> dy^2 - dz^2). There is no other way regardless any *guessing* game
> you have.
Nonsense. Apparently you don't know how to do the Lorentz
transformation for a relative velocity that isn't aligned along one of
the axes. Lorentz did. I don't see why you don't.
PD
Sue...
> > The above equation would not be able to predict the results of any
> > experiments.
>
> Sure it is. You have to play with it a little, but there is quite a
> bit that can be gleaned from the above. As an example, Einstein
> derived from this the rate of slow down of a clock in a circular
> motion around an inertial clock, right there in his 1905 paper.
> There's a *ton* that can be predicted from this simple equation.
> That's the beauty of it. If you'd like some examples, I suggest you
> work some problems, many of which have been compared to real
> measurements in real experiments, at the end of chapters in Spacetime
> Physics, by Taylor and Wheeler.
In the 1920 paper you'll find the imaginary operator is described.
http://www.bartleby.com/173/17.html
This would be about 2 years after Paul Langevin pointed out
some absrudities.
The imaginary operators shown here:
http://www.nrao.edu/~smyers/courses/astro12/speedoflight.html
are from Taylor and Wheeler.
Sue...
PD
> On Mar 3, 11:13 am, "PD" <TheDraperFam...@gmail.com> wrote:
>
> > > The above equation would not be able to predict the results of any
> > > experiments.
>
> > Sure it is. You have to play with it a little, but there is quite a
> > bit that can be gleaned from the above. As an example, Einstein
> > derived from this the rate of slow down of a clock in a circular
> > motion around an inertial clock, right there in his 1905 paper.
> > There's a *ton* that can be predicted from this simple equation.
> > That's the beauty of it. If you'd like some examples, I suggest you
> > work some problems, many of which have been compared to real
> > measurements in real experiments, at the end of chapters in Spacetime
> > Physics, by Taylor and Wheeler.
>
> In the 1920 paper you'll find the imaginary operator is described.http://www.bartleby.com/173/17.html
A complex number is not usually referred to as an "imaginary
operator."
>
> This would be about 2 years after Paul Langevin pointed out
> some absrudities.
No absurdities. What absurdities?
Sue...
> On Mar 3, 11:21 am, "Sue..." <suzysewns...@yahoo.com.au> wrote:
>
>
>
>
>
> > On Mar 3, 11:13 am, "PD" <TheDraperFam...@gmail.com> wrote:
>
> > > > The above equation would not be able to predict the results of any
> > > > experiments.
>
> > > Sure it is. You have to play with it a little, but there is quite a
> > > bit that can be gleaned from the above. As an example, Einstein
> > > derived from this the rate of slow down of a clock in a circular
> > > motion around an inertial clock, right there in his 1905 paper.
> > > There's a *ton* that can be predicted from this simple equation.
> > > That's the beauty of it. If you'd like some examples, I suggest you
> > > work some problems, many of which have been compared to real
> > > measurements in real experiments, at the end of chapters in Spacetime
> > > Physics, by Taylor and Wheeler.
>
> > In the 1920 paper you'll find the imaginary operator is described.http://www.bartleby.com/173/17.html
>
> A complex number is not usually referred to as an "imaginary
> operator."
>
>
>
> > This would be about 2 years after Paul Langevin pointed out
> > some absrudities.
>
> No absurdities. What absurdities?
The absurdity you keep repeating. A temporal clock makes
a trip and is wrong when it returns home.
That won't happen if you put the imaginares in
and transform appropriately.
http://arxiv.org/abs/physics/0204034
Sue...
>
>
>
>
>
> > The imaginary operators shown here:http://www.nrao.edu/~smyers/courses/astro12/speedoflight.html
> > are from Taylor and Wheeler.
>
> > Sue...- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
PD
> On Mar 3, 1:56 pm, "PD" <TheDraperFam...@gmail.com> wrote:
>
>
>
> > On Mar 3, 11:21 am, "Sue..." <suzysewns...@yahoo.com.au> wrote:
>
> > > On Mar 3, 11:13 am, "PD" <TheDraperFam...@gmail.com> wrote:
>
> > > > > The above equation would not be able to predict the results of any
> > > > > experiments.
>
> > > > Sure it is. You have to play with it a little, but there is quite a
> > > > bit that can be gleaned from the above. As an example, Einstein
> > > > derived from this the rate of slow down of a clock in a circular
> > > > motion around an inertial clock, right there in his 1905 paper.
> > > > There's a *ton* that can be predicted from this simple equation.
> > > > That's the beauty of it. If you'd like some examples, I suggest you
> > > > work some problems, many of which have been compared to real
> > > > measurements in real experiments, at the end of chapters in Spacetime
> > > > Physics, by Taylor and Wheeler.
>
> > > In the 1920 paper you'll find the imaginary operator is described.http://www.bartleby.com/173/17.html
>
> > A complex number is not usually referred to as an "imaginary
> > operator."
>
> > > This would be about 2 years after Paul Langevin pointed out
> > > some absrudities.
>
> > No absurdities. What absurdities?
>
> The absurdity you keep repeating. A temporal clock makes
> a trip and is wrong when it returns home.
Why is that absurd? It has been demonstrated to really happen. Why is
reality absurd?
At variance with your expectations, maybe, but not absurd.
>
> That won't happen if you put the imaginares in
> and transform appropriately.
One shouldn't have to transform reality to get it to conform to
expectations. That would be, well, wrong.
Sue...
An experiment is not acceptable if if contains mathematical
aburdities because it is impossible to interpret.
>
> At variance with your expectations, maybe, but not absurd.
I can paint a barn with a litre of paint and make
an induction motor produce more power than
it uses if I want to be loose with imaginary numbers.
Both are absurd just as your evaluation of spacetime
intervals. It is one of several mistake described here.
Abstract
Einstein addressed the twin paradox in special relativity
in a relatively unknown, unusual and rarely cited paper
written in 1918, in the form of a dialogue between a
critic and a relativist. Contrary to most textbook versions
of the resolution, Einstein admitted that the special
relativistic time dilation was symmetric for the twins,
and he had to invoke, asymmetrically, the general relativistic
gravitational time dilation during the brief periods
of acceleration to justify the asymmetrical aging.
Notably, Einstein did not use any argument related to
simultaneity or Doppler shift in his analysis. I discuss
Einstein's resolution and several conceptual issues
that arise. It is concluded that Einstein's resolution using
gravitational time dilation suffers from logical and
physical flaws, and gives incorrect answers in a general
setting. The counter examples imply the need to reconsider
many issues related to the comparison of transported
clocks. The failure of the accepted views and
resolutions is traced to the fact that the special relativity
principle formulated originally for physics in empty
space is not valid in the matter-filled universe.
C. S. Unnikrishnan
Gravitation Group,
Tata Institute of Fundamental Research,
Homi Bhabha Road, Mumbai 400 005, India
http://www.iisc.ernet.in/currsci/dec252005/2009.pdf
>
>
>
> > That won't happen if you put the imaginares in
> > and transform appropriately.
>
> One shouldn't have to transform reality to get it to conform to
> expectations. That would be, well, wrong.
We expect to use Lorenz gauge in the emitter nearfield,
Coulomb gauge in the farfield
Lorenz gauge in the absorber nearfield.
http://arxiv.org/abs/physics/0204034
Is that how you were doing it?
Sue...
http://www.aoc.nrao.edu/~smyers/courses/astro12/speedoflight.html
PD
What mathematical absurdities in the experiment? The experiment is a
*measurement* -- very straightforward.
PD
>
>
>
> > At variance with your expectations, maybe, but not absurd.
>
> I can paint a barn with a litre of paint and make
> an induction motor produce more power than
> it uses if I want to be loose with imaginary numbers.
>
> Both are absurd just as your evaluation of spacetime
> intervals. It is one of several mistake described here.
>
>
[irrelevant chaff snipped]
Sue...
The experiment doesn't matter. If I show you a bellhop
that makes dollars vanish and some calculations that predict
he will make dollars vanish, is that science?
http://mathworld.wolfram.com/MissingDollarParadox.html
http://en.wikipedia.org/wiki/Scientific_method
Sue...
>
> PD
>
>
>
> > > At variance with your expectations, maybe, but not absurd.
>
> > I can paint a barn with a litre of paint and make
> > an induction motor produce more power than
> > it uses if I want to be loose with imaginary numbers.
>
> > Both are absurd just as your evaluation of spacetime
> > intervals. It is one of several mistake described here.
>
> [irrelevant chaff snipped]- Hide quoted text -
PD
Ah, see, this is where you diverge from science, Dennis. In science,
the experiment -- the *real* experiment -- does indeed matter. It's
what settles things.
All this hooha about paradoxes and puzzling calculations are games to
flex your understanding, nothing more. They tell you very little about
reality, though they're very good at telling you that you must not
understand something very clearly.
But experiment! That tells you what is real.
>
> http://mathworld.wolfram.com/MissingDollarParadox.htmlhttp://en.wikipedia.org/wiki/Scientific_method
Koobee Wublee
> On Mar 3, 1:24 am, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
> > I am sure you stand by the Lorentz transformation. From the Lorentz
> > transformation, all you need is the following equation.
>
> > ds^2 = c^2 dt'^2 - dx'^2 = c^2 dt^2 - dx^2
>
> > The above equation would not be able to predict the results of any
> > experiments.
>
> Sure it is. You have to play with it a little, but there is quite a
> bit that can be gleaned from the above.
The mathematics is so simple to the equation above that it leaves no
room for clever matheMagical tricks.
> As an example, Einstein
> derived from this the rate of slow down of a clock in a circular
> motion around an inertial clock, right there in his 1905 paper.
No, Einstein fudged the Lorentz transform in his 1905 paper. The
derivation was totally gibberish.
The time dilation was known as one of the properties of the Lorentz
transform discovered by Larmor. This strange phenomenon of time
dilation was addressed by Larmor, discussed by Lorentz, and theorized
by Poincare. <shrug>
> There's a *ton* that can be predicted from this simple equation.
The equation does not even allow you to derive a Lagrangian that would
yield unique and well defined geodesic equations for the y and z axes.
> That's the beauty of it. If you'd like some examples, I suggest you
> work some problems, many of which have been compared to real
> measurements in real experiments, at the end of chapters in Spacetime
> Physics, by Taylor and Wheeler.
I don't have that book.
> I don't know where you got the idea that the above equation would not
> be able to predict the results of any experiments.
It does not couple into the transverse dimensions at all.
> > To *Guess* the spacetime as (ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2), it
> > takes a lot of fudging and self-hypnotism. To accept (ds^2 = c^2 dt^2
> > - dx^2 - dy^2 - dz^2), you must abandon the Lorentz transformation.
> > On the other hand, to accept the Lorentz transformation, you must
> > embrace (ds^2 = c^2 dt^2 - dx^2) and not (ds^2 = c^2 dt^2 - dx^2 -
> > dy^2 - dz^2). There is no other way regardless any *guessing* game
> > you have.
>
> Nonsense. Apparently you don't know how to do the Lorentz
> transformation for a relative velocity that isn't aligned along one of
> the axes.
Oh, yes, I do.
> Lorentz did.
It is irrelevant.
> I don't see why you don't.
You are so wrong. If the relative speed is not aligned with the x
axis, I merely just have to find another axis that does align. From
that I can write down the same equation as the following in which we
are back to square one.
ds^2 = c^2 dt^2 - du^2
Where Axis of u is aligned with the relative speed. <shrug>
Koobee Wublee
> On Mar 3, 6:20 pm, "Sue..." <suzysewns...@yahoo.com.au> wrote:
> Ah, see, this is where you diverge from science, Dennis. In science,
> the experiment -- the *real* experiment -- does indeed matter. It's
> what settles things.
Any experiments must be interpreted properly. Is that too much to
ask?
> All this hooha about paradoxes and puzzling calculations are games to
> flex your understanding, nothing more.
The paradox exists in the hypothesis and not any experiments.
> They tell you very little about
> reality, though they're very good at telling you that you must not
> understand something very clearly.
> But experiment! That tells you what is real.
If any experiment time after time does not show the paradox, the
hypothesis must be wrong. <shrug>
Sue...
I think you need a new maths teacher. One that can
calculate the betting odds that two people don't know
their own identity.
Dork usually calls me Dennis when he realises he is
trying to argue an unbalanced equation so I'll assume
you are following the same practice.
>
> All this hooha about paradoxes and puzzling calculations are games to
> flex your understanding, nothing more. They tell you very little about
> reality, though they're very good at telling you that you must not
> understand something very clearly.
>
> But experiment! That tells you what is real.
There is no relationship between absurd math
and any experiment. No experiment can
confirm absurd maths. You are hawking
(pun intended) absurd maths.
<< Note: if you know about complex numbers you will
notice that the space part enters as if it were imaginary
R2 = (ct)2 + (ix)2 + (iy)2 + (iz)2 = (ct)2 + (ir)2
where i^2 = -1 as usual. This turns out to be the
essence of the fabric (or metric) of spacetime
geometry - that space enters in with the
imaginary factor i relative to time >
http://www.nrao.edu/~smyers/courses/astro12/speedoflight.html
Imaginary numbers do not mean a free ticket
"through the looking glass". Appropriate
rigour still applies.
Free time and free energy are the same thing.
http://en.wikipedia.org/wiki/Noether's_theorem
The cranks that hawk either one exploit the same error.
Find some integrity:
http://wwwcdf.pd.infn.it/~loreti/science.html
http://www.quackwatch.org/01QuackeryRelatedTopics/pseudo.html
Sue...
PD
> On Mar 3, 6:55 pm, "PD" <TheDraperFam...@gmail.com> wrote:
>
> > On Mar 3, 6:20 pm, "Sue..." <suzysewns...@yahoo.com.au> wrote:
> > Ah, see, this is where you diverge from science, Dennis. In science,
> > the experiment -- the *real* experiment -- does indeed matter. It's
> > what settles things.
>
> Any experiments must be interpreted properly. Is that too much to
> ask?
Yes. If I *predict*, based on a model, that the *measured* velocity of
a proton with a kinetic energy 69 GeV will be 0.9999c, then it is
fairly straightforward to *measure* that velocity the same way I
measure the speed of cars on a race track. I see no need to
"interpret" the measurement to be something other than 0.9999c.
In an experiment, the *measurement* should be, and usually is, pretty
straightforward. There is no need to "interpret" it to be something
other than what it is.
>
> > All this hooha about paradoxes and puzzling calculations are games to
> > flex your understanding, nothing more.
>
> The paradox exists in the hypothesis and not any experiments.
What paradox? There is no paradox.
>
> > They tell you very little about
> > reality, though they're very good at telling you that you must not
> > understand something very clearly.
> > But experiment! That tells you what is real.
>
> If any experiment time after time does not show the paradox, the
> hypothesis must be wrong. <shrug>
There is no paradox.
PD
PD
I don't think so. Your history precedes you, independent of your self-
labeling.
>
>
> > All this hooha about paradoxes and puzzling calculations are games to
> > flex your understanding, nothing more. They tell you very little about
> > reality, though they're very good at telling you that you must not
> > understand something very clearly.
>
> > But experiment! That tells you what is real.
>
> There is no relationship between absurd math
> and any experiment. No experiment can
> confirm absurd maths. You are hawking
> (pun intended) absurd maths.
A couple of comments -
- First, you haven't pointed out what math is absurd.
- Secondly, it's bad scientific practice to pre-dismiss an experiment
if it condemns a theory you have difficulty with. The good scientist
uses
experiment to be the arbiter on the scientist's judgment whether a
theory
is sound or not.
- An experiment is an observation of *reality*. If it directly
confirms
the predictions of a theory you find difficult to swallow, then nature
has advised you that you'd better work harder to find the theory
palatable.
>
> << Note: if you know about complex numbers you will
> notice that the space part enters as if it were imaginary
>
> R2 = (ct)2 + (ix)2 + (iy)2 + (iz)2 = (ct)2 + (ir)2
> where i^2 = -1 as usual. This turns out to be the
> essence of the fabric (or metric) of spacetime
> geometry - that space enters in with the
> imaginary factor i relative to time >http://www.nrao.edu/~smyers/courses/astro12/speedoflight.html
>
> Imaginary numbers do not mean a free ticket
> "through the looking glass". Appropriate
> rigour still applies.
Complex number mathematics is *completely* rigorous.
>
> Free time and free energy are the same thing.http://en.wikipedia.org/wiki/Noether's_theorem
PD
> On Mar 3, 8:13 am, "PD" <TheDraperFam...@gmail.com> wrote:
>
> > On Mar 3, 1:24 am, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
> > > I am sure you stand by the Lorentz transformation. From the Lorentz
> > > transformation, all you need is the following equation.
>
> > > ds^2 = c^2 dt'^2 - dx'^2 = c^2 dt^2 - dx^2
>
> > > The above equation would not be able to predict the results of any
> > > experiments.
>
> > Sure it is. You have to play with it a little, but there is quite a
> > bit that can be gleaned from the above.
>
> The mathematics is so simple to the equation above that it leaves no
> room for clever matheMagical tricks.
That's simply incorrect. F=ma is quite simple, too, but you can pull
an enormous amount of information from it: Liouville's theorem,
Bernoulli's equation, Kepler's laws. The fact that you don't see how
it is possible that something so simple can have far-reaching
implications is not anyone's problem but yours.
>
> > As an example, Einstein
> > derived from this the rate of slow down of a clock in a circular
> > motion around an inertial clock, right there in his 1905 paper.
>
> No, Einstein fudged the Lorentz transform in his 1905 paper. The
> derivation was totally gibberish.
I find it rather easy to follow. I gather that you don't understand
it. That's not my problem. There are other lovely derivations. I
mentioned below the one in Taylor & Wheeler. You whine that you don't
have that. That's not my problem.
>
> The time dilation was known as one of the properties of the Lorentz
> transform discovered by Larmor. This strange phenomenon of time
> dilation was addressed by Larmor, discussed by Lorentz, and theorized
> by Poincare. <shrug>
>
> > There's a *ton* that can be predicted from this simple equation.
>
> The equation does not even allow you to derive a Lagrangian that would
> yield unique and well defined geodesic equations for the y and z axes.
Actually, yes, it does. I'm surprised you find that difficult.
>
> > That's the beauty of it. If you'd like some examples, I suggest you
> > work some problems, many of which have been compared to real
> > measurements in real experiments, at the end of chapters in Spacetime
> > Physics, by Taylor and Wheeler.
>
> I don't have that book.
Well, then, I highly recommend you spend $5 to get a used copy from
Amazon, or if you like, go to a university library where you will
likely find a copy. Your lack of resources is not my problem and does
not represent a need for me to accommodate.
>
> > I don't know where you got the idea that the above equation would not
> > be able to predict the results of any experiments.
>
> It does not couple into the transverse dimensions at all.
>
> > > To *Guess* the spacetime as (ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2), it
> > > takes a lot of fudging and self-hypnotism. To accept (ds^2 = c^2 dt^2
> > > - dx^2 - dy^2 - dz^2), you must abandon the Lorentz transformation.
> > > On the other hand, to accept the Lorentz transformation, you must
> > > embrace (ds^2 = c^2 dt^2 - dx^2) and not (ds^2 = c^2 dt^2 - dx^2 -
> > > dy^2 - dz^2). There is no other way regardless any *guessing* game
> > > you have.
>
> > Nonsense. Apparently you don't know how to do the Lorentz
> > transformation for a relative velocity that isn't aligned along one of
> > the axes.
>
> Oh, yes, I do.
Not obvious that you do. All you did below was to rotate so that the
relative velocity was along your new axis.
>
> > Lorentz did.
>
> It is irrelevant.
>
> > I don't see why you don't.
>
> You are so wrong. If the relative speed is not aligned with the x
> axis, I merely just have to find another axis that does align. From
> that I can write down the same equation as the following in which we
> are back to square one.
>
> ds^2 = c^2 dt^2 - du^2
>
> Where Axis of u is aligned with the relative speed. <shrug>
And this redefinition of axes is completely unnecessary. Do you know
how to write the Lorentz transformation in terms of x, y, z
coordinates for a boost that is NOT along one of those axes? Please
show that, or ask for help.
PD
Eric Gisse
No, NO, NO!
Why do people keep doing that?! It is ds^2 = (cdt)^2 - (dx_i)^2, there
are NO IMAGINARY NUMBERS.
If you allow the coordinate values to be complex, you have to allow
for the possibility of them being _real_. A Wick rotation has to work
both ways, in order to work at all. That means you get complex
quantities like time, position, velocity....none of which have any
meaning.
>
> Imaginary numbers do not mean a free ticket
> "through the looking glass". Appropriate
> rigour still applies.
Of course it does. Complex analysis has all the rigor of real
analysis. Just because you are unfamiliar with it, does not mean
others are so unfortunate.
[snip usual irrelevant links]
Sue...
Because our friend Dr. PD Who isn't performing a Wick
rotation or working with energy quanta. He wants to swap
temporal and spatial values and perpetuate his twins
myth.
>
> If you allow the coordinate values to be complex, you have to allow
> for the possibility of them being _real_. A Wick rotation has to work
> both ways, in order to work at all. That means you get complex
> quantities like time, position, velocity....none of which have any
> meaning.
>
>
>
> > Imaginary numbers do not mean a free ticket
> > "through the looking glass". Appropriate
> > rigour still applies.
>
> Of course it does. Complex analysis has all the rigor of real
> analysis. Just because you are unfamiliar with it, does not mean
> others are so unfortunate.
You have to get Dr. P.D. Who to teach you the semantic
operator. It is discussed here:
http://www.quackwatch.org/01QuackeryRelatedTopics/pseudo.html
http://wwwcdf.pd.infn.it/~loreti/science.html
Then you won't need any Wick rotations.
Sue...
Eric Gisse
> On Mar 3, 8:13 am, "PD" <TheDraperFam...@gmail.com> wrote:
>
> > On Mar 3, 1:24 am, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
> > > I am sure you stand by the Lorentz transformation. From the Lorentz
> > > transformation, all you need is the following equation.
>
> > > ds^2 = c^2 dt'^2 - dx'^2 = c^2 dt^2 - dx^2
>
> > > The above equation would not be able to predict the results of any
> > > experiments.
>
> > Sure it is. You have to play with it a little, but there is quite a
> > bit that can be gleaned from the above.
>
> The mathematics is so simple to the equation above that it leaves no
> room for clever matheMagical tricks.
Have you...you know...actually tried using SR in that form? Have you
ever read a textbook that shows using it in that form?
Of course not. Doing physics is far below your standing.
>
> > As an example, Einstein
> > derived from this the rate of slow down of a clock in a circular
> > motion around an inertial clock, right there in his 1905 paper.
>
> No, Einstein fudged the Lorentz transform in his 1905 paper. The
> derivation was totally gibberish.
You keep repeating that assertion, yet you always have trouble
justifying it when called on it.
>
> The time dilation was known as one of the properties of the Lorentz
> transform discovered by Larmor. This strange phenomenon of time
> dilation was addressed by Larmor, discussed by Lorentz, and theorized
> by Poincare. <shrug>
Finally, something pinned on Larmor that doesn't deal with
electromagnetism. Regardless of how true it is.
That guy was a real one-track physicist. :D
>
> > There's a *ton* that can be predicted from this simple equation.
>
> The equation does not even allow you to derive a Lagrangian that would
> yield unique and well defined geodesic equations for the y and z axes.
>
*sigh*
Poor little K-W, once again talking about a subject he has no
education in. Deriving the Lagrangian for a relativistic free particle
is eaaasssyy. It is a textbook problem, one apparently you never did.
Think carefully back to whenever you were taught [big assumption here]
about the Lagrangian . L = K - V, right? Since this is SR by itself
there is no potential energy so L = K = kinetic energy.
Think carefully again. Remember that p_u = m U_u = m dx_i / dtau ?
Remember that the time component of the four momentum is energy?
So we have p_0 = m U_0 = -mdx_0 / dtau = -gamma m = -m sqrt(1-v^2). I
assume you know how to handle dt/dtau from the metric...
Thus L = -m*sqrt(1-v^2). Or if you are real prissy, L = -mc^2 *
sqrt( 1 - (v/c)^2 )
That isn't the only way to do it. You can also use the principle of
least action to derive it, and the results match.
Generalized coordinates are cool, aren't they?
[snip remaining]
Koobee Wublee
> On Mar 4, 12:01 am, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
> > Any experiments must be interpreted properly. Is that too much to
> > ask?
>
> Yes.
Then, any physical experiments are worthless.
> If I *predict*, based on a model, that the *measured* velocity of
> a proton with a kinetic energy 69 GeV will be 0.9999c, then it is
> fairly straightforward to *measure* that velocity the same way I
> measure the speed of cars on a race track. I see no need to
> "interpret" the measurement to be something other than 0.9999c.
What does this have anything to do with what we are discussing?
> In an experiment, the *measurement* should be, and usually is, pretty
> straightforward. There is no need to "interpret" it to be something
> other than what it is.
The measurement is but to decipher the measurement is not as trivial
as you think. It is subject to influence and interpretations.
Koobee Wublee
> On Mar 3, 11:56 pm, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
> > ds^2 = c^2 dt'^2 - dx'^2 = c^2 dt^2 - dx^2
>
> > The mathematics is so simple to the equation above that it leaves no
> > room for clever matheMagical tricks.
>
> That's simply incorrect. F=ma is quite simple, too, but you can pull
> an enormous amount of information from it: Liouville's theorem,
> Bernoulli's equation, Kepler's laws. The fact that you don't see how
> it is possible that something so simple can have far-reaching
> implications is not anyone's problem but yours.
How do you justify (ds^2 = c^2 dt^2 - dx^2) to (ds^2 = c^2 dt^2 - dx^2
- dy^2 - dz^2)?
> > No, Einstein fudged the Lorentz transform in his 1905 paper. The
> > derivation was totally gibberish.
>
> I find it rather easy to follow.
Incompetents like Gisse and moortel may not, but I know you are full
of sh*t.
> I gather that you don't understand
> it.
Oh, I understand it fully. It is gibberish.
> That's not my problem.
Not mine, either.
> There are other lovely derivations.
Like Einstein's book on relativity where he pulled out the Lorentz
transform from two equations equating zero with zero.
> > The equation does not even allow you to derive a Lagrangian that would
> > yield unique and well defined geodesic equations for the y and z axes.
>
> Actually, yes, it does. I'm surprised you find that difficult.
Yes, I do. Show me how you derive the Lagrangian for the geodesic
motion from the equation below.
ds^2 = c^2 dt^2 - dx^2
> Nonsense. Apparently you don't know how to do the Lorentz
> transformation for a relative velocity that isn't aligned along one of
> the axes.
>
> > You are so wrong. If the relative speed is not aligned with the x
> > axis, I merely just have to find another axis that does align. From
> > that I can write down the same equation as the following in which we
> > are back to square one.
>
> > ds^2 = c^2 dt^2 - du^2
>
> > Where Axis of u is aligned with the relative speed. <shrug>
>
> And this redefinition of axes is completely unnecessary. Do you know
> how to write the Lorentz transformation in terms of x, y, z
> coordinates for a boost that is NOT along one of those axes? Please
> show that, or ask for help.
dS' = dS + B (B * dS / (1 + sqrt(1 - B^2)) + c dt) / sqrt(1 - B^2)
dt' = (dt + B * dS / c) / sqrt(1 - B^2)
S', S, B are all vectors.
Need me to explain what the symbols mean? Couldn't you figure it out
if you indeed understand the subject well as a professor of physics?
gdew...@gmail.com
Eric Gisse
What would your ISP say if I forwarded every one of these to them as a
complaint?
gdew...@gmail.com
They will think "gheee, this Eric Gisse really is an idiot.
http://groups.google.com/group/sci.physics/msg/9cd4a7c4e631a9fb
http://groups.google.com/group/sci.physics/msg/1db0357180750547
http://groups.google.com/group/sci.physics/msg/12d10f1bda5326cf
http://groups.google.com/group/sci.physics/msg/918d2e4193ab7580
http://groups.google.com/group/sci.physics/msg/4c7115c0ccf4ad36
On Feb 20, 9:32 pm, "Eric Gisse" <jowr...@gmail.com> wrote:
> Do I need to harass you more?
http://groups.google.com/group/sci.physics/msg/43f0d9c4922f0b6a
http://groups.google.com/group/sci.physics/msg/39537145ac68837a
http://groups.google.com/group/sci.physics/msg/92a9e764b15aa85e
http://groups.google.com/group/sci.physics/msg/d6e2b5b28aa879ca
On Feb 28, 6:17 am, "Eric Gisse" <jowr...@gmail.com> wrote:
> Since there is nothing to be gained, you can finally stop posting on a
> daily basis about it.
>
> [snip rot]
http://groups.google.com/group/sci.physics/msg/fceecaa8e211cb79
http://groups.google.com/group/sci.physics/msg/4584dbdcda4d8326
etc etc etc
Koobee Wublee
Likewise, what would your ISP say if I forward all of your posts and
moortels' as complaints?
Eric Gisse
Nothing?
PD
> On Mar 4, 11:22 am, "PD" <TheDraperFam...@gmail.com> wrote:
>
> > On Mar 4, 12:01 am, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
> > > Any experiments must be interpreted properly. Is that too much to
> > > ask?
>
> > Yes.
>
> Then, any physical experiments are worthless.
In your opinion, perhaps.
>
> > If I *predict*, based on a model, that the *measured* velocity of
> > a proton with a kinetic energy 69 GeV will be 0.9999c, then it is
> > fairly straightforward to *measure* that velocity the same way I
> > measure the speed of cars on a race track. I see no need to
> > "interpret" the measurement to be something other than 0.9999c.
>
> What does this have anything to do with what we are discussing?
It's an illustration of a straightforward *measurement* in an
experiment, and how free from "interpretation" it is.
>
> > In an experiment, the *measurement* should be, and usually is, pretty
> > straightforward. There is no need to "interpret" it to be something
> > other than what it is.
>
> The measurement is but to decipher the measurement is not as trivial
> as you think. It is subject to influence and interpretations.
No, not really. One estimates the size of potential sources of
systematic and statistical uncertainty, and that's about it. The
measurement is direct and not particularly dependent on
interpretation. That's what makes it useful as *experiment*. It is
direct consultation with nature. It is NOT nature muddled with
interpretation.
>
> > > The paradox exists in the hypothesis and not any experiments.
>
> > What paradox? There is no paradox.
>
> The paradox exists in the hypothesis and not any experiments.
What paradox? There is no paradox.
>
> > > > They tell you very little about
> > > > reality, though they're very good at telling you that you must not
> > > > understand something very clearly.
> > > > But experiment! That tells you what is real.
>
> > > If any experiment time after time does not show the paradox, the
> > > hypothesis must be wrong. <shrug>
>
> > There is no paradox.
>
> The paradox exists in the hypothesis and not any experiments.
There is no paradox.
PD
PD
> On Mar 4, 11:37 am, "PD" <TheDraperFam...@gmail.com> wrote:
>
> > On Mar 3, 11:56 pm, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
> > > ds^2 = c^2 dt'^2 - dx'^2 = c^2 dt^2 - dx^2
>
> > > The mathematics is so simple to the equation above that it leaves no
> > > room for clever matheMagical tricks.
>
> > That's simply incorrect. F=ma is quite simple, too, but you can pull
> > an enormous amount of information from it: Liouville's theorem,
> > Bernoulli's equation, Kepler's laws. The fact that you don't see how
> > it is possible that something so simple can have far-reaching
> > implications is not anyone's problem but yours.
>
> How do you justify (ds^2 = c^2 dt^2 - dx^2) to (ds^2 = c^2 dt^2 - dx^2
> - dy^2 - dz^2)?
Um... the former applies to a 2D spacetime, which doesn't seem to
apply to our universe. The latter applies to a 4D spacetime, which
does (experimentally) seem to apply to our universe. Why?
And what does this have to do with being unable to draw interesting
conclusions from it?
>
> > > No, Einstein fudged the Lorentz transform in his 1905 paper. The
> > > derivation was totally gibberish.
>
> > I find it rather easy to follow.
>
> Incompetents like Gisse and moortel may not, but I know you are full
> of sh*t.
Really? On what basis? That you can't follow Einstein's derivation,
and I can, so I must be full of sh*t?
>
> > I gather that you don't understand
> > it.
>
> Oh, I understand it fully. It is gibberish.
>
> > That's not my problem.
>
> Not mine, either.
>
> > There are other lovely derivations.
>
> Like Einstein's book on relativity where he pulled out the Lorentz
> transform from two equations equating zero with zero.
Actually, I was thinking of that Taylor and Wheeler book. Didn't I
mention that?
>
> > > The equation does not even allow you to derive a Lagrangian that would
> > > yield unique and well defined geodesic equations for the y and z axes.
>
> > Actually, yes, it does. I'm surprised you find that difficult.
>
> Yes, I do. Show me how you derive the Lagrangian for the geodesic
> motion from the equation below.
>
> ds^2 = c^2 dt^2 - dx^2
>
OK, let's start from the basics. Is there a T term? Is there a V term?
I'll help you think it through.
>
> > Nonsense. Apparently you don't know how to do the Lorentz
> > transformation for a relative velocity that isn't aligned along one of
> > the axes.
>
> > > You are so wrong. If the relative speed is not aligned with the x
> > > axis, I merely just have to find another axis that does align. From
> > > that I can write down the same equation as the following in which we
> > > are back to square one.
>
> > > ds^2 = c^2 dt^2 - du^2
>
> > > Where Axis of u is aligned with the relative speed. <shrug>
>
> > And this redefinition of axes is completely unnecessary. Do you know
> > how to write the Lorentz transformation in terms of x, y, z
> > coordinates for a boost that is NOT along one of those axes? Please
> > show that, or ask for help.
>
> dS' = dS + B (B * dS / (1 + sqrt(1 - B^2)) + c dt) / sqrt(1 - B^2)
> dt' = (dt + B * dS / c) / sqrt(1 - B^2)
>
> S', S, B are all vectors.
Then you're missing some vector operators. Is that what the * is
supposed to represent? And why do you need dS and dt? We're
transforming coordinates, not infinitesimals.
>
> Need me to explain what the symbols mean? Couldn't you figure it out
> if you indeed understand the subject well as a professor of physics?
It's not a matter of figuring it out. It's a matter of writing it down
correctly.
Denote R and R' as vectors, and V as a vector. Take g to be 1/sqrt[1-
B*B], where B is V/c and * is the dot product.
Then
t' = g(t - R*B/c)
R' = R + [(g-1)(R*B)/(cB*B) - gt)V
There ya go.
PD
John Park
> On Mar 4, 10:46 pm, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
[...]
>>
>> dS' = dS + B (B * dS / (1 + sqrt(1 - B^2)) + c dt) / sqrt(1 - B^2)
>> dt' = (dt + B * dS / c) / sqrt(1 - B^2)
>>
>> S', S, B are all vectors.
>
> Then you're missing some vector operators. Is that what the * is
> supposed to represent? And why do you need dS and dt? We're
> transforming coordinates, not infinitesimals.
[...]
Be reasonable: do you expect him to understand the stuff he copies and pastes?
--John Park
PD
No, not really, but at least he has to try to find it, and in the
process he perhaps learns that the answer isn't quite the same as what
he makes up out of thin air. It's also a demonstration to himself that
wrapping stuff that he makes up out of thin air with artfully
constructed and jargon-laden verbosity is no replacement for the real
thing. So far, this hasn't made the real thing sufficiently attractive
over what he now does to motivate a change in his habits, but perhaps
it's only a matter of time or of his keeling over.
PD
Koobee Wublee
> On Mar 4, 10:46 pm, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
> > How do you justify (ds^2 = c^2 dt^2 - dx^2) to (ds^2 = c^2 dt^2 - dx^2
> > - dy^2 - dz^2)?
>
> Um... the former applies to a 2D spacetime, which doesn't seem to
> apply to our universe. The latter applies to a 4D spacetime, which
> does (experimentally) seem to apply to our universe. Why?
http://groups.google.com/group/sci.physics.relativity/msg/ad344b28076f997d?hl=en&
* * * QUOTE
> ds^2 = c^2 dt'^2 - dx'^2 = c^2 dt^2 - dx^2
I don't know where you got the idea that the above equation would not
be able to predict the results of any experiments.
* * * UNQUOTE
The former is the direct result of the Lorentz transform. The latter
requires some justification. You said the former is able to predict
the results of any experiment. There is no other way to interpret it
as any experiments in 4-D spacetime. You are contradicting yourself
again. Androcles is right again.
> And what does this have to do with being unable to draw interesting
> conclusions from it?
This is because you have failed to understand the relationship between
the former and the Lorentz transform. <shrug>
> > No, Einstein fudged the Lorentz transform in his 1905 paper. The
> > derivation was totally gibberish.
>
> I find it rather easy to follow.
>
> > Incompetents like Gisse and moortel may not, but I know you are full
> > of sh*t.
>
> Really? On what basis? That you can't follow Einstein's derivation,
> and I can, so I must be full of sh*t?
Yes, this post is going to be left for later public review. I know
you are full of sh*t.
> > ds^2 = c^2 dt^2 - dx^2
>
> OK, let's start from the basics. Is there a T term? Is there a V term?
> I'll help you think it through.
Do you not know how the above equation is derived from the Lorentz
transform?
> > dS' = dS + B (B * dS / (1 + sqrt(1 - B^2)) + c dt) / sqrt(1 - B^2)
> > dt' = (dt + B * dS / c) / sqrt(1 - B^2)
>
> > S', S, B are all vectors.
>
> Then you're missing some vector operators. Is that what the * is
> supposed to represent?
Not at all. '*' is the dot product. B^2 = B * B.
> And why do you need dS and dt?
Because that is what the Lorentz transform is all about. <sbrug>
> We're
> transforming coordinates, not infinitesimals.
We are not transforming global coordinates (t, x, y, z) but
transforming (dt, dx, dy, dz). Do you not even know that?
> > Need me to explain what the symbols mean? Couldn't you figure it out
> > if you indeed understand the subject well as a professor of physics?
>
> It's not a matter of figuring it out. It's a matter of writing it down
> correctly.
>
> Denote R and R' as vectors, and V as a vector. Take g to be 1/sqrt[1-
> B*B], where B is V/c and * is the dot product.
> Then
> t' = g(t - R*B/c)
> R' = R + [(g-1)(R*B)/(cB*B) - gt)V
Changing the above to <dt, dx, dy, dz>, they are identical with mine.
> There ya go.
This proves that you just copied them without understanding them.
<shrug>
On Mar 5, 1:23 pm, "PD" <TheDraperFam...@gmail.com> wrote:
> On Mar 5, 12:12 pm, a...@FreeNet.Carleton.CA (John Park) wrote:
> > Be reasonable: do you expect him to understand the stuff he copies and pastes?
It looks like Mr. Park is talking about you, professor Draper.
> No, not really, but at least he has to try to find it,
So, you found it without understanding it.
> and in the
> process he perhaps learns that the answer isn't quite the same as what
> he makes up out of thin air.
They are indeed the same if referring to <dt, dS>.
> It's also a demonstration to himself that
> wrapping stuff that he makes up out of thin air with artfully
> constructed and jargon-laden verbosity is no replacement for the real
> thing.
This is exactly what you represent.
> So far, this hasn't made the real thing sufficiently attractive
> over what he now does to motivate a change in his habits, but perhaps
> it's only a matter of time or of his keeling over.
Now, I have raised the issue that Einstein's 1905 derivation of the
Lorentz tranform was a total gibberish. It will be examined by the
curious. There is nothing you can do about it. The ones claiming
they understood what Einstein was doing are full of sh*t. You should
admit not seeing the emperor's clothes for a change. Isn't that why
you get into physics for? Now, you have sold out your soul to get a
career. I pity you.
Eric Gisse
[snip junk]
How about addressing the proof that the metric admits a Lagrangian
despite your assertion that it does not? Then I can be entertained by
your other delusions.
Sue...
[...]
>
> Now, I have raised the issue that Einstein's 1905 derivation of the
> Lorentz tranform was a total gibberish.
Minkowski is better expressed in 1920.
<<In order to give due prominence to this relationship,
however, we must replace the usual time co-ordinate t
by an imaginary magnitude
sqrt ( -1)
ct proportional to it. Under these conditions, the natural
laws satisfying the demands of the (special) theory of
relativity assume mathematical forms, in which the time
co-ordinate plays exactly the same rôle as the three space
co-ordinates. >>
http://www.bartleby.com/173/17.html
Otherwise stated:
<<... if you know about complex numbers you will notice that
the space part enters as if it were imaginary
R2 = (ct)2 + (ix)2 + (iy)2 + (iz)2 = (ct)2 + (ir)2
where i^2 = -1 as usual. This turns out to be the
essence of the fabric (or metric) of spacetime
geometry - that space enters in with the imaginary
factor i relative to time. >>
http://www.aoc.nrao.edu/~smyers/courses/astro12/speedoflight.html
"proper time"
http://farside.ph.utexas.edu/teaching/em/lectures/node114.html
Sue...
Eric Gisse
[...]
Using ict or ix_i is very unphysical because it allows for the
possibility of complex time and space. It deeply obscures the
fundamental geometry of the situation: spacetime is not Euclidean, but
rather it is Minkowski - the differences are fundamental.
Sue...
So where do you use imaginary time on a light path ?
http://www.sm.luth.se/~urban/master/Theory/3.html
http://en.wikipedia.org/wiki/Apparent_power
Sue...
Eric Gisse
> On Mar 6, 12:20 am, "Eric Gisse" <jowr...@gmail.com> wrote:
>
> > On Mar 5, 7:14 pm, "Sue..." <suzysewns...@yahoo.com.au> wrote:
>
> > [...]
>
> > Using ict or ix_i is very unphysical because it allows for the
> > possibility of complex time and space. It deeply obscures the
> > fundamental geometry of the situation: spacetime is not Euclidean, but
> > rather it is Minkowski - the differences are fundamental.
>
> So where do you use imaginary time on a light path ?
You don't.
>
> http://www.sm.luth.se/~urban/master/Theory/3.html
> http://en.wikipedia.org/wiki/Apparent_power
It just isn't a Dennis post without a few irrelevant links.
Do you even understand what you cite? Do you think it is somehow
relevant?
>
> Sue...
Sue...
> On Mar 5, 9:07 pm, "Sue..." <suzysewns...@yahoo.com.au> wrote:
>
> > On Mar 6, 12:20 am, "Eric Gisse" <jowr...@gmail.com> wrote:
>
> > > On Mar 5, 7:14 pm, "Sue..." <suzysewns...@yahoo.com.au> wrote:
>
> > > [...]
>
> > > Using ict or ix_i is very unphysical because it allows for the
> > > possibility of complex time and space. It deeply obscures the
> > > fundamental geometry of the situation: spacetime is not Euclidean, but
> > > rather it is Minkowski - the differences are fundamental.
>
> > So where do you use imaginary time on a light path ?
http://www.sm.luth.se/~urban/master/Theory/3.html
http://en.wikipedia.org/wiki/Apparent_power
>
> You don't.
That is fine. It is called Coulomb gauge.
Please stop babbling about other gauges
'till you learn to evaluate their intrevals.
http://arxiv.org/abs/physics/0204034
Sue...
>
>
>
PD
> On Mar 5, 7:50 am, "PD" <TheDraperFam...@gmail.com> wrote:
>
> > On Mar 4, 10:46 pm, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
> > > How do you justify (ds^2 = c^2 dt^2 - dx^2) to (ds^2 = c^2 dt^2 - dx^2
> > > - dy^2 - dz^2)?
>
> > Um... the former applies to a 2D spacetime, which doesn't seem to
> > apply to our universe. The latter applies to a 4D spacetime, which
> > does (experimentally) seem to apply to our universe. Why?
>
>
> * * * QUOTE
>
> > ds^2 = c^2 dt'^2 - dx'^2 = c^2 dt^2 - dx^2
>
> I don't know where you got the idea that the above equation would not
> be able to predict the results of any experiments.
>
> * * * UNQUOTE
>
> The former is the direct result of the Lorentz transform. The latter
> requires some justification. You said the former is able to predict
> the results of any experiment. There is no other way to interpret it
> as any experiments in 4-D spacetime. You are contradicting yourself
> again. Androcles is right again.
No. The Lorentz transform can be derived from the metric.
And I *certainly* did not say that the former is able to predict the
results of *any* experiment -- for example, it will tell you nothing
about shock aversion in a cognitive psychology experiment with rats.
However, the metric *can* be used to predict, for example, the
*measurable* speed of an outgoing daughter particle from the decay of
another particle that is flying at some known speed in a beamline,
knowing the *measured* speed of the daughter particle when the parent
is at rest. That is certainly an example of a prediction of an
experimental result and it is in fact confirmed.
>
> > And what does this have to do with being unable to draw interesting
> > conclusions from it?
>
> This is because you have failed to understand the relationship between
> the former and the Lorentz transform. <shrug>
>
> > > No, Einstein fudged the Lorentz transform in his 1905 paper. The
> > > derivation was totally gibberish.
>
> > I find it rather easy to follow.
>
> > > Incompetents like Gisse and moortel may not, but I know you are full
> > > of sh*t.
>
> > Really? On what basis? That you can't follow Einstein's derivation,
> > and I can, so I must be full of sh*t?
>
> Yes, this post is going to be left for later public review. I know
> you are full of sh*t.
Frankly, your assessment doesn't concern me.
>
> > > ds^2 = c^2 dt^2 - dx^2
>
> > OK, let's start from the basics. Is there a T term? Is there a V term?
> > I'll help you think it through.
>
> Do you not know how the above equation is derived from the Lorentz
> transform?
What does this have to do with obtaining a Lagrangian from the metric?
I'll ask you again. Is there a T term? Is there a V term?
>
> > > dS' = dS + B (B * dS / (1 + sqrt(1 - B^2)) + c dt) / sqrt(1 - B^2)
> > > dt' = (dt + B * dS / c) / sqrt(1 - B^2)
>
> > > S', S, B are all vectors.
>
> > Then you're missing some vector operators. Is that what the * is
> > supposed to represent?
>
> Not at all. '*' is the dot product. B^2 = B * B.
>
> > And why do you need dS and dt?
>
> Because that is what the Lorentz transform is all about. <sbrug>
Don't thinks so. Lorentz transform is a *coordinate* transformation.
>
> > We're
> > transforming coordinates, not infinitesimals.
>
> We are not transforming global coordinates (t, x, y, z) but
> transforming (dt, dx, dy, dz). Do you not even know that?
Beg to differ.
http://en.wikipedia.org/wiki/Lorentz_transformation
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/ltrans.html#c2
http://farside.ph.utexas.edu/teaching/em/lectures/node109.html
http://hepth.hanyang.ac.kr/~kst/lect/relativity/x146.htm
http://mathworld.wolfram.com/LorentzTransformation.html
http://www.answers.com/topic/lorentz-transformation
http://www.cv.nrao.edu/course/astr534/LorentzTransform.html
>
> > > Need me to explain what the symbols mean? Couldn't you figure it out
> > > if you indeed understand the subject well as a professor of physics?
>
> > It's not a matter of figuring it out. It's a matter of writing it down
> > correctly.
>
> > Denote R and R' as vectors, and V as a vector. Take g to be 1/sqrt[1-
> > B*B], where B is V/c and * is the dot product.
> > Then
> > t' = g(t - R*B/c)
> > R' = R + [(g-1)(R*B)/(cB*B) - gt)V
>
> Changing the above to <dt, dx, dy, dz>, they are identical with mine.
Perhaps. Would you like to demonstrate that explicitly?
>
> > There ya go.
>
> This proves that you just copied them without understanding them.
> <shrug>
>
> On Mar 5, 1:23 pm, "PD" <TheDraperFam...@gmail.com> wrote:
>
> > On Mar 5, 12:12 pm, a...@FreeNet.Carleton.CA (John Park) wrote:
> > > Be reasonable: do you expect him to understand the stuff he copies and pastes?
>
> It looks like Mr. Park is talking about you, professor Draper.
>
> > No, not really, but at least he has to try to find it,
>
> So, you found it without understanding it.
>
> > and in the
> > process he perhaps learns that the answer isn't quite the same as what
> > he makes up out of thin air.
>
> They are indeed the same if referring to <dt, dS>.
>
> > It's also a demonstration to himself that
> > wrapping stuff that he makes up out of thin air with artfully
> > constructed and jargon-laden verbosity is no replacement for the real
> > thing.
>
> This is exactly what you represent.
>
> > So far, this hasn't made the real thing sufficiently attractive
> > over what he now does to motivate a change in his habits, but perhaps
> > it's only a matter of time or of his keeling over.
>
> Now, I have raised the issue that Einstein's 1905 derivation of the
> Lorentz tranform was a total gibberish. It will be examined by the
> curious. There is nothing you can do about it.
If you'd like to see that derivation expanded a bit, some of the links
I provided above do that. Those that are curious will perhaps find the
illumination that eludes you.
> The ones claiming
> they understood what Einstein was doing are full of sh*t.
And you base that on what? The fact that you don't understand it?
> You should
> admit not seeing the emperor's clothes for a change. Isn't that why
> you get into physics for? Now, you have sold out your soul to get a
> career. I pity you.
Well, as I said, your assessment doesn't concern me.
PD
bz
news:1173151889.2...@v33g2000cwv.googlegroups.com:
> * * * QUOTE
>
>> ds^2 = c^2 dt'^2 - dx'^2 = c^2 dt^2 - dx^2
>
> I don't know where you got the idea that the above equation would not
> be able to predict the results of any experiments.
>
> * * * UNQUOTE
>
> The former is the direct result of the Lorentz transform. The latter
> requires some justification. You said the former is able to predict
> the results of any experiment.
the first statement implies that the equation can be used to predict the
results of at least ONE experiment.
the second statement claims the first statement implies the equation can be
used to predict the results of ALL experiments.
Seems to be a bit of shift in meaning there.
--
bz
please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.
bz+...@ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap
Paul B. Andersen
> On Mar 6, 1:41 am, "Eric Gisse" <jowr...@gmail.com> wrote:
>> On Mar 5, 9:07 pm, "Sue..." <suzysewns...@yahoo.com.au> wrote:
>>
>>> On Mar 6, 12:20 am, "Eric Gisse" <jowr...@gmail.com> wrote:
>>>> Using ict or ix_i is very unphysical because it allows for the
>>>> possibility of complex time and space. It deeply obscures the
>>>> fundamental geometry of the situation: spacetime is not Euclidean, but
>>>> rather it is Minkowski - the differences are fundamental.
>>> So where do you use imaginary time on a light path ?
>> You don't.
>>> http://www.sm.luth.se/~urban/master/Theory/3.html
>>> http://en.wikipedia.org/wiki/Apparent_power
>> It just isn't a Dennis post without a few irrelevant links.
>>
>> Do you even understand what you cite? Do you think it is somehow
>> relevant?
I think the answer to that question is answered below:
> That is fine. It is called Coulomb gauge.
> Please stop babbling about other gauges
> 'till you learn to evaluate their intrevals.
>
> http://arxiv.org/abs/physics/0204034
Hilarious, no? :-)
Paul
Eric Gisse
> On Mar 6, 1:41 am, "Eric Gisse" <jowr...@gmail.com> wrote:
>
> > On Mar 5, 9:07 pm, "Sue..." <suzysewns...@yahoo.com.au> wrote:
>
> > > On Mar 6, 12:20 am, "Eric Gisse" <jowr...@gmail.com> wrote:
>
> > > > On Mar 5, 7:14 pm, "Sue..." <suzysewns...@yahoo.com.au> wrote:
>
> > > > [...]
>
> > > > Using ict or ix_i is very unphysical because it allows for the
> > > > possibility of complex time and space. It deeply obscures the
> > > > fundamental geometry of the situation: spacetime is not Euclidean, but
> > > > rather it is Minkowski - the differences are fundamental.
>
> > > So where do you use imaginary time on a light path ?
>
> http://www.sm.luth.se/~urban/master/Theory/3.html
> http://en.wikipedia.org/wiki/Apparent_power
You are repeating these links yet either of these have anything to do
with special relativity.
>
>
>
> > You don't.
>
> That is fine. It is called Coulomb gauge.
> Please stop babbling about other gauges
> 'till you learn to evaluate their intrevals.
>
> http://arxiv.org/abs/physics/0204034
See? You don't.
>
> Sue...
>
>
Sue...
> On Mar 6, 1:42 am, "Sue..." <suzysewns...@yahoo.com.au> wrote:
>
>
>
>
>
> > On Mar 6, 1:41 am, "Eric Gisse" <jowr...@gmail.com> wrote:
>
> > > On Mar 5, 9:07 pm, "Sue..." <suzysewns...@yahoo.com.au> wrote:
>
> > > > On Mar 6, 12:20 am, "Eric Gisse" <jowr...@gmail.com> wrote:
>
> > > > > On Mar 5, 7:14 pm, "Sue..." <suzysewns...@yahoo.com.au> wrote:
>
> > > > > [...]
>
> > > > > Using ict or ix_i is very unphysical because it allows for the
> > > > > possibility of complex time and space. It deeply obscures the
> > > > > fundamental geometry of the situation: spacetime is not Euclidean, but
> > > > > rather it is Minkowski - the differences are fundamental.
>
> > > > So where do you use imaginary time on a light path ?
>
> >http://www.sm.luth.se/~urban/master/Theory/3.html
> >http://en.wikipedia.org/wiki/Apparent_power
>
> You are repeating these links yet either of these have anything to do
> with special relativity.
"The [ ] Incompatibility of the Law of Propagation of
Light with the Principle of Relativity [is only] Apparent"
http://www.bartleby.com/173/7.html
Time-independent Maxwell equations
Time-dependent Maxwell's equations
Relativity and electromagnetism
http://farside.ph.utexas.edu/teaching/em/lectures/lectures.html
Maxwell's equations in classic electrodynamics
(classic field theory)_
a) Maxwell equations (no movement),
b) Maxwell equations (with moved bodies)
http://www.wolfram-stanek.de/maxwell_equations.htm#maxwell_classic_extended
http://www.sm.luth.se/~urban/master/Theory/3.html
http://en.wikipedia.org/wiki/Apparent_power
http://arxiv.org/abs/physics/0204034
Sue...
>
>
>
> > > You don't.
>
> > That is fine. It is called Coulomb gauge.
> > Please stop babbling about other gauges
> > 'till you learn to evaluate their intrevals.
>
> >http://arxiv.org/abs/physics/0204034
>
> See? You don't.
>
>
>
>
>
> > Sue...- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
Eric Gisse
[snip]
When I snip all the irrelevant links, there is nothing left. Thus the
discussion is over - you have no idea what you are talking about.
Sue...
Your are arguing that the postulates of SR are irrelevant
to SR. At least you are consistat in your absurdity.
"The [ ] Incompatibility of the Law of Propagation of
Light with the Principle of Relativity [is only] Apparent"
http://www.bartleby.com/173/7.html
Time-independent Maxwell equations
Time-dependent Maxwell's equations
Relativity and electromagnetism
http://farside.ph.utexas.edu/teaching/em/lectures/lectures.html
Maxwell's equations in classic electrodynamics
(classic field theory)_
a) Maxwell equations (no movement),
b) Maxwell equations (with moved bodies)
http://www.wolfram-stanek.de/maxwell_equations.htm#maxwell_classic_extended
Sue...
Koobee Wublee
> On Mar 5, 9:31 pm, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
> The Lorentz transform can be derived from the metric.
The Lorentz transform is one of the most difficult physical laws to
derive because it is wrong. >shrug>
> And I *certainly* did not say that the former is able to predict the
> results of *any* experiment -- for example, it will tell you nothing
> about shock aversion in a cognitive psychology experiment with rats.
Man, you are a very good BS-er.
> However, the metric *can* be used to predict, for example, the
> *measurable* speed of an outgoing daughter particle from the decay of
> another particle that is flying at some known speed in a beamline,
> knowing the *measured* speed of the daughter particle when the parent
> is at rest. That is certainly an example of a prediction of an
> experimental result and it is in fact confirmed.
It is truly amazing. You are checkmated, and you are able to BS your
way out of it. You are still checkmated. However, it is your BS
skill that I tip my hat to you.
> > No, Einstein fudged the Lorentz transform in his 1905 paper. The
> > derivation was totally gibberish.
>
> I find it rather easy to follow.
>
> > Yes, this post is going to be left for later public review. I know
> > you are full of sh*t.
>
> Frankly, your assessment doesn't concern me.
Yes, that is correct. However, I know you are a liar.
> > Do you not know how the above equation is derived from the Lorentz
> > transform?
>
> What does this have to do with obtaining a Lagrangian from the metric?
It is the basis of it.
> I'll ask you again. Is there a T term? Is there a V term?
You have asked me enough questions to avoid your own incompetence.
> We're
> transforming coordinates, not infinitesimals.
>
> > We are not transforming global coordinates (t, x, y, z) but
> > transforming (dt, dx, dy, dz). Do you not even know that?
>
> Beg to differ.http://en.wikipedia.org/wiki/Lorentz_transformationhttp://hyperphysics.phy-astr.gsu.edu/hbase/relativ/ltrans.html#c2http://farside.ph.utexas.edu/teaching/em/lectures/node109.htmlhttp://hepth.hanyang.ac.kr/~kst/lect/relativity/x146.htmhttp://mathworld.wolfram.com/LorentzTransformation.htmlhttp://www.answers.com/topic/lorentz-transformationhttp://www.cv.nrao.edu/course/astr534/LorentzTransform.html
Lorentz transform does not make sense in a global coordinate
transformation. It is all about infinitesimals. Global coordinate
transformation only applies in very special circumstances. <shrug>
> > dS' = dS + B (B * dS / (1 + sqrt(1 - B^2)) + c dt) / sqrt(1 - B^2)
> > dt' = (dt + B * dS / c) / sqrt(1 - B^2)
>
> t' = g(t - R*B/c)
> R' = R + [(g-1)(R*B)/(cB*B) - gt)V
>
> > Changing the above to <dt, dx, dy, dz>, they are identical with mine.
>
> Perhaps. Would you like to demonstrate that explicitly?
Now, I know you are totally incompetent as well. Androcles is so
right about you.
You really don't understand the mathematics of the Lorentz transform.
I do think you understand simple math either. Why did you give that
kid who had plenty of wild ideas about physics but not in any math a
tough time anyway? He was your younger self. You are what he will
become. You are not teaching physics. You are spreading the gospel
of false science with Einstein as the messiah. It makes me pewk.
Dirk Van de moortel
"Koobee Wublee"
aka Australopithecus Afarensis
aka Scholarly Fungi
aka Time Traveler
aka Lordly Amoeba
aka Ibn Battuta
aka Marco Polo
<koobee...@gmail.com> wrote in message news:1173255430....@h3g2000cwc.googlegroups.com...
[snip]
> You really don't understand the mathematics of the Lorentz transform.
How lucky we all are that a neo-nazi retired aerospace engineer
like you understands it so thoroughly:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/LorentzTale.html
Thanks for that :-)
Dirk Vdm
karand...@yahoo.com
SperM.hotmail.com> wrote:
> "Koobee Wublee"
> aka Australopithecus Afarensis
> aka Scholarly Fungi
> aka Time Traveler
> aka Lordly Amoeba
> aka Ibn Battuta
> aka Marco Polo
>
>
> [snip]
>
> > You really don't understand the mathematics of the Lorentz transform.
>
> How lucky we all are that a neo-nazi retired aerospace engineer
> like you understands it so thoroughly:
> http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/LorentzTale.html
> Thanks for that :-)
>
> Dirk Vdm
Isn't he an arab from Irvine, CA? His daily trips to the local mosque
affected his brain due to his smelling of the ass-fumes of his co-
prayers.
PD
> On Mar 6, 6:28 am, "PD" <TheDraperFam...@gmail.com> wrote:
>
> > On Mar 5, 9:31 pm, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
> > The Lorentz transform can be derived from the metric.
>
> The Lorentz transform is one of the most difficult physical laws to
> derive
For whom? Oh yes, I forgot, it's all about you.
> because it is wrong. >shrug>
Doesn't seem to find much contradiction with data. What's your
criterion? Oh yes, I forgot. What you do is science, and what
scientists do isn't science.
>
> > And I *certainly* did not say that the former is able to predict the
> > results of *any* experiment -- for example, it will tell you nothing
> > about shock aversion in a cognitive psychology experiment with rats.
>
> Man, you are a very good BS-er.
>
> > However, the metric *can* be used to predict, for example, the
> > *measurable* speed of an outgoing daughter particle from the decay of
> > another particle that is flying at some known speed in a beamline,
> > knowing the *measured* speed of the daughter particle when the parent
> > is at rest. That is certainly an example of a prediction of an
> > experimental result and it is in fact confirmed.
>
> It is truly amazing. You are checkmated, and you are able to BS your
> way out of it. You are still checkmated. However, it is your BS
> skill that I tip my hat to you.
>
Checkmated? I didn't realize you were playing a game. Oh yes, I
forgot. What you do is science and what scientists do isn't science.
> > > No, Einstein fudged the Lorentz transform in his 1905 paper. The
> > > derivation was totally gibberish.
>
> > I find it rather easy to follow.
>
> > > Yes, this post is going to be left for later public review. I know
> > > you are full of sh*t.
>
> > Frankly, your assessment doesn't concern me.
>
> Yes, that is correct. However, I know you are a liar.
How? Because you can't follow the derivation and so it MUST be bogus?
If you don't understand it, it must be wrong?
Lovely large head you have there. What do you use for supports to keep
it from toppling over?
>
> > > Do you not know how the above equation is derived from the Lorentz
> > > transform?
>
> > What does this have to do with obtaining a Lagrangian from the metric?
>
> It is the basis of it.
>
> > I'll ask you again. Is there a T term? Is there a V term?
>
> You have asked me enough questions to avoid your own incompetence.
>
So all this talk about Lagrangians and you haven't the foggiest what
the T term is or the V term?
> > We're
> > transforming coordinates, not infinitesimals.
>
> > > We are not transforming global coordinates (t, x, y, z) but
> > > transforming (dt, dx, dy, dz). Do you not even know that?
>
> > Beg to differ.http://en.wikipedia.org/wiki/Lorentz_transformationhttp://hyperphysic...
>
> Lorentz transform does not make sense in a global coordinate
> transformation. It is all about infinitesimals. Global coordinate
> transformation only applies in very special circumstances. <shrug>
>
Despite all those people that disagree with you. Very well. Have you
consulted Lorentz? Oh, I forgot. You are describing the
transformations that Lorentz *meant* to write down, not what he
*actually* wrote down, even though *both* are wrong because you don't
understand the derivation of either.
> > > dS' = dS + B (B * dS / (1 + sqrt(1 - B^2)) + c dt) / sqrt(1 - B^2)
> > > dt' = (dt + B * dS / c) / sqrt(1 - B^2)
>
> > t' = g(t - R*B/c)
> > R' = R + [(g-1)(R*B)/(cB*B) - gt)V
>
> > > Changing the above to <dt, dx, dy, dz>, they are identical with mine.
>
> > Perhaps. Would you like to demonstrate that explicitly?
>
> Now, I know you are totally incompetent as well. Androcles is so
> right about you.
Good for him. He's right about so little, that it would be good for
him to be right about something.
>
> You really don't understand the mathematics of the Lorentz transform.
> I do think you understand simple math either. Why did you give that
> kid who had plenty of wild ideas about physics but not in any math a
> tough time anyway? He was your younger self. You are what he will
> become. You are not teaching physics. You are spreading the gospel
> of false science with Einstein as the messiah. It makes me pewk.
Thanks, K-W, for this post. In it you were more revealing than you
usually are, and it gave some pretty good insights to the depths of
your crankiness. Please comb your hair, tuck in your shirt tail, and
wipe that fleck of foam off your lips, and give it another polished
go.
PD
Eric Gisse
> On Mar 6, 6:28 am, "PD" <TheDraperFam...@gmail.com> wrote:
>
> > On Mar 5, 9:31 pm, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
> > The Lorentz transform can be derived from the metric.
>
> The Lorentz transform is one of the most difficult physical laws to
> derive because it is wrong. >shrug>
(ds)^2 = -c^2(dt)^2 + (dx_i)^2 [i=1,2,3].
(ds)^2 = -c^2(dt')^2 + (dx'_i)^2 [i=1,2,3].
Ignore y and z. This only works because the squared distance is a
conserved quantity in this geometry. I'm shooting for how the t and t'
components are related.
(ds)^2 = -c^2(dt)^2 + (dx)^2
(ds)^2 = -c^2(dt')^2 + (dx')^2
-c^2(dt)^2 +(dx)^2 = c^2(dt')^2 + (dx')^2
-c^2(dt/dt')^2 + (dx/dt')^2 = -c^2 + (v')^2
(dt)^2(c^2/(dt')^2 - v^2/(dt')^2) = c^2 - (v')^2
(dt/dt')^2 = [c^2 - (v')^2] / [c^2 - v^2]
(dt/dt')^2 = [1 - (v'/c)^2] / [1 - (v/c)^2]
dt = dt' * sqrt( [1 - (v'/c)^2] / [1 - (v/c)^2] )
Any comments?
If the v' term perturbs you, consider a transformation from a
stationary frame to a moving frame. You can set one of the v's to zero
then.
[snip remaining]
Sue...
> On Mar 6, 11:17 pm, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
>
> > On Mar 6, 6:28 am, "PD" <TheDraperFam...@gmail.com> wrote:
>
> > > On Mar 5, 9:31 pm, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
> > > The Lorentz transform can be derived from the metric.
>
> > The Lorentz transform is one of the most difficult physical laws to
> > derive because it is wrong. >shrug>
>
> (ds)^2 = -c^2(dt)^2 + (dx_i)^2 [i=1,2,3].
> (ds)^2 = -c^2(dt')^2 + (dx'_i)^2 [i=1,2,3].
>
> Ignore y and z. This only works because the squared distance is a
> conserved quantity in this geometry. I'm shooting for how the t and t'
> components are related.
>
> (ds)^2 = -c^2(dt)^2 + (dx)^2
> (ds)^2 = -c^2(dt')^2 + (dx')^2
>
> -c^2(dt)^2 +(dx)^2 = c^2(dt')^2 + (dx')^2
> -c^2(dt/dt')^2 + (dx/dt')^2 = -c^2 + (v')^2
>
> (dt)^2(c^2/(dt')^2 - v^2/(dt')^2) = c^2 - (v')^2
>
> (dt/dt')^2 = [c^2 - (v')^2] / [c^2 - v^2]
>
> (dt/dt')^2 = [1 - (v'/c)^2] / [1 - (v/c)^2]
>
> dt = dt' * sqrt( [1 - (v'/c)^2] / [1 - (v/c)^2] )
>
> Any comments?
<< dt = d_tau in the particle's rest frame. Thus,
d_tau corresponds to the time difference between
two neighbouring events on the particle's world-line,
as measured by a clock attached to the particle
(hence, the name proper time). According to Eq. (1422),
the particle's clock **appears** to run slow, by a factor ,
in an **inertial** frame in which the particle is moving
with velocity . This is the celebrated time
dilation effect. >>
http://farside.ph.utexas.edu/teaching/em/lectures/node114.html
But the equivalence principle demands:
<<A Lorentz transformation or any other coordinate
transformation will convert electric or magnetic
fields into mixtures of electric and magnetic fields,
but no transformation mixes them with the
gravitational field. >>
http://www.aip.org/pt/vol-58/iss-11/p31.html
So it is naughty to mass with mother nature.
http://www.esa.int/SPECIALS/GSP/SEM0L6OVGJE_0.html
Sue...
Eric Gisse
Sue...
I knew you would see the problem.
Happy to see that you agree.
Sue...
Sue...
>
> > On Mar 6, 6:28 am, "PD" <TheDraperFam...@gmail.com> wrote:
>
> > > On Mar 5, 9:31 pm, "Koobee Wublee" <koobee.wub...@gmail.com> wrote:
> > > The Lorentz transform can be derived from the metric.
>
> > The Lorentz transform is one of the most difficult physical laws to
> > derive because it is wrong. >shrug>
>
> (ds)^2 = -c^2(dt)^2 + (dx_i)^2 [i=1,2,3].
> (ds)^2 = -c^2(dt')^2 + (dx'_i)^2 [i=1,2,3].
>
> Ignore y and z. This only works because the squared distance is a
> conserved quantity in this geometry. I'm shooting for how the t and t'
> components are related.
>
> (ds)^2 = -c^2(dt)^2 + (dx)^2
> (ds)^2 = -c^2(dt')^2 + (dx')^2
>
> -c^2(dt)^2 +(dx)^2 = c^2(dt')^2 + (dx')^2
> -c^2(dt/dt')^2 + (dx/dt')^2 = -c^2 + (v')^2
>
> (dt)^2(c^2/(dt')^2 - v^2/(dt')^2) = c^2 - (v')^2
>
> (dt/dt')^2 = [c^2 - (v')^2] / [c^2 - v^2]
>
> (dt/dt')^2 = [1 - (v'/c)^2] / [1 - (v/c)^2]
>
> dt = dt' * sqrt( [1 - (v'/c)^2] / [1 - (v/c)^2] )
>
> Any comments?
It is truly amazing how you eliminate imaginaries with
just a few keystrokes and it takes Jackson and Yang
40 pages.
http://www.physics.princeton.edu/~mcdonald/examples/EM/yang_ajp_73_742_05.pdf
http://arxiv.org/ftp/physics/papers/0204/0204034.pdf
I wonder what it is that Jackson and Yang are missing.
Sue...
bz
@t69g2000cwt.googlegroups.com:
> I wonder what it is that Jackson and
> Yang are missing.
Yin, obviously.