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| Author |
Message |
Sue...
science forum Guru
Joined: 08 May 2005
Posts: 2684
|
 Posted: Sun Jun 04, 2006 6:01 pm Post subject:
Re: Another look at the Lorentz factor.
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|
|
Spoonfed wrote:
| Quote: | Sue... wrote:
It is to be found rather in the fact of his recognition
that the four-dimensional space-time continuum of the
theory of relativity, in its most essential formal properties,
shows a pronounced relationship to the three-dimensional
continuum of Euclidean geometrical space. 1 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. Formally, these four co-ordinates
correspond exactly to the three space co-ordinates in
Euclidean geometry.
Albert Einstein (1879-1955). Relativity:
The Special and General Theory. ***1920***.
http://www.bartleby.com/173/17.html
giggle
Sue...
Did he mean this in an exact manner?
Minkowski didn't think so.
A rapidity change in the x-t plane
/ct'\ = / Cosh[A] -Sinh[A]\ /ct\
\ x'/ \-Sinh[A] Cosh[A]/ \ x/
is equal to
/ct'\ = / Cos[i A] -Sin[i A]\ /ct\
\ x'/ \-Sin[i A] Cos[i A]/ \ x/
which is similar but not quite the same as a rotation in the x-y plane
with the real-valued angle replaced with an imaginary angle.
/ y'\ = / Cos[B] -Sin[B]\ / y\
\ x'/ \ Sin[B] Cos[B]/ \ x/
But here, I am referring to treating the rapidity as an imaginary
angle, whereas Einstein was talking in terms treating the value of t as
imaginary. So considerably more futzing is needed to show whether
Einstein's statement (replace c*t with i*c*t and it's exactly the same)
was correct.
|
Likely it is not. See the difference in Lorenz fransforms.
http://farside.ph.utexas.edu/teaching/em/lectures/node113.html
Rapidity is a HEP term of art and there IS considerable
futzing between:
http://en.wikipedia.org/wiki/Lorentz_factors
--AND--
http://pcitapiww.cern.ch/asd/lhc++/clhep/manual/UserGuide/VectorDefs/node63..html
There are several accelerator professionals that post to this
group who might offer you some tutoring in its use.
Perhaps you can run this:
<<Space and Lorentz Vector and Transformation Package
Formulas and Definitions >>
http://pcitapiww.cern.ch/asd/lhc++/clhep/manual/UserGuide/VectorDefs/VectorDefs.html
Sue... |
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Spoonfed (www.spoonfedrel
science forum Guru Wannabe
Joined: 28 Apr 2006
Posts: 144
|
| Posted: Mon Jun 05, 2006 4:23 am Post subject:
Re: Another look at the Lorentz factor.
|
|
|
Sue... wrote:
| Quote: | Spoonfed wrote:
Sue... wrote:
It is to be found rather in the fact of his recognition
that the four-dimensional space-time continuum of the
theory of relativity, in its most essential formal properties,
shows a pronounced relationship to the three-dimensional
continuum of Euclidean geometrical space. 1 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. Formally, these four co-ordinates
correspond exactly to the three space co-ordinates in
Euclidean geometry.
Albert Einstein (1879-1955). Relativity:
The Special and General Theory. ***1920***.
http://www.bartleby.com/173/17.html
giggle
Sue...
Did he mean this in an exact manner?
Minkowski didn't think so.
A rapidity change in the x-t plane
/ct'\ = / Cosh[A] -Sinh[A]\ /ct\
\ x'/ \-Sinh[A] Cosh[A]/ \ x/
is equal to
/ct'\ = / Cos[i A] -Sin[i A]\ /ct\
\ x'/ \-Sin[i A] Cos[i A]/ \ x/
which is similar but not quite the same as a rotation in the x-y plane
with the real-valued angle replaced with an imaginary angle.
/ y'\ = / Cos[B] -Sin[B]\ / y\
\ x'/ \ Sin[B] Cos[B]/ \ x/
But here, I am referring to treating the rapidity as an imaginary
angle, whereas Einstein was talking in terms treating the value of t as
imaginary. So considerably more futzing is needed to show whether
Einstein's statement (replace c*t with i*c*t and it's exactly the same)
was correct.
Likely it is not. See the difference in Lorenz fransforms.
http://farside.ph.utexas.edu/teaching/em/lectures/node113.html
Rapidity is a HEP term of art and there IS considerable
futzing between:
http://en.wikipedia.org/wiki/Lorentz_factors
--AND--
http://pcitapiww.cern.ch/asd/lhc++/clhep/manual/UserGuide/VectorDefs/node63.html
There are several accelerator professionals that post to this
group who might offer you some tutoring in its use.
Perhaps you can run this:
Space and Lorentz Vector and Transformation Package
Formulas and Definitions
http://pcitapiww.cern.ch/asd/lhc++/clhep/manual/UserGuide/VectorDefs/VectorDefs.html
Sue...
|
I think Minkowski was right.
I did some more futzing around with the equations to see what would
happen if I took an ordinary rotation, and flat-out replaced t with
ict. It's close, but this ain't horseshoes. (Did I use that right,
Sue?)
If you were to do a simple rotation in the x-t plane about the origin,
it would look like this
/k'\ = / x/sqrt(x^2+t^2) t/sqrt(x^2+t^2)\/k\
\m'/ \-t/sqrt(x^2+t^2) x/sqrt(x^2+t^2)/\m/
where k is the t-coordinate and m is the x coordinate in the original
frame
and k' and t' are the coordinates in the rotated frame
Now according to Einstein, if I understand him correctly, we can
replace the time terms with themselves times i*c and it will give the
Lorentz Transformation.
/ick'\ = / x/sqrt(x^2+(ict)^2) ict/sqrt(x^2+(ict)^2)\/ick\
\ m'/ \-ict/sqrt(x^2+(ict)^2) x/sqrt(x^2+(ict)^2) /\ m/
Replacing i^2 with -1
/ick'\ = / x/sqrt(x^2-(ct)^2) ict/sqrt(x^2-(ct)^2)\/ick\
\ m'/ \-ict/sqrt(x^2-(ct)^2) x/sqrt(x^2-(ct)^2) /\ m/
Multiply top and bottom of each fraction by i/ct
x becomes: (ix/ct) = ib Where b=x/ct
ict becomes: -1
sqrt(x^2-(ct)^2) becomes: sqrt([x^2-(ct)^2] * (i/ct)^2)
=sqrt(1-(x/ct)^2)
=sqrt(1-b^2)
Where b=x/ct
So
/ick'\ = / ib/sqrt(1-b^2) -1/sqrt(1-b^2) \/ick\
\ m'/ \ 1/sqrt(1-b^2) ib/sqrt(1-b^2) /\ m/
so
Now since we've got rid of our x's and t's in the transformation, I'll
slip the more common coordinate labels in there to make it look a
little more familiar.
/ict'\ = / ib/sqrt(1-b^2) -1/sqrt(1-b^2) \/ict\
\ x'/ \ 1/sqrt(1-b^2) ib/sqrt(1-b^2) /\ x/
So this gives
ict'=(btc-x)/sqrt(1-b^2)
and x'=i(ct-bx)/sqrt(1-b^2)
Replacing b with v/c in the numerator:
ict'=(vt-x)/sqrt(1-b^2)
and x'=i(ct-vx/c)/sqrt(1-b^2)
This doesn't seem to match up with the Lorentz Transformation,
especially since the imaginary term never quite seems to settle in.
This seems to indicate that every acceleration causes the x and t
coordinates of every event to switch back and forth between real and
imaginary. |
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Sue...
science forum Guru
Joined: 08 May 2005
Posts: 2684
|
| Posted: Mon Jun 05, 2006 7:55 am Post subject:
Re: Another look at the Lorentz factor.
|
|
|
Spoonfed wrote:
| Quote: | Sue... wrote:
Spoonfed wrote:
Sue... wrote:
It is to be found rather in the fact of his recognition
that the four-dimensional space-time continuum of the
theory of relativity, in its most essential formal properties,
shows a pronounced relationship to the three-dimensional
continuum of Euclidean geometrical space. 1 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. Formally, these four co-ordinates
correspond exactly to the three space co-ordinates in
Euclidean geometry.
Albert Einstein (1879-1955). Relativity:
The Special and General Theory. ***1920***.
http://www.bartleby.com/173/17.html
giggle
Sue...
Did he mean this in an exact manner?
Minkowski didn't think so.
A rapidity change in the x-t plane
/ct'\ = / Cosh[A] -Sinh[A]\ /ct\
\ x'/ \-Sinh[A] Cosh[A]/ \ x/
is equal to
/ct'\ = / Cos[i A] -Sin[i A]\ /ct\
\ x'/ \-Sin[i A] Cos[i A]/ \ x/
which is similar but not quite the same as a rotation in the x-y plane
with the real-valued angle replaced with an imaginary angle.
/ y'\ = / Cos[B] -Sin[B]\ / y\
\ x'/ \ Sin[B] Cos[B]/ \ x/
But here, I am referring to treating the rapidity as an imaginary
angle, whereas Einstein was talking in terms treating the value of t as
imaginary. So considerably more futzing is needed to show whether
Einstein's statement (replace c*t with i*c*t and it's exactly the same)
was correct.
Likely it is not. See the difference in Lorenz fransforms.
http://farside.ph.utexas.edu/teaching/em/lectures/node113.html
Rapidity is a HEP term of art and there IS considerable
futzing between:
http://en.wikipedia.org/wiki/Lorentz_factors
--AND--
http://pcitapiww.cern.ch/asd/lhc++/clhep/manual/UserGuide/VectorDefs/node63.html
There are several accelerator professionals that post to this
group who might offer you some tutoring in its use.
Perhaps you can run this:
Space and Lorentz Vector and Transformation Package
Formulas and Definitions
http://pcitapiww.cern.ch/asd/lhc++/clhep/manual/UserGuide/VectorDefs/VectorDefs.html
Sue...
I think Minkowski was right.
I did some more futzing around with the equations to see what would
happen if I took an ordinary rotation, and flat-out replaced t with
ict. It's close, but this ain't horseshoes. (Did I use that right,
Sue?)
If you were to do a simple rotation in the x-t plane about the origin,
it would look like this
/k'\ = / x/sqrt(x^2+t^2) t/sqrt(x^2+t^2)\/k\
\m'/ \-t/sqrt(x^2+t^2) x/sqrt(x^2+t^2)/\m/
where k is the t-coordinate and m is the x coordinate in the original
frame
and k' and t' are the coordinates in the rotated frame
Now according to Einstein, if I understand him correctly, we can
replace the time terms with themselves times i*c and it will give the
Lorentz Transformation.
/ick'\ = / x/sqrt(x^2+(ict)^2) ict/sqrt(x^2+(ict)^2)\/ick\
\ m'/ \-ict/sqrt(x^2+(ict)^2) x/sqrt(x^2+(ict)^2) /\ m/
Replacing i^2 with -1
/ick'\ = / x/sqrt(x^2-(ct)^2) ict/sqrt(x^2-(ct)^2)\/ick\
\ m'/ \-ict/sqrt(x^2-(ct)^2) x/sqrt(x^2-(ct)^2) /\ m/
Multiply top and bottom of each fraction by i/ct
x becomes: (ix/ct) = ib Where b=x/ct
ict becomes: -1
sqrt(x^2-(ct)^2) becomes: sqrt([x^2-(ct)^2] * (i/ct)^2)
=sqrt(1-(x/ct)^2)
=sqrt(1-b^2)
Where b=x/ct
So
/ick'\ = / ib/sqrt(1-b^2) -1/sqrt(1-b^2) \/ick\
\ m'/ \ 1/sqrt(1-b^2) ib/sqrt(1-b^2) /\ m/
so
Now since we've got rid of our x's and t's in the transformation, I'll
slip the more common coordinate labels in there to make it look a
little more familiar.
/ict'\ = / ib/sqrt(1-b^2) -1/sqrt(1-b^2) \/ict\
\ x'/ \ 1/sqrt(1-b^2) ib/sqrt(1-b^2) /\ x/
So this gives
ict'=(btc-x)/sqrt(1-b^2)
and x'=i(ct-bx)/sqrt(1-b^2)
Replacing b with v/c in the numerator:
ict'=(vt-x)/sqrt(1-b^2)
and x'=i(ct-vx/c)/sqrt(1-b^2)
|
Ahem.... no offense, but I am not much of a math
fan even when it is pretty like this:
<<This relationship is plotted in Figure 3. The total mass of
methane released in the seven years considered in the area
of AB is the volume under the plotted surface. To calculate
this volume we need to carry out a multiple integral. >>
http://www.earthsci.ucl.ac.uk/undergrad/geomaths/5-min/mincon.htm
or
http://www.herrera.unt.edu.ar/nolineales/emmcap/images/Image4.gif
| Quote: |
This doesn't seem to match up with the Lorentz Transformation,
especially since the imaginary term never quite seems to settle in.
This seems to indicate that every acceleration causes the x and t
coordinates of every event to switch back and forth between real and
imaginary.
|
This is a hobby for me so unless one of the posters that is a
fan of ascii maths is following your line of thought you would
do better running a math package to confirm your work.
If you have a unix system I feel sure there is plenty availible
for free.
ftp5.freebsd.org/pub/FreeBSD/ports
http://sourceforge.net/
Kind regards,
Sue... |
|
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|
|
Spoonfed (www.spoonfedrel
science forum Guru Wannabe
Joined: 28 Apr 2006
Posts: 144
|
| Posted: Mon Jun 05, 2006 3:12 pm Post subject:
Re: Another look at the Lorentz factor.
|
|
|
Sue... wrote:
| Quote: | Spoonfed wrote:
Sue... wrote:
Spoonfed wrote:
Sue... wrote:
It is to be found rather in the fact of his recognition
that the four-dimensional space-time continuum of the
theory of relativity, in its most essential formal properties,
shows a pronounced relationship to the three-dimensional
continuum of Euclidean geometrical space. 1 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. Formally, these four co-ordinates
correspond exactly to the three space co-ordinates in
Euclidean geometry.
Albert Einstein (1879-1955). Relativity:
The Special and General Theory. ***1920***.
http://www.bartleby.com/173/17.html
giggle
Sue...
Did he mean this in an exact manner?
Minkowski didn't think so.
A rapidity change in the x-t plane
/ct'\ = / Cosh[A] -Sinh[A]\ /ct\
\ x'/ \-Sinh[A] Cosh[A]/ \ x/
is equal to
/ct'\ = / Cos[i A] -Sin[i A]\ /ct\
\ x'/ \-Sin[i A] Cos[i A]/ \ x/
which is similar but not quite the same as a rotation in the x-y plane
with the real-valued angle replaced with an imaginary angle.
/ y'\ = / Cos[B] -Sin[B]\ / y\
\ x'/ \ Sin[B] Cos[B]/ \ x/
But here, I am referring to treating the rapidity as an imaginary
angle, whereas Einstein was talking in terms treating the value of t as
imaginary. So considerably more futzing is needed to show whether
Einstein's statement (replace c*t with i*c*t and it's exactly the same)
was correct.
Likely it is not. See the difference in Lorenz fransforms.
http://farside.ph.utexas.edu/teaching/em/lectures/node113.html
Rapidity is a HEP term of art and there IS considerable
futzing between:
http://en.wikipedia.org/wiki/Lorentz_factors
--AND--
http://pcitapiww.cern.ch/asd/lhc++/clhep/manual/UserGuide/VectorDefs/node63.html
There are several accelerator professionals that post to this
group who might offer you some tutoring in its use.
Perhaps you can run this:
Space and Lorentz Vector and Transformation Package
Formulas and Definitions
http://pcitapiww.cern.ch/asd/lhc++/clhep/manual/UserGuide/VectorDefs/VectorDefs.html
Sue...
I think Minkowski was right.
I did some more futzing around with the equations to see what would
happen if I took an ordinary rotation, and flat-out replaced t with
ict. It's close, but this ain't horseshoes. (Did I use that right,
Sue?)
If you were to do a simple rotation in the x-t plane about the origin,
it would look like this
/k'\ = / x/sqrt(x^2+t^2) t/sqrt(x^2+t^2)\/k\
\m'/ \-t/sqrt(x^2+t^2) x/sqrt(x^2+t^2)/\m/
where k is the t-coordinate and m is the x coordinate in the original
frame
and k' and t' are the coordinates in the rotated frame
Now according to Einstein, if I understand him correctly, we can
replace the time terms with themselves times i*c and it will give the
Lorentz Transformation.
/ick'\ = / x/sqrt(x^2+(ict)^2) ict/sqrt(x^2+(ict)^2)\/ick\
\ m'/ \-ict/sqrt(x^2+(ict)^2) x/sqrt(x^2+(ict)^2) /\ m/
Replacing i^2 with -1
/ick'\ = / x/sqrt(x^2-(ct)^2) ict/sqrt(x^2-(ct)^2)\/ick\
\ m'/ \-ict/sqrt(x^2-(ct)^2) x/sqrt(x^2-(ct)^2) /\ m/
Multiply top and bottom of each fraction by i/ct
x becomes: (ix/ct) = ib Where b=x/ct
ict becomes: -1
sqrt(x^2-(ct)^2) becomes: sqrt([x^2-(ct)^2] * (i/ct)^2)
=sqrt(1-(x/ct)^2)
=sqrt(1-b^2)
Where b=x/ct
So
/ick'\ = / ib/sqrt(1-b^2) -1/sqrt(1-b^2) \/ick\
\ m'/ \ 1/sqrt(1-b^2) ib/sqrt(1-b^2) /\ m/
so
Now since we've got rid of our x's and t's in the transformation, I'll
slip the more common coordinate labels in there to make it look a
little more familiar.
/ict'\ = / ib/sqrt(1-b^2) -1/sqrt(1-b^2) \/ict\
\ x'/ \ 1/sqrt(1-b^2) ib/sqrt(1-b^2) /\ x/
So this gives
ict'=(btc-x)/sqrt(1-b^2)
and x'=i(ct-bx)/sqrt(1-b^2)
Replacing b with v/c in the numerator:
ict'=(vt-x)/sqrt(1-b^2)
and x'=i(ct-vx/c)/sqrt(1-b^2)
Ahem.... no offense, but I am not much of a math
fan
|
You're off the hook, Sue.
However, presumably, Einstein has a proof similar to this one which
shows that it can be done. Anyone who has access to that proof or
prides themselves on your understanding of tensors should be able to
run through this and verify or find errors in my assumptions, setup,
and/or work.
I doubt if anybody with that skill, though, also has the interest and
time to chase after my errors without compensation. And anyway, I
realized what I did wrong this morning.
Namely, I set up the rotation thinking in terms of {x,y} and then made
it analogous to a rotation in terms of {t,x}. I switched the order.
In effect, the last line, which reads...
| Quote: | ict'=(vt-x)/sqrt(1-b^2)
and x'=i(ct-vx/c)/sqrt(1-b^2)
|
Should have read
x'=(vt-x)/sqrt(1-b^2)
and ict'=i(ct-vx/c)/sqrt(1-b^2)
After dividing the last line by ic, you'll have
x'=(vt-x)/sqrt(1-b^2)
t'=(t-vx/cc)/sqrt(1-b^2)
| Quote: |
This doesn't seem to match up with the Lorentz Transformation,
especially since the imaginary term never quite seems to settle in.
This seems to indicate that every acceleration causes the x and t
coordinates of every event to switch back and forth between real and
imaginary.
This is a hobby for me so unless one of the posters that is a
fan of ascii maths is following your line of thought you would
do better running a math package to confirm your work.
If you have a unix system I feel sure there is plenty availible
for free.
ftp5.freebsd.org/pub/FreeBSD/ports
http://sourceforge.net/
Kind regards,
Sue...
|
I already use some math software, Sue. It was invaluable working on
Tim' polysigned math. With this problem though, the biggest difficulty
was to figure out how to set it up--keeping the x's and y's of the
coordinate system separate from the dx/dy associated with a rotation,
and keeping the x and t of spacetime separate from the dx/dt associated
with velocity change, and figuring out how to represent this
mathematically and consistently, and in such a way that the
representation didn't confuse me. It took two or three go's before I
realized (to keep from getting confused) that I had to use x and y as
components of the rotation, and x and t as components of the velocity,
while making up other symbols, (k and m) for the coordinate positions.
Software would have been perfectly happy if I'd called the variables
squawk, bleep, moo, and futz, as long as they were four distinct
variables, and setting it up would have forced me to realize that there
were four distinct variables, but the software would *not* have told me
that there were four variables. The software can help a lot once you
know how to set up the problem, but it won't necessarily detect a
problem in the setup.
So anyway, if Einstein said,
| Quote: | replace the usual time co-ordinate t by an imaginary magnitude
(sqrt -1)
ct proportional to it.
|
and...
| Quote: | Formally, these four co-ordinates
correspond exactly to the three space co-ordinates in
Euclidean geometry.
|
and I believe I have just verified Einstein's statement was correct,
but I still don't know for sure which definition of "formally" Einstein
was using.
Formally: adv. 1. following incontrovertibly from strict
mathematical manipulations. 2. not making much sense to me, but
following incontrovertibly from strict mathematical manipulations. |
|
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|
|
Sue...
science forum Guru
Joined: 08 May 2005
Posts: 2684
|
| Posted: Mon Jun 05, 2006 4:32 pm Post subject:
Re: Another look at the Lorentz factor.
|
|
|
Spoonfed wrote:
| Quote: | Sue... wrote:
Spoonfed wrote:
Sue... wrote:
Spoonfed wrote:
Sue... wrote:
It is to be found rather in the fact of his recognition
that the four-dimensional space-time continuum of the
theory of relativity, in its most essential formal properties,
shows a pronounced relationship to the three-dimensional
continuum of Euclidean geometrical space. 1 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. Formally, these four co-ordinates
correspond exactly to the three space co-ordinates in
Euclidean geometry.
Albert Einstein (1879-1955). Relativity:
The Special and General Theory. ***1920***.
http://www.bartleby.com/173/17.html
giggle
Sue...
Did he mean this in an exact manner?
Minkowski didn't think so.
A rapidity change in the x-t plane
/ct'\ = / Cosh[A] -Sinh[A]\ /ct\
\ x'/ \-Sinh[A] Cosh[A]/ \ x/
is equal to
/ct'\ = / Cos[i A] -Sin[i A]\ /ct\
\ x'/ \-Sin[i A] Cos[i A]/ \ x/
which is similar but not quite the same as a rotation in the x-y plane
with the real-valued angle replaced with an imaginary angle.
/ y'\ = / Cos[B] -Sin[B]\ / y\
\ x'/ \ Sin[B] Cos[B]/ \ x/
But here, I am referring to treating the rapidity as an imaginary
angle, whereas Einstein was talking in terms treating the value of t as
imaginary. So considerably more futzing is needed to show whether
Einstein's statement (replace c*t with i*c*t and it's exactly the same)
was correct.
Likely it is not. See the difference in Lorenz fransforms.
http://farside.ph.utexas.edu/teaching/em/lectures/node113.html
Rapidity is a HEP term of art and there IS considerable
futzing between:
http://en.wikipedia.org/wiki/Lorentz_factors
--AND--
http://pcitapiww.cern.ch/asd/lhc++/clhep/manual/UserGuide/VectorDefs/node63.html
There are several accelerator professionals that post to this
group who might offer you some tutoring in its use.
Perhaps you can run this:
Space and Lorentz Vector and Transformation Package
Formulas and Definitions
http://pcitapiww.cern.ch/asd/lhc++/clhep/manual/UserGuide/VectorDefs/VectorDefs.html
Sue...
I think Minkowski was right.
I did some more futzing around with the equations to see what would
happen if I took an ordinary rotation, and flat-out replaced t with
ict. It's close, but this ain't horseshoes. (Did I use that right,
Sue?)
If you were to do a simple rotation in the x-t plane about the origin,
it would look like this
/k'\ = / x/sqrt(x^2+t^2) t/sqrt(x^2+t^2)\/k\
\m'/ \-t/sqrt(x^2+t^2) x/sqrt(x^2+t^2)/\m/
where k is the t-coordinate and m is the x coordinate in the original
frame
and k' and t' are the coordinates in the rotated frame
Now according to Einstein, if I understand him correctly, we can
replace the time terms with themselves times i*c and it will give the
Lorentz Transformation.
/ick'\ = / x/sqrt(x^2+(ict)^2) ict/sqrt(x^2+(ict)^2)\/ick\
\ m'/ \-ict/sqrt(x^2+(ict)^2) x/sqrt(x^2+(ict)^2) /\ m/
Replacing i^2 with -1
/ick'\ = / x/sqrt(x^2-(ct)^2) ict/sqrt(x^2-(ct)^2)\/ick\
\ m'/ \-ict/sqrt(x^2-(ct)^2) x/sqrt(x^2-(ct)^2) /\ m/
Multiply top and bottom of each fraction by i/ct
x becomes: (ix/ct) = ib Where b=x/ct
ict becomes: -1
sqrt(x^2-(ct)^2) becomes: sqrt([x^2-(ct)^2] * (i/ct)^2)
=sqrt(1-(x/ct)^2)
=sqrt(1-b^2)
Where b=x/ct
So
/ick'\ = / ib/sqrt(1-b^2) -1/sqrt(1-b^2) \/ick\
\ m'/ \ 1/sqrt(1-b^2) ib/sqrt(1-b^2) /\ m/
so
Now since we've got rid of our x's and t's in the transformation, I'll
slip the more common coordinate labels in there to make it look a
little more familiar.
/ict'\ = / ib/sqrt(1-b^2) -1/sqrt(1-b^2) \/ict\
\ x'/ \ 1/sqrt(1-b^2) ib/sqrt(1-b^2) /\ x/
So this gives
ict'=(btc-x)/sqrt(1-b^2)
and x'=i(ct-bx)/sqrt(1-b^2)
Replacing b with v/c in the numerator:
ict'=(vt-x)/sqrt(1-b^2)
and x'=i(ct-vx/c)/sqrt(1-b^2)
Ahem.... no offense, but I am not much of a math
fan
You're off the hook, Sue.
However, presumably, Einstein has a proof similar to this one which
shows that it can be done. Anyone who has access to that proof or
prides themselves on your understanding of tensors should be able to
run through this and verify or find errors in my assumptions, setup,
and/or work.
|
Like this?
<<...the most general transformation between two inertial frames
consists
of a Lorentz transformation in the standard configuration plus a
translation
(this includes a translation in time) and a rotation of the coordinate
axes.
The resulting transformation is called a general Lorentz
transformation, as
opposed to a Lorentz transformation in the standard configuration which
will henceforth be termed a standard Lorentz transformation. >>
http://farside.ph.utexas.edu/teaching/jk1/lectures/node13.html
When you mix time and space you want to do it equally on all
three axes. Or I misunderstood what you were doing.
| Quote: |
I doubt if anybody with that skill, though, also has the interest and
time to chase after my errors without compensation. And anyway, I
realized what I did wrong this morning.
Namely, I set up the rotation thinking in terms of {x,y} and then made
it analogous to a rotation in terms of {t,x}. I switched the order.
In effect, the last line, which reads...
ict'=(vt-x)/sqrt(1-b^2)
and x'=i(ct-vx/c)/sqrt(1-b^2)
Should have read
x'=(vt-x)/sqrt(1-b^2)
and ict'=i(ct-vx/c)/sqrt(1-b^2)
After dividing the last line by ic, you'll have
x'=(vt-x)/sqrt(1-b^2)
t'=(t-vx/cc)/sqrt(1-b^2)
even when it is pretty like this:
This relationship is plotted in Figure 3. The total mass of
methane released in the seven years considered in the area
of AB is the volume under the plotted surface. To calculate
this volume we need to carry out a multiple integral.
http://www.earthsci.ucl.ac.uk/undergrad/geomaths/5-min/mincon.htm
or
http://www.herrera.unt.edu.ar/nolineales/emmcap/images/Image4.gif
This doesn't seem to match up with the Lorentz Transformation,
especially since the imaginary term never quite seems to settle in.
This seems to indicate that every acceleration causes the x and t
coordinates of every event to switch back and forth between real and
imaginary.
This is a hobby for me so unless one of the posters that is a
fan of ascii maths is following your line of thought you would
do better running a math package to confirm your work.
If you have a unix system I feel sure there is plenty availible
for free.
ftp5.freebsd.org/pub/FreeBSD/ports
http://sourceforge.net/
Kind regards,
Sue...
I already use some math software, Sue. It was invaluable working on
Tim' polysigned math. With this problem though, the biggest difficulty
was to figure out how to set it up--keeping the x's and y's of the
coordinate system separate from the dx/dy associated with a rotation,
and keeping the x and t of spacetime separate from the dx/dt associated
with velocity change, and figuring out how to represent this
mathematically and consistently, and in such a way that the
representation didn't confuse me. It took two or three go's before I
realized (to keep from getting confused) that I had to use x and y as
components of the rotation, and x and t as components of the velocity,
while making up other symbols, (k and m) for the coordinate positions.
Software would have been perfectly happy if I'd called the variables
squawk, bleep, moo, and futz, as long as they were four distinct
variables, and setting it up would have forced me to realize that there
were four distinct variables, but the software would *not* have told me
that there were four variables. The software can help a lot once you
know how to set up the problem, but it won't necessarily detect a
problem in the setup.
So anyway, if Einstein said,
replace the usual time co-ordinate t by an imaginary magnitude
(sqrt -1)
ct proportional to it.
and...
Formally, these four co-ordinates
correspond exactly to the three space co-ordinates in
Euclidean geometry.
and I believe I have just verified Einstein's statement was correct,
but I still don't know for sure which definition of "formally" Einstein
was using.
|
The exceptions are discussed here:
http://arxiv.org/abs/physics/0204034
| Quote: |
Formally: adv. 1. following incontrovertibly from strict
mathematical manipulations. 2. not making much sense to me, but
following incontrovertibly from strict mathematical manipulations.
|
Jackson will put some meaning to it. If not,
you missed something betweeen time independent
an time dependent maxwell's equations.
Sue... |
|
| Back to top |
|
|
Spoonfed (www.spoonfedrel
science forum Guru Wannabe
Joined: 28 Apr 2006
Posts: 144
|
| Posted: Tue Jun 06, 2006 10:50 pm Post subject:
Re: Another look at the Lorentz factor.
|
|
|
Sue... wrote:
| Quote: | Spoonfed wrote:
Sue... wrote:
Spoonfed wrote:
Sue... wrote:
Spoonfed wrote:
Sue... wrote:
It is to be found rather in the fact of his recognition
that the four-dimensional space-time continuum of the
theory of relativity, in its most essential formal properties,
shows a pronounced relationship to the three-dimensional
continuum of Euclidean geometrical space. 1 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. Formally, these four co-ordinates
correspond exactly to the three space co-ordinates in
Euclidean geometry.
Albert Einstein (1879-1955). Relativity:
The Special and General Theory. ***1920***.
http://www.bartleby.com/173/17.html
giggle
Sue...
Did he mean this in an exact manner?
Minkowski didn't think so.
A rapidity change in the x-t plane
/ct'\ = / Cosh[A] -Sinh[A]\ /ct\
\ x'/ \-Sinh[A] Cosh[A]/ \ x/
is equal to
/ct'\ = / Cos[i A] -Sin[i A]\ /ct\
\ x'/ \-Sin[i A] Cos[i A]/ \ x/
which is similar but not quite the same as a rotation in the x-y plane
with the real-valued angle replaced with an imaginary angle.
/ y'\ = / Cos[B] -Sin[B]\ / y\
\ x'/ \ Sin[B] Cos[B]/ \ x/
But here, I am referring to treating the rapidity as an imaginary
angle, whereas Einstein was talking in terms treating the value of t as
imaginary. So considerably more futzing is needed to show whether
Einstein's statement (replace c*t with i*c*t and it's exactly the same)
was correct.
Likely it is not. See the difference in Lorenz fransforms.
http://farside.ph.utexas.edu/teaching/em/lectures/node113.html
Rapidity is a HEP term of art and there IS considerable
futzing between:
http://en.wikipedia.org/wiki/Lorentz_factors
--AND--
http://pcitapiww.cern.ch/asd/lhc++/clhep/manual/UserGuide/VectorDefs/node63.html
There are several accelerator professionals that post to this
group who might offer you some tutoring in its use.
Perhaps you can run this:
Space and Lorentz Vector and Transformation Package
Formulas and Definitions
http://pcitapiww.cern.ch/asd/lhc++/clhep/manual/UserGuide/VectorDefs/VectorDefs.html
Sue...
I think Minkowski was right.
I did some more futzing around with the equations to see what would
happen if I took an ordinary rotation, and flat-out replaced t with
ict. It's close, but this ain't horseshoes. (Did I use that right,
Sue?)
If you were to do a simple rotation in the x-t plane about the origin,
it would look like this
/k'\ = / x/sqrt(x^2+t^2) t/sqrt(x^2+t^2)\/k\
\m'/ \-t/sqrt(x^2+t^2) x/sqrt(x^2+t^2)/\m/
where k is the t-coordinate and m is the x coordinate in the original
frame
and k' and t' are the coordinates in the rotated frame
Now according to Einstein, if I understand him correctly, we can
replace the time terms with themselves times i*c and it will give the
Lorentz Transformation.
/ick'\ = / x/sqrt(x^2+(ict)^2) ict/sqrt(x^2+(ict)^2)\/ick\
\ m'/ \-ict/sqrt(x^2+(ict)^2) x/sqrt(x^2+(ict)^2) /\ m/
Replacing i^2 with -1
/ick'\ = / x/sqrt(x^2-(ct)^2) ict/sqrt(x^2-(ct)^2)\/ick\
\ m'/ \-ict/sqrt(x^2-(ct)^2) x/sqrt(x^2-(ct)^2) /\ m/
Multiply top and bottom of each fraction by i/ct
x becomes: (ix/ct) = ib Where b=x/ct
ict becomes: -1
sqrt(x^2-(ct)^2) becomes: sqrt([x^2-(ct)^2] * (i/ct)^2)
=sqrt(1-(x/ct)^2)
=sqrt(1-b^2)
Where b=x/ct
So
/ick'\ = / ib/sqrt(1-b^2) -1/sqrt(1-b^2) \/ick\
\ m'/ \ 1/sqrt(1-b^2) ib/sqrt(1-b^2) /\ m/
so
Now since we've got rid of our x's and t's in the transformation, I'll
slip the more common coordinate labels in there to make it look a
little more familiar.
/ict'\ = / ib/sqrt(1-b^2) -1/sqrt(1-b^2) \/ict\
\ x'/ \ 1/sqrt(1-b^2) ib/sqrt(1-b^2) /\ x/
So this gives
ict'=(btc-x)/sqrt(1-b^2)
and x'=i(ct-bx)/sqrt(1-b^2)
Replacing b with v/c in the numerator:
ict'=(vt-x)/sqrt(1-b^2)
and x'=i(ct-vx/c)/sqrt(1-b^2)
Ahem.... no offense, but I am not much of a math
fan
You're off the hook, Sue.
However, presumably, Einstein has a proof similar to this one which
shows that it can be done. Anyone who has access to that proof or
prides themselves on your understanding of tensors should be able to
run through this and verify or find errors in my assumptions, setup,
and/or work.
Like this?
...the most general transformation between two inertial frames
consists
of a Lorentz transformation in the standard configuration plus a
translation
(this includes a translation in time) and a rotation of the coordinate
axes.
The resulting transformation is called a general Lorentz
transformation, as
opposed to a Lorentz transformation in the standard configuration which
will henceforth be termed a standard Lorentz transformation.
http://farside.ph.utexas.edu/teaching/jk1/lectures/node13.html
|
Not really. I don't see any references to imaginary numbers in there
at all. Unless you see something I missed.
| Quote: | When you mix time and space you want to do it equally on all
three axes. Or I misunderstood what you were doing.
|
If you have pre-determined your axes, and your rotation or acceleration
is not aligned with them with the origin at the point where the
rotation goes around and the time the acceleration takes place, you'd
have to consider a whole lot of information which I've not bothered
with.
But if you start with an acceleration or rotation, you can select your
origin, and your axes based on that information.
| Quote: |
I doubt if anybody with that skill, though, also has the interest and
time to chase after my errors without compensation. And anyway, I
realized what I did wrong this morning.
Namely, I set up the rotation thinking in terms of {x,y} and then made
it analogous to a rotation in terms of {t,x}. I switched the order.
In effect, the last line, which reads...
ict'=(vt-x)/sqrt(1-b^2)
and x'=i(ct-vx/c)/sqrt(1-b^2)
Should have read
x'=(vt-x)/sqrt(1-b^2)
and ict'=i(ct-vx/c)/sqrt(1-b^2)
After dividing the last line by ic, you'll have
x'=(vt-x)/sqrt(1-b^2)
t'=(t-vx/cc)/sqrt(1-b^2)
even when it is pretty like this:
This relationship is plotted in Figure 3. The total mass of
methane released in the seven years considered in the area
of AB is the volume under the plotted surface. To calculate
this volume we need to carry out a multiple integral.
http://www.earthsci.ucl.ac.uk/undergrad/geomaths/5-min/mincon.htm
or
http://www.herrera.unt.edu.ar/nolineales/emmcap/images/Image4.gif
This doesn't seem to match up with the Lorentz Transformation,
especially since the imaginary term never quite seems to settle in.
This seems to indicate that every acceleration causes the x and t
coordinates of every event to switch back and forth between real and
imaginary.
This is a hobby for me so unless one of the posters that is a
fan of ascii maths is following your line of thought you would
do better running a math package to confirm your work.
If you have a unix system I feel sure there is plenty availible
for free.
ftp5.freebsd.org/pub/FreeBSD/ports
http://sourceforge.net/
Kind regards,
Sue...
I already use some math software, Sue. It was invaluable working on
Tim' polysigned math. With this problem though, the biggest difficulty
was to figure out how to set it up--keeping the x's and y's of the
coordinate system separate from the dx/dy associated with a rotation,
and keeping the x and t of spacetime separate from the dx/dt associated
with velocity change, and figuring out how to represent this
mathematically and consistently, and in such a way that the
representation didn't confuse me. It took two or three go's before I
realized (to keep from getting confused) that I had to use x and y as
components of the rotation, and x and t as components of the velocity,
while making up other symbols, (k and m) for the coordinate positions.
Software would have been perfectly happy if I'd called the variables
squawk, bleep, moo, and futz, as long as they were four distinct
variables, and setting it up would have forced me to realize that there
were four distinct variables, but the software would *not* have told me
that there were four variables. The software can help a lot once you
know how to set up the problem, but it won't necessarily detect a
problem in the setup.
So anyway, if Einstein said,
replace the usual time co-ordinate t by an imaginary magnitude
(sqrt -1)
ct proportional to it.
and...
Formally, these four co-ordinates
correspond exactly to the three space co-ordinates in
Euclidean geometry.
and I believe I have just verified Einstein's statement was correct,
but I still don't know for sure which definition of "formally" Einstein
was using.
The exceptions are discussed here:
http://arxiv.org/abs/physics/0204034
Formally: adv. 1. following incontrovertibly from strict
mathematical manipulations. 2. not making much sense to me, but
following incontrovertibly from strict mathematical manipulations.
Jackson will put some meaning to it. If not,
you missed something betweeen time independent
an time dependent maxwell's equations.
Sue...
|
I glanced through Jackson's description of Special Relativity in
"Classical Electrodynamics" I found no mention of time having an
imaginary quality. Perhaps during some set of twenty or thirty steps
he short-handed as "Clearly" or "Obviously," "trivially," or "it is
left as an exercise to determine..." |
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