What is the "reason" C is constant to all observers?

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Topic:	Science > Physics
User:	"George Hammond"
Date:	26 Aug 2005 06:43:46 PM
Object:	What is the "reason" C is constant to all observers?

Einstein's Special Relativity is famous for AD HOC
postulating that:
"All observers regardless of their relative
(uniform) velocities will measure the same
speed of light, C."
This is of course highly COUNTER INTUITIVE (the
Galilean transformation is more intuitive).
ODDLY ENOUGH, Einstein never gave a physical
reason or explanation of WHY C should be the same for
all observers regardless of their speed... he simply
AD HOC POSTULATED the fact (which has since
been experimentally confirmed a million times).
PLEASE CORRECT ME IF I AM WRONG:
But as far as I know, to this day, there is STILL NO
"logical theoretical reason"
of WHY C should appear to be the SAME to all
moving observers!
As far as I know, Quantum Theory, QED, Quantum
Field Theory, Particle Physics...... NOTHING,
offers any logical explanation for this fact... all of
them simply AD HOC ASSUME the fact just like
Einstein did in 1905.
Is that correct?
Yes, or no?
Or is there some "physical explanation" of why C
is a constant to all (uniformly) moving observers
that I don't know about?
Mystified?
.

User: "Sue..."
Title: Re: What is the "reason" C is constant to all observers?	28 Sep 2005 09:05:58 AM

Androcles wrote:

"Sue..." <suzysewnshow@yahoo.com.au> wrote in message
news:1127913175.802599.81100@g47g2000cwa.googlegroups.com...
|
| tony fleming wrote:
| > sue thanks, i understand what you are saying. i'm not asking why
does
| > it go at 2.997 x 10^8 m/s but why constant??
|
| Because the mass of an electron is constant.

Huh?
Why does price(rice) in China have to do with mass(electron)?

Ya snookered me with those spring and weight URL's
Fill a room up with those things, and give them some kind
of mechanical coupling.
You should find the weight and tension determines how fast a
purturbation propagates across the room.
I'll repost:
When light (electromagnetism) interacts with matter,
whether the matter is emitting, receiving or incidental
to the path, charges in the matter have to move
in response to the Coulomb component.
Atoms in a piano string have to move if you want
a wave to oscillate end to end.
If you want a wave to propagate faster down a
string or rope, you use a lighter rope or remove
the winding on the big fat piano strings.
If you want light to propagte faster as it moves
from charge to charge, then you use lighter charges.
If you find that all charges weigh 0.511 MeV,
then you are stuck with this:
http://physics.nist.gov/cuu/Images/alphaeq.gif
http://physics.nist.gov/cuu/Constants/alpha.html
"You know, it would be sufficient to really
understand the electron"
--Albert Einstein
and you may have missed this:
http://physics.nist.gov/cuu/Images/alphaeq.gif
http://physics.nist.gov/cuu/Constants/alpha.html
BTW... don't forget to look at:
http://physics.nist.gov/cuu/Images/alphaeq.gif
http://physics.nist.gov/cuu/Constants/alpha.html
Sue...
PS.
http://physics.nist.gov/cuu/Images/alphaeq.gif
http://physics.nist.gov/cuu/Constants/alpha.html

| <<why not a function of distance,>>

| It isn't constant over distance.
| That is why the Hubble law.

Could be, but could be getting fat as it gets old, too. I am.
The wavelength of my belt has grown a notch.

|
| whatever, why not a variation of .0001%?
|
| Like this
|
| ========================================
| speed of light in vacuum
| Value 299 792 458 m s-1
| Standard uncertainty (exact)
| Relative standard uncertainty (exact)
| Concise form 299 792 458 m s-1
|
| http://physics.nist.gov/cgi-bin/cuu/Value?c
| =========================================
| natural unit of length
| Value 386.159 2678 x 10-15 m
| Standard uncertainty 0.000 0026 x 10-15 m
| Relative standard uncertainty 6.7 x 10-9
| Concise form 386.159 2678(26) x 10-15 m
|
| http://physics.nist.gov/cgi-bin/cuu/Value?tecomwlbar|search_for=length
| =========================================
|
|
| no it's a constant,
| > WHY???? (btw i think i have already told you the answer recently)
|
| Sorry...
| I forgot to make a note of it.
|
| Sue...
I have that trouble too. I keep thinking
[quote]
we establish by definition that the "time" required by a turtle to
travel
from A to B equals the "time" it requires to travel from B to A.
[end quote]
Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/
Oops!... Did I say 'a turtle' again? It keeps slipping
out...Sorry...'light'.
Androcles.

User: "Androcles Androcles@ MyPlace.org"
Title: Re: What is the "reason" C is constant to all observers?	28 Sep 2005 11:52:11 AM

"Sue..." <suzysewnshow@yahoo.com.au> wrote in message
news:1127916358.318142.81290@g14g2000cwa.googlegroups.com...
|
| Androcles wrote:
| > "Sue..." <suzysewnshow@yahoo.com.au> wrote in message
| > news:1127913175.802599.81100@g47g2000cwa.googlegroups.com...
| > |
| > | tony fleming wrote:
| > | > sue thanks, i understand what you are saying. i'm not asking
why
| > does
| > | > it go at 2.997 x 10^8 m/s but why constant??
| > |
| > | Because the mass of an electron is constant.
| >
| > Huh?
| > Why does price(rice) in China have to do with mass(electron)?
|
| Ya snookered me with those spring and weight URL's
| Fill a room up with those things, and give them some kind
| of mechanical coupling.
|
Boing, boing...
Kong-Fu-see-us say to honolable Zu, One hand washee other, rectrickery
field is magnets makee, magnet field is rectrickery makee.
Mass bobbee on spling makee spling bob on mass, you no care how fast
snooker ball spin when flown rike snook swim.
http://indian-river.fl.us/fishing/fish/snookcom.jpg
You roll snooker ball on tabrle, then you care how fast.
You catchee snook, snook wiggle like mass on spring.
| You should find the weight and tension determines how fast a
| purturbation propagates across the room.
|
You no clap one hand makee noisee, need two hand.
Here is two hand makee noisee in snookered eyeball.
http://campus.umr.edu/physics/alab/polar/circular_wave.gif
One hand clap sideways, other hand clap up-down makee plopagation
noisee.
| I'll repost:
|
| When light (electromagnetism) interacts with matter,
| whether the matter is emitting, receiving or incidental
| to the path, charges in the matter have to move
| in response to the Coulomb component.
|
| Atoms in a piano string have to move if you want
| a wave to oscillate end to end.
| If you want a wave to propagate faster down a
| string or rope, you use a lighter rope or remove
| the winding on the big fat piano strings.
|
|
| If you want light to propagte faster as it moves
| from charge to charge, then you use lighter charges.
|
|
| If you find that all charges weigh 0.511 MeV,
Kong-Fu-see-us say MeV not kiloglams, Zu talkee funny, not makee pi-anny
righter pray music. Makee pi-anny hot.
| then you are stuck with this:
|
|
| http://physics.nist.gov/cuu/Images/alphaeq.gif
| http://physics.nist.gov/cuu/Constants/alpha.html
Zu think awr corour same red brue gleen, like note on pi-anny awr same?
Makee noisee, "na na na nah na" awr same pitch, Zu not have music.
|
| "You know, it would be sufficient to really
| understand the electron"
| --Albert Einstein
| and you may have missed this:
| http://physics.nist.gov/cuu/Images/alphaeq.gif
| http://physics.nist.gov/cuu/Constants/alpha.html
|
"The hardest thing in the world to understand
is the income tax." -- Albert Einstein
|
| BTW... don't forget to look at:
| http://physics.nist.gov/cuu/Images/alphaeq.gif
| http://physics.nist.gov/cuu/Constants/alpha.html
|
|
| Sue...
| PS.
| http://physics.nist.gov/cuu/Images/alphaeq.gif
| http://physics.nist.gov/cuu/Constants/alpha.html
|
Kong-Fu-see-us no rooky and no rookee.
Zu rookee brue sky, gleen glass, led poppy, yerrow sand, maybe catchee
Frolida snook. Catchee snook with glasshopper on hook, eh, glasshopper?
Catchee inner peace too.
No catchee snook with gov, catchee snooker tax.
Androcles.
.

User: "Sue..."
Title: Re: What is the "reason" C is constant to all observers?	28 Sep 2005 01:42:17 PM

Androcles wrote:

The photo is a fake.
No one has ever seen a snook. They are too wiley and too
mean if you do get one a hook. ;-)

You roll snooker ball on tabrle, then you care how fast.
You catchee snook, snook wiggle like mass on spring.

Snook wiggle like tasmanian devil on spider silk.

| You should find the weight and tension determines how fast a
| purturbation propagates across the room.
|
You no clap one hand makee noisee, need two hand.
Here is two hand makee noisee in snookered eyeball.
http://campus.umr.edu/physics/alab/polar/circular_wave.gif

Yes .... a fair representation of the E plane potential, assuming
the length of the arrows corresponds to intensity.

One hand clap sideways, other hand clap up-down makee plopagation
noisee.

Nope. This way:
http://web.mit.edu/8.02t/www/802TEAL3D/visualizations/light/EBlight/EBlight=
..htm
Look at Two Dimensional Waves and Radiation Pattern of a Quarter Wave
Antenna (interactive) and you can imagine why the CP is hard to
illustrate for both E and H planes.
http://web.mit.edu/8.02t/www/802TEAL3D/visualizations/light/index.htm

| I'll repost:
|
| When light (electromagnetism) interacts with matter,
| whether the matter is emitting, receiving or incidental
| to the path, charges in the matter have to move
| in response to the Coulomb component.
|
| Atoms in a piano string have to move if you want
| a wave to oscillate end to end.
| If you want a wave to propagate faster down a
| string or rope, you use a lighter rope or remove
| the winding on the big fat piano strings.
|
|
| If you want light to propagte faster as it moves
| from charge to charge, then you use lighter charges.
|
|
| If you find that all charges weigh 0.511 MeV,

Kong-Fu-see-us say MeV not kiloglams, Zu talkee funny, not makee pi-anny
righter pray music. Makee pi-anny hot.

| then you are stuck with this:
|
|
| http://physics.nist.gov/cuu/Images/alphaeq.gif
| http://physics.nist.gov/cuu/Constants/alpha.html

Zu think awr corour same red brue gleen, like note on pi-anny awr same?
Makee noisee, "na na na nah na" awr same pitch, Zu not have music.

|
| "You know, it would be sufficient to really
| understand the electron"
| --Albert Einstein

| and you may have missed this:
| http://physics.nist.gov/cuu/Images/alphaeq.gif
| http://physics.nist.gov/cuu/Constants/alpha.html
|
"The hardest thing in the world to understand
is the income tax." -- Albert Einstein

Then stop voting for the labour party and ask the Queen to
adopt you.

|
| BTW... don't forget to look at:
| http://physics.nist.gov/cuu/Images/alphaeq.gif
| http://physics.nist.gov/cuu/Constants/alpha.html
|
|
| Sue...
| PS.
| http://physics.nist.gov/cuu/Images/alphaeq.gif
| http://physics.nist.gov/cuu/Constants/alpha.html
|

Kong-Fu-see-us no rooky and no rookee.
Zu rookee brue sky, gleen glass, led poppy, yerrow sand, maybe catchee
Frolida snook. Catchee snook with glasshopper on hook, eh, glasshopper?
Catchee inner peace too.

No catchee snook with gov, catchee snooker tax.

Mucho dinero! No habla english espanol senora.
? =BF? =BF? =BF
http://www.google.com/search?hl=3Den&lr=3D&safe=3Doff&q=3Dflorida+creel+lim=
it+snook&btnG=3DSearch
Sue...

Androcles.

User: "Androcles Androcles@ MyPlace.org"
Title: Re: What is the "reason" C is constant to all observers?	28 Sep 2005 03:10:52 PM

"Sue..." <suzysewnshow@yahoo.com.au> wrote in message
news:1127932937.122835.138550@g44g2000cwa.googlegroups.com...
Androcles wrote:

| The photo is a fake.
| No one has ever seen a snook. They are too wiley and too
| mean if you do get one a hook. ;-)
Wiggly wriggly...like you dodging the electric field made by the moving
magnet
flux.

You roll snooker ball on tabrle, then you care how fast.
You catchee snook, snook wiggle like mass on spring.

| Snook wiggle like tasmanian devil on spider silk.
Or Soo the snook.

| Yes .... a fair representation of the E plane potential, assuming
| the length of the arrows corresponds to intensity.
Length of arrows is constant. cos^2+sin^2 = 1. Energy is conserved.
It flips from moving mass to compressed/stretched spring and back again,
from E-field to B-field and back again.

One hand clap sideways, other hand clap up-down makee plopagation
noisee.

Nope. This way:
http://web.mit.edu/8.02t/www/802TEAL3D/visualizations/light/EBlight/EBlight.htm
Definitely and emphatically NOT. Energy zero, created from nothing?
Sorry, you can NOT have a null B-field and a null E-field at the same
time, which is what that diagram shows.
Where's the spring? Where's the moving mass? If the spring is not
stretched or compressed, how is it going to reverse the momentum of the
mass?
http://www.kettering.edu/~drussell/Demos/SHO/damp.html
Aether was the "spring" and "mass" of the old clockwork models that
Maxwell thought in terms of, and a very bad analogy. Nor can you have
fairy dust to make it work either. What works for sound doesn't work for
light or radio, they are different
models even though both are sine waves.
Look at Two Dimensional Waves and Radiation Pattern of a Quarter Wave
Antenna (interactive) and you can imagine why the CP is hard to
illustrate for both E and H planes.
http://web.mit.edu/8.02t/www/802TEAL3D/visualizations/light/index.htm
READ CAREFULLY and acknowledge:
Look, all I want is for you to accept is a 90 degree phase shift between
the E and B fields. That's all that is necessary and sufficent, you can
have all the other diagrams, I'm not arguing about them. Once you do,
the complex plane with "real" and "imaginary" numbers becomes a simple
way to describe AC electrodynamics. Math is a tool, that's all. It's
shorthand. I've assisted in building analogue computers for Concorde
with it, and they worked for 30 years. All that is required is that we
speak the same language.
If you can just accept that the old money that sends kids to MIT and
Princeton
are not the best brains in the world of physics... Didn't GW Bush go to
MIT?
Nah...Yale. same difference, old money. Is he an engineer physicist? Is
he even a lowly can't do anything physicist like Roberts or Andersen? Of
course not. He's a loud mouth.

| I'll repost:
|
| When light (electromagnetism) interacts with matter,
| whether the matter is emitting, receiving or incidental
| to the path, charges in the matter have to move
| in response to the Coulomb component.
|
| Atoms in a piano string have to move if you want
| a wave to oscillate end to end.
| If you want a wave to propagate faster down a
| string or rope, you use a lighter rope or remove
| the winding on the big fat piano strings.
|
|
| If you want light to propagte faster as it moves
| from charge to charge, then you use lighter charges.
|
|
| If you find that all charges weigh 0.511 MeV,

Kong-Fu-see-us say MeV not kiloglams, Zu talkee funny, not makee
pi-anny
righter pray music. Makee pi-anny hot.

| then you are stuck with this:
|
|
| http://physics.nist.gov/cuu/Images/alphaeq.gif
| http://physics.nist.gov/cuu/Constants/alpha.html

Zu think awr corour same red brue gleen, like note on pi-anny awr
same?
Makee noisee, "na na na nah na" awr same pitch, Zu not have music.

|
| "You know, it would be sufficient to really
| understand the electron"
| --Albert Einstein

| and you may have missed this:
| http://physics.nist.gov/cuu/Images/alphaeq.gif
| http://physics.nist.gov/cuu/Constants/alpha.html
|
"The hardest thing in the world to understand
is the income tax." -- Albert Einstein

Then stop voting for the labour party and ask the Queen to
adopt you.
I voted AGAINST the labour party, the queen won't even invite me for
afternoon tea and bickies, and I DID wave back.
http://www.thoroughbrednews.co.nz/imgmgr/imagefiles/star_craft_head4-279118.jpg

|
| BTW... don't forget to look at:
| http://physics.nist.gov/cuu/Images/alphaeq.gif
| http://physics.nist.gov/cuu/Constants/alpha.html
|
|
| Sue...
| PS.
| http://physics.nist.gov/cuu/Images/alphaeq.gif
| http://physics.nist.gov/cuu/Constants/alpha.html
|

Kong-Fu-see-us no rooky and no rookee.
Zu rookee brue sky, gleen glass, led poppy, yerrow sand, maybe catchee
Frolida snook. Catchee snook with glasshopper on hook, eh,
glasshopper?
Catchee inner peace too.

No catchee snook with gov, catchee snooker tax.

Mucho dinero! No habla english espanol senora.
? �? �? �
http://www.google.com/search?hl=en&lr=&safe=off&q=florida+creel+limit+snook&btnG=Search
Sue...
http://fwie.fw.vt.edu/WWW/macsis/lists/M010008.htm
(guvmint snooker tax)

Androcles.

User: "brian a m stuckless"
Title: Re: What is the "reason" C is constant to all observers?	28 Sep 2005 02:57:38 PM

A REDUCED MEDiUM, "in vacuum" actually refers to, THERE.
Find WHAT the PRESSURE and/or PARTiCLE COUNT was, Sue.
VERY sincerely u c,
```Brian

<> >><> >><> >><> >><>

Sue... wrote:

tony fleming wrote:

sue thanks, i understand what you are saying. i'm not asking why does
it go at 2.997 x 10^8 m/s but why constant??

Because the mass of an electron is constant.

<<why not a function of distance,>>
It isn't constant over distance.
That is why the Hubble law.

whatever, why not a variation of .0001%?

Like this

========================================
speed of light in vacuum

insert ..see top.

Value 299 792 458 m s-1
Standard uncertainty (exact)
Relative standard uncertainty (exact)
Concise form 299 792 458 m s-1

http://physics.nist.gov/cgi-bin/cuu/Value?c
=========================================
natural unit of length
Value 386.159 2678 x 10-15 m
Standard uncertainty 0.000 0026 x 10-15 m
Relative standard uncertainty 6.7 x 10-9

Assumed ( i.e. "PEGGED" ) unMEASURED data.!!

Concise form 386.159 2678(26) x 10-15 m

Assumed ( i.e. "PEGGED" ) unMEASURED data.!!

http://physics.nist.gov/cgi-bin/cuu/Value?tecomwlbar|search_for=length
=========================================

no it's a constant,

Assumed ( i.e. "PEGGED" ) unMEASURED data.!!

WHY???? (btw i think i have already told you the answer recently)

Sorry...
I forgot to make a note of it.

Show LEAST-Squares-Adjustment for ANY MEASURED c.
(For any measured EVER in history). Show that, Sue.
Don't FORGET to verify what the PARTiCLE COUNT was.
REALLY, appreciate your posts.
Thank you SO much. ```Brian

Sue...

User: "Sue..."
Title: Re: What is the "reason" C is constant to all observers?	28 Sep 2005 04:58:55 PM

brian a m stuckless wrote:

A REDUCED MEDiUM, "in vacuum" actually refers to, THERE.
Find WHAT the PRESSURE and/or PARTiCLE COUNT was, Sue.
VERY sincerely u c,
```Brian

<> >><> >><> >><> >><>

Sue... wrote:

tony fleming wrote:

sue thanks, i understand what you are saying. i'm not asking why does
it go at 2.997 x 10^8 m/s but why constant??

insert ..see top.

Assumed ( i.e. "PEGGED" ) unMEASURED data.!!

Concise form 386.159 2678(26) x 10-15 m

Assumed ( i.e. "PEGGED" ) unMEASURED data.!!

http://physics.nist.gov/cgi-bin/cuu/Value?tecomwlbar|search_for=length
=========================================

no it's a constant,

Assumed ( i.e. "PEGGED" ) unMEASURED data.!!

WHY???? (btw i think i have already told you the answer recently)

Sorry...
I forgot to make a note of it.

Show LEAST-Squares-Adjustment for ANY MEASURED c.
(For any measured EVER in history). Show that, Sue.

Don't FORGET to verify what the PARTiCLE COUNT was.

REALLY, appreciate your posts.
Thank you SO much. ```Brian

Ya Welcomed :o)
http://nvl.nist.gov/pub/nistpubs/sp958-lide/html/191-193.html
http://www.wisegeek.com/how-can-i-become-a-better-writer.htm?referrer=adwords_campaign=betterwriter_ad=900609&_search_kw=writing
:o)
Sue...

Sue...

User: "brian a m stuckless"
Title: Re: What is the "reason" C is constant to all observers?	29 Sep 2005 06:00:03 AM

The TOPiC's NOT about "ABSORBED" photon VELOCiTY, Sue.!!!!
SHOW least-squares adjustment #s for MEASURED velocity c.!
SHOW the MEASURED particle counts of those REDUCED media.!
Or SHOW the MEASURED reduced media PRESSURE during TESTs.!
Your references were COMPLETE DUDDs in that respect, Sue.!
Thanks for the writing tips ..that we can all use.
VERY sincerely u seem to TRY to c,
```Brian

<> >><> >><> >><> >><>

Sue... wrote:

brian a m stuckless wrote:

A REDUCED MEDiUM, "in vacuum" actually refers to, THERE.
Find WHAT the PRESSURE and/or PARTiCLE COUNT was, Sue.
VERY sincerely u c,
```Brian

<> >><> >><> >><> >><>

Sue... wrote:

tony fleming wrote:

sue thanks, i understand what you are saying. i'm not asking why does
it go at 2.997 x 10^8 m/s but why constant??

insert ..see top.

Assumed ( i.e. "PEGGED" ) unMEASURED data.!!

Concise form 386.159 2678(26) x 10-15 m

Assumed ( i.e. "PEGGED" ) unMEASURED data.!!

http://physics.nist.gov/cgi-bin/cuu/Value?tecomwlbar|search_for=length
=========================================

no it's a constant,

Assumed ( i.e. "PEGGED" ) unMEASURED data.!!

WHY???? (btw i think i have already told you the answer recently)

Sorry...
I forgot to make a note of it.

Show LEAST-Squares-Adjustment for ANY MEASURED c.
(For any measured EVER in history). Show that, Sue.

Don't FORGET to verify what the PARTiCLE COUNT was.

Don't FORGET to verify what the PARTiCLE COUNT was, Sue.

REALLY, appreciate your posts.
Thank you SO much. ```Brian

Sue...

User: "sue jahn suzysewnshow"
Title: Re: What is the "reason" C is constant to all observers?	25 Sep 2005 11:40:42 AM

On Sun, 25 Sep 2005 02:47:53 -0400, Androcles MyPlace.org> <<Androcles>
wrote:

"Sue..." <suzysewnshow@yahoo.com.au> wrote in message
news:1127622628.671755.192770@g43g2000cwa.googlegroups.com...
|
| George Hammond wrote:
| > Einstein's Special Relativity is famous for AD HOC
| > postulating that:
| >
| > "All observers regardless of their relative
| > (uniform) velocities will measure the same
| > speed of light, C."
| >
| > This is of course highly COUNTER INTUITIVE (the
| > Galilean transformation is more intuitive).
| >
| > ODDLY ENOUGH, Einstein never gave a physical
| > reason or explanation of WHY C should be the same for
| > all observers regardless of their speed... he simply
| > AD HOC POSTULATED the fact (which has since
| > been experimentally confirmed a million times).
| >
| > PLEASE CORRECT ME IF I AM WRONG:
| > But as far as I know, to this day, there is STILL NO
| > "logical theoretical reason"
| > of WHY C should appear to be the SAME to all
| > moving observers!
|
| Then you have been sleeping for about 50 years:

And you are still asleep and dreaming.
George is correct, there is no logical theoretical reason why
the speed of light should be the same for all observers.

If the observers are comprised of matter there is.
http://www.conformity.com/0102reflections.html
http://physics.nist.gov/cuu/Constants/alpha.html
Sue...

| Near and Far Fields - From Statics to Radiation... operating
frequency,
| measurement distance, and wave impedance. Wave impedance is the ratio
| of the electric field magnitude, E, to that of the magnetic
| http://www.conformity.com/0102reflections.html
| Look at the *MEASURED* E/H near far field ratios.
| also:
|
http://www.google.com/search?hl=en&q=%22wave+impedance%22&btnG=Google+Search

That's not a logical theoretical reason why
the speed of light should be the same for all observers.
Androcles.

--
Using Opera's revolutionary e-mail client: http://www.opera.com/mail/
.

User: "Androcles Androcles@ MyPlace.org"
Title: Re: What is the "reason" C is constant to all observers?	25 Sep 2005 12:29:20 PM

"sue jahn" <suzysewnshow> wrote in message
news:op.sxn278fk1v3b7x@maxwell.dbknest.net...
| On Sun, 25 Sep 2005 02:47:53 -0400, Androcles MyPlace.org>
<<Androcles>
| wrote:
|
| >
| > "Sue..." <suzysewnshow@yahoo.com.au> wrote in message
| > news:1127622628.671755.192770@g43g2000cwa.googlegroups.com...
| > |
| > | George Hammond wrote:
| > | > Einstein's Special Relativity is famous for AD HOC
| > | > postulating that:
| > | >
| > | > "All observers regardless of their relative
| > | > (uniform) velocities will measure the same
| > | > speed of light, C."
| > | >
| > | > This is of course highly COUNTER INTUITIVE (the
| > | > Galilean transformation is more intuitive).
| > | >
| > | > ODDLY ENOUGH, Einstein never gave a physical
| > | > reason or explanation of WHY C should be the same for
| > | > all observers regardless of their speed... he simply
| > | > AD HOC POSTULATED the fact (which has since
| > | > been experimentally confirmed a million times).
| > | >
| > | > PLEASE CORRECT ME IF I AM WRONG:
| > | > But as far as I know, to this day, there is STILL NO
| > | > "logical theoretical reason"
| > | > of WHY C should appear to be the SAME to all
| > | > moving observers!
| > |
| > | Then you have been sleeping for about 50 years:
| >
| > And you are still asleep and dreaming.
| > George is correct, there is no logical theoretical reason why
| > the speed of light should be the same for all observers.
|
| If the observers are comprised of matter there is.
And you are still asleep and dreaming.
George is correct, there is no logical theoretical reason why
the speed of light should be the same for all observers.
Androcles
.

User: "Sue..."
Title: Re: What is the "reason" C is constant to all observers?	25 Sep 2005 01:29:14 PM

Androcles wrote:

"sue jahn" <suzysewnshow> wrote in message
news:op.sxn278fk1v3b7x@maxwell.dbknest.net...
| On Sun, 25 Sep 2005 02:47:53 -0400, Androcles MyPlace.org>
<<Androcles>
| wrote:
|
| >
| > "Sue..." <suzysewnshow@yahoo.com.au> wrote in message
| > news:1127622628.671755.192770@g43g2000cwa.googlegroups.com...
| > |
| > | George Hammond wrote:
| > | > Einstein's Special Relativity is famous for AD HOC
| > | > postulating that:
| > | >
| > | > "All observers regardless of their relative
| > | > (uniform) velocities will measure the same
| > | > speed of light, C."
| > | >
| > | > This is of course highly COUNTER INTUITIVE (the
| > | > Galilean transformation is more intuitive).
| > | >
| > | > ODDLY ENOUGH, Einstein never gave a physical
| > | > reason or explanation of WHY C should be the same for
| > | > all observers regardless of their speed... he simply
| > | > AD HOC POSTULATED the fact (which has since
| > | > been experimentally confirmed a million times).
| > | >
| > | > PLEASE CORRECT ME IF I AM WRONG:
| > | > But as far as I know, to this day, there is STILL NO
| > | > "logical theoretical reason"
| > | > of WHY C should appear to be the SAME to all
| > | > moving observers!
| > |
| > | Then you have been sleeping for about 50 years:
| >
| > And you are still asleep and dreaming.
| > George is correct, there is no logical theoretical reason why
| > the speed of light should be the same for all observers.
|
| If the observers are comprised of matter there is.

And you are still asleep and dreaming.
George is correct, there is no logical theoretical reason why
the speed of light should be the same for all observers.

Then there must be reason to dispute this relation:
http://physics.nist.gov/cuu/Images/alphaeq.gif
<< where e is the elementary charge, =3D h/2 where h is the Planck
constant, =3D 1/=B50c2 is the electric constant (permitivity of vacuum)
and =B50 is the magnetic constant (permeability of vacuum). In the
International System of Units (SI), c, , and =B50 are exactly known
constants. >>
http://physics.nist.gov/cuu/Constants/alpha.html
..=2E. or reason for the constants changing.
Sue...

=20
Androcles

User: "Androcles Androcles@ MyPlace.org"
Title: Re: What is the "reason" C is constant to all observers?	25 Sep 2005 01:43:58 PM

"Sue..." <suzysewnshow@yahoo.com.au> wrote in message
news:1127672953.999515.30150@g47g2000cwa.googlegroups.com...
Androcles wrote:

"sue jahn" <suzysewnshow> wrote in message
news:op.sxn278fk1v3b7x@maxwell.dbknest.net...
| On Sun, 25 Sep 2005 02:47:53 -0400, Androcles MyPlace.org>
<<Androcles>
| wrote:
|
| >
| > "Sue..." <suzysewnshow@yahoo.com.au> wrote in message
| > news:1127622628.671755.192770@g43g2000cwa.googlegroups.com...
| > |
| > | George Hammond wrote:
| > | > Einstein's Special Relativity is famous for AD HOC
| > | > postulating that:
| > | >
| > | > "All observers regardless of their relative
| > | > (uniform) velocities will measure the same
| > | > speed of light, C."
| > | >
| > | > This is of course highly COUNTER INTUITIVE (the
| > | > Galilean transformation is more intuitive).
| > | >
| > | > ODDLY ENOUGH, Einstein never gave a physical
| > | > reason or explanation of WHY C should be the same for
| > | > all observers regardless of their speed... he simply
| > | > AD HOC POSTULATED the fact (which has since
| > | > been experimentally confirmed a million times).
| > | >
| > | > PLEASE CORRECT ME IF I AM WRONG:
| > | > But as far as I know, to this day, there is STILL NO
| > | > "logical theoretical reason"
| > | > of WHY C should appear to be the SAME to all
| > | > moving observers!
| > |
| > | Then you have been sleeping for about 50 years:
| >
| > And you are still asleep and dreaming.
| > George is correct, there is no logical theoretical reason why
| > the speed of light should be the same for all observers.
|
| If the observers are comprised of matter there is.

And you are still asleep and dreaming.
George is correct, there is no logical theoretical reason why
the speed of light should be the same for all observers.

| Then there must be reason to dispute this relation:
In case you haven't noticed, I'm simply denying
anything you say without waiting for the explanation.
I learned that from you.
Androcles
.

User: "Sue..."
Title: Re: What is the "reason" C is constant to all observers?	25 Sep 2005 02:14:48 PM

Androcles wrote:

"Sue..." <suzysewnshow@yahoo.com.au> wrote in message
news:1127672953.999515.30150@g47g2000cwa.googlegroups.com...

Androcles wrote:

"sue jahn" <suzysewnshow> wrote in message
news:op.sxn278fk1v3b7x@maxwell.dbknest.net...
| On Sun, 25 Sep 2005 02:47:53 -0400, Androcles MyPlace.org>
<<Androcles>
| wrote:
|
| >
| > "Sue..." <suzysewnshow@yahoo.com.au> wrote in message
| > news:1127622628.671755.192770@g43g2000cwa.googlegroups.com...
| > |
| > | George Hammond wrote:
| > | > Einstein's Special Relativity is famous for AD HOC
| > | > postulating that:
| > | >
| > | > "All observers regardless of their relative
| > | > (uniform) velocities will measure the same
| > | > speed of light, C."
| > | >
| > | > This is of course highly COUNTER INTUITIVE (the
| > | > Galilean transformation is more intuitive).
| > | >
| > | > ODDLY ENOUGH, Einstein never gave a physical
| > | > reason or explanation of WHY C should be the same for
| > | > all observers regardless of their speed... he simply
| > | > AD HOC POSTULATED the fact (which has since
| > | > been experimentally confirmed a million times).
| > | >
| > | > PLEASE CORRECT ME IF I AM WRONG:
| > | > But as far as I know, to this day, there is STILL NO
| > | > "logical theoretical reason"
| > | > of WHY C should appear to be the SAME to all
| > | > moving observers!
| > |
| > | Then you have been sleeping for about 50 years:
| >
| > And you are still asleep and dreaming.
| > George is correct, there is no logical theoretical reason why
| > the speed of light should be the same for all observers.
|
| If the observers are comprised of matter there is.

And you are still asleep and dreaming.
George is correct, there is no logical theoretical reason why
the speed of light should be the same for all observers.

| Then there must be reason to dispute this relation:

In case you haven't noticed, I'm simply denying
anything you say without waiting for the explanation.
I learned that from you.

There is quiet some difference in denying what I say
and denying what a standards laboratory says.
<<Re: What is the "reason" C is constant to all observers? >>
http://physics.nist.gov/cuu/Images/alphaeq.gif
You'll forgive me, I trust, for not following along with
political, metaphysical or religious arguments why that
should not be the case.
=======================
The essential elements of a scientific method are iterations and
recursions of the following four steps:
1 Characterization (Quantification, observation and measurement)
2 Hypothesis (a theoretical, hypothetical explanation of the
observations and measurements)
3 Prediction (logical deduction from the hypothesis)
4 Experiment (test of all of the above)
http://en.wikipedia.org/wiki/Scientific_method
=======================
Empirical evidence:
http://www.google.com/search?hl=en&safe=off&q=+electromagnetic+reciprocal+principle&spell=1
Sue...

Androcles

User: "Androcles Androcles@ MyPlace.org"
Title: Re: What is the "reason" C is constant to all observers?	25 Sep 2005 02:56:45 PM

"Sue..." <suzysewnshow@yahoo.com.au> wrote in message
news:1127675688.302174.273870@g47g2000cwa.googlegroups.com...
|
| Androcles wrote:
| > "Sue..." <suzysewnshow@yahoo.com.au> wrote in message
| > news:1127672953.999515.30150@g47g2000cwa.googlegroups.com...
| >
| > Androcles wrote:
| > > "sue jahn" <suzysewnshow> wrote in message
| > > news:op.sxn278fk1v3b7x@maxwell.dbknest.net...
| > > | On Sun, 25 Sep 2005 02:47:53 -0400, Androcles MyPlace.org>
| > > <<Androcles>
| > > | wrote:
| > > |
| > > | >
| > > | > "Sue..." <suzysewnshow@yahoo.com.au> wrote in message
| > > | > news:1127622628.671755.192770@g43g2000cwa.googlegroups.com...
| > > | > |
| > > | > | George Hammond wrote:
| > > | > | > Einstein's Special Relativity is famous for AD HOC
| > > | > | > postulating that:
| > > | > | >
| > > | > | > "All observers regardless of their relative
| > > | > | > (uniform) velocities will measure the same
| > > | > | > speed of light, C."
| > > | > | >
| > > | > | > This is of course highly COUNTER INTUITIVE (the
| > > | > | > Galilean transformation is more intuitive).
| > > | > | >
| > > | > | > ODDLY ENOUGH, Einstein never gave a physical
| > > | > | > reason or explanation of WHY C should be the same for
| > > | > | > all observers regardless of their speed... he simply
| > > | > | > AD HOC POSTULATED the fact (which has since
| > > | > | > been experimentally confirmed a million times).
| > > | > | >
| > > | > | > PLEASE CORRECT ME IF I AM WRONG:
| > > | > | > But as far as I know, to this day, there is STILL NO
| > > | > | > "logical theoretical reason"
| > > | > | > of WHY C should appear to be the SAME to all
| > > | > | > moving observers!
| > > | > |
| > > | > | Then you have been sleeping for about 50 years:
| > > | >
| > > | > And you are still asleep and dreaming.
| > > | > George is correct, there is no logical theoretical reason why
| > > | > the speed of light should be the same for all observers.
| > > |
| > > | If the observers are comprised of matter there is.
| > >
| > > And you are still asleep and dreaming.
| > > George is correct, there is no logical theoretical reason why
| > > the speed of light should be the same for all observers.
| >
| > | Then there must be reason to dispute this relation:
| >
| > In case you haven't noticed, I'm simply denying
| > anything you say without waiting for the explanation.
| > I learned that from you.
|
| There is quiet some difference in denying what I say
| and denying what a standards laboratory says.
You are wrong.
Androcles
.

User: ""
Title: Re: What is the "reason" C is constant to all observers?	06 Sep 2005 06:20:53 AM

Light is the specific frequency-spectrum that the eye reacts on and the
brain interprets as light.
When the observer moves relative to the source (c-v) his light change
wavelength-spectrum but the frequency-spectrum is invariant.
The speed of light is constant relative to the source. But the emitted
radiation's waves are increasing in length and velocity caused of the
entropy-effect that forces the radiation,s energy to equilibrium. So,
every individual wave moves faster than light and increases
accelerating proportional to the increasing wavelengths.
So, the big-bang's expansion is a misinterpretation of the accelerating
radiation's redshift.
Ingvar Astrand, Sweden
http://www.theuniphysics.info
.

User: "Androcles Androcles@ MyPlace.org"
Title: Re: What is the "reason" C is constant to all observers?	06 Sep 2005 06:52:01 AM

<ingvar_astrand@yahoo.se> wrote in message
news:1126005653.523348.98890@g43g2000cwa.googlegroups.com...
| Light is the specific frequency-spectrum that the eye reacts on and
the
| brain interprets as light.
| When the observer moves relative to the source (c-v) his light change
| wavelength-spectrum but the frequency-spectrum is invariant.
Ok up to there.
| The speed of light is constant relative to the source. But the emitted
| radiation's waves are increasing in length and velocity caused of the
| entropy-effect that forces the radiation,s energy to equilibrium.
Now you are guessing.
Androcles
So,
| every individual wave moves faster than light and increases
| accelerating proportional to the increasing wavelengths.
| So, the big-bang's expansion is a misinterpretation of the
accelerating
| radiation's redshift.
|
| Ingvar Astrand, Sweden
| http://www.theuniphysics.info
|
.

User: "Hatunen"
Title: Re: What is the "reason" C is constant to all observers?	07 Sep 2005 12:43:54 PM

On 6 Sep 2005 04:20:53 -0700,

wrote:

In science, "light" is the entire electrmagnetic spectrum but
this being rather sweeping, scientists will call different parts
of the spectrum things like "x-rays" and "gamm rays" and radio
waves". But the term "light" does include the near and far
infrared and ultraviolet, which the brain never interprets as
"light".

The speed of light is constant relative to the source. But the emitted
radiation's waves are increasing in length and velocity caused of the
entropy-effect that forces the radiation,s energy to equilibrium. So,
every individual wave moves faster than light and increases
accelerating proportional to the increasing wavelengths.

That is so impressive sounding and so wrong.

So, the big-bang's expansion is a misinterpretation of the accelerating
radiation's redshift.

Ah. And you hvae an agenda for not having the red-shift be due to
expansion of the universe?
************* DAVE HATUNEN (hatunen@cox.net) *************
* Tucson Arizona, out where the cacti grow *
* My typos & mispellings are intentional copyright traps *
.

User: ""
Title: Re: What is the "reason" C is constant to all observers?	19 Sep 2005 11:00:50 AM

The standard reasoning is given at:
http://en.wikipedia.org/wiki/Special_relativity_for_beginners
Why does Pythagoras' theorem apply to 3D space?
Best Wishes
Alex Green
.

User: "Hatunen"
Title: Re: What is the "reason" C is constant to all observers?	19 Sep 2005 05:19:23 PM

On 19 Sep 2005 09:00:50 -0700,

wrote:

The standard reasoning is given at:

http://en.wikipedia.org/wiki/Special_relativity_for_beginners

Why does Pythagoras' theorem apply to 3D space?

Maybe I don't understand the question, but it's a fundamental of
three dimensional calculations that d^2 = a^2 + b^2 + c^2,
usually expressed in Cartesian form as R^2 = X^2 + Y^2 + Z^2
************* DAVE HATUNEN (hatunen@cox.net) *************
* Tucson Arizona, out where the cacti grow *
* My typos & mispellings are intentional copyright traps *
.

User: "Bilge"
Title: Re: What is the "reason" C is constant to all observers?	21 Sep 2005 02:34:28 AM

Hatunen:

On 19 Sep 2005 09:00:50 -0700,

wrote:

The standard reasoning is given at:

http://en.wikipedia.org/wiki/Special_relativity_for_beginners

Why does Pythagoras' theorem apply to 3D space?

Maybe I don't understand the question, but it's a fundamental of
three dimensional calculations that d^2 = a^2 + b^2 + c^2,

More precisely, that is what you get if you assume euclid's postulates.
The length of an arc connecting two points, could be anything.

usually expressed in Cartesian form as R^2 = X^2 + Y^2 + Z^2

The geometric distance between the origin and the point (x, y, z)
is given by the sqare root of the scalar product of the vector
from the origin to (x, y, z). The scalar product of two vectors
is defined by,
A.B = g_ij A^i B^j
For the vector dr = (dx, dy, dz), then the expresion is,
dr^2 = g_11 dx^2 + g_22 dy^2 + g_33 dz^2
+ (g_12 + g_21) dxdy + (g_13 + g_31)dxdz + (g_23 + g_32)dydz
Euclid's postulates and a cartesian basis imply,
g_11 = 1, g_22 = 1, g_33 = 1, and g_ij = 0, i != j.

Which defines what we normally call a straight line. But, we could choose
to define a different geometry, by not insisting on euclid's postulates.
We would still arrive at a ``pythagorean theorem,'' since the pythagorean
theorem is a measurement of distance and need not be restricted to
any particular geometry. In four dimensions, you could have a euclidean
geometry and the vector dr
dr^2 = g_ij dx^i dx^j
with i,j = (0,1,2,3) and g_ij = 1, for i=j, and 0 for i != j.
The geometry in relativity is given by,
g_00 = -1 g_11 = g_22 = g_33 = 1, g_ij = 0 for i != j
Then the length of a line is,
dr^2 = -(dx^0)^2 + (dx^1)^2 + (dx^2)^2 + (dx^3)^2
Since we are measuring distances in all four directions using the
same units, there's no reason for even mentioning the quantity, `c'.
The only reason that `c' ever appears is because we typically use
different units for time and `c' converts meters to seconds.
We would have the same thing happen in the 3-d if we measured the
length of the z-axis in cm and the other axes in meters:
dr^2 = dx^2 + dy^2 + (0.1)^2 dz^2
b = 0.1 m/cm
There's no meaning to the constant b = 0.1m/cm. It's just the choice
to use different units. The same applies to relativity.
.

User: "brian a m stuckless"
Title: Re: What is the "reason" C is constant to all observers?	21 Sep 2005 06:41:22 AM

Permeability Uo & Permittivity Eo cannot BOTH be UNiTY ..at once,
in any COHERENT SYSTEM of STANDARD UNiTs.!! This is WHY the old
HEAVYSiDE/LORENTZ (double) STANDARDs is ONLY used "in-house" now
by ONLY the most dire GR-Tivity gtr GR Cracked-pots (crackedpots).
[ i.e. Like "PEGGiNG" various PLANCK units and c all to ONE (1).]

<> >><> >><> >><> >><>

Bilge wrote:

The length of an arc connecting two points, could be anything.

NOTE:
-[ This is NOT *TRUE* for all ADjACENT points, on ANY PATH.!! ]-
Sincerely c,
```Brian
p.s.
MOTiON happens *BETWEEN* ADjACENT POiNTs.!!
Also note..
ANGULAR momentum pA = mass*radius*velocity:
WHERE is the QUANTiTY for a CENTRAL MASS.?!
OBViOUSLY, pA is > QUANTiTATiVELY free <.!!
(As far as any CENTRAL mass is concerned.)!
Please EXPLAiN this CONTRADiCTiON, Bilge.!!

<> >><> >><> >><> >><>

Which defines what we normally call a straight line. But,
The only reason that `c' ever appears is because we typically
use different units for time and `c' converts meters to seconds.

Whereas YOU "PEG" c = 1; The subsequent 1 / Uo*Eo ..CANNOT = 1.!!
This,of course, means you are a spiFFY GR gtr Tivity Dimwit.!!

The same applies to relativity.

<> >><> >><> >><> >><>

User: "tadchem"
Title: Re: What is the "reason" C is constant to all observers?	20 Sep 2005 04:18:01 PM

"Hatunen" <hatuunen@cox.net> wrote in message
news:i7eui15e88ivn3os3sonnoe863t5v3o31q@4ax.com...
<snip>

Maybe I don't understand the question, but it's a fundamental of
three dimensional calculations that d^2 = a^2 + b^2 + c^2,
usually expressed in Cartesian form as R^2 = X^2 + Y^2 + Z^2

The Pythagorean Theorem (and its generalization) applies only in Euclidean
space of 2 or more dimensions.
It looks vastly different if one uses a different orthonormal coordinate
basis set.
If one introduces curvature to space (such as relativistic gravitation) then
the space is no longer Euclidean.
See for example the links here:
http://scienceworld.wolfram.com/physics/topics/Metrics.html

************* DAVE HATUNEN (hatunen@cox.net) *************
* Tucson Arizona, out where the cacti grow *

Heard of any thunderbirds out there lately?

* My typos & mispellings are intentional copyright traps *

Mine are honest errors...
"Only one thing is impossible for God: To find any sense in any copyright
law on the planet." - Mark Twain
Tom Davidson
Richmond, VA
.

User: ""
Title: Re: What is the "reason" C is constant to all observers?	20 Sep 2005 04:43:43 PM

In article <uvednSEqB6wj463eRVn-vA@comcast.com>, "tadchem" <tadchemNOSPAM@comcast.net> writes:

"Hatunen" <hatuunen@cox.net> wrote in message
news:i7eui15e88ivn3os3sonnoe863t5v3o31q@4ax.com...

<snip>

Maybe I don't understand the question, but it's a fundamental of
three dimensional calculations that d^2 = a^2 + b^2 + c^2,
usually expressed in Cartesian form as R^2 = X^2 + Y^2 + Z^2

The Pythagorean Theorem (and its generalization) applies only in Euclidean
space of 2 or more dimensions.

It looks vastly different if one uses a different orthonormal coordinate
basis set.

In Euclidean space? Thhere is nothing about the Pythagorean Theorem
that is coordinate dependent. Of course ...

If one introduces curvature to space (such as relativistic gravitation) then
the space is no longer Euclidean.

.... for sure. But that's separate from the issue of what coordinates
are used.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
.

User: "tadchem"
Title: Re: What is the "reason" C is constant to all observers?	21 Sep 2005 04:42:10 AM

<mmeron@cars3.uchicago.edu> wrote in message
news:jo%Xe.19$55.3084@news.uchicago.edu...

In article <uvednSEqB6wj463eRVn-vA@comcast.com>, "tadchem"

<tadchemNOSPAM@comcast.net> writes:

"Hatunen" <hatuunen@cox.net> wrote in message
news:i7eui15e88ivn3os3sonnoe863t5v3o31q@4ax.com...

<snip>

Maybe I don't understand the question, but it's a fundamental of
three dimensional calculations that d^2 = a^2 + b^2 + c^2,
usually expressed in Cartesian form as R^2 = X^2 + Y^2 + Z^2

The Pythagorean Theorem (and its generalization) applies only in

Euclidean

space of 2 or more dimensions.

It looks vastly different if one uses a different orthonormal coordinate
basis set.

In Euclidean space? Thhere is nothing about the Pythagorean Theorem
that is coordinate dependent. Of course ...

Think so? Try rewriting Pythagoras in spherical coordinates [r,theta,phi].
X = r*sin(theta)*cos(phi)
Y = r*sin(theta)*sin(phi)
Z = r*cos(theta)
What are the "X", "Y", and "Z" if not coordinates?

If one introduces curvature to space (such as relativistic gravitation)

then

the space is no longer Euclidean.

... for sure. But that's separate from the issue of what coordinates
are used.

Regardless of the coordinates chosen, Pythagoras fails in non-Euclidean
space.
Tom Davidson
Richmond, VA
.

User: "Bilge"
Title: Re: What is the "reason" C is constant to all observers?	21 Sep 2005 12:07:07 PM

tadchem:

Think so? Try rewriting Pythagoras in spherical coordinates [r,theta,phi].

OK. I'll do precisely that. Start with the pythagorean theore,
ds^2 = g_ij dx^i dx^j
The metric coefficients, g_ij for a cartesian basis are,
g_ij = 1, i = j, and zero otherwise.
Now, I'll assume that i,j = r,\theta,\phi and that e_r, e_\theta and
e_\phi are unit vectors along r, \theta and \phi, respectively with a
generic vector given by,
du_i = (dr, d\theta, d\phi)
What I have to do is calculate the metric coefficients, g_ij.

X = r*sin(theta)*cos(phi)
Y = r*sin(theta)*sin(phi)
Z = r*cos(theta)

What are the "X", "Y", and "Z" if not coordinates?

For the coordinates given,
dx/dr = sin(\theta) cos(\phi)
dy/dr = sin(\theta) sin(\phi)
dz/dr = cos(\theta)
with a unit vector along r given by,
e_r = (dx/dr)e_x + (dy/dr)e_y + (dz/dr)e_z
The coefficient g_rr for the induced metric is then <e_r|e_r>,
[sin(\theta) cos(\phi)]^2 <e_x|e_x>

+ [sin(\theta) sin(\phi)]^2 <e_y|e_y> + [cos(\theta)]^2 <e_z|e_z>
Since <e_i|e_j> = \delta_ij, for i,j = x,y,z,
g_rr = [sin(\theta) cos(\phi)]^2 + [sin(\theta) sin(\phi)]^2

+ [cos(\theta)]^2 <e_z|e_z>
= [sin(\theta)]^2 [(cos(\phi)^2 + sin(\phi)^2] + [cos(\theta)]^2
= 1
Similarly, for \theta,
dx/d(\theta) = r cos(\theta) cos(\phi)
dy/d(\theta) = r cos(\theta) sin(\phi)

dz/d(\theta) = -r sin(\theta)
e_\theta = (r cos(\theta) cos(\phi)) e_x

+ (r cos(\theta) sin(\phi)) e_y - r sin(\theta)
g_\theta\theta follows from <e_\theta|\e_\theta> = r^2
and for \phi
dx/d(\phi) = -r cos(\theta) sin(\phi)
dx/d(\phi) = r cos(\theta) cos(\phi)

dz/d(\phi) = 0
e_\phi = -r cos(\theta) sin(\phi) e_x + r cos(\theta) cos(\phi) e_y
and
g_\phi\phi = (r sin(\theta))^2
You can check to see that the rest of the coefficients like g_r\theta,
etc. are zero.
So, a vector du^i = (dr, d\theta, d\phi) has a length given by,

ds^2 = g_ij du^i du^j
= (du^r)^2 + r^2 (d^\theta)^2 + (r sin\theta)^2 (d\phi)^2
The coordinate chart r,\theta,\phi do not cover the sphere, since the
transformation is singular at the poles. However, you can construct
an atlas with 2 different charts that does cover the sphere.

Regardless of the coordinates chosen, Pythagoras fails in non-Euclidean
space.

That isn't true. The pythagorean theorem is nothing more than the
definition of a line element. Pythagoreas obviously did not consider
the generic definition of a line, but that doesn't prevent us from
recognizing that he imposed an artificial restriction due to a lack
of knowledge, which we have since acquired.
.

User: ""
Title: Re: What is the "reason" C is constant to all observers?	21 Sep 2005 10:24:57 AM

In article <U_OdnbLds6xJsKzeRVn-rg@comcast.com>, "tadchem" <tadchemNOSPAM@comcast.net> writes:

<mmeron@cars3.uchicago.edu> wrote in message
news:jo%Xe.19$55.3084@news.uchicago.edu...

In article <uvednSEqB6wj463eRVn-vA@comcast.com>, "tadchem"

<tadchemNOSPAM@comcast.net> writes:

"Hatunen" <hatuunen@cox.net> wrote in message
news:i7eui15e88ivn3os3sonnoe863t5v3o31q@4ax.com...

<snip>

Maybe I don't understand the question, but it's a fundamental of
three dimensional calculations that d^2 = a^2 + b^2 + c^2,
usually expressed in Cartesian form as R^2 = X^2 + Y^2 + Z^2

The Pythagorean Theorem (and its generalization) applies only in

Euclidean

space of 2 or more dimensions.

It looks vastly different if one uses a different orthonormal coordinate
basis set.

In Euclidean space? Thhere is nothing about the Pythagorean Theorem
that is coordinate dependent. Of course ...

Think so? Try rewriting Pythagoras in spherical coordinates [r,theta,phi].

X = r*sin(theta)*cos(phi)
Y = r*sin(theta)*sin(phi)
Z = r*cos(theta)

What are the "X", "Y", and "Z" if not coordinates?

Sigh. You're confused. I repeat, the Pythagorean theorem has
***nothing*** to do with coordinates. Find out a Euclidean geometry
textbook and read the exact formulation of the theorem.

If one introduces curvature to space (such as relativistic gravitation)

then

the space is no longer Euclidean.

... for sure. But that's separate from the issue of what coordinates
are used.

Regardless of the coordinates chosen, Pythagoras fails in non-Euclidean
space.

If by "fails" you mean "doesn't maintain its original form" then yes,
certainly. And, *regardless* of the coordinates chosen the
Pythagorean theorem stands, in its original form, in a Euclidean
space.
Parametrizing a Euclidean space using spherical coordinates does not
make it non-Euclidean.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
.

User: "tadchem"
Title: Re: What is the "reason" C is constant to all observers?	21 Sep 2005 12:18:06 PM

wrote:
<snp>
If you count carefully, I cited *TWO* instances in which the
Pythagorean Theorem does not take the form "A^2 + B^2 = C^2"
(regardless of the names you give the variables).
One is when non-Cartesian coordinates are used.
For example, X^2 + Y^2 + Z^2 in spherical coordinates becomes
R^2*sin^2(theta)*cos^2(phi) + R^2*sin^2(theta)*sin^2(phi) +
R^2*cos^2(theta). The *distances* are unchanged, but the *form* is
completely different.
The other is when non-Euclidean space is used.

Regardless of the coordinates chosen, Pythagoras fails in non-Euclidean
space.

You counter my statement regarding PT in *non-Euclidean* space with an
explicit reference to *Euclidean* space??? Can you stay focussed for a
sentence or two, please? Or do you have a factory somewhere that
manufactures straw men for you wholesale?
http://en.wikipedia.org/wiki/Pythagorean_theorem
The properties of the space are inextricable from any metric used to
determine distances within the space.

Parametrizing a Euclidean space using spherical coordinates does not
make it non-Euclidean.

http://en.wikipedia.org/wiki/Pythagorean_theorem#The_Pythagorean_theorem_in_non-Euclidean_geometry
Note particularly the forms here for spherical triangles (curvature >
0) and hyperbolic triangles (curvature < 0).
Tom Davidson
Richmond, VA
.

User: "Bilge"
Title: Re: What is the "reason" C is constant to all observers?	22 Sep 2005 07:05:38 PM

tadchem:

The other is when non-Euclidean space is used.

Ok, then essentially, you are excluding the pythagorean theorem
from physics. That pretty much exludes the possibility of discussing
any physics (even newtonian physics) with anyone other than physicists
or mathematicians. Frankly, I see no reason to tell someone that
he formally erred by employing an abstraction like a right triangle
to build a dog house. That sort of thing gives people the idea that
they aren't learning anything useful in most high school mathematics
courses.
[...]

OK, if you want to limit the scope of the pythagorean theorem to
the abstract concept of a right triangle in an abstract euclidean
space, how about providing a better term that allows me to connect
a measurement of distance to the length of a vector.
.

User: "brian a m stuckless"
Title: Re: What is the "reason" C is constant to all observers?	21 Sep 2005 01:18:49 PM

Dimwit HiGHLY DETAiLs answer to UNasked QUESTiON:
ANGULAR momentum pA = mass*radius*velocity
Q. WHERE is the QUANTiTY for a CENTRAL MASS here?
..can ANYone answer this, please.?!!
Sincerely c,
```Brian

<> >><> >><> >><> >><>

tadchem wrote:

mmeron@cars3.uchicago.edu wrote:

<snp>

If you count carefully, I cited *TWO* instances in which the
Pythagorean Theorem does not take the form "A^2 + B^2 = C^2"
(regardless of the names you give the variables).

One is when non-Cartesian coordinates are used.

For example, X^2 + Y^2 + Z^2 in spherical coordinates becomes
R^2*sin^2(theta)*cos^2(phi) + R^2*sin^2(theta)*sin^2(phi) +
R^2*cos^2(theta). The *distances* are unchanged, but the *form* is
completely different.

The other is when non-Euclidean space is used.

Regardless of the coordinates chosen, Pythagoras fails in non-Euclidean
space.

insert ..see top.!!

http://en.wikipedia.org/wiki/Pythagorean_theorem

The properties of the space are inextricable from any metric used to
determine distances within the space.

Parametrizing a Euclidean space using spherical coordinates does not
make it non-Euclidean.

http://en.wikipedia.org/wiki/Pythagorean_theorem#The_Pythagorean_theorem_in_non-Euclidean_geometry

Note particularly the forms here for spherical triangles (curvature >
0) and hyperbolic triangles (curvature < 0).

Tom Davidson
Richmond, VA

User: ""
Title: Re: What is the "reason" C is constant to all observers?	21 Sep 2005 01:32:48 PM

In article <1127323086.437884.146290@o13g2000cwo.googlegroups.com>, "tadchem" <thomas.davidson@dla.mil> writes:

Gee whiz. Look at at the momentum of point mass. We have p = mv, not
p^2 = m^2 + v^2. Obvious violation og the Pythagorean theorem, ain't
it?:-)
This is getting silly. The A, B and C in the Pythagorean theorem are
not just any "names of variables. They are *lenghts*. To be
specific, they're the lenghts of the three sides of a right angle
triangle. And the Pythagorean theorem says that the square of the length
of the side oppoite to the right angle (which we call C in the above)
is equal to the sum of the squares of the other two lengths. That's
all. Now, if for whatever perverse reasons I choose to introduce the
parametrization
A = rs^2 - t
B = rs^2 + t
C = 2t + sr^2
Then, written in terms of the parameters r, s, t the relationship
between A, B and C will become
2*r^2*s^4 + 2*t^2 = r^4*s^2 + 4*r^2*s*t + 4*t^2
Which is not at all in the form of the original Pythagorean theorem
but constitutes no violation suince r, s, t *are not* the lengths of
the sides of a right angle triangle, just some arbitrary parameters
which were used to express said lengths.
I repeat again, there are ***no*** coordinates in the Pythagorean
theorem and the relationship in the theorem is not between
coordinates. Don't confuse lengths and coordinates. Coordinates are
just "addresses". You employ some sufficient set of parameters such
that to every point in the space you're interested in there
corresponds a unique set of coordinate values. Very convenient for
many applications but, as I said, the coordinates do not have to have
anything to do with lengths.
When students are first introduced to analytical geometry, this is
done using the *simplest* coordinate system possible, where the
coordinates of a point are just the orthogonal projections of said
point on a set of orthogonal axes. In this case coordinates (or, more
exactly, coordinate differences) are indeed lengths/distances. So, if
you'll take (for simplicity) a 2D space, with two points with
orthogonal coordinates (x1,y1) and (x2,y2) then x2-x1 and y2-y1 are
indeed projections of the length of the line joining the two points on
the x and y axes. And in this, very special case, if you'll pick
three points being the vertices of a right angle triangle then
rperesenting the lenghts in the Pythagorean theorem by the coordinates
of the vertices yeilds a relationship closely resembling in form the
original relationship. But this is a special case, one specific
coordinate system out of an infinity of possible ones, and not at all
representative.
So, if you'll instead use polar coordiantes in this 2D plane, so that
the two points will have the "addresses" of (r1,phi1) and (r2,phi2)
then it is most certainly *not true* that either r2-r2 or phi2-phi1
represent any specific lengths or distances. Thus, the distance
between the two ponts will *not* be (r2-r1)^2 + (phi2-phi1)^2 and this
is not because the Pythagorean theorem is no longer valid (it very
much is) but becasue you're not plugging in the lenghts you should've.
To give another example, consider a city map (and not necessarily
amodern city built on a grid but just any city). Now, the coordinates
of a house are (street name, house number). Those are indeed
coordinates since given the address of the house, in the form of
(street name, number), you can locate the house. But this in no way
means that if you live on 184 Sycamore St. and your friend lives on
39 Lakefront Ave., then the distnace between your houses is given by
distance^2 = (184-39)^2 + (Sycamore St. - Lakefront Ave.)^2
And if you consider this one more ridiculous then the one above, it is
only because you attribute to coordinates a meaning beyond of wwhat's
there. And I doubt that you'll consider the above to be an instance
of a "failure of the Pythagorean theorem".
Bilge already provided the proper picture in a differential form.
I'll do something similar now in an non-differential form (the
differential one is more convenient application wise but perhaps this
one will make matters clearer.
Assume you've an n-dimensional space, with a completely arbitrary
coordinate system emploing tn paremeters s1,s2,...sn. All this meanes
is that an arbitrary point P has a unique address of the form of
(sP1, sP2, ...sPn)
Now (we're assuming that the space is metric and all sorts of other
details that a mathematician will be better qualified than I to fill
in here), given two points P and Q, the distance between them is given
by some function of the 2*n coordinates, i.e.
D(P,Q) = D(sP1,...sPn,sQ1,...,sQn)
That's all. Note that there is nothing here about the function being
quadratic or whatever. The form of the function depends on the
coordinate system chosen, that's all).
OK, so lets now assume that you've three such points, P,Q and R, suche
that they form a right angle triangle, with R being the vertex of the
right angle. Then, if the space is Euclidean, the Pythagorean theorem
tells you that
D(P,R)^2 + D(Q,R)^2 = D(P,Q)^2
(again, there is nothing here about any squares or sums of suares
of individual coordinates being involved). And if the space is not
Euclidean, then the relationship between the three distances is
different than the above, in general.

One is when non-Cartesian coordinates are used.
...

The other is when non-Euclidean space is used.

Regardless of the coordinates chosen, Pythagoras fails in non-Euclidean
space.

All I'm trying to do is to keep you from confusing two completely
differnt issues. One is the the matter of the metric of the space,
the other is the matter of the coordiantes used. These two are
independent. The Pythagorean theorem stands in its original form in a
Euclidean space, *regardless* of the coordinates being used. It
doesn't stand in