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This is your guru

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  This is your guru mlwozniak@wp.pl - 2016-11-16 00:09 -0800
    Re: This is your guru Melzzzzz <mel@zzzzz.com> - 2016-11-16 10:52 +0100
      Re: This is your guru mlwozniak@wp.pl - 2016-11-16 02:41 -0800
        Re: This is your guru Melzzzzz <mel@zzzzz.com> - 2016-11-16 11:43 +0100
          Re: This is your guru mlwozniak@wp.pl - 2016-11-16 03:49 -0800
            Re: This is your guru Melzzzzz <mel@zzzzz.com> - 2016-11-16 13:18 +0100
              Re: This is your guru mlwozniak@wp.pl - 2016-11-16 04:48 -0800
                Re: This is your guru JanPB <filmart@gmail.com> - 2016-11-16 21:54 -0800
                  Re: This is your guru mlwozniak@wp.pl - 2016-11-16 23:19 -0800
                    Re: This is your guru JanPB <filmart@gmail.com> - 2016-11-16 23:32 -0800
                      Re: This is your guru mlwozniak@wp.pl - 2016-11-16 23:38 -0800
                        Re: This is your guru JanPB <filmart@gmail.com> - 2016-11-17 00:14 -0800
                          Re: This is your guru mlwozniak@wp.pl - 2016-11-17 00:38 -0800
                            Re: This is your guru JanPB <filmart@gmail.com> - 2016-11-17 10:37 -0800
                              Re: This is your guru mlwozniak@wp.pl - 2016-11-17 22:36 -0800
      Re: This is your guru Tom Roberts <tjroberts137@sbcglobal.net> - 2016-11-16 10:26 -0600
        Re: This is your guru Kieth Staudt <KiethS@StaudtInstitute.info> - 2016-11-16 18:12 +0000
        Re: This is your guru Melzzzzz <mel@zzzzz.com> - 2016-11-16 19:28 +0100
        Re: This is your guru JanPB <filmart@gmail.com> - 2016-11-16 21:57 -0800
          Re: This is your guru mlwozniak@wp.pl - 2016-11-16 23:34 -0800
            Re: This is your guru JanPB <filmart@gmail.com> - 2016-11-17 00:23 -0800
              Re: This is your guru mlwozniak@wp.pl - 2016-11-17 03:37 -0800
                Re: This is your guru JanPB <filmart@gmail.com> - 2016-11-17 10:39 -0800
                  Re: This is your guru mlwozniak@wp.pl - 2016-11-17 22:50 -0800
                    Re: This is your guru JanPB <filmart@gmail.com> - 2016-11-17 23:52 -0800
                      Re: This is your guru mlwozniak@wp.pl - 2016-11-18 00:53 -0800
                        Re: This is your guru JanPB <filmart@gmail.com> - 2016-11-18 12:28 -0800
                          Re: This is your guru Maciej Woźniak <mlwozniak@wp.pl> - 2016-11-19 10:07 +0100
                            Re: This is your guru Marion Vanriper <Marion@Vanriper.info> - 2016-11-26 09:30 +0000
          Re: This is your guru Tom Roberts <tjroberts137@sbcglobal.net> - 2016-11-17 10:09 -0600
    Re: This is your guru JanPB <filmart@gmail.com> - 2016-11-16 12:11 -0800
      Re: This is your guru mlwozniak@wp.pl - 2016-11-16 23:14 -0800
        Re: This is your guru JanPB <filmart@gmail.com> - 2016-11-16 23:35 -0800
          Re: This is your guru mlwozniak@wp.pl - 2016-11-16 23:52 -0800
            Re: This is your guru JanPB <filmart@gmail.com> - 2016-11-17 00:25 -0800
              Re: This is your guru paparios <paparios@gmail.com> - 2016-11-17 03:59 -0800
                Re: This is your guru "Dan S. MacAbre" <no@way.com> - 2016-11-17 12:12 +0000
                  Re: This is your guru Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-11-18 05:07 +0100
                    Re: This is your guru "Dan S. MacAbre" <no@way.com> - 2016-11-18 10:02 +0000
                      Re: This is your guru Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-11-18 22:13 +0100
                      Re: This is your guru "David (Time Lord) Fuller" <fuller.david@hotmail.com> - 2016-11-18 21:46 -0800
                      Re: This is your guru "Paul B. Andersen" <relativity@paulba.no> - 2016-11-19 09:30 +0100
                  Re: This is your guru Tom Roberts <tjroberts137@sbcglobal.net> - 2016-11-19 10:01 -0600
                    Re: This is your guru "David (Time Lord) Fuller" <fuller.david@hotmail.com> - 2016-11-19 08:07 -0800
                    Re: This is your guru Norberto Haman <norberman@norberman.info> - 2016-11-19 16:37 +0000
                      Re: This is your guru Steven Carlip <carlip@physics.ucdavis.edu> - 2016-11-20 09:37 -0800
                        Re: This is your guru Norberto Haman <norberman@norberman.info> - 2016-11-20 20:32 +0000
                          Re: This is your guru Steven Carlip <carlip@physics.ucdavis.edu> - 2016-11-20 15:54 -0800
                            Re: This is your guru Maryann Tonn <libradaz@libradazert.info> - 2016-11-21 15:00 +0000
                              Re: This is your guru Steven Carlip <carlip@physics.ucdavis.edu> - 2016-11-21 19:16 -0800
                                Re: This is your guru Maryann Tonn <libradaz@libradazert.info> - 2016-11-22 10:39 +0000
                                  Re: This is your guru paparios <paparios@gmail.com> - 2016-11-22 03:16 -0800
                                    Re: This is your guru Maryann Tonn <libradaz@libradazert.info> - 2016-11-22 11:24 +0000
                                      Re: This is your guru Ned Latham <nedlatham@woden.valhalla.oz> - 2016-11-22 12:13 +0000
                                  Re: This is your guru benj <benj@nobody.net> - 2016-11-22 17:37 -0500
                                  Re: This is your guru Steven Carlip <carlip@physics.ucdavis.edu> - 2016-11-22 22:57 -0800
                                    Re: This is your guru Dusty Kollman <uberdriver@uberdriver.info> - 2016-11-23 13:37 +0000
                                      Re: This is your guru paparios <paparios@gmail.com> - 2016-11-23 06:04 -0800
                                        Re: This is your guru Dusty Kollman <uberdriver@uberdriver.info> - 2016-11-23 14:27 +0000
                                          Re: This is your guru paparios <paparios@gmail.com> - 2016-11-23 06:44 -0800
                                            Re: This is your guru Dusty Kollman <uberdriver@uberdriver.info> - 2016-11-23 14:56 +0000
                                      Re: This is your guru Steven Carlip <carlip@physics.ucdavis.edu> - 2016-11-23 10:46 -0800
                                        Re: This is your guru Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-11-23 23:21 +0100
                                          Re: This is your guru Bob Goodman <bobg@boobgodman.info> - 2016-11-23 22:58 +0000
                                            Re: This is your guru Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-11-25 20:00 +0100
                                              Re: This is your guru Bernardo Dearmond <bnard@dearmond.org> - 2016-11-25 21:46 +0000
                                                Re: This is your guru Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-11-25 22:56 +0100
                                        Re: This is your guru "David (Lord Kronos) Fuller" <fuller.david@hotmail.com> - 2016-11-23 20:13 -0800
                                    Re: This is your guru Odd Bodkin <bodkinodd@gmail.com> - 2016-11-23 09:24 -0600
                                      Re: This is your guru Dusty Kollman <uberdriver@uberdriver.info> - 2016-11-23 15:33 +0000
                                      Re: This is your guru Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-11-23 23:23 +0100
                                        Re: This is your guru "David (Lord Kronos) Fuller" <fuller.david@hotmail.com> - 2016-11-23 20:15 -0800
                                          Re: This is your guru Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-11-25 20:05 +0100
                                            Re: This is your guru "David (Lord Kronos) Fuller" <fuller.david@hotmail.com> - 2016-11-25 14:57 -0800
                                    Re: This is your guru Dusty Kollman <uberdriver@uberdriver.info> - 2016-11-23 18:28 +0000
                                      Re: This is your guru Poutnik <poutnik4nntp@gmail.com> - 2016-11-23 19:37 +0100
                                        Re: This is your guru Dusty Kollman <uberdriver@uberdriver.info> - 2016-11-23 18:39 +0000
                              Re: This is your guru "David (Lord Kronos) Fuller" <fuller.david@hotmail.com> - 2016-11-21 22:40 -0800
                    Re: This is your guru Tom Roberts <tjroberts137@sbcglobal.net> - 2016-11-19 10:47 -0600
                      Re: This is your guru Norberto Haman <norberman@norberman.info> - 2016-11-19 17:20 +0000
                Re: This is your guru mlwozniak@wp.pl - 2016-11-17 04:37 -0800

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#398592 — This is your guru

> > 
> > >     ds^2 = dx^2/x^4 + dy^2
> > > 
> > > ...is a square, then please write the entity it's the square _of_. Good luck.
> > 
> > So I have my answer: 2)you don't know what ds is .
> 
> This is not the answer. You claimed that "ds^2" in my question was
> a square. Just WRITE the thing this "ds^2" is the square of. Is this
> some sort of a secret now?
> 
> BTW, this is now the SECOND mistake you're making (the first being
> not answering the test question in the first place). Care to continue?
> 

This is your guru: he screams, he waves his arms
and he doesn't know, what a differential is.

Jan is an extreme case. But others are not much
better. Really.

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#398593

On Wed, 16 Nov 2016 00:09:53 -0800 (PST)
mlwozniak@wp.pl wrote:

> > >   
> > > >     ds^2 = dx^2/x^4 + dy^2
> > > > 
> > > > ...is a square, then please write the entity it's the square
> > > > _of_. Good luck.  
> > > 
> > > So I have my answer: 2)you don't know what ds is .  
> > 
> > This is not the answer. You claimed that "ds^2" in my question was
> > a square. Just WRITE the thing this "ds^2" is the square of. Is this
> > some sort of a secret now?
> > 
> > BTW, this is now the SECOND mistake you're making (the first being
> > not answering the test question in the first place). Care to
> > continue? 
> 
> This is your guru: he screams, he waves his arms
> and he doesn't know, what a differential is.
> 
> Jan is an extreme case. But others are not much
> better. Really.

Well, following formula from wikipedia:
ds^2 = Edx^2 + 2Fdxdy+Gdy^2
E = 1/x^4, F = 0 , G = 1
and transformation tensor is:
E F
F G

so metric tensor is:
1/x^4  0
0      1

-- 
press any key to continue or any other to quit

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#398594

W dniu środa, 16 listopada 2016 10:52:09 UTC+1 użytkownik Melzzzzz napisał:
> On Wed, 16 Nov 2016 00:09:53 -0800 (PST)
> mlwozniak@wp.pl wrote:
> 
> > > >   
> > > > >     ds^2 = dx^2/x^4 + dy^2
> > > > > 
> > > > > ...is a square, then please write the entity it's the square
> > > > > _of_. Good luck.  
> > > > 
> > > > So I have my answer: 2)you don't know what ds is .  
> > > 
> > > This is not the answer. You claimed that "ds^2" in my question was
> > > a square. Just WRITE the thing this "ds^2" is the square of. Is this
> > > some sort of a secret now?
> > > 
> > > BTW, this is now the SECOND mistake you're making (the first being
> > > not answering the test question in the first place). Care to
> > > continue? 
> > 
> > This is your guru: he screams, he waves his arms
> > and he doesn't know, what a differential is.
> > 
> > Jan is an extreme case. But others are not much
> > better. Really.
> 
> Well, following formula from wikipedia:
> ds^2 = Edx^2 + 2Fdxdy+Gdy^2
> E = 1/x^4, F = 0 , G = 1
> and transformation tensor is:
> E F
> F G
> 
> so metric tensor is:
> 1/x^4  0
> 0      1

And - does it prevent somehow ds to be the differential of s?

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#398595

On Wed, 16 Nov 2016 02:41:12 -0800 (PST)
mlwozniak@wp.pl wrote:

> W dniu środa, 16 listopada 2016 10:52:09 UTC+1 użytkownik Melzzzzz
> napisał:
> > On Wed, 16 Nov 2016 00:09:53 -0800 (PST)
> > mlwozniak@wp.pl wrote:
> >   
> > > > >     
> > > > > >     ds^2 = dx^2/x^4 + dy^2
> > > > > > 
> > > > > > ...is a square, then please write the entity it's the square
> > > > > > _of_. Good luck.    
> > > > > 
> > > > > So I have my answer: 2)you don't know what ds is .    
> > > > 
> > > > This is not the answer. You claimed that "ds^2" in my question
> > > > was a square. Just WRITE the thing this "ds^2" is the square
> > > > of. Is this some sort of a secret now?
> > > > 
> > > > BTW, this is now the SECOND mistake you're making (the first
> > > > being not answering the test question in the first place). Care
> > > > to continue?   
> > > 
> > > This is your guru: he screams, he waves his arms
> > > and he doesn't know, what a differential is.
> > > 
> > > Jan is an extreme case. But others are not much
> > > better. Really.  
> > 
> > Well, following formula from wikipedia:
> > ds^2 = Edx^2 + 2Fdxdy+Gdy^2
> > E = 1/x^4, F = 0 , G = 1
> > and transformation tensor is:
> > E F
> > F G
> > 
> > so metric tensor is:
> > 1/x^4  0
> > 0      1  
> 
> And - does it prevent somehow ds to be the differential of s?
> 

Well strictly speaking, formula is not complete without explanation.
One can't surely come up with tensor without knowing original
formula.

-- 
press any key to continue or any other to quit

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#398597

W dniu środa, 16 listopada 2016 11:43:43 UTC+1 użytkownik Melzzzzz napisał:
> On Wed, 16 Nov 2016 02:41:12 -0800 (PST)
> mlwozniak@wp.pl wrote:
> 
> > W dniu środa, 16 listopada 2016 10:52:09 UTC+1 użytkownik Melzzzzz
> > napisał:
> > > On Wed, 16 Nov 2016 00:09:53 -0800 (PST)
> > > mlwozniak@wp.pl wrote:
> > >   
> > > > > >     
> > > > > > >     ds^2 = dx^2/x^4 + dy^2
> > > > > > > 
> > > > > > > ...is a square, then please write the entity it's the square
> > > > > > > _of_. Good luck.    
> > > > > > 
> > > > > > So I have my answer: 2)you don't know what ds is .    
> > > > > 
> > > > > This is not the answer. You claimed that "ds^2" in my question
> > > > > was a square. Just WRITE the thing this "ds^2" is the square
> > > > > of. Is this some sort of a secret now?
> > > > > 
> > > > > BTW, this is now the SECOND mistake you're making (the first
> > > > > being not answering the test question in the first place). Care
> > > > > to continue?   
> > > > 
> > > > This is your guru: he screams, he waves his arms
> > > > and he doesn't know, what a differential is.
> > > > 
> > > > Jan is an extreme case. But others are not much
> > > > better. Really.  
> > > 
> > > Well, following formula from wikipedia:
> > > ds^2 = Edx^2 + 2Fdxdy+Gdy^2
> > > E = 1/x^4, F = 0 , G = 1
> > > and transformation tensor is:
> > > E F
> > > F G
> > > 
> > > so metric tensor is:
> > > 1/x^4  0
> > > 0      1  
> > 
> > And - does it prevent somehow ds to be the differential of s?
> > 
> 
> Well strictly speaking, formula is not complete without explanation.
> One can't surely come up with tensor without knowing original
> formula.

However, the thread is not about the tensor.
It is about ds - starting from Jan's doubt
if there is any ds in ds^2.

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#398598

On Wed, 16 Nov 2016 03:49:41 -0800 (PST)
mlwozniak@wp.pl wrote:

> W dniu środa, 16 listopada 2016 11:43:43 UTC+1 użytkownik Melzzzzz
> napisał:
> > On Wed, 16 Nov 2016 02:41:12 -0800 (PST)
> > mlwozniak@wp.pl wrote:
> >   
> > > W dniu środa, 16 listopada 2016 10:52:09 UTC+1 użytkownik Melzzzzz
> > > napisał:  
> > > > On Wed, 16 Nov 2016 00:09:53 -0800 (PST)
> > > > mlwozniak@wp.pl wrote:
> > > >     
> > > > > > >       
> > > > > > > >     ds^2 = dx^2/x^4 + dy^2
> > > > > > > > 
> > > > > > > > ...is a square, then please write the entity it's the
> > > > > > > > square _of_. Good luck.      
> > > > > > > 
> > > > > > > So I have my answer: 2)you don't know what ds is .      
> > > > > > 
> > > > > > This is not the answer. You claimed that "ds^2" in my
> > > > > > question was a square. Just WRITE the thing this "ds^2" is
> > > > > > the square of. Is this some sort of a secret now?
> > > > > > 
> > > > > > BTW, this is now the SECOND mistake you're making (the first
> > > > > > being not answering the test question in the first place).
> > > > > > Care to continue?     
> > > > > 
> > > > > This is your guru: he screams, he waves his arms
> > > > > and he doesn't know, what a differential is.
> > > > > 
> > > > > Jan is an extreme case. But others are not much
> > > > > better. Really.    
> > > > 
> > > > Well, following formula from wikipedia:
> > > > ds^2 = Edx^2 + 2Fdxdy+Gdy^2
> > > > E = 1/x^4, F = 0 , G = 1
> > > > and transformation tensor is:
> > > > E F
> > > > F G
> > > > 
> > > > so metric tensor is:
> > > > 1/x^4  0
> > > > 0      1    
> > > 
> > > And - does it prevent somehow ds to be the differential of s?
> > >   
> > 
> > Well strictly speaking, formula is not complete without explanation.
> > One can't surely come up with tensor without knowing original
> > formula.  
> 
> However, the thread is not about the tensor.
> It is about ds - starting from Jan's doubt
> if there is any ds in ds^2.
> 

Well, sure, ds is differential of s. It is derivation of original
formula.

-- 
press any key to continue or any other to quit

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#398599

W dniu środa, 16 listopada 2016 13:18:46 UTC+1 użytkownik Melzzzzz napisał:
> On Wed, 16 Nov 2016 03:49:41 -0800 (PST)
> mlwozniak@wp.pl wrote:
> 
> > W dniu środa, 16 listopada 2016 11:43:43 UTC+1 użytkownik Melzzzzz
> > napisał:
> > > On Wed, 16 Nov 2016 02:41:12 -0800 (PST)
> > > mlwozniak@wp.pl wrote:
> > >   
> > > > W dniu środa, 16 listopada 2016 10:52:09 UTC+1 użytkownik Melzzzzz
> > > > napisał:  
> > > > > On Wed, 16 Nov 2016 00:09:53 -0800 (PST)
> > > > > mlwozniak@wp.pl wrote:
> > > > >     
> > > > > > > >       
> > > > > > > > >     ds^2 = dx^2/x^4 + dy^2
> > > > > > > > > 
> > > > > > > > > ...is a square, then please write the entity it's the
> > > > > > > > > square _of_. Good luck.      
> > > > > > > > 
> > > > > > > > So I have my answer: 2)you don't know what ds is .      
> > > > > > > 
> > > > > > > This is not the answer. You claimed that "ds^2" in my
> > > > > > > question was a square. Just WRITE the thing this "ds^2" is
> > > > > > > the square of. Is this some sort of a secret now?
> > > > > > > 
> > > > > > > BTW, this is now the SECOND mistake you're making (the first
> > > > > > > being not answering the test question in the first place).
> > > > > > > Care to continue?     
> > > > > > 
> > > > > > This is your guru: he screams, he waves his arms
> > > > > > and he doesn't know, what a differential is.
> > > > > > 
> > > > > > Jan is an extreme case. But others are not much
> > > > > > better. Really.    
> > > > > 
> > > > > Well, following formula from wikipedia:
> > > > > ds^2 = Edx^2 + 2Fdxdy+Gdy^2
> > > > > E = 1/x^4, F = 0 , G = 1
> > > > > and transformation tensor is:
> > > > > E F
> > > > > F G
> > > > > 
> > > > > so metric tensor is:
> > > > > 1/x^4  0
> > > > > 0      1    
> > > > 
> > > > And - does it prevent somehow ds to be the differential of s?
> > > >   
> > > 
> > > Well strictly speaking, formula is not complete without explanation.
> > > One can't surely come up with tensor without knowing original
> > > formula.  
> > 
> > However, the thread is not about the tensor.
> > It is about ds - starting from Jan's doubt
> > if there is any ds in ds^2.
> > 
> 
> Well, sure, ds is differential of s. It is derivation of original
> formula.

Yes. Now look above. Or below:

> > > > > > > You claimed that "ds^2" in my
> > > > > > > question was a square. Just WRITE the thing this "ds^2" is
> > > > > > > the square of. Is this some sort of a secret now?

That's Jan's text. The chief expert of differential
geometry here. He doesn't know, what ds^2 is square of.
Unbelievable, isn't it? 
He is an extreme case, true. But, seriously: relativity
is promoted by these like him. They're juggling with
symbols OK, they just forgot what they mean.

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#398653

On Wednesday, November 16, 2016 at 4:48:04 AM UTC-8, mlwo...@wp.pl wrote:
> 
> Yes. Now look above. Or below:
> 
> > > > > > > > You claimed that "ds^2" in my
> > > > > > > > question was a square. Just WRITE the thing this "ds^2" is
> > > > > > > > the square of. Is this some sort of a secret now?
> 
> That's Jan's text. The chief expert of differential
> geometry here. He doesn't know, what ds^2 is square of.

Grasping for rhetorical straws, aren't we? Haha. ds^2 (metric tensor) is not a square
of anything. Instead of telling everyone what I "don't know" you can win this argument
very easily: by writing down what exactly is "ds^2" the square of. (Saying "ds" doesn't
count as answer, sorry.)

> Unbelievable, isn't it? 

To you, yes.

> He is an extreme case, true.

No, I am just an expert on that stuff. 

> But, seriously: relativity
> is promoted by these like him. They're juggling with
> symbols OK, they just forgot what they mean.

Oh stop this pathetic poetry already and simply tell everyone at long last: what is "ds^2"
the square of? Yes, I know you've already said "ds" but this is just word play. What
_is_ this "ds"?

If you knew what "ds" was, you'd see immediately where your mistake lies. The part that
confuses you is that "ds" and "ds^2" are related in a way.

--
Jan

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#398659

W dniu czwartek, 17 listopada 2016 06:54:36 UTC+1 użytkownik JanPB napisał:
> On Wednesday, November 16, 2016 at 4:48:04 AM UTC-8, mlwo...@wp.pl wrote:
> > 
> > Yes. Now look above. Or below:
> > 
> > > > > > > > > You claimed that "ds^2" in my
> > > > > > > > > question was a square. Just WRITE the thing this "ds^2" is
> > > > > > > > > the square of. Is this some sort of a secret now?
> > 
> > That's Jan's text. The chief expert of differential
> > geometry here. He doesn't know, what ds^2 is square of.
> 
> Grasping for rhetorical straws, aren't we? Haha. ds^2 (metric tensor) is not a square
> of anything. Instead of telling everyone what I "don't know" you can win this argument
> very easily: by writing down what exactly is "ds^2" the square of. (Saying "ds" doesn't
> count as answer, sorry.)

I wrote it 3 or 4 times, Melzzz wrote it too.
But - talk to an idiot.

> If you knew what "ds" was, you'd see immediately where your mistake lies. The part that
> confuses you is that "ds" and "ds^2" are related in a way.
 
I wrote it 3 or 4 times, Melzzz wrote it too.
But - talk to an idiot.

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#398660

On Wednesday, November 16, 2016 at 11:19:11 PM UTC-8, mlwo...@wp.pl wrote:
> W dniu czwartek, 17 listopada 2016 06:54:36 UTC+1 użytkownik JanPB napisał:
> > On Wednesday, November 16, 2016 at 4:48:04 AM UTC-8, mlwo...@wp.pl wrote:
> > > 
> > > Yes. Now look above. Or below:
> > > 
> > > > > > > > > > You claimed that "ds^2" in my
> > > > > > > > > > question was a square. Just WRITE the thing this "ds^2" is
> > > > > > > > > > the square of. Is this some sort of a secret now?
> > > 
> > > That's Jan's text. The chief expert of differential
> > > geometry here. He doesn't know, what ds^2 is square of.
> > 
> > Grasping for rhetorical straws, aren't we? Haha. ds^2 (metric tensor) is not a square
> > of anything. Instead of telling everyone what I "don't know" you can win this argument
> > very easily: by writing down what exactly is "ds^2" the square of. (Saying "ds" doesn't
> > count as answer, sorry.)
> 
> I wrote it 3 or 4 times, Melzzz wrote it too.

No. All you wrote was "ds". You claimed "ds^2" was a square of something. Here is
my ds^2 again:

    ds^2 = dx^2/x^4 + dy^2

Just write the mathematical entity, EXPLICITLY,  which when squared yields
dx^2/x^4 + dy^2.

I'm still waiting.

> But - talk to an idiot.

And spare us the rhetoric. It'll get you nowhere.

> > If you knew what "ds" was, you'd see immediately where your mistake lies. The part that
> > confuses you is that "ds" and "ds^2" are related in a way.
>  
> I wrote it 3 or 4 times, Melzzz wrote it too.
> But - talk to an idiot.

Nope, you haven't written anything except poetry (as usual) and some Wikipedia clippings
in English. The question is precise, you've made a claim - prove it: write down what quantity
exactly yields dx^2/x^4 + dy^2 when squared.

I know (and Tom knows) EXACTLY what sort of trap you've fallen into but I'm not helping.

--
Jan

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#398663

W dniu czwartek, 17 listopada 2016 08:32:09 UTC+1 użytkownik JanPB napisał:
> On Wednesday, November 16, 2016 at 11:19:11 PM UTC-8, mlwo...@wp.pl wrote:
> > W dniu czwartek, 17 listopada 2016 06:54:36 UTC+1 użytkownik JanPB napisał:
> > > On Wednesday, November 16, 2016 at 4:48:04 AM UTC-8, mlwo...@wp.pl wrote:
> > > > 
> > > > Yes. Now look above. Or below:
> > > > 
> > > > > > > > > > > You claimed that "ds^2" in my
> > > > > > > > > > > question was a square. Just WRITE the thing this "ds^2" is
> > > > > > > > > > > the square of. Is this some sort of a secret now?
> > > > 
> > > > That's Jan's text. The chief expert of differential
> > > > geometry here. He doesn't know, what ds^2 is square of.
> > > 
> > > Grasping for rhetorical straws, aren't we? Haha. ds^2 (metric tensor) is not a square
> > > of anything. Instead of telling everyone what I "don't know" you can win this argument
> > > very easily: by writing down what exactly is "ds^2" the square of. (Saying "ds" doesn't
> > > count as answer, sorry.)
> > 
> > I wrote it 3 or 4 times, Melzzz wrote it too.
> 
> No. All you wrote was "ds". You claimed "ds^2" was a square of something. 

A lie. As expected from relativistic trash.
But, ok, once again. This time I can even quote
Tom. 

If, on the other hand, ds^2 is the line element, then there is no nomenclature
convention, no bold text, no tensor-product symbols, and ds, dx, and dy are all
ordinary differentials along some (unspecified) path of integration.

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#398667

On Wednesday, November 16, 2016 at 11:38:35 PM UTC-8, mlwo...@wp.pl wrote:
> W dniu czwartek, 17 listopada 2016 08:32:09 UTC+1 użytkownik JanPB napisał:
> > On Wednesday, November 16, 2016 at 11:19:11 PM UTC-8, mlwo...@wp.pl wrote:
> > > W dniu czwartek, 17 listopada 2016 06:54:36 UTC+1 użytkownik JanPB napisał:
> > > > On Wednesday, November 16, 2016 at 4:48:04 AM UTC-8, mlwo...@wp.pl wrote:
> > > > > 
> > > > > Yes. Now look above. Or below:
> > > > > 
> > > > > > > > > > > > You claimed that "ds^2" in my
> > > > > > > > > > > > question was a square. Just WRITE the thing this "ds^2" is
> > > > > > > > > > > > the square of. Is this some sort of a secret now?
> > > > > 
> > > > > That's Jan's text. The chief expert of differential
> > > > > geometry here. He doesn't know, what ds^2 is square of.
> > > > 
> > > > Grasping for rhetorical straws, aren't we? Haha. ds^2 (metric tensor) is not a square
> > > > of anything. Instead of telling everyone what I "don't know" you can win this argument
> > > > very easily: by writing down what exactly is "ds^2" the square of. (Saying "ds" doesn't
> > > > count as answer, sorry.)
> > > 
> > > I wrote it 3 or 4 times, Melzzz wrote it too.
> > 
> > No. All you wrote was "ds". You claimed "ds^2" was a square of something. 
> 
> A lie. As expected from relativistic trash.

Oh, so I've FINALLY made you retract your nonsense. It was like pulling teeth.

> But, ok, once again. This time I can even quote
> Tom. 

> If, on the other hand, ds^2 is the line element, then there is no nomenclature
> convention, no bold text, no tensor-product symbols, and ds, dx, and dy are all
> ordinary differentials along some (unspecified) path of integration.

Yes, Tom let the very cat out of the bag that I was trying to get you to see. Can you
catch it? It's the words "along some (unspecified) path of integration".

THIS is what you were supposed to say if you knew this: (1) "ds^2' as a metric tensor
is not a square of anything (i.e., there is no covector which would produce ds^2 when
tensored with itself). (2) OTOH in a context in which a _specific fixed curve is being
considered_ [which was not the context of the original dx^2/x^4+dy^2 question],
THEN one can consider the arc length along that curve (aka. the 1D volume form on it,
aka. "ds")  whose coefficient at a curve point, when squared, is equal to the tensor _ds^2 
evaluated on the vector tangent to the curve at that point_. In this context the equation
for ds^2 has only a symbolic meaning as differentials are undefined unless one does
Abraham's non-standard analysis. OTOH ds^2 considered as a tensor (symmetric,
rank-2) is well-defined, with dx and dy being specific covectors (namely, exterior
derivatives of the coordinate functions x and y, respectively).

The difference between "dx as a covector/exterior derivative" and "dx as an undefined
symbol with an intutive meaning presumed known by experience and tradition" is what
Tom called "dx in bold text" vs. "dx not in bold text".

--
Jan

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#398672

W dniu czwartek, 17 listopada 2016 09:14:25 UTC+1 użytkownik JanPB napisał:

> > But, ok, once again. This time I can even quote
> > Tom. 
> 
> > If, on the other hand, ds^2 is the line element, then there is no nomenclature
> > convention, no bold text, no tensor-product symbols, and ds, dx, and dy are all
> > ordinary differentials along some (unspecified) path of integration.
> 
> Yes, Tom let the very cat out of the bag that I was trying to get you to see. Can you
> catch it? It's the words "along some (unspecified) path of integration".

You're a stupid, arrogant, lying shit, Jan.
It's possible, however, these others you wrote it to
will believe.

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#398706

On Thursday, November 17, 2016 at 12:38:18 AM UTC-8, mlwo...@wp.pl wrote:
> W dniu czwartek, 17 listopada 2016 09:14:25 UTC+1 użytkownik JanPB napisał:
> 
> > > But, ok, once again. This time I can even quote
> > > Tom. 
> > 
> > > If, on the other hand, ds^2 is the line element, then there is no nomenclature
> > > convention, no bold text, no tensor-product symbols, and ds, dx, and dy are all
> > > ordinary differentials along some (unspecified) path of integration.
> > 
> > Yes, Tom let the very cat out of the bag that I was trying to get you to see. Can you
> > catch it? It's the words "along some (unspecified) path of integration".
> 
> You're a stupid, arrogant, lying shit, Jan.
> It's possible, however, these others you wrote it to
> will believe.

Finally you are at a loss of words. Next step for you is to be able to
admit you've made a mistake (actually, two).

And cut cursing, it makes you look feeble-minded.

_
Jan

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#398757

W dniu czwartek, 17 listopada 2016 19:37:44 UTC+1 użytkownik JanPB napisał:
> On Thursday, November 17, 2016 at 12:38:18 AM UTC-8, mlwo...@wp.pl wrote:
> > W dniu czwartek, 17 listopada 2016 09:14:25 UTC+1 użytkownik JanPB napisał:
> > 
> > > > But, ok, once again. This time I can even quote
> > > > Tom. 
> > > 
> > > > If, on the other hand, ds^2 is the line element, then there is no nomenclature
> > > > convention, no bold text, no tensor-product symbols, and ds, dx, and dy are all
> > > > ordinary differentials along some (unspecified) path of integration.
> > > 
> > > Yes, Tom let the very cat out of the bag that I was trying to get you to see. Can you
> > > catch it? It's the words "along some (unspecified) path of integration".
> > 
> > You're a stupid, arrogant, lying shit, Jan.
> > It's possible, however, these others you wrote it to
> > will believe.
> 
> Finally you are at a loss of words. Next step for you is to be able to
> admit you've made a mistake (actually, two).
> And cut cursing, it makes you look feeble-minded.

It's not cursing. It's a fact.

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#398612

On 11/16/16 11/16/16 - 3:52 AM, Melzzzzz wrote:
> Well, following formula from wikipedia:
> ds^2 = Edx^2 + 2Fdxdy+Gdy^2
> E = 1/x^4, F = 0 , G = 1
> and transformation tensor is:
> E F
> F G
>
> so metric tensor is:
> 1/x^4  0
> 0      1

Actually not. That is the COMPONENTS of the metric tensor projected onto the 
coordinates {x,y}.

Note there is a nomenclature ambiguity here: ds^2 could be the metric tensor, or 
it could be the line element -- both meanings are used, but in an ASCII 
newsgroup they are indistinguishable. (In either case one can directly read the 
components of the metric for these coordinates.)

If ds^2 is the metric tensor, then ds^2, dx, and dy should be printed in bold 
text, and the tensor-product symbol X has been omitted by convention; using more 
standard nomenclature, the equation is implicitly understood to be:
	ds^2 = E dx X dx + 2 F dx X dy + G dy X dy     [ds^2,dx,dy bold]
Here dx and dy are the usual 1-forms, and ds^2 is explicitly a symmetric rank-2 
tensor (with a funny symbol -- feel free to replace ds^2 with g).

If, on the other hand, ds^2 is the line element, then there is no nomenclature 
convention, no bold text, no tensor-product symbols, and ds, dx, and dy are all 
ordinary differentials along some (unspecified) path of integration.

In only the second case is ds^2 the square of some quantity.

> Well strictly speaking, formula is not complete without explanation.

YES! All too many people around here forget this important point, presenting 
meaningless formulas without explanations. Your recognition of this is why I 
provided the above explanation.


JanPB knows all this, but you other guys apparently do not. Of course none of 
you have actually answered his challenge question. (I can do so, and he knows I 
can, but why should I spoil the circus....)

Tom Roberts

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#398622

Tom Roberts wrote:

>> Well strictly speaking, formula is not complete without explanation.
> 
> YES! All too many people around here forget this important point,
> presenting meaningless formulas without explanations. Your recognition
> of this is why I provided the above explanation.
> JanPB knows all this, but you other guys apparently do not. Of course
> none of you have actually answered his challenge question. (I can do so,
> and he knows I can, but why should I spoil the circus....) Tom Roberts

Hmm yes, as solving Einstein's non-linear differential equations of 
gravity is very difficult. Not even Einstein himself thought that this 
would ever be possible. However, Karl Schwarzschild was able to solve the 
equations for a spherical symmetric static body. Static means that the 
mass is kept unchanged (or independent over time.) Today, the Schwarzschild 
solution is used for most simple scenarios. Einstein himself did not use 
the methods of Schwarzschild but instead used direct approximation 
techniques. Still he was able to conclude that planet Mercury had a 
precession of 43 arc seconds (43/3600 degrees) per century, which was in 
stunning agreement with observation. :)

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#398624

On Wed, 16 Nov 2016 10:26:43 -0600
Tom Roberts <tjroberts137@sbcglobal.net> wrote:

> On 11/16/16 11/16/16 - 3:52 AM, Melzzzzz wrote:
> > Well, following formula from wikipedia:
> > ds^2 = Edx^2 + 2Fdxdy+Gdy^2
> > E = 1/x^4, F = 0 , G = 1
> > and transformation tensor is:
> > E F
> > F G
> >
> > so metric tensor is:
> > 1/x^4  0
> > 0      1  
> 
> Actually not. That is the COMPONENTS of the metric tensor projected
> onto the coordinates {x,y}.
> 
> Note there is a nomenclature ambiguity here: ds^2 could be the metric
> tensor, or it could be the line element -- both meanings are used,
> but in an ASCII newsgroup they are indistinguishable. (In either case
> one can directly read the components of the metric for these
> coordinates.)
> 
> If ds^2 is the metric tensor, then ds^2, dx, and dy should be printed
> in bold text, and the tensor-product symbol X has been omitted by
> convention; using more standard nomenclature, the equation is
> implicitly understood to be: ds^2 = E dx X dx + 2 F dx X dy + G dy X
> dy     [ds^2,dx,dy bold] Here dx and dy are the usual 1-forms, and
> ds^2 is explicitly a symmetric rank-2 tensor (with a funny symbol --
> feel free to replace ds^2 with g).
> 
> If, on the other hand, ds^2 is the line element, then there is no
> nomenclature convention, no bold text, no tensor-product symbols, and
> ds, dx, and dy are all ordinary differentials along some
> (unspecified) path of integration.
> 
> In only the second case is ds^2 the square of some quantity.
> 
> > Well strictly speaking, formula is not complete without
> > explanation.  
> 
> YES! All too many people around here forget this important point,
> presenting meaningless formulas without explanations. Your
> recognition of this is why I provided the above explanation.
> 
> 
> JanPB knows all this, but you other guys apparently do not. Of course
> none of you have actually answered his challenge question. (I can do
> so, and he knows I can, but why should I spoil the circus....)
> 
> Tom Roberts

What was challenge question? ;)
I just read about metric tensor and this ;)
I can't probably answer as I am not physician. I am programmer with
some mathematical background...
I can somewhat understand physics but too rusty, too many years of
coding and not doing math actually ...

-- 
press any key to continue or any other to quit

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#398654

On Wednesday, November 16, 2016 at 8:26:50 AM UTC-8, tjrob137 wrote:
> 
> If, on the other hand, ds^2 is the line element, then there is no nomenclature 
> convention, no bold text, no tensor-product symbols, and ds, dx, and dy are all 
> ordinary differentials along some (unspecified) path of integration.

Yes.

> In only the second case is ds^2 the square of some quantity.

Yes.

Of course all this will go right over Maciej's head.

--
Jan

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#398661

W dniu czwartek, 17 listopada 2016 06:57:49 UTC+1 użytkownik JanPB napisał:
> On Wednesday, November 16, 2016 at 8:26:50 AM UTC-8, tjrob137 wrote:
> > 
> > If, on the other hand, ds^2 is the line element, then there is no nomenclature 
> > convention, no bold text, no tensor-product symbols, and ds, dx, and dy are all 
> > ordinary differentials along some (unspecified) path of integration.
> 
> Yes.
> 
> > In only the second case is ds^2 the square of some quantity.
> 
> Yes.

So, now, poor idiot - take a look at your 
problem. No bold text, right? no tensor-product 
symbols, right? Too bad. 
Stop pretending. Even your fellow idiot knows,
what ds is and what ds^2 is.

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