The Road to Reality
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Chapter 10 - Problem with Xi
http://www.roadtoreality.info/viewtopic.php?f=22&t=1978
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Author:  fruity [ 31 Jul 2012, 18:55 ]
Post subject:  Chapter 10 - Problem with Xi

Hi,
At page 187, \xi is introduced:
\xi = A \frac{\partial}{\partial X} + B \frac{\partial}{\partial Y}
But I'm not sure what is meant by this.
Somewhere is written that \xi = \frac{\partial}{\partial x} but I simply don't understand this section. For example, \frac{\partial}{\partial x} doesn't return a vector, right? Why is \xi then called a 'vector field'?
Furthermore, I still wonder what is 'dx' and the '1-form'. I think, I miss anything to understand this section.
Does anyone know how it works?
Greetings,
fruity

Author:  vasco [ 31 Jul 2012, 20:31 ]
Post subject:  Re: Chapter 10 - Problem with Xi

You may or may not find it useful to read this discussion in the forum:
viewtopic.php?f=4&t=1439

Author:  vasco [ 31 Jul 2012, 20:39 ]
Post subject:  Re: Chapter 10 - Problem with Xi

\xi here is an operator, so \xi F(x,y)=[A \frac{\partial }{\partial X} + B \frac{\partial }{\partial Y}]F=A \frac{\partial F}{\partial X} + B \frac{\partial F}{\partial Y}

Author:  vasco [ 31 Jul 2012, 20:58 ]
Post subject:  Re: Chapter 10 - Problem with Xi

\frac{\partial}{\partial X} can be thought of as a vector since it is the derivative in the direction of the x axis.

Author:  fruity [ 31 Jul 2012, 22:31 ]
Post subject:  Re: Chapter 10 - Problem with Xi

Thanks... I will have to think about it (,too).

Author:  fruity [ 01 Aug 2012, 07:47 ]
Post subject:  Re: Chapter 10 - Problem with Xi

Ok, I what I've understood now:
\xi is the direction we want to calculate the slope in.
d\Phi is the covector to the vector along the contour, but I'm not sure. There is the equation \Phi \bullet \xi = 0, which says me that the slope vanishes when the vector \xi points in the right direction (along the contour). \Phi \bullet \xi = 0 also says me that \Phi has to be the covector of \xi, right? I think I still miss something related to d\Phi...

vasco wrote:
\xi here is an operator, so \xi F(x,y)=[A \frac{\partial }{\partial X} + B \frac{\partial }{\partial Y}]F=A \frac{\partial F}{\partial X} + B \frac{\partial F}{\partial Y}

So this equation in the book didn't mean the vector \xi but the operator \xi \Phi or \xi F, which returns a number...
Furthermore, \xi is called a vector field. Is \xi now an operator or a vector or a vector field?

Author:  vasco [ 01 Aug 2012, 11:50 ]
Post subject:  Re: Chapter 10 - Problem with Xi

fruity wrote:
Thanks... I will have to think about it (,too).
fruity wrote:
Ok, I what I've understood now:
\xi is the direction we want to calculate the slope in.
d\Phi is the covector to the vector along the contour, but I'm not sure. There is the equation \Phi \bullet \xi = 0, which says me that the slope vanishes when the vector \xi points in the right direction (along the contour). \Phi \bullet \xi = 0 also says me that \Phi has to be the covector of \xi, right? I think I still miss something related to d\Phi...

\Phi is not a vector, but a scalar field on S (see page 183 section 10.2 near the top), so you can't write \Phi\bullet\xi.
\xi(\Phi) is also not a vector.

Author:  fruity [ 01 Aug 2012, 18:18 ]
Post subject:  Re: Chapter 10 - Problem with Xi

Sorry, I meant
d\Phi \bullet \xi
and
d\Phi
. What I still don't understand is why d\Phi should "give the direction of the contours" (Fig. 10.8). Isn't it the same as \nabla \Phi?

Author:  vasco [ 02 Aug 2012, 07:09 ]
Post subject:  Re: Chapter 10 - Problem with Xi

Hi fruity
By the way I don't understand this topic either, I'm just hoping that by discussing it with someone else I might get there.

Author:  fruity [ 02 Aug 2012, 09:12 ]
Post subject:  Re: Chapter 10 - Problem with Xi

Thanks for your help. I think what I know now is enough for reading the next chapters...
Maybe the use of these things in later chapters will help, too.

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