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 Exercise [12.10] 
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Joined: 07 May 2009, 16:45
Posts: 62
Post Exercise [12.10]
Let G = \int_{-\infty}^{\infty} e^{-x^{2}} dx. Explain why G^{2} = \int_{\Re^{2}} e^{-(x^{2}+y^{2})} dx \wedge dy and evaluate this by chaning to polar coordinates(r,\theta) .

Hence prove G = \sqrt{\pi}.
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This, e^{-x^{2}} dx, is a 1-form.

So is this e^{-y^{2}} dy.

If G = \int_{-\infty}^{\infty} e^{-x^{2}} dx , then also G = \int_{-\infty}^{\infty} e^{-y^{2}} dy because the x or y is a dummy variable that disappears with integration.

So multiplying these two ways of expressing G together would give us G^2

G^2 =  \int_{-\infty}^{\infty} e^{-x^{2}} dx   \int_{-\infty}^{\infty} e^{-y^{2}} dy

This part is a little confusing to me. We don't multiply 1-forms, we "wedge product" them. Is that justification enough to go to the next step as follows,

G^2 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}  e^{-x^{2}} dx \wedge e^{-y^{2}} dy ?

e^{-x^{2}} dx \wedge  e^{-y^{2}} dy = e^{-x^{2}} e^{-y^{2}} dx \wedge dy = e^{-(x^{2}+y^{2})} dx \wedge dy is a 2-form, appropriate for integrating over a plane, as Penrose tells us (See page 231 and note 12.9 on page 245). A little confusion remains on my part.

G^2 = \int_{-\infty}^{\infty}   \int_{-\infty}^{\infty}  e^{-x^{2}} e^{-y^{2}} dx \wedge dy

G^2 = \int_{\Re^{2}} e^{-(x^{2}+y^{2})} dx \wedge dy

Changing to polar coordinates:

r = \sqrt{x^2 +y^2}
\theta = \sin^{-1} \left( \frac{y}{\sqrt{x^2+y^2}} \right)

x = r \cos \theta
y = r \sin \theta
dx = \cos \theta{  }dr - r \sin\theta{  } d\theta
dy = \sin \theta{  } dr + r \cos\theta{  }d\theta

dx\wedge dy =
\left(r \cos^2 \theta \right)dr \wedge d\theta - \left(r \sin^2 \theta \right) d\theta \wedge dr
= r\left( \cos^2 \theta + \sin^2 \theta \right) dr \wedge d\theta
= r  dr \wedge d \theta

\int_{-\infty}^{\infty}   \int_{-\infty}^{\infty} e^{-(x^{2}+y^{2})} dx \wedge dy

= \int_{-\infty}^{\infty} \int_{0}^{\infty}   e^{-(x^{2}+y^{2})} dx \wedge dy + \int_{-\infty}^{\infty} \int_{-\infty}^{0}   e^{-(x^{2}+y^{2})} dx \wedge dy

= \int_{-\infty}^{\infty} \int_{0}^{\infty}   e^{-(x^{2}+y^{2})} dx \wedge dy + \int_{-\infty}^{\infty} \int_{0}^{\infty}   e^{-(x^{2}+y^{2})} dx \wedge dy

= \int_{-\infty}^{\infty} \int_{0}^{\infty}  2 e^{-(x^{2}+y^{2})} dx \wedge dy

= \int_{\theta (y = -\infty) }^{\theta (y  = \infty)} \int_{r(x =0)}^{r(x = \infty)}  2r e^{-r^{2}} dr \wedge d\theta


= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{\infty}  2r e^{-r^{2}} dr \wedge d\theta

The change in integration limits justified because for \theta

\sin^{-1} \left( \frac {-\infty}{\sqrt{\left(-\infty\right)^2 + x^2}} \right) \rightarrow -\frac{\pi}{2} (OK, I mean this is the limit as y approaches -\infty)

\sin^{-1} \left( \frac { \infty}{\sqrt{ \infty^2 + x^2}} \right) \rightarrow \frac{\pi}{2} ( for y going to \infty )

and for r

\sqrt {\left(\infty^2 + y\right)} \rightarrow \infty (for x going to \infty )

What about the lower limit of r corresponding to when x=0? We are integrating along the half line from a point where x=0 to x=+\infty in the rintegration. I think integrating along any line to infinity (except along the y-axis in which case x would not change) would do the trick, just so long as the angle \theta remains constant in this integration. If we started at any other point than the origin on the y-axis, x=0 at the start of this integration, but the angle \theta would have to change as x increased towards \infty. That is about as good a reason as I can come up with as to why x=0 for a lower limit meansr=0 for a lower limit with the variable change.

Continuing, and substituting u = -r^2, du = -2rdr, r=\infty \rightarrow u = -\infty
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{\infty}  2r e^{-r^{2}} dr \wedge d\theta
= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{-\infty}  -e^{u} du \wedge d\theta

= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( -e^{-\infty} - -e^{0} \right) d\theta (I am a little uncertain about this step as well. I mean losing the wedge. Is it justified?)

= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 d\theta = \frac{\pi}{2} - -\frac{\pi}{2} = \pi

So

G^2 = \pi,
and
G = \sqrt\pi


Last edited by DimBulb on 30 Jul 2009, 06:02, edited 5 times in total.

29 Jul 2009, 19:14

Joined: 07 May 2009, 16:45
Posts: 62
Post Re: Exercise [12.10]
Also a little confusing is this notion that "The wedge product in Cartan's differential-form notation takes care of everything if we choose to change our coordinates" Seems to me we are still doing quite a bit of work with coordinate change, and while fairly simple in this case, we had to add anr to the wedge product when we went fromdx \wedge dy to rdr \wedge d\theta.

And then there is this notion of a 2-form. I mean, a 1-form spans all the dimensions in the space except one. In our case the space is two dimensional so a 1-form represents a line or a "line density". A 2-form spans all the dimensions in a space except two. Does that means in our case a 2-form represents a point, or a "point density", which I guess is the same as an "area density"?


30 Jul 2009, 03:48

Joined: 07 May 2009, 16:45
Posts: 62
Post Re: Exercise [12.10]
The breaking up of

\int_{-\infty}^{\infty}   \int_{-\infty}^{\infty} e^{-(x^{2}+y^{2})} dx \wedge dy

into

\int_{-\infty}^{\infty} \int_{0}^{\infty}   e^{-(x^{2}+y^{2})} dx \wedge dy + \int_{-\infty}^{\infty} \int_{-\infty}^{0}   e^{-(x^{2}+y^{2})} dx \wedge dy

was necessitated by the fact that the transfer function

r = \sqrt{x^2 +y^2}

is not one-to-one, that is both -x and x will give the same r. So x=-\infty \rightarrow r=\infty and x = \infty \rightarrow r=\infty which would make the starting and ending point of the integration the same, making the integration 0, which would be wrong.


30 Jul 2009, 19:56

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Re: Exercise [12.10]
Dimbulb, I don't think you need to split your integrals up the way you have.

It's sufficient to note that:

\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} (\cdots) dx \wedge dy

and

\int_{-\pi}^{\pi}\int_0^{\infty} (\cdots) dr \wedge d\theta

both represent area integrals over the same region (manifold), namely the \Re^2 plane.

Also, you seem to be trying to change x into r and y into \theta, which isn't kosher - you can't (generally) change coordinates pairwise like this.

If I recall my calculus correctly (and that's definitely an if!), to change coordinates in a regular area (or volume, etc.) integral, you need to (a) ensure that the integration is occurring over the same exact region, (b) change the variables in the term being integrated, and (c) insert det(J) into the integral, where J is the Jacobian matrix of the coordinate transformation.

i.e. \int_{Region} f(\boldsymbol{x}).d\boldsymbol{x} \ = \ \int_{Region} f(\boldsymbol{x'}).\vbar\boldsymbol{J}\vbar.d\boldsymbol{x'}

where

\boldsymbol{J} = \left(\begin{array}{cccc}\frac{\partial x_0}{\partial {x_0}'}&\frac{\partial x_0}{\partial {x_1}'}&\cdots&\frac{\partial x_0}{\partial {x_n}'}\\\\\frac{\partial x_1}{\partial {x_0}'}&\frac{\partial x_1}{\partial {x_1}'}&\cdots&\frac{\partial x_1}{\partial {x_n}'}\\\\\vdots&\vdots&\ddots&\vdots\\\\\frac{\partial x_n}{\partial {x_0}'}&\frac{\partial x_n}{\partial {x_1}'}&\cdots&\frac{\partial x_n}{\partial {x_n}'}\end{array}\right)


Intuitively \vbar\boldsymbol{J}\vbar can be understood as the ratio of the infinitesimal areas/volumes dx : dx'. For example, in the case of changing from cartesian coordinates dx dy into polar coordinates dr d\theta, dr has the same scale as dx or dy, but d\theta scales proportionally to r, so at any given point \vbar dx dy\vbar = r\vbar dr d\theta\vbar.

It appears that this is what the Grassman product notation provides "for free", in that the rules for coordinate transforms of 1-forms automatically involve calculation of the elements of J, and then calculating out the wedge-product automatically yeilds its determinant, so when changing coordinates in an integral involving p-forms, you don't need to worry about step (c).

Like Dimbulb, though, I'll confess I'm a little unclear about the step involving the combining two regular integrals to get a "wedge" integral, and then the later step of integrating out \theta from the wedge product to get a regular integral again... it seems OK, and obviously works in this case, but it's not clear if there are caveats in more general cases...


17 Jul 2010, 19:14

Joined: 23 Aug 2010, 13:12
Posts: 33
Post Re: Exercise [12.10]
A comment on how the "r" in r dr is automatically generated by the wedge product. See PDF file.


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24 Aug 2010, 00:05

Joined: 07 May 2009, 16:45
Posts: 62
Post Re: Exercise [12.10]
I think I understand what you all are saying. Submitted a more correct version Exercise [12.10]b


29 Nov 2010, 18:46
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