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Exercise [12.11]
Author	Message
Sameed Zahoor Joined: 12 Mar 2008, 10:57 Posts: 69 Location: India	Exercise [12.11] $d(Adx+Bdy)=dA\wedge\ dx+dB\wedge\ dy$ $=(\frac{\partial{A}}{\partial{x}}dx+\frac{\partial{A}}{\partial{y}}dy)\wedge\ dx+(\frac{\partial{B}}{\partial{x}}dx+\frac{\partial{B}}{\partial{y}}dy)\wedge\ dy$ $=\frac{\partial{A}}{\partial{y}}dy\wedge\ dx+\frac{\partial{B}}{\partial{x}}dx\wedge\ dy$ $=(\frac{\partial{B}}{\partial{x}}-\frac{\partial{A}}{\partial{y}})dx\wedge\ dy$
26 May 2008, 12:04

DimBulb Joined: 07 May 2009, 16:45 Posts: 54	Re: Exercise [12.11] I am confused by your symbology. I did this way: $\boldsymbol\alpha = A dx + B dy$ substituting $\alpha_1 = A$ $\alpha_2 = B$ gives $\boldsymbol\alpha = \sum_{r=1}^2 \alpha_r dx^r$ (like the equation near bottom of page 232, but for a 1-form) $d\boldsymbol\alpha = \sum_{q=1}^2 \sum_{r=1}^2 {\left(d \boldsymbol \alpha \right)}_{qr} dx^q \wedge dy^r$ $= \sum_{q=1}^2 \sum_{r=1}^2 \frac {\partial}{\partial x^{[q}} \alpha_{r]} dx^q \wedge dy^r$ (with "component" from equation near middle of page 233) where $\frac {\partial}{\partial x^{[q}} \alpha_{r]} = \frac{1}{2}\left( \frac {\partial}{\partial x^q} \alpha_r - \frac {\partial}{\partial x^r} \alpha_q \right)$ Writing out the summations, ignoring the components where because $dx^1 \wedge dx^1 = dx^2 \wedge dx^2 = 0$ $d\boldsymbol\alpha = \frac{1}{2}\left( \frac {\partial}{\partial x^1} \alpha_2 - \frac {\partial}{\partial x^2} \alpha_1 \right) dx^1 \wedge dx^2 + \frac{1}{2}\left( \frac {\partial}{\partial x^2} \alpha_1 - \frac {\partial}{\partial x^1} \alpha_2 \right) dx^2 \wedge dx^1$ $= \left( \frac {\partial}{\partial x^1} \alpha_2 - \frac {\partial}{\partial x^2} \alpha_1 \right) dx^1 \wedge dx^2$ $= \left( \frac {\partial}{\partial x} B - \frac {\partial}{\partial y} A \right) dx \wedge dy$ (undoing the A,B,x and y substitutions from above) Is this the same thing?
31 Jul 2009, 21:39

DimBulb Joined: 07 May 2009, 16:45 Posts: 54	Re: Exercise [12.11] Maybe I am using Penrose statement of how $d\boldsymbol\alpha$ has components $\frac {\partial}{\partial x^{[q}}\alpha_{r...t]}$ and using that to show how it works in this specific case, and you are proving it in this specific case?
01 Aug 2009, 06:40

vasco Supporter Joined: 07 Jun 2008, 08:21 Posts: 182	Re: Exercise [12.11] DimBulb I could be wrong, but I think you will be lucky to get a reply from Sameed. If you look you will see that he has not visited the forum since March. You could try sending him a private message and if he has the options set to be informed on his private email address, then he will get a message. Good Luck and keep posting. I am not as far on in the book as you, so I cannot help you very much myself. Vasco
01 Aug 2009, 07:12

DimBulb Joined: 07 May 2009, 16:45 Posts: 54	Re: Exercise [12.11] That's OK. I did not really expect an answer from Sameed Zahoor. I am trying to puzzle these things out on-line. If somebody can help, great. If not, maybe my ramblings will help somebody else following after. Or maybe it will just confuse them as well. After further investigation and thinking, I think Sameed Zahoor is correct in his proof, showing in the process how the operator acts on a 1-form. What I did, which I think is also instructive, was show how the general notation of the components of $d\boldsymbol\alpha$ , $\boldsymbol\alpha$ being a p-form, works out in the specific case of the exercise.
02 Aug 2009, 02:04

flashbang Joined: 13 Aug 2009, 00:08 Posts: 13	Re: Exercise [12.11] Sorry to be asking you to go back in time, but I am really struggling with Ch 10 & 12 so I am making sure I understand the answers to the exercises. Firstly, DimBulb, I understand your solution apart from a small typo (I think) near the top: you say: $\boldsymbol\alpha = \sum_{r=1}^2 \alpha_r dx^r$ but I think you mean: $\boldsymbol\alpha = \sum \alpha_{rs} dx^r \wedge dx^s$ to be in keeping with the equation near the bottom of p232 for a 1-form -- even though it is a 1-form it still has two coordinates. Secondly, Sameer's solution looks good and neat to me but he has made one intellectual leap that is not obvious to me. Right at the start he says: $d(Adx+Bdy)=dA\wedge\ dx+dB\wedge\ dy$ thereby seemingly assuming that $d(Adx)=dA\wedge\ dx$ Is that an obvious jump? Where does Penrose say we can do that? Or is it so simple that he didn't need to? Apart from that his solution makes sense to me. Any thoughts?
22 Aug 2009, 13:35

flashbang Joined: 13 Aug 2009, 00:08 Posts: 13	Re: Exercise [12.11] I'm wondering whether this is one of those things that is so obvious it cannot be explained. What do you think of this: By "ordinary" derivative ideas ("Leibniz Law" on RTR p115, section 6.5) we have the chain rule: $d(A\alpha)=Ad\alpha + dA d\alpha$ Now with $\alpha = dx$ then we have The first term on the right disappears because of Penrose's third rule that , so we are left with on the right. And I suppose you have to ask how you multiply two p-forms like this, and the answer is to use a wedge, so we have $d(Adx) = dA \wedge dx$ and Sameer's solution follows fairly easily if you remember that $dx \wedge dx = 0$ although I notice that Penrose said nothing about the nature of A and B -- they must be functions of x and y (or constants in the degenerate case) for the question to make sense, I suppose. So! Is my step from to $dA \wedge dx$ valid? Has Penrose said we could do that? I guess that's the only way to multiply p-forms, and that's why we start with which seems to be two items multiplied together, and then when multiplying and we get $dA \wedge dx$ . I'm about ready to give up here and just assume it's right. I think there was a famous mathematician who said something along the lines of "There are some parts of Mathematics you don't understand, you simply get used to them".
22 Aug 2009, 19:59

Universus Joined: 19 Oct 2009, 20:10 Posts: 2	Re: Exercise [12.11] Hi, I haven't checked yet your solution Dimbulb in details, but let me please try to make Sameed Zahoor's solution a bit clearer (which is the way I almost did it myself). This solution has the advantage not to be based on the explicit definition of the exterior derivative of a form, which would ideally necessitate to solve problem 12.14 first (I don't know if you did). $\mathrm{d} (A \mathrm{d}x + B \mathrm{d}y) = \mathrm{d} (A \mathrm{d}x) + \mathrm{d} (B \mathrm{d}y)$ (from the first 'axiom') $= \mathrm{d}A \wedge\mathrm{d}x + A \mathrm{d}^2 x + \mathrm{d}B \wedge \mathrm{d}y + B \mathrm{d}^2y$ (from the second 'axiom') $= \mathrm{d}A \wedge \mathrm{d}x +\mathrm{d}B \wedge \mathrm{d}y$ (from the third 'axiom') $= \left( \frac {\partial A}{\partial x } \mathrm{d}x + \frac {\partial A}{\partial y } \mathrm{d}y \right) \wedge \mathrm{d}x + \left( \frac {\partial B}{\partial x } \mathrm{d}x + \frac {\partial B}{\partial y } \mathrm{d}y \right) \wedge \mathrm{d}y$ (from the definition of the gradient given at p.186) $= \frac {\partial A}{\partial y } \mathrm{d}y \wedge \mathrm{d}x + \frac {\partial B}{\partial x } \mathrm{d}x \wedge \mathrm{d}y$ (since the wedge product of a form by itself is 0 ) $= \left( \frac {\partial B}{\partial x } - \frac {\partial A}{\partial y } \right) \mathrm{d}x \wedge \mathrm{d}y$ (from the anticommutation of the wedge product ) This is exactly what Sameed and flashbang did. To answer flashbang, the second line is obtained from the fact that A and B are scalars, so 0-forms ; you can then use the second 'axiom' since it holds, as an axiom, for any pair of forms (even if they don't have the same 'rank').
22 Oct 2009, 19:12

Jayroberts Joined: 29 May 2010, 23:31 Posts: 5	Re: Exercise [12.11] I am an undergrad in mathematics, and am getting into the nitty gritty of multivariable calculus, So would I be wrong in saying that this is a way of proving Green's Theorem without any geometrical interpretation. If so that is incredible. And I think grassman products should be taught alongside linear algebra since they seem so useful.
30 May 2010, 04:16

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Exercise [12.11]

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