I am not subscribed to the forum but I had a thought that complements the posted solution to 12.10 -- namely what thinking of the dxdy as dx^dy gets you: x=rcos(t), y=rsin(t) so dx^dy =(cos(t)dr -rsin(t)dt)^(sin(t)dr +rcos(t)dt) = rcos(t)cos(t)dr^dt -rsin(t)sint(t) dt^dr = rdr^dt. The antisymmetry generates the Jacobean, which is a determinant, ie an antisymnmetric sum of products
In exercise [12.12], the conclusion d(dT) = 0 for T of degree 0 can be proved directly: d(dT) = d(S(dT/dxi)dxi) = SS(d2T/dxidxj)dxi^dxj =0 because dxi^dxj is antisymmetric but the second partial derivative (d2T/dxidxj) is symmetric. (Forgive my orthography, SS stands for the double summation on i and j. The indices i and j must be understood as raised or lowered as appropriate.) Establishing this result directly for degree zero permits one to establish it for all degrees by induction using the second of the two defining relations given immediately above in RtR.
Unfortunately the owner of the site has not replied yet, so I do not know what is happening. I have contacted him by email and also via Twitter. If it becomes clear that this situation is permanent then we could consider starting all over again. It would be a pity to lose all those previous solutions though. Hopefully some previous contributors would resubmit their solutions.
Thank you very much. I find book very nice, but I do not have sufficient knowledge to solve all the exercises... The site was very useful for me. Hi Vasco
The RTR site is being overhauled and revamped. The guy who maintains the site is sorting it out, but it may take some time before it's up and running again.
The RTR site is working again as an archive site. If you use the link http://legacy.roadtoreality.info/ you will be able to access all parts of the old site. The links from my site have not yet been updated.
All links to exercise solutions in all chapters should now be working. If you find any links which are not working please let me know on this blog. Thanks Vasco
Thank you so much for rehosting this website. I am struggling through the book and I find it invaluable to see solutions. I am currently on Chapter 10.
I have a few solutions to problems no one else has uploaded to any of the sites. In addition I have caught errors and omissions in several posts and have corrected/extended those solutions. I have some solutions that are different and sometimes simpler than ones posted. I also have a different interpretation of what Penrose was asking for certain problems. Are you seeking additions posts?
Full disclosure - my solutions are hand written and would have to be uploaded as pdf files. My handwriting is pretty clear but not as good as if I had used an equation editor.
Hi Bud Thanks for the positive feedback. I am always happy to add new posts. Handwritten solutions are fine. If you want to upload them to google drive, share them by allowing anyone with the link to read and download them, and then post the link here, I will put a link on the website and your solutions will then be available to all our visitors.
I'm having a problem understanding the 1st equation on p. 239. I don't understand how if alpha is a p-form there can be an expression with a component of alpha that has q subscripts when p≠q. Are you permitting questions like this on your site and, if so, is there a different thread where I should post it?
I have created a Geometric Algebra package for Mathematica. It allows one to compute geometric products, wedge products, etc., either numerically or symbolically so that we don't have to do these lengthy calculations by hand. The package is available for download at https://github.com/matrixbud/Geometric-Algebra.
I am not subscribed to the forum but I had a thought that complements the posted solution to 12.10 -- namely what thinking of the dxdy as dx^dy gets you: x=rcos(t), y=rsin(t) so dx^dy =(cos(t)dr -rsin(t)dt)^(sin(t)dr +rcos(t)dt) = rcos(t)cos(t)dr^dt -rsin(t)sint(t) dt^dr = rdr^dt.
ReplyDeleteThe antisymmetry generates the Jacobean, which is a determinant, ie an antisymnmetric sum of products
In exercise [12.12], the conclusion d(dT) = 0 for T of degree 0 can be proved directly: d(dT) =
ReplyDeleted(S(dT/dxi)dxi) = SS(d2T/dxidxj)dxi^dxj =0 because dxi^dxj is antisymmetric but the second partial derivative (d2T/dxidxj) is symmetric. (Forgive my orthography, SS stands for the double summation on i and j. The indices i and j must be understood as raised or lowered as appropriate.) Establishing this result directly for degree zero permits one to establish it for all degrees by induction using the second of the two defining relations given immediately above in RtR.
I really found your solutions to the exercises of Road to REALITY helpful. :)
ReplyDeleteThe RTR site seems to be down again. I have contacted the owner of the site.
ReplyDeleteUnfortunately the owner of the site has not replied yet, so I do not know what is happening. I have contacted him by email and also via Twitter. If it becomes clear that this situation is permanent then we could consider starting all over again. It would be a pity to lose all those previous solutions though. Hopefully some previous contributors would resubmit their solutions.
ReplyDeleteI have had a reply from the owner of the site and he says he will fix things this coming week.
ReplyDeleteHi Vasco... Site is not available... I really found your solutions to the exercises of Road to REALITY helpful. :)
ReplyDeleteI hope that the website with the solutions of the exercises go back to work. The solutions help me a lot
ReplyDeleteYes, I will try and sort this out. I will get in touch with the guy who maintains the site.
ReplyDeleteThank you very much. I find book very nice, but I do not have sufficient knowledge to solve all the exercises... The site was very useful for me. Hi Vasco
ReplyDeleteThe RTR site is being overhauled and revamped. The guy who maintains the site is sorting it out, but it may take some time before it's up and running again.
ReplyDeleteThanks Vasco, you are very kind.
DeleteThe RTR site is PARTIALLY working again. Attachments are not yet working and links from this site need to be updated, which is quite a long job.
ReplyDeleteThe RTR site is working again as an archive site. If you use the link http://legacy.roadtoreality.info/ you will be able to access all parts of the old site.
ReplyDeleteThe links from my site have not yet been updated.
Vasco, thanks for this work. Link for Ex 3.2 gives File Not Found :-(
ReplyDeleteYes, I know about these link problems and I am slowly correcting them. All the links for chapter 2 now work. I will do chapter 3 next and so on
ReplyDeleteChapter 3 links now OK
ReplyDeleteChapter 4 links now OK
ReplyDeleteChapters 5 & 6 now have working links.
ReplyDeleteChapter 7,8 & 9 now have working links.
ReplyDeleteChapters 1-16 now have working links.
ReplyDeleteAll links to exercise solutions in all chapters should now be working. If you find any links which are not working please let me know on this blog. Thanks
ReplyDeleteVasco
Vasco,
DeleteThank you so much for rehosting this website. I am struggling through the book and I find it invaluable to see solutions. I am currently on Chapter 10.
I have a few solutions to problems no one else has uploaded to any of the sites. In addition I have caught errors and omissions in several posts and have corrected/extended those solutions. I have some solutions that are different and sometimes simpler than ones posted. I also have a different interpretation of what Penrose was asking for certain problems. Are you seeking additions posts?
Full disclosure - my solutions are hand written and would have to be uploaded as pdf files. My handwriting is pretty clear but not as good as if I had used an equation editor.
Bud Simrin
Hi Bud
DeleteThanks for the positive feedback. I am always happy to add new posts. Handwritten solutions are fine. If you want to upload them to google drive, share them by allowing anyone with the link to read and download them, and then post the link here, I will put a link on the website and your solutions will then be available to all our visitors.
Hi Vasco,
ReplyDeleteI'm having a problem understanding the 1st equation on p. 239. I don't understand how if alpha is a p-form there can be an expression with a component of alpha that has q subscripts when p≠q. Are you permitting questions like this on your site and, if so, is there a different thread where I should post it?
Bud
Hi Bud
DeleteYes I allow these kinds of posts. I have only just seen your post, as I don't check this blog every day. I will have a look at it
Vasco
I have created a Geometric Algebra package for Mathematica. It allows one to compute geometric products, wedge products, etc., either numerically or symbolically so that we don't have to do these lengthy calculations by hand. The package is available for download at https://github.com/matrixbud/Geometric-Algebra.
ReplyDeleteBud Simrin