tag:blogger.com,1999:blog-2269351477810212131.post6829604728023765376..comments2017-06-10T07:53:37.316-07:00Comments on Copasetic Flow: Cool Math Tricks: Deriving the Divergence, (Del or Nabla) into New (Cylindrical) Coordinate SystemsHamilton Carterhttps://plus.google.com/108242372478733707643noreply@blogger.comBlogger16125tag:blogger.com,1999:blog-2269351477810212131.post-81714074455711818512016-07-18T23:21:29.281-07:002016-07-18T23:21:29.281-07:00Thank you a lot man!!!Thank you a lot man!!!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2269351477810212131.post-33072229743310223762016-06-22T04:48:17.164-07:002016-06-22T04:48:17.164-07:00thanxs bro.. great helpthanxs bro.. great helpAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2269351477810212131.post-1060460037639325182016-04-20T10:27:23.653-07:002016-04-20T10:27:23.653-07:00Awesome stuff man!
Your work saved me a lot of ti...Awesome stuff man! <br />Your work saved me a lot of time<br />Thanks a lotAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2269351477810212131.post-78319543368070362762015-11-15T10:23:31.601-08:002015-11-15T10:23:31.601-08:00Thanks a lot. I was searching for this material an...Thanks a lot. I was searching for this material and I found great your explanation !!<br /><br />Alexandre Santiagohttps://www.blogger.com/profile/17969382846880515900noreply@blogger.comtag:blogger.com,1999:blog-2269351477810212131.post-77065012287534185782015-11-15T09:43:52.392-08:002015-11-15T09:43:52.392-08:00Thanks a lot. I was searching for this material an...Thanks a lot. I was searching for this material and I found great your explanation !!<br /><br />Unknownhttps://www.blogger.com/profile/17969382846880515900noreply@blogger.comtag:blogger.com,1999:blog-2269351477810212131.post-26275074856960237672015-03-28T18:17:37.938-07:002015-03-28T18:17:37.938-07:00awesome awesome Amal Biswashttps://www.blogger.com/profile/08574313797070079074noreply@blogger.comtag:blogger.com,1999:blog-2269351477810212131.post-60094211648354131592012-09-10T10:29:36.218-07:002012-09-10T10:29:36.218-07:00Hi Doughboy,
Is the following for spherical coordi...Hi Doughboy,<br />Is the following for spherical coordinates about what you had in mind?<br /><a href="http://copaseticflow.blogspot.com/2012/09/an-intuitive-way-to-spherical-gradient.html" rel="nofollow">http://copaseticflow.blogspot.com/2012/09/an-intuitive-way-to-spherical-gradient.html</a><br /><br />Thanks again! You made my homework this week a lot easier and much more of a learning experience, (in a good way).Hamiltonhttps://www.blogger.com/profile/08273077706643157078noreply@blogger.comtag:blogger.com,1999:blog-2269351477810212131.post-83001005309956608592012-09-10T10:28:31.052-07:002012-09-10T10:28:31.052-07:00This comment has been removed by the author.Hamiltonhttps://www.blogger.com/profile/08273077706643157078noreply@blogger.comtag:blogger.com,1999:blog-2269351477810212131.post-16906578882513825012012-09-09T05:37:15.384-07:002012-09-09T05:37:15.384-07:00Thank you Doughboy! That's an awesome pointer...Thank you Doughboy! That's an awesome pointer! It took me awhile to work through all the details of what you said, but in the end it's far easier and left me with a physical understanding of the derivation instead of just a mechanical list of steps. Expect to see a post on deriving the spherical laplacian in this way soon. Thanks again!Hamiltonhttps://www.blogger.com/profile/08273077706643157078noreply@blogger.comtag:blogger.com,1999:blog-2269351477810212131.post-3201528278924983802012-09-07T14:06:18.822-07:002012-09-07T14:06:18.822-07:00Your derivation is great, but it's too lengthy...Your derivation is great, but it's too lengthy. You can actually derive it starting with the definition of divergence, which is the limit as the volume of a closed surface approaches zero of the flux through that surface. Taking the surface as the standard differential volume unit in cylindrical coordinates, you can directly derive it much more quickly than going through the chain rule and coordinate transformations.Doughboyhttps://www.blogger.com/profile/15559242691697477711noreply@blogger.comtag:blogger.com,1999:blog-2269351477810212131.post-56168254890922755382012-03-23T08:50:06.433-07:002012-03-23T08:50:06.433-07:00hi,
i'm from india and i found your explanatio...hi,<br />i'm from india and i found your explanation extremely helpful.i was able to derive expressions for divergence and curl on my own by following a method similar to yours.<br />it was enlightening. <br />thanks.manihttps://www.blogger.com/profile/03129239496565707962noreply@blogger.comtag:blogger.com,1999:blog-2269351477810212131.post-10750248491244518632010-05-18T14:17:34.033-07:002010-05-18T14:17:34.033-07:00Firstly, thanks for this, I've been trying in ...Firstly, thanks for this, I've been trying in vain to derive something similar and I've finally noticed that I've been using the wrong transformation of unit vectors!<br /><br />Similarly to previous comments, this particular result falls out fairly quickly when you note that r_x = cos(phi); r_y = sin(phi); phi_x = -sin(phi)/r and phi_y=cos(phi)/r. Otherwise a good exercise in the chain rule!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2269351477810212131.post-70770031939925812002010-03-01T11:00:05.184-08:002010-03-01T11:00:05.184-08:00Found your tutorial on google. Very nicely done. T...Found your tutorial on google. Very nicely done. Thank you!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2269351477810212131.post-13298614007379249252009-12-06T06:15:06.709-08:002009-12-06T06:15:06.709-08:00Hi,
Good article. I noticed a shortcut for d phi /...Hi,<br />Good article. I noticed a shortcut for d phi / dx and d phi / dy. If you use phi = arctan(y/x), <br />d phi / dx = 1/(1+(y/x)^2).d/dx(y/x)<br />=-1/(1+(y/x)^2).y/x^2<br />=-y/(x^2+y^2)<br />=-sin(phi)/r<br />similar for d phi/dy, and avoids a lot of messy algebra.John F.noreply@blogger.comtag:blogger.com,1999:blog-2269351477810212131.post-46336816229516260862009-11-05T13:58:06.035-08:002009-11-05T13:58:06.035-08:00Thank you....
great help!
--SuThank you....<br /><br />great help!<br /><br />--SuAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2269351477810212131.post-48439110465982494182009-10-27T09:26:57.924-07:002009-10-27T09:26:57.924-07:00Hi Hamilton,
very nice tutorial. Following your r...Hi Hamilton,<br /><br />very nice tutorial. Following your reasoning I was able to derive the Nabla from Cartesian coordinate system to my own defined non-standard spherical coordinate system. And it worked perfectly!<br /><br />Thanks again.<br /><br />HernanAnonymousnoreply@blogger.com