tag:blogger.com,1999:blog-8030944689053965636.post2781577220943501445..comments2023-09-25T03:59:32.505-05:00Comments on $\hbar = c = 1$ : Why is the Lorentz transformation linear?Unknownnoreply@blogger.comBlogger7125tag:blogger.com,1999:blog-8030944689053965636.post-27969844960817884352017-12-07T06:13:01.742-06:002017-12-07T06:13:01.742-06:00Why can't they of an exponential form which is...Why can't they of an exponential form which is still linear?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8030944689053965636.post-69431630719027717232017-12-07T06:10:32.923-06:002017-12-07T06:10:32.923-06:00Why can't the transformation be in an exponent...Why can't the transformation be in an exponential form? It is still linear.ArimaKouseihttps://www.blogger.com/profile/12487770871455000864noreply@blogger.comtag:blogger.com,1999:blog-8030944689053965636.post-22873607493548510742014-04-11T14:24:02.778-05:002014-04-11T14:24:02.778-05:00In your example, $\frac{\partial^2 y}{\partial x_1...In your example, $\frac{\partial^2 y}{\partial x_1 \partial x_1} = -\frac{1}{4}\frac{1}{\sqrt{x_1^3}} \ne 0$. aichihttps://www.blogger.com/profile/10170976534690871504noreply@blogger.comtag:blogger.com,1999:blog-8030944689053965636.post-19533591207743283812014-04-03T00:09:20.600-05:002014-04-03T00:09:20.600-05:00If y = square(x1) + square (x2), it is non-linear
...If y = square(x1) + square (x2), it is non-linear<br /><br />but its second-order derivatives, y'' = zero.<br /><br />so, y''=0 does not imply linearity, right or wrong?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8030944689053965636.post-60024677467271601792014-02-09T14:12:28.427-06:002014-02-09T14:12:28.427-06:00you forgot the dt/dt' term in your first red t...you forgot the dt/dt' term in your first red term.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8030944689053965636.post-74019249273595401062014-01-06T17:55:36.013-06:002014-01-06T17:55:36.013-06:00This comment has been removed by the author.aichihttps://www.blogger.com/profile/10170976534690871504noreply@blogger.comtag:blogger.com,1999:blog-8030944689053965636.post-69722802524640719612013-12-28T22:19:02.042-06:002013-12-28T22:19:02.042-06:00Thanks!
But what happened with the (d^2 t / dt'...Thanks!<br />But what happened with the (d^2 t / dt'^2) - terms?<br />I couldn't find out how to type formula here, so here's a screenshot<br />http://www.psicolor.de/derivate.pngAnonymousnoreply@blogger.com