tag:blogger.com,1999:blog-14574087.post112527525000466668..comments2023-08-25T09:44:42.886-04:00Comments on The Lyceum: Time Travel, Part TwoDanielhttp://www.blogger.com/profile/10190478379984737926noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-14574087.post-46403643794281505232022-06-03T01:47:02.729-04:002022-06-03T01:47:02.729-04:00Hello maate nice postHello maate nice postBobby Mhttps://www.bobbymatthews.com/noreply@blogger.comtag:blogger.com,1999:blog-14574087.post-1125693911501059862005-09-02T16:45:00.000-04:002005-09-02T16:45:00.000-04:00Ash Sere, please let me know when your new story i...Ash Sere, please let me know when your new story is ready. It's nice to inspire something other than rolled eyes at philosophy parties.Danielhttps://www.blogger.com/profile/10190478379984737926noreply@blogger.comtag:blogger.com,1999:blog-14574087.post-1125438301888886152005-08-30T17:45:00.000-04:002005-08-30T17:45:00.000-04:00Ah, got it, there's a lack of bracketing going on ...Ah, got it, there's a lack of bracketing going on here! <BR/><BR/>(Having said that, it is my fault as division takes preference over subtraction.)<BR/><BR/>It's t = 1 /(1-(v^2/c^2))½.<BR/><BR/>Now it all makes sense. Sorry for my random ramblings... Save that it did raise the interesting sub topic of the mighty zero.<BR/><BR/>I like 0/0.<BR/><BR/>Ash.Matt McGrathhttps://www.blogger.com/profile/06076825151483605719noreply@blogger.comtag:blogger.com,1999:blog-14574087.post-1125437812675640622005-08-30T17:36:00.000-04:002005-08-30T17:36:00.000-04:00That 10 year loop suggestion in the middle gives m...That 10 year loop suggestion in the middle gives me the single greatest idea for a new story in about 6 months. Thanks! I'll have to work on it at some point when I'm through writing the two I'm presently writing...!<BR/><BR/>Anyway, more to the point, I was thinking about you're post of yesterday as I walked across the Rochester bridge this morning and I knew the maths didn't quite work, but I couldn't work out how.<BR/><BR/>Here it is:<BR/><BR/>You stated the Lorentz Equation as t=1/(1-v^2/c^2)½.<BR/><BR/>Then you said, as v approaches c then the denominator approaches zero. Yet this is not the case, surely? c = speed of light > 1. Therefore as v tends to c the denominator tends to (-1)½ which is an imaginary number. <BR/><BR/>I can only imagine it's meant to read something like t=1/(c^2-v^2/c^2)½. Though that would tend to 0/c^2 which is undefined (since you can't divide zero by anything.)<BR/><BR/>I guess I should Google it, or perhaps I'm getting something wrong.Matt McGrathhttps://www.blogger.com/profile/06076825151483605719noreply@blogger.comtag:blogger.com,1999:blog-14574087.post-1125409034126470042005-08-30T09:37:00.000-04:002005-08-30T09:37:00.000-04:00Thanks for the comments.Doh. I changed Verne to W...Thanks for the comments.<BR/><BR/>Doh. I changed Verne to Wells. Thanks.Danielhttps://www.blogger.com/profile/10190478379984737926noreply@blogger.comtag:blogger.com,1999:blog-14574087.post-1125366510594269352005-08-29T21:48:00.000-04:002005-08-29T21:48:00.000-04:00Verne? Surely you're thinking of Wells.Verne? Surely you're thinking of Wells.Brandonhttps://www.blogger.com/profile/06698839146562734910noreply@blogger.com