Message Boards Message Boards

0
|
3471 Views
|
2 Replies
|
0 Total Likes
View groups...
Share
Share this post:

Help me get Function N running to get large gammas in the Twin Paradox?

Posted 11 years ago

I've written this twice, months apart from scratch. Failed both times. I'll bet this is a trivial mistake! The attachment is .nb, identical to this paste, below.

(Twin Paradox)

(* dilation =tprime/time *)

(natural units, c=1 ; v in fraction of c )

(* At v = .999 c, etc. *)

v = 0.999; dilation = 1/ ((1 - (v^2))^0.5)

v = 0.999999; dilation = 1/ ((1 - (v^2))^0.5)

v = 0.999999999999; dilation = 1/ ((1 - (v^2))^0.5)

v = 0.999999999999999999999999; dilation = 1/ ((1 - (v^2))^0.5)

22.3663

707.107

707115.

Power::infy: Infinite expression 1/0. encountered. >>

ComplexInfinity

FullForm[dilation]

\!( TagBox[ StyleBox[ RowBox[{"DirectedInfinity", "[", "]"}], ShowSpecialCharacters->False, ShowStringCharacters->True, NumberMarks->True], FullForm])

dilation = N[1.000000000000000000000000/ ((1.000000000000000000000000 - (0.999999999999999999999999^2))^(1/ 2)), 1000]

Power::infy: Infinite expression 1/0.*10^-12 encountered. >>

ComplexInfinity

Attachments:
POSTED BY: Douglas Youvan
2 Replies

Marco,

Thanks. 0."24 nines" of c turns a billion years into about 12 hours!

10^9 / dilation 365 24

Doug

POSTED BY: Douglas Youvan

Hi,

this should work:

dilation = N[1/((1 - (SetPrecision[0.999999999999999999999999^2, Infinity]))^(1/2)), 1000]

It was just a little numerical problem. The term

((1.000000000000000000000000 - (0.999999999999999999999999^2))^(1/ 2))

is numerically zero:

0.*10^-12

Cheers,

Marco

POSTED BY: Marco Thiel
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Reply Preview
Attachments
Remove
or Discard

Group Abstract Group Abstract