Greg Egan
Apr7-04, 08:46 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resize=yes,status=no,wi dth=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In article <jOQnrrEir8aAFwO3@btopenworld.com>, Oz\n<acoohdb@btopenworld.com> wrote:\n\n> In discussion with my son I said a body in orbit followed a geodesic.\n>\n> He disagreed, and a $5 bet was taken.\n>\n> To me this seems a perfectly simple question, with a yes-no answer.\n>\n> However he has asked me to pose two questions, with which he hopes to\n> wriggle out, I guess:\n>\n> In a Schwarzschild metric (but not necessarily a black hole):\n>\n> 1) Is a geodesic the path of minimum proper time.\n\nNo, a geodesic in spacetime is a path of *maximum* proper time. That\'s\none of the crucial differences between spatial geometry and spacetime\ngeometry, which comes from the different signature of the metric (the\nfact that there\'s a minus sign for the time coordinate). To put it\nanother way, the proper-time hypotenuse T of a right triangle in flat\nspacetime (or a small enough right triangle in any spacetime) obeys the\nnot-quite-Pythagorean formula:\n\nT^2 = t^2 - x^2\n\nmaking the hypotenuse shorter than the timelike side.\n\nIf you stick two such triangles together to make a small detour, you get:\n\nB\n|\\\n| \\\nt | \\ T\n| \\\n|_x__\\\n| /\n| /\nt | / T\n| /\n|/\nA\n\nHere someone following the geodesic from event A to B experiences proper\ntime 2t, whereas someone who takes the detour experiences a shorter\nproper time 2T.\n\n> 2) Does an orbital geodesic have a greater or lesser proper time than a\n> radial one.\n>\n> I take no responsibility for the latter question making sense, or not.\n> As far as I can see they are solutions with different boundary\n> conditions.\n\nQuite right. It\'s beside the point to compare the lengths of geodesics\nwith different endpoints.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <jOQnrrEir8aAFwO3@btopenworld.com>, Oz
<acoohdb@btopenworld.com> wrote:
> In discussion with my son I said a body in orbit followed a geodesic.
>
> He disagreed, and a $5 bet was taken.
>
> To me this seems a perfectly simple question, with a yes-no answer.
>
> However he has asked me to pose two questions, with which he hopes to
> wriggle out, I guess:
>
> In a Schwarzschild metric (but not necessarily a black hole):
>
> 1) Is a geodesic the path of minimum proper time.
No, a geodesic in spacetime is a path of *maximum* proper time. That's
one of the crucial differences between spatial geometry and spacetime
geometry, which comes from the different signature of the metric (the
fact that there's a minus sign for the time coordinate). To put it
another way, the proper-time hypotenuse T of a right triangle in flat
spacetime (or a small enough right triangle in any spacetime) obeys the
not-quite-Pythagorean formula:
T^2 = t^2 - x^2
making the hypotenuse shorter than the timelike side.
If you stick two such triangles together to make a small detour, you get:
B
|\
| \
t | \ T
| \
|_x__\
| /
| /
t | / T
| /
|/
A
Here someone following the geodesic from event A to B experiences proper
time 2t, whereas someone who takes the detour experiences a shorter
proper time 2T.
> 2) Does an orbital geodesic have a greater or lesser proper time than a
> radial one.
>
> I take no responsibility for the latter question making sense, or not.
> As far as I can see they are solutions with different boundary
> conditions.
Quite right. It's beside the point to compare the lengths of geodesics
with different endpoints.
<acoohdb@btopenworld.com> wrote:
> In discussion with my son I said a body in orbit followed a geodesic.
>
> He disagreed, and a $5 bet was taken.
>
> To me this seems a perfectly simple question, with a yes-no answer.
>
> However he has asked me to pose two questions, with which he hopes to
> wriggle out, I guess:
>
> In a Schwarzschild metric (but not necessarily a black hole):
>
> 1) Is a geodesic the path of minimum proper time.
No, a geodesic in spacetime is a path of *maximum* proper time. That's
one of the crucial differences between spatial geometry and spacetime
geometry, which comes from the different signature of the metric (the
fact that there's a minus sign for the time coordinate). To put it
another way, the proper-time hypotenuse T of a right triangle in flat
spacetime (or a small enough right triangle in any spacetime) obeys the
not-quite-Pythagorean formula:
T^2 = t^2 - x^2
making the hypotenuse shorter than the timelike side.
If you stick two such triangles together to make a small detour, you get:
B
|\
| \
t | \ T
| \
|_x__\
| /
| /
t | / T
| /
|/
A
Here someone following the geodesic from event A to B experiences proper
time 2t, whereas someone who takes the detour experiences a shorter
proper time 2T.
> 2) Does an orbital geodesic have a greater or lesser proper time than a
> radial one.
>
> I take no responsibility for the latter question making sense, or not.
> As far as I can see they are solutions with different boundary
> conditions.
Quite right. It's beside the point to compare the lengths of geodesics
with different endpoints.