One way to see that spacetime is curved is to try and draw a "rectangle" in spacetime (see the figure in the Feynman lectures, ch 42.7): If I wait for 100 seconds and then move upwards on earth, I end up at a different point in spacetime than when I first move upwards and then wait for 100 seconds.(adsbygoogle = window.adsbygoogle || []).push({});

As long as height changes are small, the time dilation on earth is linear with height,

[tex] t(h) = (1+gh/c^2) t(0)[/tex]

The total distance in time between the two points I reach using the two different ways is thus t(h)-t(0), proportional to the time I wait (waiting 200s instead of 100 will double it) and proportional to the height.

This means that if I draw a small "square" with edges in the r- and t-direction and edgelength ε, the distance of the two end points to reach the opposite corner is proportional to ε².

However, both Penroses Road to reality (ch 14) as well as this page:

http://math.ucr.edu/home/baez/gr/torsion.html

state that in a torsion-free space, the opposite corners of a square constructed this way should touch up to order ε³.

So, obviously, I'm making a stupid mistake. Can anybody tell me where it is?

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# I Confusion about parallelogram in curved space-time

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