# I Confusion about parallelogram in curved space-time

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1. Apr 4, 2017

### Sonderval

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.
As long as height changes are small, the time dilation on earth is linear with height,
$$t(h) = (1+gh/c^2) t(0)$$
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.

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?

2. Apr 4, 2017

### stevendaryl

Staff Emeritus
I'm not confident that this is the complete answer, but I think that John Baez is talking about making a rectangle whose sides are geodesics. In contrast, it sounds like Feynman is talking about a rectangle whose sides are not geodesics. If you wait 100 seconds at a constant height $h$ above the surface of the Earth, then you aren't following a geodesic. So that's not following a geodesic.

On the other hand, the reason I'm not confident in my answer is because my first intuition would be that the procedure that Feynman is talking about would give a coordinate-dependent quantity, not spacetime curvature.

3. Apr 4, 2017

### Sonderval

@stevendaryl
Good idea, I did not think of that.
I'm not sure it helps, though. Let's replace the vertical (time-oriented) edges of the square with lightcones (null geodesics), and the horizontal part with the world line of a free-falling particle that is initially at rest. (Or am I not allowed to use null geodesics here because they have zero four-length? Alternatively, I could use worldlines of a particle moving with v close to c.)
The distance in free falling for the horizontal part is proportional to t². Both horizontal lines will "fall" by the same amount of r=(1/2)gt², so this does not seem to change their distance.
I've tried to calculate this using a square with side lengths 1ns and 0,3m (to have meaningful ε).
The vertical fall is then 5E-18m; and this gets divided by c² in the formula for the time dilation I wrote down above; so I do not see how this can compensate for the quadratic dependence.