Torsion & Non-Closed Rectangle in Feynman & Penrose

[Sonderval]

In the Feynman Lectures on Physics, Feynman explains the curvature of spacetime by drawing a rectangle in spacetime, see
http://www.feynmanlectures.caltech.edu/II_42.html Fig. 42.18
First waiting 100 sec and then moving 100 feet in height on Earth's surface results in a different situation than first moving 100 feet and then waiting 100 sec due to the time dilation that depends on height.
The time dilation effect is (h as height, g as gravitational acceleration, c=1, t(0) the time waiting at the lower height)
$$t(h) = (1+ gh) t(0)$$
So the difference between the end points in the diagram is given by $$g h t(0)$$. It is thus proportional to the area of the rectangle.
However, in Road To Realty, Fig 14.9 b and c, Penrose says that in a torsion-free space, a small rectangle with sidelength ε is closed up to an order ε³.
To me this seems like a contradiction - so obviously I'm making a mistake somewhere.
Can anybody tell me where I'm going off?

 

[Orodruin]

What Feynman describes is not a rectangle or even a paralellogram. The world-lines described are not geodesics.

 

[Sonderval]

@Orodruin
You are right, at least for the horizontal lines (the vertical lines are straight lines from one height to another, so they are spacelike geodesics in the Schwarzschild metric, are they not?). The horizontal lines connect two points at the same height and different times. If I replace them with geodesics (parabolas), I do not see how this changes the numbers since the parabola at h=0 and at h=100ft would differ only extremely slightly. I don't really see how this would change the order of the deviation.

 

[Orodruin]

the vertical lines are straight lines from one height to another, so they are spacelike geodesics in the Schwarzschild metric, are they not?

Feynman seems to be describing actually lifting something physically. To do that it needs to move on a time-like path. The assumption must be that the two lifts are related by a time translation (generated by the global time-like Killing field $\partial_t$. Without further specification it is impossible to know exactly what he meant. The parts where the object is sitting still at the surface or at h = 100 ft are definitely not a geodesics as the proper acceleration in those cases is the gravitational acceleration.

 

[Sonderval]

@Orodruin
I don't think Feynman supposes to lift anything - he says that we have two objects and follow the worldline of each of them.
And yes, as I said in my previous comments, to make to horizontal lines geodesics, we should and could use a parabola. If the waiting time is, for example 2 seconds, I could throw each object 5 Meters high (from its starting position) and have it follow its geodesics to reach the same height after 2 seconds.
But even in this case, I do not see how this changes the basic picture that the difference between the two times is proportional to ght(0).

 

[Orodruin]

This is still not sufficient. In order to get the statement of the torsion you need to make sure that the direction vectors of the second parallel transports are parallel transported along the first transports. In addition, the torsion statement is a local statement and a priori only needs to hold for infinitesimal transports (in other words, the relative displacement that is linear in both parallel transport distances vanish). The global difference is a different cup of tea.

 

[Sonderval]

@Orodruin
Thanks, I wil have to think about that.
I'm not quite sure why making the Feynman rectangle infinitesimally small would change anything and how exactly the transport of vectors along the geodesics would change the picture (After all, both the vertical spacelike geodesics and the two parabolas are identical), but I probably need to think about it a bit more...

 

[Sonderval]

So, after thinking a bit more about it, I think I finally understand it:
Consider the lower side of the rectangle and replace it with a parabola to make it a geodesic.
The vertical line on the left has to be parallel transported along this geodesic.
The geodesic starts with a vector that points upwards and to the right (object thrown upwards, initial four-velocity) and ends with a vector that points downwards and to the right (object falling down now).
Since the four-velocity vector rotates, the vertical line has to rotate accordingly during parallel transport.
Due to the minus-sign in the metric between space and time components, the vertical line rotates counter-clockwise when the four-velocity along the parabola rotates clockwise.
Therefore, on the right side of the graph, the line that was vertical before now points slightly to the left.
Thus, following this line now, we end up to the left of the graph, i.e. at a smaller value of time at height h than with a vertical line drawn.

Thanks again for helping me understanding this - it has bothered me for quite a while now.