H1    Gravitation and the Curvature of Space-Time

Below is a space-time diagram, as we learned to draw with STR in section C:

Clocks are placed at the positions xA and xB and both have been set to zero in A and B. Through the projection of the space-time position of the clocks on the time-axis, we can determine what time the clocks show at that point. One can use the time-axis to read the elapsed proper time for each object since the start of an operation.

Let the gravity at position xB be stronger than at xA. What can be done so that at point xB along the path BB'  time runs a bit less quickly than at point xA along the path AA' , which is of the same length in space-time? Epstein shows us the simple yet elegant solution: we only need to bend the time-axis a bit away from B to produce the desired effect!

Objects that linger at point xA move on a circle centered at Z with radius ZO + xA. The corresponding applies to stationary objects at xB. It is important that the two arc segments AA' and BB' are exactly the same length. This arc length specifies how long the process lasted for an observer in the OFF. We can read the elapsed proper time for A or B, by projecting A' or B' vertically on the now circular time-axis. We need only link A' and B' with Z. The clock at B' apparently shows less time than the one at A'!

The curvature of the time-axis, i.e., the inverse of the radius ZA = ZA', is critical for the strength of the effect at point xA. This curvature is a function of the distance r from the center of the central mass. This is to the right of xB, because gravity is stronger at xB than xA. If the curvature is zero, i.e., the radius ZO is infinitely large as in the first Epstein diagram above, then there is no effect from gravity and we are back in the STR.

We have now described the behavior of static objects in a gravitational field. As examples of such objects, imagine an apple in the Netherlands or a pine cone in the Swiss Alps each of which is hanging fixed on its branch. But what happens when the connection between the stem and the fruit breaks? The Epstein diagram holds the answer to this question:

Apple and branch are located at time zero at point A in space-time. Just then the connection between the two is cut. The branch moves up as fixed from A to C, while the apple moves along the tangent to arc AC, i.e., along the red dotted line and arrives at B! For the freely falling apple, there is no longer gravity. For the apple it is rather as if the tree is being prevented by the ground from following an inertial trajectory. After a short time, a distance Δx opens between the apple and the branch, whereby the apple has moved in the direction of the center of mass. The arc length AC and the segment length AB are again exactly the same length.

The branch traverses the arc AC proportional to time. At point P on the arc, the angle between the tangent to the circle at P and the red line AB is equal to the angle PZO. This angle indicates, however, how quickly the apple is moving away from the branch at a given moment, that is, the relative velocity of the apple and branch. Thus we can even read Galileo’s law for falling bodies from the diagram: Regardless of its mass the apple has a speed proportional to the elapsed time! Recall, however, that this law has a limited scope: v must be much smaller than c!

The angle between two curves in the space-time diagram still indicates a relative velocity. More precisely: sin(φ) = v/c ! The maximum angle is still 90º. Light moves, as usual, perpendicular to the time-axis. However, in the case of apples falling at the earth's surface, a quantitative analysis of the intermediate angle is tricky, as section H3 will show. But before doing so we would like to use our bent space-time diagrams to present a new variant of the twin paradox.