01 January 1970 23 9K Report

Another thread here highlighted the need for a solid counterargument to Dingle's objection to Relativity Theory. It seems that previous participants were unable to resolve the issue convincingly. Therefore, we would like to present a straightforward and intuitive solution, providing easier access to understanding the problem.

We apply the theory of segmented spacetime to this problem. If you are not familiar with the theory yet, here is an illustrative paper. Please don’t dismiss it simply because it also addresses singularities - that’s only part of the discussion. The paper is straightforward and engaging to read.

Article Segmented Spacetime and the Natural Boundary of Black Holes:...

Essentially, the segmented spacetime theory proposes that gravitation segments space, slowing down objects, while moving objects are segmented themselves. This concept is expanded upon in a subsequent paper:

Preprint Segmented Spacetime - A New Perspective on Light, Gravity an...

Now, consider two clocks in different gravitational fields. In this scenario, space is segmented differently in each field, meaning these are distinct environments with varying segmentations. If we take a clock from environment A and place it near a clock in environment B, the time difference between them remains observable due to these distinct segmentations.

However, in the example with two clocks, where one moves and the other remains stationary, both exist within the same environment, with the same spatial segmentation. Here, any time difference arises from the moving clock itself: as it accelerates, its internal segmentation increases, causing it to run slower. Upon negative acceleration, these segments are removed, allowing the clock to speed up again.

This is why a clock accelerating and negative accelerating within the same environment will not retain a permanent time difference. It contrasts with moving a clock from environment A to environment B, where the segmentation is tied to the environment, not the clock itself.

We invite you to discuss this concept.

More Carmen Wrede's questions See All
Similar questions and discussions