Say I have a massive object:

enter image description here

This object as we know causes spacetime to bend and curve. The "maximum curve" which I would define as the line in space time that runs directly through the center of the object has the maximum displacement from the space time of zero. This "maximum curve" would take the shape similar to a upside down bell curve or a upside down first derivative of a logistic function:

enter image description here enter image description here

What would the derivative of the above functions mean? (either the derivative of the bell curve or the second derivative of the logistic function). What would each individual point on the derivative mean: acceleration at that point? Velocity?

I apologize if my wording is poor. I am simply a curious high school student who has completed only the most basic physics and calculus courses. Any help in simplifying my questions or my descriptions would be welcome.


That picture is only an analogy of spacetime, but it is not how spacetime works. The curvature of spacetime is sadly much more complicated than a bent sheet, so the function that describes this curve is not really very meaningful. It's just an illustration, not a description of how gravity actually is.

  • $\begingroup$ While true, I don't see how this answers the question posed. $\endgroup$
    – Kyle Kanos
    Jun 21 '18 at 10:01
  • 4
    $\begingroup$ @Kyle sadly, sometimes the only answer is that the question is wrong. I could have tried to relate the shape of the sheet to some curvature-related function, but I didn't want to because I could have given the wrong impression that this "analogy" actually works. $\endgroup$
    – Javier
    Jun 21 '18 at 10:59
  • $\begingroup$ even if wrong, this a comment on the post (better also to include relevant questions, such as those that Qmechanic links) and should have been posted as such than posting a non-answer. $\endgroup$
    – Kyle Kanos
    Jun 21 '18 at 11:07
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    $\begingroup$ @Kyle respectfully, no it's not. It's about the derivative of the curve given by the curved sheet, which is not the metric tensor. This is exactly my point: I don't want to lead OP into thinking that the picture is somehow correct. $\endgroup$
    – Javier
    Jun 21 '18 at 11:35
  • 1
    $\begingroup$ @Nat there is no physical interpretation of the gradient of the sheet. How would you answer the question if not like this? $\endgroup$
    – Javier
    Jun 21 '18 at 15:11

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