# If the space-time curve was quantified and has a mathematical function what would the derivative of the function mean? [closed]

Say I have a massive object: 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:

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.

## closed as unclear what you're asking by AccidentalFourierTransform, WillO, Jon Custer, Sebastian Riese, Qmechanic♦Jun 29 '18 at 16:24

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• Your definition of "maximum curve" is ambiguous. Even confined to 3-D space, there are an infinity of curves that pass through any point. Further complicating the issue, in 4-D spacetime, an object's center is no longer a single point but a worldline. – enumaris Jun 20 '18 at 16:16
• – Qmechanic Jun 20 '18 at 17:51
• Ahoy! If you'd like to learn about the math necessary for understanding general relativity I'd highly recommend MathTheBeautiful's tensor calculus series on YouTube! – Eben Cowley Jun 27 '18 at 22:06

## 1 Answer

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.

• While true, I don't see how this answers the question posed. – Kyle Kanos Jun 21 '18 at 10:01
• @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. – Javier Jun 21 '18 at 10:59
• 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. – Kyle Kanos Jun 21 '18 at 11:07
• @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. – Javier Jun 21 '18 at 11:35
• @Nat there is no physical interpretation of the gradient of the sheet. How would you answer the question if not like this? – Javier Jun 21 '18 at 15:11