-3

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.

Qmechanic
  • 201,751
user15836
  • 11
  • 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
  • 1
    Related: https://physics.stackexchange.com/q/155328/2451, https://physics.stackexchange.com/q/90592/2451 , http://physics.stackexchange.com/q/7781/2451 and http://physics.stackexchange.com/a/13839/2451 and links therein. – 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 Kadile Jun 27 '18 at 22:06

1 Answers1

5

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.

Javier
  • 28,106
  • While true, I don't see how this answers the question posed. – Kyle Kanos Jun 21 '18 at 10:01
  • 4
    @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 I disagree. OP had some doubts and I (hopefully) clarified them. This should count as an answer, even if it's not very satisfying. To put it another way: if this shouldn't be an answer, what should? Should the question be closed? "False premises" is not AFAIK a closing reason. – Javier Jun 21 '18 at 11:11
  • the question is about the meaning of the derivative of the metric tensor. You could have addressed what meaning there could arise in doing such a thing. – Kyle Kanos Jun 21 '18 at 11:19
  • 2
    @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
  • This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. – Nat Jun 21 '18 at 15:08
  • Have to agree with @KyleKanos in that this looks like a comment on the nature of the curved sheet illustration rather than an answer to the question about how to interpret the "derivative" (gradient, I'd think) of it. – Nat Jun 21 '18 at 15:10
  • 1
    @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
  • Honestly my mind's elsewhere right now, so dunno what I'd like to say if I got to think this out. As a knee-jerk, I think that those graphics are kinda like gravitational potential energy wells, and then the gradient would point at the steepest slope out of 'em, so it'd basically be the negative of gravitational force. Then I'd probably connect that to Einstein's elevator thought experiment, I'd guess? – Nat Jun 21 '18 at 15:30
  • Here we go, from Wikipedia: "Given the universality of free fall, there is no observable distinction between inertial motion and motion under the influence of the gravitational force. This suggests the definition of a new class of inertial motion, namely that of objects in free fall under the influence of gravity. This new class of preferred motions, too, defines a geometry of space and time—in mathematical terms, it is the geodesic motion associated with a specific connection which depends on the gradient of the gravitational potential.". – Nat Jun 21 '18 at 15:37