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posted by Fnord666 on Wednesday March 01 2017, @05:21PM   Printer-friendly
from the seemed-like-a-hack-anyway dept.

The dominant Lambda-CDM model is the standard model of physical cosmology, and it has proved reasonably successful. It does, however, have problems, such as dark matter, whose true nature remains elusive. Dutch physicist Erik Verlinde has, in a recent paper, proposed that gravity might not actually be a fundamental interaction at all, but rather an emergent property of spacetime itself, and as such, what current cosmological theory considers dark matter is really an emergent gravity phenomenon. Sabine Hossenfelder has an article about several recent tests of Verlinde's theory, which show that the idea might have promise.

Physicists today describe the gravitational interaction through Einstein's Theory of General Relativity, which dictates the effects of gravity are due to the curvature of space-time. But it's already been 20 years since Ted Jacobson demonstrated that General Relativity resembles thermodynamics, which is a framework to describe how very large numbers of individual, constituent particles behave. Since then, physicists have tried to figure out whether this similarity is a formal coincidence or hints at a deeper truth: that space-time is made of small elements whose collective motion gives rise to the force we call gravity. In this case, gravity would not be a truly fundamental phenomenon, but an emergent one.

[...] Verlinde pointed out that emergent gravity in a universe with a positive cosmological constant – like the one we live in – would only approximately reproduce General Relativity. The microscopic constituents of space-time, Verlinde claims, also react to the presence of matter in a way that General Relativity does not capture: they push inwards on matter. This creates an effect similar to that ascribed to particle dark matter, which pulls normal matter in by its gravitational attraction.

[...] So, it's a promising idea and it has recently been put to test in a number of papers.

[...] Another paper that appeared two weeks ago tested the predictions from Verlinde's model against the rotation curves of a sample of 152 galaxies. Emergent gravity gets away with being barely compatible with the data – it systematically results in too high an acceleration to explain the observations.

A trio of other papers show that Verlinde's model is broadly speaking compatible with the data, though it doesn't particularly excel at anything or explain anything novel.

[...] The real challenge for emergent gravity, I think, is not galactic rotation curves. That is the one domain where we already know that modified gravity – at last some variants thereof – work well. The real challenge is to also explain structure formation in the early universe, or any gravitational phenomena on larger (tens of millions of light years or more) scales.

Particle dark matter is essential to obtain the correct predictions for the temperature fluctuations in the cosmic microwave background. That's a remarkable achievement, and no alternative for dark matter can be taken seriously so long as it cannot do at least as well. Unfortunately, Verlinde's emergent gravity model does not allow the necessary analysis – at least not yet.

Previously:
Emergent Gravity and the Dark Universe


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  • (Score: 0) by Anonymous Coward on Thursday March 02 2017, @12:17AM (11 children)

    by Anonymous Coward on Thursday March 02 2017, @12:17AM (#473634)

    "Now go read some [..] as we need to get you out of short trousers first."

    How thoughtful, I ll take that under consideration. Now, if you are done with your childish nonsense and laughable patronizing attempts, please direct your attention here as I got something special for you too: it is called a free-falling spring, and will help you understand how you have been meticulously deluding yourself into believing that 'expanding the mind' means 'quoting shit others said or wrote'.

    The challenge is simple: a spring of length x and stiffness k is hanging at rest inside a gravity field g, and is released. Describe (math model) the motion of this free-falling spring.

    Good Luck (you are going to need it)

  • (Score: 2) by HiThere on Thursday March 02 2017, @12:45AM (7 children)

    by HiThere (866) Subscriber Badge on Thursday March 02 2017, @12:45AM (#473640) Journal

    That's a very simple problem. You just model the "free falling spring", and stop before it collides with anything. Of course, the shape of the spring, and how it's rotating will introduce a few complexities even before the first collision. ... When, among other things, you need to start considering the elasticity of to two surfaces (as well as the reaction of the rest of the spring). So stop first.

    A fairer example would be considering how the air moves past the blades of a rotating fan. There are lots of places you can look things about that up without needing to rent time on a supercomputer.

    --
    Javascript is what you use to allow unknown third parties to run software you have no idea about on your computer.
    • (Score: 0) by Anonymous Coward on Thursday March 02 2017, @01:22AM (6 children)

      by Anonymous Coward on Thursday March 02 2017, @01:22AM (#473656)

      That's a very simple problem.

      Oh really? Solve it then, and report back with your solution.

      Or don't, and just keep believing you know "how simple" it is.

      • (Score: 1) by khallow on Thursday March 02 2017, @12:57PM (5 children)

        by khallow (3766) Subscriber Badge on Thursday March 02 2017, @12:57PM (#473828) Journal

        Oh really? Solve it then, and report back with your solution.

        The other poster already did.

        • (Score: 0) by Anonymous Coward on Friday March 03 2017, @12:08PM (4 children)

          by Anonymous Coward on Friday March 03 2017, @12:08PM (#474326)

          The other poster already did.

          Hi khallow, no, he did not, perhaps my fault for not explaining the problem better. I welcome you to give it a go: a spring of mass m and stiffness k is at rest hanging vertically inside a gravitational field g and is released. The question is to derive the equation of motion for the bottom tip of the spring as a function of the vertical distance from the ground against time, from the moment the spring is released to the moment it contacts the ground, and the answer is to be expressed this in terms of m, k and g, or any other parameter pleases you as long as you formulate it in math (differentials, a polynomial or a plot are all good) as long as you can justify your derivation.

          • (Score: 1) by khallow on Friday March 03 2017, @03:08PM (3 children)

            by khallow (3766) Subscriber Badge on Friday March 03 2017, @03:08PM (#474370) Journal

            Hi khallow, no, he did not, perhaps my fault for not explaining the problem better. I welcome you to give it a go: a spring of mass m and stiffness k is at rest hanging vertically inside a gravitational field g and is released. The question is to derive the equation of motion for the bottom tip of the spring as a function of the vertical distance from the ground against time, from the moment the spring is released to the moment it contacts the ground, and the answer is to be expressed this in terms of m, k and g, or any other parameter pleases you as long as you formulate it in math (differentials, a polynomial or a plot are all good) as long as you can justify your derivation.

            You should have said that in your original demand. But even so, the process for deriving the solution was laid out. But having said that, I'd use the Lagrange approach:

            L = K - V.

            Here, K and V are the kinetic and potential energy of the spring. There are a variety of possibilities depending on how the mass and potential energy are distributed. In general, you can always find a solution computationally, even with complex three dimensional dynamics or if the spring can break or other sort of "memory" (where the current behavior of the spring depends on what happened to it in the past).

            But let's suppose the simplest case where the mass of the frictionless spring is all concentrated in a single point with height x - larger means higher, the potential energy of the spring follows Hooke's law, and the gravitational field is constant. Then
            K = 1/2 mv^2 (v = dx/dt, m is the constant mass of the spring concentrated at point x) and
            V = 1/2 k(x_0 - x)^2 + mgx where k is a stiffness coefficient (not necessarily your stiffness coefficient, g is the acceleration due to the gravity field). The potential energy of the spring is just the sum of the potential energy of the spring and the potential energy of the mass in the gravitational field.

            The key trick to using the Lagrangian L is that the partial derivative of L with respect to x is equal to the normal derivative with respect to time of the partial derivative of L with respect to v = dx/dt. Hmmm, let's call it

            \dee L/ \dee x = d/dt( \dee L/ \dee v).

            Then

            k(x - x_0) + mg = m dv/dt.

            Still looks pretty simple to me.

            • (Score: 0) by Anonymous Coward on Friday March 03 2017, @05:12PM (2 children)

              by Anonymous Coward on Friday March 03 2017, @05:12PM (#474448)

              The question is for the bottom of the spring, so the spring has to be considered at least as a one-dimensional 'physical' object; no point masses.

              k(x - x_0) + mg = m dv/dt.

              Again, it is the bottom of the spring that is concerned as a function of time: does x here represent the bottom of the spring?

              • (Score: 1) by khallow on Friday March 03 2017, @06:34PM (1 child)

                by khallow (3766) Subscriber Badge on Friday March 03 2017, @06:34PM (#474487) Journal

                The question is for the bottom of the spring, so the spring has to be considered at least as a one-dimensional 'physical' object; no point masses.

                If you're assuming that the mass distribution is distributed along the spring, then we'll have to do something different. Let's suppose uniform distribution of mass. x(a) (actually x(a) := x(a;t)) is the position of the spring where fraction a mass is on one side (as parameterized by the coefficient a) at given time t. Overlapping of the spring is allowed. Kinetic energy and the potential energy due to gravity do not change much:

                K = 1/2 m integral{a=0 to 1} dx(a)/dt da, and
                P_{grav} = mg integral {a=0 to 1} x(a) da.

                The potential energy due to the spring is now proportional along the length of the spring to the square of change in parameter a from a default rest state spring with no force acting on it. P_{spring} = 1/2 k integral{a=0 to 1} (dx(A)/da - b)^2 da. And fix the initial parameters of the spring (x(a; t=0). b happens to be the at rest length of the spring without tension (so b is your "x"). Now, we have a variational equation. We still can use the Lagrangian:

                d/dt \delta L/ \delta (dx(a)/dt) = \delta L / \delta x(a). Solving the math becomes either a matter of a computer model, or getting all your integrals in terms of the variation of x(a), not various derivatives of the variation of x(a) (there's a second derivative of time t, which needs to be reexpressed in terms of a Kronecker delta function, and a first derivative of parameter a, which can be transformed via some integration by parts manipulation, as integrals of some function of x(a) and its derivatives collectively times a single counter-factor, the variation of x. The first factor thus has to be identically zero resulting in a differential equation. I feel there's little point to going that far though. Notice that while there is a lot of it, the math has come through via a variety of simple rules.

                And as HiThere noted, once you have the model, you just use the model to compute the position of the spring. That remains simple as expected. Once we have the basic equation of motion, we can then add other effects such as your interaction with a solid surface, internal friction of the spring, etc. While that may look very hard from our very limited viewpoint it doesn't actually change the computational complexity of the problem that much.

                • (Score: 0) by Anonymous Coward on Friday March 03 2017, @07:37PM

                  by Anonymous Coward on Friday March 03 2017, @07:37PM (#474524)

                  If you're assuming that the mass distribution is distributed along the spring

                  I am,

                  And as HiThere noted, once you have the model, you just use the model to compute the position of the spring.

                  No dispute there: only I do not have such a model, and it is not trivial to produce. Your honest analysis may change that, so I guess I can treat it as homework, implement a model and see what happens.

  • (Score: 2) by FatPhil on Thursday March 02 2017, @10:27AM (2 children)

    by FatPhil (863) <pc-soylentNO@SPAMasdf.fi> on Thursday March 02 2017, @10:27AM (#473791) Homepage
    The top of the spring falls under both gravity and the tension on the spring. The bottom of the spring, and any other part of the spring that's not the top, was under gravity/tension equilibrium before it was dropped, and therefore remains stationary until the tension disappears, i.e. the spring collapses to its neutral extension. Assuming a slinky-like spring, where the neutral extension is totally collapsed, the spring will simply collapse, base unmoving. This is independent of the massiness or loadedness of the spring, as it depends purely on the initial state being tension/gravity equilibrium.

    Why are you asking a pure mathematician an applied maths question? How will that help you grasp metaphysics?
    --
    Great minds discuss ideas; average minds discuss events; small minds discuss people; the smallest discuss themselves
    • (Score: 1) by curril on Thursday March 02 2017, @11:20PM

      by curril (5717) on Thursday March 02 2017, @11:20PM (#474175)

      Sorry, that's not correct. The instant the spring is released, the top the spring will start moving down, releasing tension and the bottom of the spring will start falling (albeit slower than the top). The center of mass of the spring will fall strictly by the acceleration of gravity g at that point, while the ends of the spring will oscillate about the center of mass according to Hook's Law and simple harmonic oscillation. Of course, if the length of the spring is such that g can't be assumed to the same at the ends of the spring, then spring will also experience tidal forces that change the frequency of oscillation over time.

    • (Score: 0) by Anonymous Coward on Friday March 03 2017, @12:38PM

      by Anonymous Coward on Friday March 03 2017, @12:38PM (#474332)

      was under gravity/tension equilibrium before it was dropped, and therefore remains stationary until the tension disappears [..] where the neutral extension is totally collapsed, the spring will simply collapse, base unmoving

      Are you proclaiming the slingy a special case spring that exactly cancels out it stiffness with gravity so its bottom always stays put until the top collapses, even when its mass or stiffness is messed with? If so, please math it up (express it as a function of time)

      Why are you asking a pure mathematician an applied maths question?

      Sorry, you lost me: who is the applied mathematician?

      How will that help you grasp metaphysics?

      Where did that one come from? Would you mind if we settle the math part of the discussion first?