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We know the size of observable universe, that it is expanding, that expansion is accelerating. Does sufficiently precise data exist to determine at least the SIGN of the next (the 3rd) derivative? If Dark Energy is causing acceleration, shouldn't expansion dilute the repulsive force of dark energy? While continuing to accelerate, shouldn't the rate of acceleration diminish over time? Shouldn't $\frac{{d^3}s}{dt^3}$ be negative? If were again surprised and determine that the acceleration itself is increasing, then maybe we're not exploding, maybe we're being pulled apart.

I'll leave theory to others to ponder. I just want to know when we can do some more High School math to determine the next, the 3rd, derivative of the size of the universe? Does the data exist?

Wouldn't it be nice to know?

Kyle Kanos
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William Schwartz
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    There is no experimental evidence for the third derivative. Not that having some would help, because derivatives only give rise to local (in this case short time) polynomial approximations. The long term dynamic could be periodic, chaotic or even catastrophic (big rip) and there would be absolutely no information about that in the local derivatives. – CuriousOne Oct 13 '14 at 20:47
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    If you know $a(t)$, you can use simple high school math to take the 3rd derivative. Also, we're not exploding or being pulled apart. Space is expanding, not matter. – HDE 226868 Oct 13 '14 at 23:00
  • @CuriousOne 1: I think, analyzing the data of enough 1a supernovas were enough. And we have currently much more 1a data as the acceleration was determined. 2: Did you ever heard about Taylor series? As more derivative you know, as better forecasts can you make from the future. – peterh Oct 14 '14 at 16:59
  • @HDE226868 Yes, it is true - although I think practically he wants to know the actualy best available experimental data to the 3rd derivative. IMHO, the conclusion has only secondary priority in his question. – peterh Oct 14 '14 at 17:01
  • @PeterHorvath: Do you have a citation for a paper that does the analysis? I would love to read it. – CuriousOne Oct 15 '14 at 06:09
  • @CuriousOne No, sorry. For me is only unclear, what were the problem with a simple 3rd grade polynomial regression to the latest 1a database. As I know, before around 4mrd years ago the acceleration of the size of the universe was yet negative. Since then it is positive. It is a clear sign, there is a jerk. – peterh Oct 15 '14 at 09:47
  • @PeterHorvath: I would like to see the error calculus on that. Having said that, if the signal is there, it's a clear indication that knowing the higher polynomial orders at any given time without knowing the underlying dynamics is pretty low in information content. If there was a jerk, and we don't know its physics, what does that tell us about the future? – CuriousOne Oct 15 '14 at 17:37
  • @CuriousOne 1: with them, we had better predictions. 2: knowing the jerk could maybe show some interesting about the underlying physics, too. – peterh Oct 16 '14 at 08:53
  • @PeterHorvath: The prediction would always be model dependent... and if the model is wrong (which we can't know until it's the far future), then the prediction is worthless. Even the weather report is scientifically on better footings than the future of the universe, I am afraid. – CuriousOne Oct 16 '14 at 09:20
  • @CuriousOne Exactly on this logic you could even deny the importance of the calculation of the 2nd derivative soon. – peterh Oct 16 '14 at 09:38
  • @CuriousOne "There is too much error" (proved with statistical data), or "there is no visible difference from the prediction" (proved with statistical data), these all were valid answers. But these all needed calculation, and such, paper on arxiv. For example, earlier with 6e+9 years, it were probably possible to predict the current accelerating expansion, but only based on the jerk. – peterh Oct 16 '14 at 10:53
  • @PeterHorvath: You have to excuse me. I am an experimental physicist who doesn't care about irrelevant calculations. One can construct cosmological models all day long and fit data to them (statistically relevant, or not), that still doesn't allow us to predict things in the future that haven't been measured. Case in point: We have "known" about the Higgs for something like 40 years. If that was enough in science, we didn't have to build LHC to observe it. And as far as cosmology is concerned: the models about the future of the universe have changed three times in my lifetime, alone. – CuriousOne Oct 16 '14 at 14:27

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If we take the simple approach of determining the state of the "jerk" today by assuming an exponential expansion (e.g., $a(t)\sim\exp(H_0 t)$), then $$ \dot a=H_0a\tag{1} $$ The derivative of this is then, $$ \frac{d^2a}{dt^2}=H_0\dot{a}=H^2_0a $$ And now for the "jerk," $$ \frac{d^3a}{dt^3}=H^2_0\dot{a}=H^3_0a\tag{2} $$ The Hubble constant is already pretty small at about 70 km/s/Mpc (2.26$\cdot$10$^{-18}$ 1/s), so taking the cube of this makes for a very small number and is probably a difficult value to tack down observationally. Since the scale factor is non-negative, then we can conclude that (2) is also non-negative, at least for now

Adding the cosmological constants (and neglecting the expected 0 values) such that (1) becomes $$ \dot{a}=H_0a\sqrt{\Omega_{M,0}a^{-3} + \Omega_{\Lambda,0}}\tag{3} $$ probably won't change the sign of the result, given that both $\Omega_{M,0}$ and $\Omega_{\Lambda,0}$ are positive constants.

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Kyle Kanos
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    Your definition (1) is not correct. $\dot{a}=H(t)a$, by definition. $H_0$ only gives you the present-day value of $\dot{a}$, so you cannot use it to calculate higher-order derivatives. – Pulsar Oct 14 '14 at 18:03
  • @Pulsar: Did you miss the part where I said that I'm taking the simple approach? Perhaps I should clarify that this would be the appropriate value for today and not all time that you are mistaking it for. – Kyle Kanos Oct 14 '14 at 18:08
  • In your simple approach you haven't omitted cosmological constants, in fact you assumed that the energy content of the universe is a cosmological constant, so that the expansion is exponential. But fair enough, your conclusion that the third derivative is positive is correct. I'll retract my downvote. – Pulsar Oct 14 '14 at 18:16
  • @Pulsar: You are correct, it is an exponential expansion. I've amended the opening statement to reflect that. – Kyle Kanos Oct 14 '14 at 18:20