3
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – CuriousOne Oct 13 '14 at 20:47
  • 2
    $\begingroup$ 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. $\endgroup$ – HDE 226868 Oct 13 '14 at 23:00
  • $\begingroup$ @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. $\endgroup$ – peterh - Reinstate Monica Oct 14 '14 at 16:59
  • $\begingroup$ @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. $\endgroup$ – peterh - Reinstate Monica Oct 14 '14 at 17:01
  • $\begingroup$ @PeterHorvath: Do you have a citation for a paper that does the analysis? I would love to read it. $\endgroup$ – CuriousOne Oct 15 '14 at 6:09
3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Pulsar Oct 14 '14 at 18:03
  • $\begingroup$ @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. $\endgroup$ – Kyle Kanos Oct 14 '14 at 18:08
  • $\begingroup$ 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. $\endgroup$ – Pulsar Oct 14 '14 at 18:16
  • $\begingroup$ @Pulsar: You are correct, it is an exponential expansion. I've amended the opening statement to reflect that. $\endgroup$ – Kyle Kanos Oct 14 '14 at 18:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.