I've found that the speed of expansion is about $70\frac{km}{s\cdot 10^{6}pc}$.

But what is the acceleration of expansion in terms of $\frac{km}{s^{2}\cdot 10^{6}pc}$ ?

I want to know how fast this value of "70" is changing, ie what is $\mathrm{d}H_0/\mathrm{d}t$?

  • $\begingroup$ Did you read the WIkipedia page on this subject and if so, can you be more specific about what you're having trouble with ? $\endgroup$
    – StephenG
    Sep 7 '17 at 10:58
  • $\begingroup$ I've read, I didn't found an answer, I've corrected my question to be a bit more specific $\endgroup$ Sep 7 '17 at 12:00
  • 1
    $\begingroup$ Not clear to me we've had enough observation time to answer this question. Let's let the visible universe expand to double its current size, then remeasure the Hubble constant to see if it's still around 70, or gone up to 75. That'd allow your calculation, but I don't think 10 or 20 years worth of data is going to give much accuracy. $\endgroup$ Sep 8 '17 at 13:58
  • $\begingroup$ Double is too much and not necessary, obviously... So you've meant, we can not estimate acceleration rate with any reasonable error yet? Then I should learn how do we know, it it even exist :) $\endgroup$ Sep 9 '17 at 16:17
  • 3
    $\begingroup$ @WayfaringStranger You don't have to wait for the universe to expand in the future to measure the rate of change of $H_o$. You have 13.8 billion years worth of measurements you can make to determine $dH_o/dt$. $\endgroup$
    – zephyr
    Sep 10 '17 at 22:54

The figure below shows the evolution of the Hubble parameter $H$ from 10 billion years (Gyr) ago, to 10 Gyr into the future:


As you can see, the change in $H$ is modest nowadays, compared to the past. The "acceleration" of the expansion at any point in time is given by the tangent to the curve at that time. Today, the change is $dH/dt \simeq -1.2\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}\,\mathrm{Gyr}^{-1}$, or roughly $-10^{-17}\,\mathrm{km}\,\mathrm{s}^{-2}\,\mathrm{Mpc}^{-1}$.

In other words, it will take roughly a billion years before the expansion rate has decreased by $1\,\mathrm{km}\,\mathrm{s}^{-1}$, and it asymptotically approaches some $57\,\mathrm{km}\,\mathrm{s}^{-1}$.

Since $H$ is defined as $(da/dt)\,/\,a$, where $a$ is the scale factor (the "size" of the Universe), a constant $H$ implies that $a\propto e^{Ht}$; that is, the size of the Universe increases exponentially.

  • $\begingroup$ Wait I thought this is backwards, and expansion is accelerating... $\endgroup$ Sep 11 '17 at 17:09
  • $\begingroup$ @ЕвгенийАртеменко: The expansion rate at a given distance approaches a constant value, 55 km/s/Mpc. But the velocity with which a given galaxy recedes from us increases, because it increases its distance, and hence lies at a distance where the expansion rate is larger. A galaxy that is today 1 Mpc away, recedes at 68 km/s/Mpc. In a billion years, it will be at a distance of 1.1 Mpc, where the expansion rate is 73 km/s/Mpc, and so on. $\endgroup$
    – pela
    Sep 11 '17 at 18:45
  • $\begingroup$ Aw man, second time today! @RobJeffries. $\endgroup$
    – pela
    Sep 11 '17 at 20:10
  • $\begingroup$ @ЕвгенийАртеменко I added a paragraph on this. $\endgroup$
    – pela
    Sep 11 '17 at 20:25
  • $\begingroup$ The acceleration of the expansion is surely $d(Ha)/dt = H\dot{a} + a\dot{H}$? $\endgroup$
    – ProfRob
    Apr 19 '20 at 17:27

Pela's answer gives the numerical value, I thought I would just explain the difference between $\dot{H}$, which is negative, and an acceleration, which is positive.

The Friedmann acceleration equation is given by $$\frac{\ddot{a}}{a} = -\frac{4 \pi G}{3}\left(\rho+\frac{3p}{c^2}\right) + \frac{\Lambda c^2}{3}, \tag*{(1)}$$ whilst the Hubble parameter is defined as $$H = \frac{\dot{a}}{a} \tag*{(2)}$$

In equation (1) the first term gets smaller with time since both matter density $\rho$ and pressure $p$ due to matter and radiation become smaller. The second term due to the cosmological constant $\Lambda$ is positive and eventually dominates and we are in that regime now. This is what is meant by an accelerating universe, where the second derivative of the scale factor $\ddot{a}$ is positive.

However, you have modified your question to ask what the time derivative of the Hubble parameter (not constant!) is. Differentiating equation (2) with respect to time, we have $$\dot{H} = \frac{\ddot{a}}{a} - H^2$$ $$\dot{H} = -\frac{4 \pi G}{3}\left(\rho+\frac{3p}{c^2}\right) + \frac{\Lambda c^2}{3} - H^2 \tag*{(3)}\, .$$

$\dot{H}$ is the gradient of the plot shown in Pela's answer and it is negative. That is $H$ is getting smaller. As the universe gets bigger, $H$ reaches its asymptotic value when $\ddot{a}/a = H^2$.


You have 1pc=3.26ly=3.26*300000*365.25*24*3600 km = ...


So you could easily find H0 :)


Sorry I've read to quickly the question. The acceleration rate of the universe is given by second equation of Friedmann Lemaître.

$$\frac{\ddot{a}}{a} = -\frac{4 \pi G}{3}\left(\rho+\frac{3p}{c^2}\right) + \frac{\Lambda c^2}{3}$$

with $\rho$ and p density of the fluids (matter, dark energy...) you consider in the cosmological (standard) model.



  • $\begingroup$ The OP is not asking about expansion rate $H_0$ (which he actually gives in the question), but about the acceleration of this rate. $\endgroup$
    – pela
    Sep 8 '17 at 9:30
  • 1
    $\begingroup$ This still doesn't answer the question because you don't show how this equation relates to $dH_o/dt$ or what the actual numerical value works out to, which is what the OP was asking for. $\endgroup$
    – zephyr
    Sep 10 '17 at 22:52
  • $\begingroup$ @zephyr I started to answer the questions along the same line. It is not clear what is meant by the "acceleration rate of the Universe"; $\ddot{a}/a$ might be as good a definition as any, as might $\dot{H_0}$, but they are not the same thing. $\endgroup$
    – ProfRob
    Sep 11 '17 at 9:16
  • 1
    $\begingroup$ What the OP is wanting is $\dot{H_0}$. You can relate this to the equation you give by the fact that $$\dot{H} + H^2 = \frac{\ddot{a}}{a}$$ $\endgroup$
    – Beta Decay
    Sep 11 '17 at 16:50
  • $\begingroup$ @BetaDecay well yes, that is now clear, though you might like to explain in your answer how a decreasing value of H corresponds to an "acceleration rate"... $\endgroup$
    – ProfRob
    Sep 11 '17 at 19:25

The acceleration of the universe is written as $\dot{H}$, where $H$ is the Hubble Rate (today, the Hubble Rate is given the name the Hubble Constant, $H_0$, and is given the value of 70 km/s/106pc). The Friedmann Equation can be simplified to the form:

$$\dot{H} + H^2 = -\frac{4\pi G}{3}\left(\rho + \frac{3p}{c^2}\right)$$

Where $\rho$ is the density, and $p$ is the pressure. $G$ is the gravitational constant, and $c^2$ is the speed of light.

Now, to find the acceleration, simply substitute the relavent values into the equation below:

$$\dot{H_0} = -\frac{4\pi G}{3}\left(\rho_0 + \frac{3p_0}{c^2}\right) - H^2_0$$

I will try to find the relavent values, but as the previous answer says, the value is roughly $-10^{-17}\,\mathrm{km}\,\mathrm{s}^{-2}\,\mathrm{Mpc}^{-1}$

  • $\begingroup$ $\dot{H}$ is negative. $\endgroup$
    – ProfRob
    Apr 19 '20 at 17:28

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.