# Are objects in the universe moving away from each other at the same acceleration?

This is a follow up question to Does the universe expand at the same rate everywhere in the universe?.

If the universe is expanding, then it would seem necessary for each object (ie galaxy) to be accelerating away from every other object at the same rate of acceleration which is constant (or not?). If that is true, what is the acceleration?

In other words, relative to the Milky Way, does it seem that all the other galaxies are moving directly away from us at the same rate of acceleration? If so, is it possible for us to prove that for any other galaxy that we know, the same observation would be true, were we able to view the universe from that galaxy?

• Note that this acceleration applies to objects that are not gravitationally bound: superclusters of galaxies. Two neighbor galaxies don't yield to space expansion because they are gravitationally bound to each other; they may move in relation to each other, including moving apart, even accelerating due to some other galaxies, but they are not accelerating due to space expansion. Two superclusters of galaxies do though.
– SF.
Nov 20 '15 at 10:08
• I highly recommend you look at hubble's law, Edwin Hubble was the scientist who first discovered that the universe was expanding, he also defined a constant and equation to prove it. en.wikipedia.org/wiki/Hubble%27s_law Nov 26 '15 at 19:09

I assume you mean is $\frac{dv}{dt}$ the same for all objects and/or is it constant for all objects, where $v$ is recessional velocity (and $t$ is cosmic time)? The answer is no it is not the same for all objects and it is not constant for all objects.

$$v=\frac{dD}{dt}=H(t)D$$

Where $D$ is proper distance and $H(t)$ is the Hubble parameter. If we take the Hubble parameter as a constant, then:

$$D = D_0e^{Ht}$$

Where $D_0$ is the proper distance at the present time. So,

$$D''(t) = \frac{dv}{dt} = D_0H^2e^{Ht}$$

So the "recessional acceleration" of an object depends on its present proper distance and is exponentially increasing with time.

Now the Hubble parameter isn't a constant and in LCDM cosmology is currently asymptotically decreasing to a constant. In the current epoch though it is fair to say that the "recessional acceleration" of an object depends on its proper distance and is increasing with time.

• So the farther from the Milky Way the object is, the faster its acceleration away from us is increasing? Is there a term for acceleration of accleration? Nov 11 '15 at 6:59
• Yes, "recessional acceleration" increases with distance. Do you mean is there a name for the first time derivative of acceleration? If so that is sometimes called "jerk", the first time derivative of jerk is sometimes called "jounce". Nov 11 '15 at 19:06
• jerk and jounce. got it. Nov 11 '15 at 19:15

Here's a non-mathematical answer. Something you can think about in your head. The further away a galaxy is from us the faster it will move away from us. There is no exact way to measure the exact speed because the distance is phenomenal, however, we can use red-shift to work out that this is actually happening. A way to prove that this happens is to get a balloon and draw dots on it. Blow it up a little bit, maybe half way. Focus on one dot and blow it up even more. You will notice that the other dots move away from the one you focus on faster than the dot you are looking at. This is a good way of explaining what is actually happening in the universe - and why things further away from us move away faster.

This applies to whatever galaxy we are in. If we were in the andromeda galaxy, we would still experience every other galaxy moving away from us faster.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}