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We know space is expanding at a rate of roughly 432 miles/light-year/year. Since Einstein showed that time was intrinsically linked into the 4 dimensional structure of spacetime one would logically wonder if space expansion is linked to the passage of time. Due to the cosmic microwave background we can infer that the universe is 13.5-13.9 billion years old. 1 light-year is ~5.88 trillion miles or $5.8786 × 10^{12}$ miles if we divide $$5.8786 * 10^{12} /432$$ we find that the universe is expanding by a factor of 1/13607870370.4 every year so every year any given piece of space increases in size by about a 13.6 billionth of its size. Looking at the fact that every year the universe also gets a ~13.6 billionth of its age older. Looking at that link made me wonder whether the passage of time, and cosmological expansion are the same cosmological constant type effect but one on space and the other on time. Does space expand proportionally to the age of the universe?

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    $\begingroup$ Is time inflating… Physicists use the word “inflation” to mean something much more dramatic than the kind of cosmological expansion happening now. (I’m not suggesting that time is doing either.) $\endgroup$
    – Ghoster
    Commented Mar 21 at 23:46
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    $\begingroup$ The next-to-the-last equation here gives the spatial scale factor as a (non-linear) function of time. $\endgroup$
    – Ghoster
    Commented Mar 21 at 23:58
  • $\begingroup$ It's not clear what you are asking. Note that passage of time is not a physical concept, and neither is expansion of space. $\endgroup$
    – Sten
    Commented Mar 22 at 2:54
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    $\begingroup$ Does this answer your question? Does time expand with space? (or contract) $\endgroup$ Commented Mar 22 at 5:57
  • $\begingroup$ Voting to reopen. A perfectly clear question - the questioner wants to know if the cosmological constant has changed over time,. $\endgroup$
    – gandalf61
    Commented Mar 22 at 8:26

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In the FLRW model of the expanding universe, the metric is given in terms of the co-moving coordinates: $$ds^2=dt^2-a(t)^2\bigg[{dr^2\over1-kr^2}+r^2\sin^2(\theta)d\theta^2+r^2d\phi^2\bigg]$$ where the function $a(t)$ is the scale factor of the universe that satisfies the Friedmann equations: $$\bigg({\dot a\over a}\bigg)^2+{kc^2\over a^2}-{\Lambda c^2\over 3}={kc^4\over 3}\rho,$$ and $$2{\ddot a\over a}+\bigg({\dot a\over a}\bigg)^2+{kc^2\over a^2}-\Lambda c^2=-kc^2p.$$ These equations were known some years prior to Hubble's discovery of the expanding universe, at any rate, however, they show the expansion rate of the universe as a function of time, e.g. Hubble's constant is written: $$H_0={\dot a(t)\over a(t)}\bigg|_{t={\text{now}}}.$$ So we can see the explicit dependence of Hubble's constant on matter/energy, including the cosmological constant: $$\bigg(H(t)\bigg)^2=-{kc^2\over a^2}+{\Lambda c^2\over 3}+{kc^4\over 3}\rho.$$

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    $\begingroup$ the metric seems wrong, there should be no a(t) multiplying dt $\endgroup$ Commented Mar 22 at 2:29
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    $\begingroup$ It's correct only if $t$ is conformal time $\endgroup$
    – Sten
    Commented Mar 22 at 2:51
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    $\begingroup$ @Sten But aren’t the stated Friedmann equations for cosmological, not conformal, time? $\endgroup$
    – Ghoster
    Commented Mar 22 at 6:00
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    $\begingroup$ @Ghoster indeed the overdots are derivatives with respect to the proper time. $\endgroup$
    – Sten
    Commented Mar 22 at 6:58
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    $\begingroup$ Thanks, I dropped an $a(t)$ where I shouldn't have. $\endgroup$ Commented Mar 22 at 13:25
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Yes. New time is created (it "expands") as the age of the universe increases. The big bang is at the center. The age of the universe is the radius of the hypersphere. The past universe is a smaller hypersphere inside the present one.

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  • $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Physics Meta, or in Physics Chat. Comments continuing discussion may be removed. $\endgroup$
    – Buzz
    Commented Mar 22 at 15:26

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