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Q & A: Hubble constant age of universe

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Most recent answer: 03/15/2013
Given Hubble's law $v = H_0 x$ and the differential equation $dx/dt = H_0 x$ that describes the expansion of space, why is the age of the universe $t = H_0^{-1}$ instead of $t = H_0^{-1} \ln x$? What am I doing wrong?
- Peter Franusic (age 61)
El Paso, Texas
Nice question.

The Hubble "constant" H0, the ratio of the typical speed a distant galaxy is receding from us to its distance, actually changes over time.  Think of the simplest picture of the expansion, where after some initial complicated events, everything moves away from everything else at a roughly constant velocity. After time t the distance between two things moving apart at velocity v is just x=vt.  So v/x = 1/t, regardless of v, and that's why we call that ratio a constant. However, at earlier times it was obviously larger.

Interesting things happen if we leave that simple picture. The speeds of separation look like they're increasing. There  seems to be some constant density of "dark energy" driving that acceleration. In the future the density of ordinary matter will become very low, and the dark energy effects will become even more important than they are now. To a good approximation, the acceleration rather than the velocity will be proportional to the distance.  In that case you get exponential expansion, called "inflation": x=x0e(t/ τ). Taking the derivative gives just the equation you were considering: v=x/ τ, where  τ is a fixed constant, no longer changing. (Experts will understand that these particular equations apply only for things close enough to let us describe everything in a special relativistic frame.)

It will take some time before the current weak inflationary expansion comes to be more important than the earlier expansion for the visible universe. So your idea is right, just a few billion years ahead of its time.

Mike W.

(published on 03/15/2013)

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