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If I have a formula for velocity with respect to distance, like:

$73 (km / s / megaparsec)$

And I want to convert it to a formula for velocity (or any of its derivatives or its integral) with respect to time, how would I go about it?

My approach thus far is as follows:

$v(s) = 73s(t)$

By the chain rule:

$\frac{ds'}{ds} = \frac{ds'}{dt} * \frac{dt}{ds}$

$\frac{ds'}{ds} = s''(t) * \frac{1}{s'(t)}$ (because of $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$)

I can solve this differential equation with the characteristic equation:

$s^2 - 73s = 0$

I arrive at:

$s(t) = e^{74 * t}$

And this doesn't work because substituting values for $t$ here does not agree with what we would expect given $v(s)$

David Bandel
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  • There seems to be a problem with units. Velocity is distance divided by time, so how can we have velocity km/hr/??. – André Nicolas Jul 21 '16 at 21:36
  • It's velocity with respect to distance. For example, if the distance was 1 megaparsec, then the speed would be 1 km/sec. v(s) not v(t) kind of weird I know but it should still be valid – David Bandel Jul 21 '16 at 21:39

2 Answers2

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Your value $73\text{ (km/hr)/megaparsec}$ has units of inverse seconds. It is one value of the Hubble constant relating the speed an object is receding from us to the distance to the object. You can naively take the inverse and get $4.227\cdot 10^{17}$ seconds or $13.39$ billion years. It is roughly the age of the universe in that if you reverse the expansion and do not let things accelerate under gravity this is the time until everything collapses. It is an observed constant at the present epoch. If you want to solve a differential equation for distance as a function of time, you need to consider the time rate of change of the Hubble constant. Ignoring gravitational slowing and dark energy expansion, you would expect the Hubble "constant" to decrease as $\frac 1t$ because when the time since the big bang doubles, things are twice as far away but moving at the same speed relative to us. The constant speed lets you say $s=vt$, where the $v$ is the current speed.

Ross Millikan
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  • Yep that is why I'm doing this. I was hoping to come up with some rough and weak arguments for the universe's size and some conjecture about primordial expansion rates. Purely 100% amateur work and mostly a mathematical and conceptual exercise – David Bandel Jul 21 '16 at 23:34
  • is using s = vt this way really useful though? all it'd give you is the current size of the universe. i was hoping to do some integrals with it and look at the size and speed at different points in time – David Bandel Jul 21 '16 at 23:37
  • It gives you the distance to any given galaxy at any time based on its current distance from us. That is how I interpreted "distance with respect to time" – Ross Millikan Jul 22 '16 at 01:16
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You are making it too complicated. We have:

$$\frac{ds}{dt} = v = 73s$$

which immediately gives us

$$s=Ae^{73t}$$

for some constant $A$. Your differential equation looks highly suspicious to me: what exactly does $\dfrac{ds'}{ds}$ mean?

TonyK
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  • I think ds'/ds is reasonable. See here - http://math.stackexchange.com/questions/926010/derivative-of-velocity-with-respect-to-to-distance-and-vice-versa/926275 Though I think it practically will evaluate to 1? But can be expressed as arbitrary independent functions toward the goal of applying the chain rule – David Bandel Jul 21 '16 at 23:31
  • And thank you. My way really was too complicated. – David Bandel Jul 21 '16 at 23:38
  • As your post shows, $ds'/ds$ got you into a mess. You can think of $s'$ as a function of $t$, which in turn is a function of $s$, and you get $ds'/ds = ds'/dt \cdot dt/ds = s''(t)/(ds/dt)$, but this is not a standard differential equation $-$ it has three differentials in it ($ds',ds,dt$) instead of the usual two. – TonyK Jul 22 '16 at 12:26