# Converting Hubble constant to GR Units

I'm working my way through Thomas Moore's General Relativity Workbook and having a problem with Exercise 24.4.1 (p. 288), which asks you to verify Hubble constant values (with errors) for a variety of units. Given $H = 70.4 \pm 1.5\, {\rm km\,s^{-1}Mpc^{-1}}$, they ask for H in inverse meters, inverse seconds and billions of years. I have no problem with inverse seconds:

$$70.4 {\rm(km/s)(1/Mpc)} = (70.4)(10^3 {\rm m/km})/(30.857 \times 10^21 {\rm m/Mpc}) {\rm s} = 2.28 \times 10^{-18} {\rm s^{-1}}$$

the error is $(1.5)(10^3 {\rm m/km})/(30.857 \times 10^21 {\rm m/Mpc}) {\rm s} = \pm 0.0486 \times 10^{-18} {\rm s^{-1}}$

which verifies the values in 24.2 p. 281.

Likewise, for $\rm m^{-1}$:

$$(2.28 \times 10^{-18} {\rm s}^{-1})(3.336 \times 10^{-9} {\rm s/m}) = 7.61 \times 10^{-27} {\rm m}^{-1}$$

with an error of $\pm(0.0486 \times 10^{-18} {\rm s}^{-1})(3.336 \times 10^{-9} {\rm s/m}) = 0.162 \times 10^{-27} {\rm m}^{-1}$

which also verifies the value in 24.2.

But using the same method for Gy:

\begin{align}(2.28 \times 10^{-18} {\rm s}^{-1})(3.1556 \times 10^7 {\rm s/y})& = 7.1948 \times 10^{-11} {\rm y}^{-1} \\&= (\frac{1}{7.1948}) \times 10^{11} {\rm y} \\&= 13.9 \times 10^9 {\rm y} = 13.9 {\rm Gy}\end{align}

which is correct, but for the error

\begin{align}(0.0486 \times 10^{-18} {\rm s}^{-1})(3.1556 \times 10^7\, {\rm s/y}) &= 0.1534 \times 10^{-11} {\rm y}^{-1} \\&= (\frac{1}{0.1534}) \times 10^{11} {\rm y} \\&= 652 \times 10^9 {\rm y} = 652 {\rm Gy}\end{align}

which is obviously wrong. But I don't see why (which makes me feel pretty stupid, considering that the harder stuff, like the tensor calculus, has gotten pretty natural). I apologize for the lack of subscripts and superscripts, but this is my first post. Please, someone show me the error of my ways.

• Hey and welcome to Physics SE! Posts here support Latex, so you can format equations properly. Here's a link to a tutorial math.meta.stackexchange.com/questions/5020/… – CDCM Oct 8 '17 at 0:28
• Hi. I typeset parts of your question in Latex (will be visible pending edit approval), but please do so yourself in the future. It is hard to follow equations otherwise. Please see that some of your steps are contradictory. – Sayan Mandal Oct 8 '17 at 0:43

The problem lies in how you're calculating your error. We can estimate the error on a function $f(x)$ where $x$ has an error $\Delta x$ as $$\Delta f = \lvert f(x+\Delta x)-f(x)\rvert. \tag{1}$$
When $f(x)=a x$, this simplifies to $\Delta f = f(\Delta x)$. This is the case for the first two questions, and is what you've used. However for the third part, $f(x)\neq ax$, so you can't use that simplification anymore. You now have $f(x)=\frac{a}{x}$, so the error using equation 1 is: $$\Delta f = \lvert \frac{a \Delta x}{x^2 + x \Delta x} \rvert.$$