# Chapter One

On page 11, I wrote "c is for the speed of light, which is 299,792,458 meters per second, or (approximately) 186,282 miles per second."  The first figure is exact: it's the way we define a meter.  Here's how the conversion goes to get miles per second.  One inch is 2.54 centimeters, by definition.  So

$$1\,{\rm mile} = 1 {\rm mile} \times {5280\, {\rm ft} \over 1 {\rm mile}} \times {12\,{\rm inches} \over 1\,{\rm ft}} \times {2.54\,{\rm cm} \over 1\,{\rm inch}} \times {1\,{\rm m} \over 100\,{\rm cm}} = 1609.344\,{\rm m} \,.$$

That implies that the speed of light is

$$c = 299,792,458\,{{\rm m} \over {\rm s}} \times {1 {\rm mile} \over 1609.344\,{\rm m}} = 186,282.39705122\ldots {{\rm miles} \over {\rm s}}$$

I quoted the final result to more digits than one could ever need.  In principle this is an exact number.  But 186,282 miles per second is plenty of accuracy for most purposes.

On p. 12, I wrote "My mass is about 75 kilograms.  Rounding up a bit, that's about 50,000,000,000,000,000,000,000,000,000 nucleons."  The way to get this is to remember that the mass of a mole of hydrogen is approximately one gram, and hydrogen's mass comes mostly from the proton that comprises its nucleus.  (Both of these statements are true to better than a part in a hundred.)  So we convert

$$75\,{\rm kg} = 75\,{\rm kg} \times {1000\,{\rm g} \over 1\,{\rm kg}} \times {1 {\rm mole} \over 1 {\rm g}} \times {6.02 \times 10^{23} {\rm protons} \over 1 {\rm mole}} = 4.52 \times 10^{28} \, {\rm protons} \,,$$

where I've remembered that a mole, by definition, is Avagadro's number of whatever it is you're counting (in this case protons). Also I recall that protons and neutrons weight approximately the same.  Because I was aiming at only a rough estimate, I rounded up to $5 \times 10^{28}$ nucleons.

On p. 13,  I wrote, "Suppose you could [run a mile at a speed of $10\,{\rm m/s}$.]...  Time would run ... slower (for you) by one part in about $10^{15}$. In general, time runs slower by a factor of

$$\gamma = {1 \over \sqrt{1 - v^2/c^2}} \approx 1 - {v^2 \over 2c^2} \,,$$

where in the approximate equality I've expanded to leading non-trivial order in $v/c$.  So the one part in $10^{15}$ is supposed to be $v^2 / (2c^2)$.  Let's work it out:

$${v^2 \over 2c^2} = {1 \over 2} {(10\,{\rm m} / {\rm s})^2 \over (3 \times 10^8\,{\rm m}/{\rm s})^2} = {1 \over 2} {1 \over 9 \times 10^{14}} = {1 \over 1.8 \times 10^{15}} \,.$$

I "rounded" $1.8$ down to $1$ to get the one part in $10^{15}$ figure. Ordinarily I might care a lot about a factor of $1.8$, but when quoting an extremely small effect, it is common practice to give the "order of magnitude", which is the closest power of $10$.

On p. 14,  I wrote, "Time for [particles whirling around modern accelerators] runs about 1000 times slower than for a proton at rest."  The time dilation factor can be conveniently calculated as

$$\gamma = {E \over m} \,.$$

At time of writing (2009), the Tevatron can accelerate protons up to energies of $980\,{\rm GeV}$.  The proton's mass is $0.938\,{\rm GeV}$.  So $\gamma \approx 1044$.

On p. 14, I wrote, "Light can go all the way around the equator of the earth in about 0.1 seconds."  The earth's equatorial radius is $r=6378.1\,{\rm km}$.  So the time it takes light to get once around is

$$t = {2\pi r \over c} = {2 \times 3.1415927 \times 6378.1\,{\rm km} \over 299,792,458\,{\rm m/s}} \times {1000\,{\rm m} \over 1\,{\rm km}} = 0.134\,{\rm s} \,.$$

Likewise, I wrote that the moon is about 1.3 light-seconds away.  This is because its average distance is $384,403\,{\rm km}$: a similar conversion to the one I just gave gives $1.282\,{\rm s}$.  Earth's distance from the sun is $1.496 \times 10^{11}\,{\rm m}$ away, which is $499\,{\rm s}$.

On p. 16, I wrote that the energy released in fission of Uranium-235 is "a trillion times bigger as a fraciton of rest energy than the fraction that an Olympic springer can call up in the form of kinetic energy."  The fraction of energy released in nuclear fission is about one part in a thousand, which is $10^{-3}$.  The kinetic energy of an Olympic sprinter going $10\,{\rm m/s}$ divided by his rest energy is

$${{1 \over 2} mv^2 \over mc^2} = {v^2 \over 2c^2} = {1 \over 2} \left( {10\,{\rm m/s} \over 3 \times 10^8\,{\rm m/s}} \right)^2 = {1 \over 1.8 \times 10^{15}} \approx 10^{-15} \,.$$

Now the point is that $10^{-15}$ is a trillion times smaller than $10^{-3}$.  Because I replaced the $1.8$ by $1$ in quoting $10^{-15}$, this should be understood as an order of magnitude estimate.