The infinitesimal length interval between two events in spacetime $ds$ is defined by
$$ds^2=c^2 dt^2 - dx^2 - dy^2 - dz^2$$
The creature is dimensionally consistent, because time is multiplied with a speed. You can think of $(t,x,y,z)$ as the four coordinates of spacetime $(x^0,x^1,x^2,x^3)$ and $c$ appears naturally in the equations. However, the usual approach is setting $x^0=ct$ and so all the four coordinates have units of length. The definition of infinitesimal interval is then
$$ds^2=(dx^0)^2-(dx^1)^2-(dx^2)^2-(dx^3)^2$$
By doing that, $c$ vanishes from equations. For convenience, $x^0$ is named $t$ and, since it is actually proportional to time, it is called time or the time coordinate, but it is not really time. It is a distance. The original idea of Minkowski was even more bizarre, since he originally did $x^0=ict$, an imaginary (complex) distance.
The quantity that matters is a distance. It is proportional to time, but it is nevertheless a distance. The whole conceptual business is quickly (but not very enlightening) explained by saying "we work in units such that $c=1$" and the beginner is lost...
So, when you see $t$ you must remember that you are calling "time" to a quantity that is actually a distance. This trick affects also other physical quantities, so that you will be calling energy to something that is NOT energy, but energy divided by $c^2$. You will call speed to an adimensional quantity that must be multiplied by c to recover the "real" speed... It is not so unfamiliar to you: a supersonic plane may fly at "match 2.5", which really means 2.5 times the speed of sound.
Look at this list you may derive and check by yourself as an exercise:
Things that happen when you call "time" to the distance $ct$:
What you call LENGTH is still a length.
What you call TIME is a distance.
What you call MASS is still a mass.
What you call SPEED is something adimensional.
What you call ACCELERATION is an acceleration divided by $c^2$
What you call MOMENTUM is a momentum divided by $c$.
What you call ENERGY is energy divided by $c^2$
What you call ELECTRIC CURRENT is an electric current divided by $c$
By looking at this list, you now know how to "undo" the definitions so that, for instance, you have to multiply by $c^2$ when you want to recover the energy from the E that appears in the equations (e.g. the famous Einstein equation relating mass and energy is written $E=m$).
Don't worry, you will get used very soon. If you think this is bizarre, then look at what particle physicists do: They rename things so that not only the speed of light vanishes from the equations, but also the Planck constant and the electron charge, all at once. Don't ask me how to "undo" that...
EDIT: Some users keep asking in the comments so, for the ease of beguinners, here is how to reproduce that list:
Decompose every physical quantity into the basic dimensions length, mass, time. For instance:
Energy=work (dimensionally)=
[Force]x[length]=
[mass]x[acceleration]x[length]=
[mass]x[length]x[length]/([time]x[time])
Now, take every [time] factor you see and change it into [length]:
Energy=
[mass]x[length]x[length]/([time]x[time]) turns into
[mass]x[length]x[length]/([length]x[length]) =
[mass]
That is, by pretending that time is a length, we are pretending too that Energy is mass. And so on with the other quantities.
SECOND EDIT, About the metric signature:
One user was puzzled about the sign of the interval $ds^2$. This edit attempts to make this point clear:
Einstein defined in the 1921 Princeton lectures the interval as:
$$ds^2= + c^2 dt^2 - dx^2 - dy^2 - dz^2 $$
this sign convention is referred to as $(+---)$ as a shorthand notation. It is used by many authors, both in the field of Relativity and Particle Physics, like Weinberg, Peskin & Schröder or Zee.
However, there are too very good, canonical books in both fields with the opposite convention $(-+++)$, that is
$$ds^2= - c^2 dt^2 + dx^2 + dy^2 + dz^2 $$
For instance Schutz himself, Wald (except in the spinors chapter), MTW or Srednicki. It is also the convention used in Wikipedia.
The first thing you have to look at, when consulting a book, is which convention is the author using, because usually there are sign differences in the same equation when based upon one convention or the other.
The "simplest" classical explanation I know is the van der Waals interaction described by Keesom between two permanent dipoles. Let us consider two permanent dipoles $\vec{p}_1$ (located at $O_1$) and $\vec{p}_2$ located at $O_2$. Their potential energy of interaction is:
\begin{equation}
U(\vec{p}_1,\vec{p}_2,\vec{O_1 O_2}) = -\vec{p}_1\cdot \vec{E}_2 = -\vec{p}_2 \cdot \vec{E}_1
= -\frac{[3(\vec{p}_1\cdot\vec{u})(\vec{p}_2\cdot\vec{u})-\vec{p}_1\cdot \vec{p}_2]}{4\pi \epsilon_0 ||\vec{O_1 O_2}||^3}
\end{equation}
where $\vec{u}= \vec{O_1 O_2}/||\vec{O_1O_2}||$ and $\vec{O_1 O_2}$ is a vector going from $O_1$ to $O_2$.
Now, if the system is subject to thermal fluctuations then the dipoles can change their orientations according to a Boltzmann weight. If the time scale associated to the motion of the particles is much longer than the one associated to the dipole orientations then for each distance $r = ||\vec{O_1O_2}||$ on can trace over all the possible orientations.
The effective interaction between the dipoles is then characterized by the free energy of the system i.e. by:
\begin{equation}
\mathcal{F}(r|p_1,p_2) \equiv -k_B T \ln \int \frac{d\Omega_1 d\Omega_2}{(4\pi)^2} \:e^{-\beta U(\vec{p}_1,\vec{p}_2,\vec{O_1 O_2})}
\end{equation}
where $d\Omega_i$ is the solid angle element associated with the direction of dipole $\vec{p_i}$ and $\beta=1/k_B T$.
This quantity is super difficult to compute. However, if the dipoles are sufficiently far apart, their interaction energy is small and one may expand the exponential in powers of $\beta U$:
\begin{eqnarray}
\mathcal{F}(r|p_1,p_2) \approx -k_B T \ln \int \frac{d\Omega_1 d\Omega_2}{(4\pi)^2} && \times\\
&& \:\left(1 \right.\\
&& -\beta U(\vec{p}_1,\vec{p}_2,\vec{O_1 O_2}) \\
&& \left.+\frac{1}{2}\beta^2 U(\vec{p}_1,\vec{p}_2,\vec{O_1 O_2})^2\right)
\end{eqnarray}
The first integral $\int \frac{d\Omega_1 d\Omega_2}{(4\pi)^2} 1 $ is trivial and equal to $1$. The second integral of the potential is zero by symmetry (each dipole is as likely to be at either a positive or negative angle from $\vec{u}$ and the integral of a scalar product of two independent vectors is zero). The only remaining term is of order 2 in $\beta U$ which decays with $1/r^6$. At the end of the day we end up with something like:
\begin{equation}
\mathcal{F}(r|p_1,p_2) \approx -k_B T \ln \left(1+\beta^2\frac{C}{r^6} \right) \approx -\frac{C}{k_B T}\frac{1}{r^6}
\end{equation}
where
\begin{equation}
C = \int \frac{d\Omega_1 d\Omega_2}{(4\pi)^2}\left( \frac{[3(\vec{p}_1\cdot\vec{u})(\vec{p}_2\cdot\vec{u})-\vec{p}_1\cdot \vec{p}_2]}{4\pi \epsilon_0}\right)^2
\end{equation}
is a number not important in terms of physical insight.
What matters here is that the average value of the dipole-dipole interaction is zero while the first non zero term not to be zero is owing to fluctuations. This is roughly the rule with van der Waals like interactions which is that they are fluctuation driven (the same is true in the quantum case).
Remark: the previous derivation assumes that the interactions travel at infinite speed which is not true. When the limitation of the speed of light is introduced then at large distances, the van der Waals interaction decays as $1/r^7$.
Best Answer
All of the dimensionless quantities in the chart are derived using simple dimensional analysis. The basic premise is as follows: you are given a length $\sigma$, an energy $\epsilon$, and a mass $m$. Different combinations of these quantities will give additional constants with different units. In order to convert a parameter into its dimensionless form, you divide by the particular combination of these constants which has the same units.
Therefore, to get the reduced time, we must divide the time by a combination of $\sigma$, $\epsilon$, and $m$ that has units of time. Suppose we take a product of arbitrary powers of these three constants: $\sigma^a \epsilon^b m^c$. The units of this combination are:
$$\textrm{m}^a\cdot\textrm{J}^b\cdot\textrm{kg}^c=\textrm{kg}^{b+c}\cdot\textrm{m}^{a+2b}\cdot \textrm{s}^{-2b}$$
Finding a combination which has units of seconds is equivalent to solving the equation:
$$\textrm{kg}^{b+c}\cdot\textrm{m}^{a+2b}\cdot\textrm{s}^{-2b}=\textrm{s}^1$$
It should be clear that $-2b=1$, so $b=-1/2$. We also have that $b+c=0$, which means that $c=1/2$, and $a+2b=0$, which means that $a=1$. Plugging these powers back into our original product, we see that we can make a new constant $\tau$ which has units of time:
$$\tau=\sigma^1\epsilon^{-1/2} m^{1/2}=\sigma\sqrt{\frac{m}{\epsilon}}=\sqrt{\frac{\sigma^2 m}{\epsilon}}$$
So our reduced time $t^*$ is
$$t^*=\frac{t}{\tau}=t\sqrt{\frac{\epsilon}{\sigma^2 m}}$$
This approach is quite general, and can be used in any situation when you need to derive a constant in certain units from a collection of constants with different units. In fact, this procedure is how the Planck length, mass, energy, etc. are defined, where the given collection of constants is $G$, $\hbar$, and $c$.