An invariant for transformations of Lorentz Exist a physical demonstration why 
$$E^2- p^2c^2 =m^2c^4=E'^2- p'^2c^2 $$
is an invariant for transformations of Lorentz?
N.B.: $m$ is mass; $E$ is the energy and $p$ is momentum in the frame $\Sigma$. Similary $E'$ is the energy and $p'$ is momentum in the frame $\Sigma'$.
 A: The threshold beam energies for particle production experiments depend on the way you set up the experiment (mostly fixed-target or symmetric collider systems but there are asymmetric colliders as well).1
However, those experimental thresholds are deducible on the basis on the invariant mass of the final-state system.
I suppose the simplest possible choice of reactions to discuss here would be
$$ e^- + e^+ \to \mu^- \mu^+ \;.$$
If the two muons are at relative rest then the (invariant) mass of the final state is $2m_\mu$.
In symmetric collider mode, you have net zero momentum so you can produce the pair at rest in the lab frame. Accordingly each beam needs to have a kinetic energy $T_\text{collider} = m_\mu - m_e \approx 105\,\mathrm{MeV}$ which is a lot but not a lot. 
Let's figure what you need in fixed target mode.2
You need $|\mathbf{p}_{+} + \mathbf{p}_-| = 2m_\mu \approx 211\,\mathrm{MeV}$. But we know $\mathbf{p}_- = (m_e,0)$. We write $\mathbf{p}_+ = (T+m_e,T+m_e)$ because we are very much in the ultra-relativistic regime.3
This leads to
\begin{align}
(\mathbf{p}_{+} + \mathbf{p}_-)^2
&= (T+2m_e,T+m_e)^2 \\
&= (T+2m_e)^2 - (T+m_e)^2 \\
&= 2Tm_e + 3m_e^2 \;,
\end{align}
and as you will see $T \gg m_e$ to so a very good approximation 
$$ \sqrt{2Tm_e} = 211\,\mathrm{MeV} \;.$$
But of course the electron mass is $m_e = 0.511\,\mathrm{MeV}$, so this comes to 
$$ T_\text{fixed target} \approx 43.6\,\mathrm{GeV} \;,$$
which is a lot!4
And these kinds of predictions are borne out in practice.5

1 There is a wrinkle here in that the production rate at the exact threshold is essentially zero. So we don't find the threshold by slowly dialing up the beam energy until we see the first particle, we do it by measuring the way the rate varies not much above the threshold and then working backwards from the theory.
2 Better use a positron beam as it is hard to find a positron target...
3 The boldface symbols are four-momenta and are broken down $\mathbf{p} = (E,p)$ becase we need care only about the beam direction for this calculation. In principle it ought to be $\mathbf{p} = (E,\vec{p})$. Also note that I'm working in natural units here. 
4 You should see right away that conservation of momentum requires that the products be in motion. That's where the extra energy goes: the kinetic energy of the products. The perfectly symmetric collider reaction has the products at rest at threshold.
5 This is also why we build colliders when they are considerable harder to engineer than fixed-target machines.
