We do. The LHC accelerates two protons, each with 3.5 TeV of energy, giving a total of 7 TeV in the CoM frame (The energies are from the initial phase of the previous LHC run. Later in the run this was increased to 8 TeV and the combination of the two dataset was what discovered the Higgs boson. The energies are roughly doubling now for Run II, to 13 TeV).
The main reason for this is as you mentioned, the energy involved. In any frame, we have the following invariant quantity, $s = (p_1 + p_2)^2$ which its square root, $\sqrt{s}$, gives the Centre of Mass (CoM) energy for the experiment, and here $p_i$ represents the momentum four-vector for each particle i. In a collision where two particles are moving in opposite directions with equal energy we have the following:
$$ s \equiv (p_1 + p_2) \cdot (p_1 + p_2) = (E + E, \mathbf{p}_1 − \mathbf{p}_2 ) \cdot (E + E, \mathbf{p}_1 − \mathbf{p}_2 ) = (2E , 0 ) \cdot (2E, 0) = 4E^2$$
and now the CoM energy is given by the square root of this quantity, $$\to E_{CoM} = \sqrt{s} = 2E$$
In a an experiment where one of the particles is at rest (has mass $m_t$) and the other is travelling with momentum $\mathbf{p}$ (and has mass $m_b$) we have the following:
$$ s \equiv (p_1 + p_2) \cdot (p_1 + p_2) = (E_b + m_t, \mathbf{p}_b) \cdot (E_b + m_t, \mathbf{p}_b) = E_b^2 + m_t^2 + 2E_b m_t − p_b^2 = m_t^2 + m_b^2 + 2E_b m_t $$
Assuming the masses are negligible, we have the fixed target (FT) CoM energy,
$$\to E^{\text{FT}}_{CoM} = \sqrt{s} = \sqrt{2E_b m_t}$$
Thus we would need much more energy input in a fixed target experiment to achieve the same energies as in the case with two co-moving particles.
EDIT: Regarding a comment below which I think arises from confusion of what the CoM frame is. $\sqrt{s}$ gives the CoM energy in both cases. This is useful because we can now compare between a fixed target experiment and an experiment where both particles are accelerated at the same speed but in different directions.
So, say my collider has the capability to produce a magnetic field which at its maximum, can accelerate a charged particle so as to have energy of 3.5 TeV. Now in the case that we have two particles with the same energy going in opposite directions, we will give a total CoM energy of 7 Tev, following the result above. In the second case though, theres only one accelerating particle, hence $\sqrt{s} = \sqrt{2 \times 3.5 \times m_t}$ and since E $\ll$ m, this is always less than in the first case.
So be careful, because both experiments can be transformed into a CoM frame. In the CoM frame $|\mathbf{p_1}| = -|\mathbf{p_2}|$. Note this is true in both experiments, even in the second case where one of the particles is stationary. Well, the whole point is that we can use the above formulas so we can skip transforming to the CoM frame; we can compute this quantity directly.
There are many competing limits on the maximum energy an accelerator like the LHC (i.e. a synchrotron, a type of circular accelerator) can reach. The main two are energy loss due to bremsstrahlung (also called synchrotron radiation in this context, but that's a much less fun name to say) and the bending power of the magnets.
The bending power of the magnets isn't that interesting. There's a maximum magnetic field that we can acquire with current technology, and the strength of it fundamentally limits how small the circle can be. Larger magnetic fields means the particles curve more and let you build a collider at higher energy with the same size. Unfortunately, superconducting magnets are limited in field: a given material has a maximum achievable field strength. You can't just make a larger one to get a larger field - you need to develop a whole new material to make them from.
Bremsstrahlung
Bremsstrahlung is German for "braking radiation." Whenever a charged particle is accelerated, it emits some radiation. For acceleration perpendicular to the path (for instance, if its traveling in a circle), the power loss is given by:
$$P=\frac{q^2 a^2\gamma^4}{6\pi\epsilon_0c^3}$$
$q$ is the charge, $a$ is the acceleration, $\gamma$ is the Lorentz factor, $\epsilon_0$ is the permittivity of free space, and $c$ is the speed of light.
In high energy, we usually simplify things by setting various constants equal to one. In those units, this is
$$ P=\frac{2\alpha a^2\gamma^4}{3}$$
This is instantaneous power loss. We're usually more interested in power loss over a whole cycle around the detector. The particles are going essentially at the speed of light, so the time to go around once is just $\frac{2\pi r}{c}$. We can simplify some more: $\gamma=\frac{E}{m}$, and $a=\frac{v^2}{r}$. All together, this gives:
$$ E_{\rm loop} = \frac{4\pi\alpha E^4}{3m^4r}$$
The main things to note from this are:
- As we increase the energy, the power loss increases very quickly
- Increasing the mass of the particles is very effective at decreasing the power loss
- Increasing the radius of the accelerator helps, but not as much as increasing the energy hurts.
To put these numbers in perspective, if the LHC were running with electrons and positrons instead of protons, at the same energy and everything, each $6.5~\rm TeV$ electron would need to have $37\,000~\rm TeV$ of energy added per loop. All told, assuming perfect efficiency in the accelerator part, the LHC would consume about $20~\rm PW$, or about 1000 times the world's energy usage just to keep the particles in a circle (this isn't even including the actually accelerating them part). Needless to say, this is not practical. (And of course, even if we had the energy, we don't have the technology.)
Anyway, this is the main reason particle colliders need to be large: the smaller we make them, the more energy they burn just to stay on. Naturally, the cost of a collider goes up with size. So this becomes a relatively simple optimization problem: larger means higher-up front costs but lower operating costs. For any target energy, there is an optimal size that costs the least over the long run.
This is also why the LHC is a hadron collider. Protons are much heavier than electrons, and so the loss is much less. Electrons are so light that circular colliders are out of the question entirely on the energy frontier. If the next collider were to be another synchrotron, it would probably either collide protons or possibly muons.
The problem with using protons is that they're composite particles, which makes the collisions much messier than using a lepton collider. It also makes the effective energy available less than it would be for an equivalent lepton collider.
The next collider
There are several different proposals for future colliders floating around in the high-energy physics community. A sample of them follows.
One is a linear electron-positron collider. This would have allow us to make very high-precision measurements of Higgs physics, like previous experiments did for electroweak physics, and open up other precision physics as well. This collider would need to be a linear accelerator for the reasons described above. A linear accelerator has some significant downsides to it: in particular, you only have one chance to accelerate the particles, as they don't come around again. So they tend to need to be pretty long. And once you accelerate them, most of them miss each other and are lost. You don't get many chances to collide them like you do at the LHC.
Another proposal is basically "the LHC, but bigger." A $100~\rm TeV$ or so proton collider synchrotron.
One very interesting proposal is a muon collider. Muons have the advantage of being leptons, so they have clean collisions, but they are much heavier than electrons, so you can reasonably put them in a synchrotron. As an added bonus, muon collisions have a much higher chance of producing Higgs bosons than electrons do. The main difficulty here is that muons are fairly short-lived (around $2.2~\rm\mu s$), so they would need to be accelerated very quickly before they decay. But very cool, if it can be done!
The Future
If we want to explore the highest energies, there's really no way around bigger colliders:
- For a fixed "strongest magnet," synchrotrons fundamentally need to be bigger to get to higher energy. And even assuming we could get magnets of unlimited strength, as we increase the energy there's a point where it's cheaper to just scrap the whole thing and build a bigger one.
- Linear accelerators are limited in the energy they can reach by their size and available accelerator technology. There is research into better acceleration techniques (such as plasma wakefield accelerators), but getting them much better will require a fundamental change in the technology.
There is interesting research that can be done into precision measurements of particle physics at low energy, but for discovering new particles higher energy accelerators will probably always be desirable.
Best Answer
The experiments at LHC hit protons on each other at total energy of 7 TeV.
In comparison, a flying moscquito has a kinetic energy of
Could one collide two flying bees of 3.5TeV kinetic energy and get a proton proton collision? i.e. give the total energy of the bee on a proton of each bee's proboscis?
The answer is "no" because macroscopic objects as the bee's proboscis have a surface composed by a very large number of atoms and the contact of the two proboscises will disperse the energy among those atoms much before it can reach one proton.
So it is true that macroscopic objects have a lot of kinetic energy in TeV, but to get it on an elementary particle level is a problem that has not been solved. Actually, maybe, with nanotechnology something might be devised, with magnetic fields and electric fields and God knows what, but certainly it will not be with pistons.