Can the Unruh effect be confirmed by the LHC? Two short questions regarding the Unruh effect. There are related answers on this forum and on wikipedia,  but I am looking for confirmation of my own intuitive assumptions, so a straightfoward yes or no answer would suffice.
My source is Smolin's "The Trouble With Physics" in which he mentions the effect occurring due to accelerated motion through a vacuum producing a temperature increase, although he does not give any detailed explanation, simply referring to  the effect in general terms.
My questions follow this chain of reasoning:
There exist quantum fluctuations due to the uncertainty principle relating energy and  time. To a observer either stationary or moving at a constant velocity, these energy fluctuations are, on average, zero over a timescale of say, arbitrarily, 1 second. 
If we use the crude analogy of water waves, the crests and troughs of the wave combine to produce an overall flat surface.
However,  to an accelerating observer, on average they will not observe the overall energy average of zero. Sticking to my analogy, the crest and troughs will not average out as the observer accelerates,  therefore the accelerating observer will gain energy, which can be linked, through the Boltzmann constant, to a temperature increase. 
My questions are:


*

*Is this crude intuitive picture correct, at least in principle?


As far as I am aware, the gain in energy/increase in temperature due to this effect  is extremely small, involving terms incorporating h over c,   implying that a velocity very close to that of light, would be required to detect it.


*Is a particle accelerator, such as the LHC, even remotely capable of confirming this effect by measuring, however indirectly, the increase in mass of a proton, given the high velocity required plus low energy gain involved? 

 A: From scholarpedia:

The Unruh effect is a surprising prediction of quantum field theory: From the point of view of an accelerating observer or detector, empty space contains a gas of particles at a temperature proportional to the acceleration. Direct experimental confirmation is difficult because the linear acceleration needed to reach a temperature 1 K is of order 10^20 m/s**2, but it is believed that an analog under centripetal acceleration is observed in the spin polarization of electrons in circular accelerators.

Handwaving one can give a bound on the acceleration of the proton beam at LHC taking the injection speed to the final speed as a delta(v)

The particles in the LHC are ultra-relativistic and move at 0.999997828 times the speed of light at injection  and  0.999999991 the speed of light at top energy

which is 2.1*10^-6 of the speed of light (c= about 3 10^8 m/s) so delta(v)~6*10^2m/sec
Now how long does it take to reach the final speed? 

Then our proton has to wait up to 20 minutes on the LHC 450 GeV injection plateau before the 25 minutes ramp to high energy, and these 45 minutes dominates the transit time.

So delta(t) is at best 25 minutes, in no way can one reach the 10^20m/s^2 to even see microwave radiation .
In conclusion one can be sure that the LHC protons cannot see particles in the vacuum as their acceleration is too low.
A: No definite answer to this question, the effect in some sources is accepted and other sources dispute it.
From Wikipedia:

The hypothetical Unruh effect (or sometimes Fulling–Davies–Unruh
  effect) is the prediction that an accelerating observer will
  observe black-body radiation where an inertial observer would observe
  none. In other words, the background appears to be warm from an
  accelerating reference frame; in layman's terms, a thermometer waved
  around in empty space, subtracting any other contribution to its
  temperature, will record a non-zero temperature. The ground state for
  an inertial observer is seen as in thermodynamic equilibrium with a
  non-zero temperature by the uniformly accelerating observer.The Unruh
  effect was first described by Stephen Fulling in 1973, Paul Davies in
  1975 and W. G. Unruh in 1976.[1][2][3] It is currently not clear
  whether the Unruh effect has actually been observed, since the claimed
  observations are under dispute. There is also some doubt about whether
  the Unruh effect implies the existence of Unruh radiation

I will attempt to find out if the power output of the LHC is comparable with that needed to verify or refute the effect and edit this answer accordingly with the results.
A: anna v's nice answer didn't go where I expected: the big acceleration at the LHC isn't in the accelerator, it's in the collisions. 
Let's suppose we have a proton in the LHC that undergoes an elastic, billiard-ball type collision and ends up with its original momentum in the opposite direction:
\begin{align}
\vec p_\text{initial} &= +7\,\mathrm{TeV}/c \\
\vec p_\text{final}   &= -7\,\mathrm{TeV}/c
\end{align}
The relativistic version of Newton's second law is
\begin{align}
\vec F = \gamma^3 m \vec a_\parallel + \gamma m \vec a_\perp = \frac{d\vec p}{dt}.
\end{align}
For our backscattered proton we have $\Delta \vec p = 14\,\mathrm{TeV}/c$. Let's assume the impulse acts over the time it takes light to cross the proton, $$\Delta t = 1\,\mathrm{fm}/c = 0.3\times10^{-23}\,\mathrm s.$$
Our LHC proton has $\gamma m = 7\,\mathrm{TeV}/c^2$, or 
$\gamma = \frac{\gamma m}{m} \approx 7\times10^3,$ or $\gamma^2 \approx 50\times10^6.$
Since it's backscattered it has $\vec a_\perp = 0$, giving us parallel acceleration of
\begin{align}
\vec a_\parallel &= \frac{\Delta\vec p/\Delta t}{\gamma^3 m}\\
&= \frac{14\,\mathrm{TeV}}{c} \frac{c}{1\,\rm fm} \frac{1}{50\times10^6} \frac{c^2}{7\,\mathrm{TeV}} 
= \frac{2c^2}{50\times10^6\,\rm fm } \\
&= 4\times10^{23}\,\mathrm{{m}/{s^2}}
\end{align}
The Unruh temperature is linear in the acceleration, so this corresponds to about 2000 kelvin --- pretty wimpy on the scale of what happens in an LHC collision. Even if you mumbled about using proper time for $\Delta t$ and threw in another factor of $\gamma$, that would still only takes you up to kilo-eV temperatures.
Counterintuitively, an impulse with the same order of magnitude in the direction perpendicular to the beam corresponds to a much larger acceleration:\begin{align}
\vec a_\perp &= \frac{\Delta\vec p_\perp/\Delta t}{\gamma m}\\
&= \frac{7\,\mathrm{TeV}}{c} \frac{c}{1\,\rm fm} \frac{c^2}{7\,\mathrm{TeV}} 
= \frac{c^2}{\rm fm } \\
&= 10^{32}\,\mathrm{{m}/{s^2}}
\end{align}
That corresponds to about $0.4\times 10^{12}\,\rm K$, or a typical thermal energy of about 30 MeV; still a zero-temperature background compared the actual LHC center-of-momentum collision energy (though it would turn into a percent-level correction if you were to mumble again about a missing factor of $\gamma$).
My relativistic dynamics is a little rusty, so please comment or edit if I am making some error beyond using handwavy order-of-magnitude arithmetic.
