# How can we speak use the notion of “particle” in LHC, given that we live in a curved spacetime?

I understood from lectures that the metric of a spacetime was absolute: It does not depend upon the test charge we put inside. Indeed, all the calculation our professor carried out were independent of the type of particle that "traveled" over that curved metric.

But then, when one tries to define the notion of particle in general curved backgrounds we notice that, since Poincaré invariance doesn't hold in general spacetimes, we cannot really define such a notion in an observer-independent way. So, the question is: Since we live in curved spacetime, how can one speak about "particles" at CERN-LHC?

Possible answer: for the particles themselves, metric is not curved. But then the metric does depends upon the kind of particle that you put inside, so the whole formulas are wrong, or incomplete, because they need a more term to indicate the particle mass, so.. are there some logical mistakes anywhere? In my reasoning or in physics?

• You're theoretically right to have these doubts but it's a question of extent of curvature. Over the spacetime regions that are the scenes of almost any particle interaction we are likely to observe on Earth, the curvature is utterly negligible, thus Minkowski for all practical purposes and the Poincaré group is the correct group of isometries. – WetSavannaAnimal Oct 22 '15 at 11:18
• For the Schwarschild metric, the curvature invariant $\Psi_2$ is $\Psi_2 = -GM/c^2r^3$. At the LHC, $\Psi_2 \approx 1.7\cdot 10^{-23}\, \text{m}^{-2}$, while the LHC detectors have dimensions on the order of $L = 10^1\, \text{m}$. Then $\Psi_2 L^2$, measuring the effect of curvature, is something like $10^{-20}$. Even if we took $L$ as the size of entire LHC, we would get something like $10^{-16}$. – Robin Ekman Oct 22 '15 at 11:39

1. The length scales involved in most processes on Earth (and particularly in the high energy, i.e. short lengths scale, experiments at LHC) are very small compared to the characteristic curvature scale of spacetime, so it can be ignored in practical calculations. In fact, the quantum field theories that we use to predict outcomes for LHC experiments is completely based on the Minkowski metric $\eta_{\mu\nu}=\operatorname{diag}(-1,1,1,1)$. Effects from general relativity are simply too small to have a measurable impact. This also means that we don't worry, in practical calculations, about the ill-definedness of the notion of a particle in general curved backgrounds.
• @Henry, in principle the metric is determined by all the energy and mass present in the spacetime (represented by the energy-momentum tensor $T_{\mu\nu}$). But, as I pointed out in my answer, when we speak of "test particles", we talk mean particles that have a negligible impact on the curvature of spacetime, so that we can ignore it completely while studying the metric: This is an idealized situation which nonetheless holds approximately true in many situations. – Danu Oct 22 '15 at 21:40