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?

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    $\begingroup$ 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. $\endgroup$ Oct 22, 2015 at 11:18
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    $\begingroup$ 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 $10^{-16}$. $\endgroup$ Oct 22, 2015 at 11:39

1 Answer 1


You seem to be having a few things mixed up, and I hope I manage to address all your confusions. Here are some points to keep in mind, in no particular order.

  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.

  2. When one does calculations involving "test particles" in general relativity, one tacitly assumes that the backreaction of the metric due to the motion of the particle can be ignored because it is negligibly small. Thus, one uses a fixed background metric: In particular, it doesn't depend on the type of test particle that you use.

  3. In a perfect treatment of QFT in curved spacetimes, one should of course take into account the backreaction of the particles (or, much more generally, the quantum fields that give rise to these particles), including all their properties, in all calculations. However, nobody has really figured out how to do that yet, so that's something that has simply not been done. When people talk about quantum field theory in curved spacetimes, they usually mean doing QFT on a fixed background metric (and praying that the backreaction isn't too large to be ignored... Or do some checks to see if this is the case).

  • $\begingroup$ Uhm I have another question: I was thinking about the possibility that very different masses looked at metric in a very different way: if the metric depends upon the mass, and I write some equations (like Einstein Equations) I might expect that those would depends in some way upon the mass test. But this doesn't happen. Why?? @Danu $\endgroup$
    – Les Adieux
    Oct 22, 2015 at 21:23
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    $\begingroup$ @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. $\endgroup$
    – Danu
    Oct 22, 2015 at 21:40

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