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Why particles only make sense in flat spacetime?

Physics Asked by wasnik on April 22, 2021

I don’t understand why exactly the notion of particles only makes sense in flat spacetime. From explanations I have read, it says that in general curved spacetime a unique vacuum couldn’t be talked about as different coordinate systems used to describe the spacetime would imply different vacuum, messing up covariance.

Stationary spacetimes have a global timelike killing vector which could be used to define a time coordinate. But then how does Rindler spacetime which possesses a timelike Killing vector have a vacuum different from Minkowski spacetime, when Rindler spacetime locally is just a change of coordinates from Minkowski spacetime.

One Answer

The concept of particles also makes sense for non-inertial observers and for observers in curved spacetime, as long as we remember that real observers are local and that the particle concept is only approximate.

(In this answer, "local/localized" doesn't mean localized at a point. It only means localized in some small neighborhood.)

The traditional global approach

Recall the usual approach to defining particles in flat spacetime:

  1. We define the energy observable to be the operator that generates time-translations whose integral curves are timelike geodesics.

  2. We define the vacuum state to be the state of lowest energy, and we notice that this state is invariant under Lorentz boosts, so it doesn't depend on which time-translation symmetry we used to define energy.

  3. We define particles with respect to the vacuum state. The key attribute of particles is that they can be counted and that the vacuum state has none of them.

That's all so familiar that it might feel necessary, but it isn't. In quantum field theory, observables are tied to spacetime, not to particles, so we don't need to worry if the familiar particle concept turns out to be only approximately meaningful.

A local approach

Real observers are localized: any given observer only has direct access to observables localized in some small neighborhood of the observer's worldline. Before we worry about how to generalize the preceding definitions 1,2,3 to non-inertial observers or to curved spacetime, we should think about how to replace the definitions 1,2,3 with something more local, because that's more realistic anyway.

Consider some local observer $O$. It could be a uniformly accelerating observer in flat spacetime, or a free-falling observer in curved spacetime, or whatever. The important thing is that $O$ is localized. What state $|0rangle$ should $O$ designate as the effective vacuum state?

Before trying to answer this, let's recall some basics:

  1. The state is supposed to account for whatever information we have about how the system was prepared, so that we can make predictions about subsequent measurements. A localized observer only has access to nearby local observables, and lots of different states all lead to the same predictions for those nearby local observables. This is true even for the familiar case of an inertial observer in flat spacetime.

  2. The conventional energy operator (Hamiltonian) $H$ is not a local observable. No local observer can actually measure $H$. What observable should we use instead of $H$ for defining the effective vacuum state? Any quantum field theory formulated using a background spacetime metric $g_{ab}$ has an associated stress-energy tensor $T^{ab}(x)$. In flat spacetime, integrating $T^{00}(x)$ over all of space gives the usual Hamiltonian $H$. More generally, we can consider the local observable $$ H(R)equivint_R d^3x T^{00}(x) $$ where the integration region $R$ is any finite region of space, which we can take to be the neighborhood whose observables are accessible to the observer $O$, and where the "time" components are understood to be with respect to a timelike vector field that has the observer's worldline as an integral curve.

Now we can see that for all practical purposes, any state $|0rangle$ that minimizes the expectation value of $H(R)$ is an equally good candidate for the effective vacuum state for an observer $O$ who is localized within $R$. Any local observable that (almost) annihilates $|0rangle$ is a candidate for a (slightly noisy) particle-detecting observable, so we have what we wanted: a generalization of the particle-concept that works for any local observer, whether inertial or non-inertial, and in any spacetime, whether flat or curved.

Caveats

Most things in physics are only approximate, including most of the things that we like to pretend are exact. I'll finish this answer by acknowledging a few of the ways in which the approach described above is only approximate, and I'll explain why the approximation is good enough.

  1. In flat spacetime, the Reeh-Schlieder thereom implies that the vacuum state (the lowest-energy state of the global Hamiltonian $H$) cannot be annihilated by any local observable. This means that perfectly noiseless particle-detecting observables cannot exist in any strictly finite region of space, as I explained in more detail in my answer to What's the physical meaning of the statement that "photons don't have positions"?. The Reeh-Schlieder property is expected (and often postulated) in curved spacetime, too. This isn't a problem in practice, because for a region $R$ of any reasonable macroscopic size, this fundamental noisiness is negligible compared to other practical sources of noise in real detectors.

  2. The spectrum of the operator $H(R)$ can be made arbitrarily negative by making $R$ arbitrarily small. This is easy to prove in the case of a free scalar field in flat spacetime, and I cited a review paper here: The positive-energy condition in quantum field theory for Hamiltonians associated with different timelike Killing vectors. The approach described above only makes sense if the region $R$ is large enough so that the lower bound of the spectrum of $H(R)$ is relatively insensitive to the precise size of $R$. That's okay, because any region $R$ of reasonably-macroscopic size should satisfy this condition.

  3. Even for an inertial observer in flat spacetime, state that minimizes the expectation value of $H(R)$ isn't necessarily the traditional vacuum state (which minimizes the expectation value of the full Hamiltonian $H$). That's okay, because if $R$ has reasonably-macroscopic size, then the traditional vacuum state should be among the many states that approximately minimize the expectation value of $H(R)$. Since localized particle-detecting observables are slightly noisy anyway, any such state should be good enough to use as the effective vacuum state for an observer localized in $R$. For a non-inertial observer, the traditional vacuum state might not be among those that approximately minimize the expectation value of $H(R)$, at least if the observer's acceleration is extreme. This leads to the Unruh effect, which illustrates the observer-dependence of the particle concept.

  4. As time passes, things that were previously unaccessible to the local observer can have effects that eventually propagate to the observer's location. If the effective vacuum state is chosen based on $H(R)$ at some instant in time, then it won't be able to make good predictions about those later-arriving effects that originated outside of $R$. This is no different than the situation we face every day in every real experiment: we don't control and don't even know what's happening very far away, and there's always a chance that some of those unknown far-away events (like earthquakes, solar storms, asteroids, etc) might have effects that eventually propagate into our laboratory. Only on paper can we pretend to know the global state of the system, and the main message of this answer is that when we adopt a more realistic local viewpoint, the obstacles to defining particles in curved spacetime go away. This generalized particle concept is only approximate, and that's okay. Most things in physics are like that.

Answered by Chiral Anomaly on April 22, 2021

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