Within the formalism of QFT in curved space, we find negative energies to be unavoidable (theoretically). Even worse, all proposed Energy Conditions fail (in curved spacetime), not only within the context of "toy" field theories e.g. Klein Gordon Field, but in full blown QED for example. The interesting part about negative energies is the association with superluminal phenomena from faster-than-light photons to wormholes due to the repulsive gravity sourced from these negative energies (vaguely of course, what I am referring to is the associated "reverse" Shapiro delay due to the "widening" of the lightcones). Even within Loop Quantum Gravity, there seem to exist repulsive gravitational fields under extreme conditions, not necessarily just Planck Stars and all that, but topologically non-trival spacetimes such as wormholes.

I know for a fact that negative energies are verified to exist through effects such as the Casimir Effect, the Dynamical Schwinger Effect, and even through the mass defects of nuclei or bound systems in general, but none of these effects probe the gravitational consequences of negative energies.

My question is, if negative energies are so pervasive theoretically, where are they...experimentally? Are there any experiments which have probed how such "negative" energies gravitate? Just as a sweetener, does anybody know of any significant energy condition violations within QCD?

  • $\begingroup$ There are negative frequencies, but no negative energies. $\endgroup$
    – my2cts
    Apr 7, 2021 at 8:31
  • $\begingroup$ I am talking Hamiltonian densities $\endgroup$
    – Joeseph123
    Apr 7, 2021 at 8:33
  • $\begingroup$ I like your answer to that question (it was actually me who asked it). However, this question is from an experimental point of view...since we these negative energies are unavoidable theoretically, how do they gravitate? Negative energies should gravitate repulsively, and I am looking for experimental verification of this effect. $\endgroup$
    – Joeseph123
    Apr 7, 2021 at 22:41
  • $\begingroup$ @my2cts "There are negative frequencies, but no negative energies" Really ? Explain that to dark energy due to which universe space has propulsion force. This force is only due to negative mass density, which is manifestation of negative energy. Or Do you have your own-other explanation why universe accelerates in expansion ? $\endgroup$ Apr 9, 2021 at 8:51
  • 1
    $\begingroup$ @my2cts Energy conditions are not a part of Einstein's general theory of relativity. They're extra things added on top of it. $\endgroup$
    – user76284
    Apr 16, 2021 at 3:35

3 Answers 3


I don't know of any experiments that directly observed the gravitational consequences of QFT's negative energy densities. Observing such effects would be very difficult at best.

To see why it would be very difficult, remember that in relativistic QFT, the total energy of any state must be greater than the total energy of the vacuum state, which we conventionally take to be zero. The energy density can be less than the vacuum energy density within a localized region (I cited some references in another answer), but that requires preparing an appropriate non-vacuum state. Since negative energy density is a quantum effect, we can expect that experimental realizations of negative energy density would be restricted to very small regions of space, and the energy of the whole setup (including the mass-energy of the experimental equipment) would necessarily still be positive. See ref 1 for a related analysis.

The challenge is to observe gravitational effects in that small region where the energy density is negative, and to separate them from the gravitational effects of the nearby positive-mass equipment that we're using to prepare the quantum state in that small region. That's why I said that observing such effects would be very difficult at best.

To quantify this, we need to use a model. What model should we use? Ideally, we should use a fully quantum model of gravity, but I don't understand quantum gravity well enough to do that.

Another option is to use semi-classical gravity, which interprets the right-hand side of the classical gravitational field equation as $\langle T_{ab}\rangle$, the expectation value of the stress-energy tensor $T_{ab}$ in the given quantum state. This model is good enough for most practical purposes, like for describing the gravitational interaction between the earth and moon, but we should think twice before applying it to the question about negative energy density. Negative energy density a quantum effect (it requires preparing a special quantum state), and when quantum effects are significant, the semi-classical model is not self-consistent.

A classical system can influence a quantum system in a self-consistent model, and we use such models all the time. In particular, we can use a self-consistent model to calculate the effect of earth's gravity (treated as a classical field) on quantum objects. But to observe the gravitational effects of negative energy density, the influence needs to go in the other direction: we need to know how the quantum system affects gravity, and this can't be self-consistent if gravity is treated classically. In a self-consistent model, a quantum system cannot influence a classical system. (That's why the old idea of trying to define measurement as an interaction between a quantum system and a classical system is not viable. By definition, measurement is a phenomenon in which the thing being measured influences the thing doing the measurement.)

For now, let's ignore that warning and suppose that the semi-classical approximation is good enough for back-of-the-envelope estimates. As a short-cut, I'll use experimental observations of the Casimir effect as a proxy. Standard Casimir-effect calculations give an energy per unit area that scales like $\sim \hbar c/a^3$, where $a$ is the distance between the plates. As far as I know, the largest values of $a$ for which the Casimir effect has been measured are $a\sim$ a few micrometers (refs 2,3,4,5), which gives $E\sim -10^{-8}$ Joule per square meter. In contrast, as far as I know, the smallest mass-energies whose gravitational influences have been observed are vastly larger than this. Even the quantum-gravity experiment proposed (but not yet done) in ref 6 requires a minimum mass of $\sim 10^{-15}$ kg, which corresponds to $E=mc^2\sim 10^2$ Joules. This all indicates that observing the gravitational effects of negative energy density is currently beyond reach. Experimental physicists can be very clever, so I won't say it's impossible, but I'm pretty sure it hasn't been done yet.

  1. Bekenstein (2013), "If vacuum energy can be negative, why is mass always positive? Uses of the subdominant trace energy condition" Physical Review D 88, 125005

  2. Lamoreaux (1998), Demonstration of the Casimir Force in the 0.6 to 6$\mu$m Range (https://doi.org/10.1103/PhysRevLett.78.5)

  3. Mohideen and Roy, Precision Measurement of the Casimir Force from 0.1 to 0.9 microns (https://arxiv.org/abs/physics/9805038)

  4. Bressi et al, Measurement of the Casimir force between parallel metallic surfaces (https://arxiv.org/abs/quant-ph/0203002)

  5. Zou et al, Casimir forces on a silicon micromechanical chip (https://arxiv.org/abs/1207.6163)

  6. van de Kamp et al, Quantum Gravity Witness via Entanglement of Masses: Casimir Screening (https://arxiv.org/abs/2006.06931)


This answer will confirm that QFT does propose that quantum fields can be in configurations that break the energy bounds believed to hold for ordinary matter, such as the weak energy condition and dominant energy condition. I think it is misleading, however, to use the terminology "negative mass" for this situation, at least in many examples, as I shall explain. But to answer the main point of the question: the way to relate these aspects of QFT to gravitation is at present unknown, and they are not experimentally observable as yet because no experiment is anywhere near sensitive enough.

Now some more general comments on mass, tension, and gravitation.

In general relativity gravitation is sourced by the stress-energy tensor, whose diagonal elements are energy density $\rho c^2$ and pressure $p$. For ordinary matter pressure can be positive or negative (negative pressure is tension) and energy density is positive.

If one has a small spherical shell of test particles in some spherically symmetric situation, then the volume $V$ of the shell (in freefall) varies as $$ \frac{d^2 V}{dt^2} = -4 \pi G(\rho + 3 p /c^2) V. $$ The minus sign shows that gravitation is normally attractive (the mass density $\rho$ causes the shell to collapse) but if $p$ is large enough and negative then one can have an overall gravitational repulsion (the shell expands with accelerating volume $V$). This repulsion does not indicate negative $\rho$; it indicates negative (and large) $p$. However I note that at least one wikipedia article refers to this situation as "negative mass", which is misleading and I hope it will be corrected.

Coming now to quantum field theory, the Casimir effect is an effect in which conducting plates attract one another. This attraction can be seen as arising from dipoles in one plate interacting with induced dipoles in the other, but another nice way to see it is in terms of the available states of the electromagnetic field between the plates. The plates reflect and therefore confine low-frequency modes, and they transmit high-frequency modes. Because of this they exclude some low-frequency modes (those whose wavelength is longer than twice the plate separation, for example) and if the plate separation is increased fewer modes are excluded so the zero-point energy in the field increases. This increasing energy as separation is increased will manifest as an attractive force between the plates, which can be seen as a tension in the field in its lowest energy state. Note that we do not encounter negative energy in this discussion. So again, any wiki article that says that we do is misleading.

I suppose the notion of negative energy here is that one assigns the zero of energy to the situation where the plates are far apart, and then the field energy is negative when the plates are closer together. However, the plates have to be very wide, or else form a closed cavity, so such an approach seeks to assign the zero of energy to the case where one has a huge cavity with conducting walls. It is, at the least, debatable whether one should assign $\rho = 0$ to that physical system.

This does not rule out that there could be other, more exotic, field configurations where something more like negative mass might appear. But the Casimir effect, at least, is not such a configuration. The Casimir effect exhibits tension, not negative mass.

Overall, the present state of knowledge (as I mentioned at the start) is that it is not yet clear how to make the connection between quantum field theory and gravitation, and this is especially true of the vacuum energy, sometimes called zero-point energy. Since the effects we are talking about are very small at laboratory scales, no experiment is sensitive enough to probe the gravitational aspect. The only 'laboratory' with that capability is, at present, the cosmos at large, and the hint offered by evidence for accelerating expansion.


Are there any experiments which have probed how such "negative" energies gravitate?

There is The Archimedes Experiment

Archimedes is an INFN-funded pathfinder experiment aimed at verifying the feasibility of measuring the interaction of vacuum fluctuations with gravity. The final experiment will measure the force exerted by the gravitational field on a Casimir cavity whose vacuum energy is modulated with a superconductive transition, by using a balance as a small force detector.
Archimedes is two-year project devoted to test the most critical experimental aspects, in particular the balance resonance frequency and quality factor, the thermal modulation efficiency
and the superconductive sample realization.

  • $\begingroup$ Hello! It is preferable to include the relevant information from links instead of posting just them – in case the link breaks, the answer would otherwise become useless. Maybe you can edit it to sum up the linked article? Thanks! $\endgroup$
    – jng224
    Aug 9, 2021 at 20:46

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