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If the vacuum were "emptiness" it were easy to accept that there is no "preferred frame of coordinates" and the light velocity is the same in any inertial frame of coordinates. But the vacuum is an ocean of "virtual" particles (that carry energy with themselves, and quickly can disappear). Essentially, the vacuum is not empty.

Now, all these particles don't provide, at least in some regions of the space, a PREFERRED frame of coordinates? Can't we talk of a frame AT REST with respect to these particles and frames IN MOVEMENT with respect to them? If, suppose, we can say that there is a frame at rest with respect to these particles, why doesn't the light velocity differ in this frame from its value in other frames? Isn't the light scattered by these particles? A beam prepared as monochromatic, won't suffer a dispersion in its wavelength?

To be clear, the principle of the relativity that says that the light velocity is the same in any inertial frame, seems to me questionable vis-à-vis possible interactions of a beam of light with such particles.

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  • $\begingroup$ I don't know if virtual particles even work like this, but nevermind. What makes you think they would be at rest with respect to each other? $\endgroup$ – Javier Nov 13 '14 at 22:45
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    $\begingroup$ To Javier Badia: Of course they won't be at rest with respect to each other. In a galaxy (which is another scale of dimensions) the stars are not at rest with respect to each other, though, the galaxy has a movement as a whole, and there exists a frame in which its center of mass is at rest. $\endgroup$ – Sofia Nov 13 '14 at 22:50
  • $\begingroup$ Yes, but the galaxy is a bounded object, it has a center of mass. Where is the center of mass of an infinite universe filled with particles popping in and out of existence? $\endgroup$ – Javier Nov 13 '14 at 22:59
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    $\begingroup$ To Javier Badia: it's written in the text of the question. Notice the words "at least in some regions of the space". $\endgroup$ – Sofia Nov 13 '14 at 23:02
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The commonly accepted interpretation of Special Relativity is that it's impossible to determine an (inertial) frame of references absolute motion (by performing experiments within that frame).

So the default answer (given what we know) would be no, you can't determine your absolute motion relative to the background sea of virtual particles.

However this doesn't mean that new physics or information couldn't prove this wrong.

The things to consider would be:

  • What you're referring to as 'Virtual particles' should really be called Vacuum Energy (or vacuum fluctuations). They are more like energy fluctuations than actual particles.

  • You can't bounce light off the Vacuum Energy so the experiment as you state it won't work, however...

  • Have a look at Hawking Radiation where the conditions at a black holes event horizon are predicted to give rise to actual particles. I don't think any have been detected yet though.

  • Also see the Casimir Effect which is supposedly due to Vacuum Energy (so it's a method of detecting it). Also see this link for an article on how 'virtual particles' can become real photons.

  • Relativity states that physics experiments performed in one (inertial) frame of reference will be the same in another (so your suggested experiment should be the same in all frames of reference). However you could consider the Vacuum Energy to be a different frame of reference (you're just travelling through it rather than past it) and you can compare one frame relative to another.

  • You have an implicit assumption that the 'virtual particles' or Vacuum Energy is 'tied' to 'stationary' space. I don't think we really know enough about what causes Vacuum Energy to say this for sure. It is believed that it is due to quantum fluctuations in space (and the uncertainty principle) but you can also imagine a scenario where it is not tided to space, e.g. space is filled with a sea of dark matter particles which continuously collide and create these 'virtual particles' which quickly disappear again. I'm not advocating that this is the case, simply that you've made an assumption and we don't know enough about it to state that it is true.

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Virtual particles aren't real. It's in the name.

The reason people say there are "vacuum fluctuations" in form of virtual particle-pairs forming and annihilating again is because they misunderstood Feynman diagrams. In Feynman diagrams, internal lines are called virtual particles, since the external lines (i.e. the open-ended lines) correspond to real particles when computing scattering amplitudes perturbatively with them.1

Now, a priori, if we sum over Feynman diagrams, they can contain disconnected pieces with no external legs. The quantum partition function is, in the path integral formalism, simply the path integral of the exponential of the action, which is also the propagator for the vacuum onto itself, and turns more or less surprisingly out to be given by the sum of all diagrams without any external legs - sometimes called "vacuum bubbles" - in the diagram picture. Furthermore, one can show that this partition function is essentially the energy of the vacuum of the fully interacting theory.

So, we've got the partition function, which is the sum over all "purely virtual" diagrams, which is the vacuum energy. Some therefore say that the "vacuum is filled with virtual particles". But these virtual particles aren't particles, they are mere computational crutches in a crude perturbative expansion of the full vacuum energy, and they do not correspond to any detectable particle. Therefore, any argument that relies on the existence, however short and localized, of actual particles fails if it cannot be formulated in the more precise language of diagrams or, better even, in a non-perturbative way.


1In hindsight, this is one of the worst naming desicions ever.

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    $\begingroup$ To the CuriousMind: I never learned in a systematic way about the "vacuum energy", and "propagator for the vacuum", etc. So, I apologize for my ignorance. But these "virtual particles" are only on the paper? I was told that the fact that two distant charges "feel" the presence of one another, is DUE to the virtual particles. And they should form a VERY DENSE population, because "action-at-a-distance doesn't exist. So, the virtual particles should be somehow REAL. Please correct me if I miss something $\endgroup$ – Sofia Nov 13 '14 at 23:42
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    $\begingroup$ @Sofia: These are "basic" quantum field theory concepts. If you are not familiar with QFT, then it is very difficult to explain what these concepts actually mean, especially the virtual particles, but yes, they are only "on the paper" in the sense that you cannot measure them. They are really just lines in a diagram, helping us to calculate other, measurable results, but nothing in these results demands that you think of them as anything more than computational aids. $\endgroup$ – ACuriousMind Nov 13 '14 at 23:46
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    $\begingroup$ To the CuriousMind: I am not a very good mathematician, I incline to phenomenology. Now, measurement means in quantum theory the use of an apparatus. Please see my precedent comment about two distant charges. THESE CHARGES that "feel" one another and get accelerated, seem to "measure" (represent a testimony) of the EXISTENCE of the virtual particles. I repeat, the physics doesn't admit "action-at-a-distance". $\endgroup$ – Sofia Nov 13 '14 at 23:52
  • $\begingroup$ (continuation) To be clear, it is not necessary that we, human beings, detect the presence of an object. Other bodies, e.g. the two distant charges, can do the test for us and show, by their behavior, that the object has an influence on them. $\endgroup$ – Sofia Nov 14 '14 at 0:03
  • $\begingroup$ @Sofia: That many abhor the idea of "action at a distance" is probably the very reason why thinking of forces as exchange of virtual particles is so popular. But the force falls out in the (tree-level, non-relativistic limit) of QFT without ever saying that the virtual particle is, in any sense, real. The theory predicts it purely from the diagram, which does not assign reality to the internal lines by itself, and predicts the force without ever talking about what "transmits" it. There is nothing that forbids the force from acting at a distance (as long as it stays causal, which is the case) $\endgroup$ – ACuriousMind Nov 14 '14 at 0:05

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