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  1. The Higgs Field and Higgs Boson are theorized to be responsible for mass, or $m$. If these could be manipulated, altered, or controlled for, if the Higgs Field could somehow be parted, would this result in $m=0$ in all physics equations?

  2. Since the units of $c$ are just time and distance, would $c$ be affected?

This is NOT asking about redefining and applying the standard model universally, this is about a specific localized instance. The standard model with non-zero Higgs VEV still applies universally, so discussion of global effects is not relevant. This is a 'bubble', if you will, where there is a 'wave' (for a real groaner, a 'disturbance') in the Higgs field, such that the Higgs boson inside does not interact with the Higgs field outside. The field is manipulated, NOT made zero.

The Alcubierre drive has been accepted as theoretically plausible, this is related. Would the Higgs field be enclosed in this space-time bubble, or would it remain outside?

Yes, this would seem to break Lorentz symmetry, but the same rules would apply inside as outside, it's just that the outside is not the inside. That is, the Higgs field inside appears to move through the Higgs filed outside. Thus, m inside the bubble would appear to be zero from outside.

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closed as off-topic by JamalS, ZeroTheHero, sammy gerbil, Rococo, John Rennie Sep 1 '17 at 16:35

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "We deal with mainstream physics here. Questions about the general correctness of unpublished personal theories are off topic, although specific questions evaluating new theories in the context of established science are usually allowed. For more information, see Is non mainstream physics appropriate for this site?." – ZeroTheHero, sammy gerbil, John Rennie
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ I don't think this is a bad question, or that it deserves closing. After all, the answer is more or less "yes", so OP is not way off base. $\endgroup$ – Javier Aug 31 '17 at 20:53
  • $\begingroup$ -1. Hypothetical questions which ask "What would happen if we changed the laws of nature?" are not part of mainstream physics. $\endgroup$ – sammy gerbil Aug 31 '17 at 23:13
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    $\begingroup$ Possible duplicate of What happens to matter in a standard model with zero Higgs VEV? $\endgroup$ – Rococo Sep 1 '17 at 4:54
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    $\begingroup$ As an aside, while I do think that this is probably a duplicate of the above question or other similar ones (albeit at a lower technical level), it is unfortunate otherwise that a question that in the past got many upvotes and thoughtful responses is now being met with complaints that it doesn't belong on this site. $\endgroup$ – Rococo Sep 1 '17 at 4:57
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    $\begingroup$ You misunderstand. I was defending you, and the validity of this question. (I don't think I agree that "Higgs himself came to the conclusion about the existence of the Higgs field from a very simplistic thought", but that is neither here nor there). $\endgroup$ – Rococo Sep 1 '17 at 21:47
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If in theory you could change the Higgs potential so the vacuum expectation value of the Highs field became zero, some elementary particles would become massless but hadrons would not (almost all their mass is due to the color force). This would prevent atomic electrons (the Dirac equation gives the electron's energy in hydrogen as this, which becomes $0$ for all quantum numbers in the massless limit), but $c$ would be unchanged.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Sep 1 '17 at 11:57
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I am replying at the level of a highschool student.

I will use your last comment:

The fact that c is the speed of SOMETHING is derived from working through the basic units of the equation e=mc^2, until you are left with c being distance over time, once all the other units are cancelled out.

This is a complicated way of looking at a dated equation, which is no longer used in particle physics since it is not a fundamental property but result of the mathematics. Fundamental are the four vectors in the pseudoeuclidean space of special relativity, space and time and energy and momentum. The inertial form can be derived from these simple assumptions

That is speed, but does it HAVE to be the speed of light? That it is the same value as the speed of light in a vacuum could be just an artifact.

The speed calculated from E=mc^2 is for an effective inertial mass of a massive macroscopic body. It is useful to show how a massive body behaves under large accelerations, but irrelevant for basic physics. The speed of light in vacuum is an axiom of special relativity, and special relativity has been validated innumerable times in nuclear and particle physics.

Does the Higgs field have a 'speed', or is it absolutely stationary?

Fields are mathematical constructs that fill up all space, all the elementary particles have a field over all space. A photon field, a neutrino field etc are hypothesized mathematically over all space. It is the creation and annihilation operators acting on these fields that have measurable meaning. Higgs bosons appear with the creation operators on the Higgs field. Fields are stationary by construction, like extra coordinate systems.

If there were no Higgs boson, EVERYTHING would travel at the speed of c.

If there were no Higgs boson a new theory would be needed. The fact that the Higgs boson has been found at LHC is a validation of the present standard model of particle physics.

At the universe at present the Higgs field gives mass only to the elementary particles of the table. The atoms and molecules we observe would still have an invariant mass due to the strong force in the nucleus and the rules of special relativity. Whether bound states would form is a moot question.

The only laboratory we have is the cosmological history of the universe, where due to the recovery of the symmetry, which is broken at the low energies we live in, the elementary particles have zero mass up to ~10^-12 seconds after the big bang. The particles bouncing around there still have invariant masses when in pairs and systems.

To answer your question, no the Higgs field cannot be manipulated in our present universe, and it certainly has nothing to do with the speed of light. In this sense the symbol "m" is miss used in E=mc^2, it is a variable, it is just a mathematical result of the special relativity equations, which do assign an invariant mass to each four vector and thus to each particle.

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  • $\begingroup$ But the other fields are not equal throughout space. Apparently, only the Higgs field is homologous. Does it HAVE to be? Gravity has a 'speed', but it also has waves. Yet it is also a 'gravitational field'. ALL post-Einstein (actually, post-Planck physics - Einstein and Planck were very close associates, Einstein just got the better press) is just a construct, but the most widely-held construct. e = mc^2 is what started it all. Are you saying the c in this equation is NOT the c widely used? Are you saying the equation is no longer valid, just because it is no longer used? $\endgroup$ – Justin Thyme Sep 1 '17 at 15:24
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    $\begingroup$ The photon, electron higgs fields are quantum field theory fields,not classical(general relativity or newtonian gravity or classical electromagnetism). At present, we are at low energy and the Higgs field has undergone the symmetry breaking and its vacuum expectation value is over 200 GeV, whereas all the other fields of the particle table have zero VEV. c is the c widely used, it is m that is a variable coming from the equations of special relativity. now one uses invariant mass, a fixed for a four vector, not relativistic mass because it is a function. You have to look at the links. $\endgroup$ – anna v Sep 1 '17 at 17:08
  • $\begingroup$ So what is the answer - is it theoretically possible to envision a condition where a body could be put into a scenario where it is not influenced by an external Higgs field, like a bubble, so that externally the body does not appear to have a mass, but internally it does? The links all pertain to the standard model globally, not the standard model under manipulation, along the lines of the Alcubierre drive? My question about c is that it has the units distance/time, but just by itself, these units have zero value. If they remain zero in any equation do they take on a value in c? $\endgroup$ – Justin Thyme Sep 1 '17 at 17:33
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    $\begingroup$ The answer is no. The standard model is local, at all x,y,z,t. it cannot be manipulated except by interactions that are within the standard model. The Higgs field at our energies cannot have a zero VEV. all the rest is science fiction , not science $\endgroup$ – anna v Sep 1 '17 at 18:03
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    $\begingroup$ This is a classical general relativity concept and has nothing to do with quantization, as general relativity is not yet quantized and it is not possible to guess if this en.wikipedia.org/wiki/Alcubierre_drive speculation would work. Higgs fields are in the quantum mechanical frame. $\endgroup$ – anna v Sep 1 '17 at 18:20

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