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JG1009

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- Thread starter JG1009
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In summary: The author suggests that the relationship between velocity \vec{v} and \vec{k} can be either \vec{v} = \pm \vec{k}/\omega, but I don't understand that explanation, either.K-G eq is a wave eq. and its solutions must be plane waves. The above sols. are incoming and outgoing plane waves and that is the difference of 1 & 2. say.Sol 3 & 4 are negative energy sols. they that must not be there but they are the concequence of the K-G eq.

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JG1009

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bhobba

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JG1009 said:

The existence of actual zero point energy is open to question. In QFT its usually eliminated by something called normal ordering.

But if you want to read a detailed discussion of it see:

http://users.ox.ac.uk/~lina0174/vacuum.pdf

I don't entirely agree with the above - for example I believe you can explain the Casmir effect without ZPE.

Thanks

Bill

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WannabeNewton

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JG1009 said:

The zero-point energy is due to quantum fluctuations so it doesn't show up in the average energy.

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Zarqon

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JG1009 said:

I'm not an expert on this, but have been reading a bit about it lately, and my conclusion is this: The reason the field (in amplitude) averages to zero is because fluctuations over time are not phase coherent, but random in phase. Of course, if you time average fields over random phases you get zero, but that does not mean that the field is zero for any given moment in time. On the other hand, if you instead look at the average of the energy/intensity you square the field first and then there is no longer any cancellation in the time average. In fact, if you look at the absolute value of the field, you will find that the average value for this is also non-zero.

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Xertese

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Space is described as a void but it is anything but.

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jerromyjon

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Only when there is stuff in it, if you take away the stuff you have no "space" to describe. Space and the zero point vacuum state are a geometry.Xertese said:Space is described as a void but it is anything but.

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andresB

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(Mechanism for Vanishing Zero Point Energy)

http://arxiv.org/abs/astro-ph/0309679

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When you go to General Relativity, energy does make a difference, because a constant, nonzero energy density will cause spacetime curvature. But, there, I think that it's true that any constant energy density can be absorbed into the cosmological constant. So the issue really becomes: Is there some deeper theory unifying GR and QM that would allow us to compute the cosmological constant?

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andresB said:

(Mechanism for Vanishing Zero Point Energy)

http://arxiv.org/abs/astro-ph/0309679

That paper makes a side-point about the solutions to the Klein-Gordon equation which makes no sense to me. Could somebody explain it?

The author says that the KG equation:

[itex](\partial_\mu \partial^\mu + \mu^2) \phi= 0[/itex]

has 4 different classes of solutions:

- [itex]\phi = e^{i \omega t - i \vec{k} \cdot \vec{x}}[/itex]
- [itex]\phi = e^{i \omega t + i \vec{k} \cdot \vec{x}}[/itex]
- [itex]\phi = e^{-i \omega t - i \vec{k} \cdot \vec{x}}[/itex]
- [itex]\phi = e^{-i \omega t + i \vec{k} \cdot \vec{x}}[/itex]

In the appendix, the author suggests that the relationship between velocity [itex]\vec{v}[/itex] and [itex]\vec{k}[/itex] can be either [itex]\vec{v} = \pm \vec{k}/\omega[/itex], but I don't understand that explanation, either.

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moss

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Sol 3 & 4 are negative energy sols. they that must not be there but they are the concequence of the K-G eq.

but your text is correct as it is the short coming of K-G theory that it accepts the negative energy sol.

Note that the +/- signs in the exponent depends on the metric (-+++) or (+- - -). From 4-vector k goto 4-vector p then you will see the negative energy easily.

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Bob108

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Regarding the question of Science Author’s solutions 1 & 2 seeming to be the same thing, since 3-momentum

There is a difference between the sign before a quantity in an algebraic relation and the sign that quantity itself may have.

If one does the traditional QFT analysis (see Chap. 3 of my text

The negative energy particles, if they exist, do not appear in our universe as real particles, but they might well play a role as virtual particles.

As an aside, I devote all of Chap. 10 in the book to the vacuum.

Bob Klauber

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andresB

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Bob108

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The Euler-Heisenberg effective Lagrangian has not been used much for decades (no modern QFT texts I am aware of discuss it), and frankly, I know little about it. I do believe, however, that it pertains to pair production in an electromagnetic field, rather than a vacuum, and that the infinite quantity is of the sort one finds before renormalization in usual QFT (i.e., arising from divergent integrals that need to be regularized in the process of renormalization.) See Itzykson and Zuber, Quantum Field Theory (McGraw-Hill 1980), pgs. 195-197.

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So how do you overcome this problem with the "negative-probability states" when keeping the "supplemental modes" as physical modes?

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Bob108

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The concept clearly does not work if the supplemental particles are coupled to the traditional particles (within the standard model – exclusive of gravity). Two non-interacting (except for gravity) sectors are envisioned, so unitarity within the SM should be OK. Interaction probabilities for the SM traditional particles remain just as they are, positive and of identical magnitudes.

If by probability, you mean, as in NRQM, the square of the absolute value of the state wave function integrated over space (norms), the supplementals have alternating positive and negative norms, as the particle number of the state increases, i.e., an indefinite metric in Fock space. I address, and believe resolve, this in my second paper on the subject (available at www.quantumfieldtheory.info link near bottom of home page).

If by probability, you mean, as in RQM and QFT, the relation derived from manipulating the relativistic wave equation and its conjugate (or equivalently from Noether’s theorem), then the antiparticle expressions for probability of traditional SM antiparticles are negative anyway. This led to interpretation of the probability density as charge density. Exactly parallel reasoning holds for supplemental particles. Relativistic “probability” (charge, really) density is opposite sign for supplemental particles vs supplemental anti-particles.

Yes, I agree normal ordering has advantages when it comes to streamlining the theory development, for tadpoles and elsewhere. But I am not aware of anyone using renormalization to remove the infinite (or at least ginormous) half quanta vacuum contributions that one gets without the ad hoc introduction of normal ordering (that happens to violate the basic commutation principle upon which the entire theory is based). Lots and lots of people talk about why this disagrees with empirical observation (by a factor of 10^120 or so). A number of papers have been published on possible solutions.

I submit the supplemental solution approach is the simplest mathematically of all of them. The troublesome half quanta simply drop out straightforwardly by including all of the mathematically possible solutions to the wave equation in the analysis.

What does this all mean physically? I’m working on it. As Dirac said,

“ .. it [is] an easier matter to discover the mathematical formalism needed for a fundamental physical theory than its interpretation … with the interpretation most unexpected things may turn up.”

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Of course, you have to renormalize tadpoles. Naive normal ordering can be problematic in gauge theories, and thus usually one uses the path-integral formalism, where one needs Weyl ordering and not normal ordering and renormalizes the tadpole diagrams. An example, where this becomes clear in the most easy way, is in scalar QED, where you have a tadpole matter-particle loop in the leading-order self-energy correction of the photon. Only taking it into account together with the usual one-loop diagram you get a transverse regularized photon self-energy when using a regularization scheme that is compatible with gauge invariance (e.g., dimensional regularization). The same holds true for the gluon self-energy in QCD (or pure Yang-Mills). There you also need the gluon-tadpole loop go get a transverse self-energy.

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Bob108

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Zero-point energy is the lowest possible energy state that a quantum mechanical physical system can have. It is the energy that remains in a system even at its lowest possible temperature (absolute zero). This concept arises from the uncertainty principle in quantum mechanics, which states that even in a vacuum, there is still some residual energy present.

The QED vacuum state, also known as the quantum vacuum state or the vacuum of quantum chromodynamics (QCD), is the lowest energy state of a quantum field theory. It is considered to be the state of the universe at its lowest possible energy level, and it is characterized by the presence of virtual particles constantly popping in and out of existence.

The concept of zero-point energy is closely related to the QED vacuum state. In quantum field theory, the vacuum state is not truly empty, as it contains fluctuations and virtual particles. These fluctuations contribute to the zero-point energy of the system, which is the minimum energy that remains even at absolute zero. Therefore, the QED vacuum state plays a crucial role in the existence of zero-point energy.

There is ongoing research and debate about the potential practical applications of zero-point energy. Some theories propose that it could be harnessed for energy generation, propulsion systems, and even as a potential source for the creation of matter. However, there is currently no conclusive evidence or technology that can efficiently harness zero-point energy for practical use.

The existence of zero-point energy and the QED vacuum state has significant implications for our understanding of the universe, particularly in cosmology. These concepts play a crucial role in theories such as the inflationary model of the universe, which explains the rapid expansion of the universe in its early stages. They also help us understand the large-scale structure of the universe and the effects of vacuum energy on the expansion of the universe.

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