A solution to cosmological constant problem?

[Mordred]

I don't particularly have a problem with any chosen particle. I mentioned that numerous times.

If you look back though my issue is regardless of any chosen particle or particle field you should still apply Maxwell Boltzmann and not simply use volumes.

Secondly all quantum fields has an inherent quantum uncertainty regardless of temperature. I also showed that the calculations for a QCD vacuum is distinctive to a QED vacuum.

I also includes peer reviewed links describing dual Meissner for QCD. Not just a single Meissner for QED.

This is the details the author didn't include or didn't examine. Let me ask you how many formulas has the author posted showing the numerous amplitudes contained within a proton ?

Each field within that proton has inherent uncertainty.

So how precisely does that match up to a single vector field calculation for the vacuum catastrophe when not even the electric charges match between quarks and electrons ?

The amplitudes mediating the electric charges between protons and electrons don't match each other either. That was part of that examination I did earlier.

If the author had applied those missing details I wouldn't have any real problem however he didn't looked deep enough ie into the mathematical proofs of the theories he tries to put together. He doesn't show the first second third and fourth NLO (next leading order integrals involved)

In essence he's ignoring a huge set of amplitudes with regards to protons/neutrons etc. Every time you use a Greens Function with regards to any Hamilton has uncertainty and that's every single wavefunction in QFT or QM. You have additional uncertainty adding to a total sum .

 

 

Edited 8 hours ago by Mordred

[MJ kihara]

3 hours ago, Mordred said:

I don't particularly have a problem with any chosen particle. I mentioned that numerous times.

If you look back though my issue is regardless of any chosen particle or particle field you should still apply Maxwell Boltzmann and not simply use volumes.

Secondly all quantum fields has an inherent quantum uncertainty regardless of temperature. I also showed that the calculations for a QCD vacuum is distinctive to a QED vacuum.

I also includes peer reviewed links describing dual Meissner for QCD. Not just a single Meissner for QED.

This is the details the author didn't include or didn't examine. Let me ask you how many formulas has the author posted showing the numerous amplitudes contained within a proton ?

Each field within that proton has inherent uncertainty.

So how precisely does that match up to a single vector field calculation for the vacuum catastrophe when not even the electric charges match between quarks and electrons ?

The amplitudes mediating the electric charges between protons and electrons don't match each other either. That was part of that examination I did earlier.

If the author had applied those missing details I wouldn't have any real problem however he didn't looked deep enough ie into the mathematical proofs of the theories he tries to put together. He doesn't show the first second third and fourth NLO (next leading order integrals involved)

In essence he's ignoring a huge set of amplitudes with regards to protons/neutrons etc. Every time you use a Greens Function with regards to any Hamilton has uncertainty and that's every single wavefunction in QFT or QM. You have additional uncertainty adding to a total sum .

 

 

What would make you conclude that cosmological constant problem has been solved in a precise way?

[MJ kihara]

5 hours ago, Mordred said:

The amplitudes mediating the electric charges between protons and electrons don't match each other either. That was part of that examination I did earlier.

How is this related to the vacuum inside a proton?

5 hours ago, Mordred said:

Secondly all quantum fields has an inherent quantum uncertainty regardless of temperature. I also showed that the calculations for a QCD vacuum is distinctive to a QED vacuum.

Putting cut off at planck scale doesn't it  help?

[Mordred]

The amplitudes are directly related to the anplitudes inside a proton. Recall All particles are field excitations.

Not little balls of matter.

1 hour ago, MJ kihara said:

 

Putting cut off at planck scale doesn't it  help?

Great idea take 936 MeV and multiply it by 10^{123} atoms how much energy does that give ?

One doesn't need to be a mathematician to see it will exceed 10^19 GeV which is the total energy density at BB.

Exceeding total energy/mass of the universe.

(Ps 10^19 GeV is the Planck temp cutoff when you convert to Kelvin)

Lol you could for example assume each SU(3) atom has exactly 1 quanta of energy and do the same calculation above just looking at the powers indicate it will exceed also.

 

Edited 1 hour ago by Mordred

[joigus]

4 hours ago, MJ kihara said:

What would make you conclude that cosmological constant problem has been solved in a precise way?

1st: Find the reason for the monumental overcount in QFT

Example: The exactly supersymmetric Hamiltonian gives zero for the expectation value of energy of the vacuum.

2nd: Find the reason why the actual energy is not exactly zero, but a little positive correction to that

Example: Postulate a mechanism to break SUSY ever so slightly that the expectation value of vacuum energy is slightly above zero. Then solve for the values of symmetry-breaking parameters for different models. Then go to the lab.

Something like that.

Edited 54 minutes ago by joigus
minor correction

[MJ kihara]

2 hours ago, Mordred said:

The amplitudes are directly related to the anplitudes inside a proton. Recall All particles are field excitations.

Not little balls of matter.

I think there is a lot of misunderstanding going around here...a proton is a proton because the fields inside it( quark fields) behave in a certain way( the way those quark combine)...by restricting those fields you get a proton otherwise we could have one proton filling the whole universe.when we measure a proton,I assume sum of this 'restricted'fields within that 'volume' give a result consistent with a proton.

Do you mean a proton is just a mathematical object?

2 hours ago, Mordred said:

Great idea take 936 MeV and multiply it by 10^{123} atoms how much energy does that give ?

You are getting me wrong,am talking about the formula used by the author to derive zero point energy..what's wrong with that formula? and yet it's clear they are talking of summing up all available quantum including for gravitons.