GR deals with what is Real

[MultiSingularity]

Somewhere around 1 x 10-18g in a vacuum is a new constant.

The diffraction gets too small to identify fringes. It's a natural size for the object to be physical and to never be in superposition.

Uncollapsed(stateless | unphysical | virtual) Quantum Waves + State(Matter Field or wave collapse or decoherence) + zero Diffraction showing fringes = Physical Matter (Real)

If an object is too large to display fringes, it is automatically physical. The question now is if auto-physical objects have a physical state or maybe being naturally physical doesn't require it.

Do I need to claim there is a physical state in the first place for even quantum sized objects if it is the same thing as: wave collapse, decoherence, and zero diffraction?

There has to be something that causes a particle to be physical or not before it even starts moving. If it is to only be a wave in flight, duality doesn't apply. But if physical, duality is allowed. Maybe I need a different term for "physical state".

If I started using "Real" instead of "physical state" would that get physicists off my back about mass meaning a physical property?

GR deals with what is Real.
Wave Collapse, Decoherence, and Zero Diffraction cause something to be Real.

We just need GR to handle duality for Unification.
GR for reality
QM wave function for unreal (only probabilities)

Does this mean Diffraction is directly related to Superposition?

It is curious to me that spacetime limits the speed of light and gravity to the same speed.
https://www.forbes.com/sites/startswithabang/2019/10/24/this-is-why-the-speed-of-gravity-must-equal-the-speed-of-light/#737e30e62fc0

The "unobserved" isn't real/physical and therefore not involved with spacetime. It doesn't have gravitons.
Unobserved/Unreal quantum waves do not follow the laws set by spacetime.

Maybe it would help if I explained what is happening in three famous quantum experiments. The Double Slit, Delay Choice Quantum Eraser, Which Way Quantum Eraser.

The Double Slit
Layered, Unobserved quantum fields begin to combine to assemble a new particle. The Dimension of the Unreal is able to know if the physical state of the particle will be requested in it's path. Something we know is capable of doing a state change is called a detector. But there are other more natural means of causing it. A particle with a physical state going through a double slit will only go through one slit. An unreal, quantum wave will go through both slits and display interference/fringes on a final panel. The final panel does cause a wave collapse but does not give the particle a physical state while in flight.

Delay Choice Quantum Eraser
Shows us the entire path of the particle is known before it starts moving. Entangled particles hold the same state while in flight. When the first particle hits it's final panel in a shorten path, it knows if its entangled brother will ever be physical or not in its path.

Which Way Quantum Eraser
Something very interesting happens when you cause two state changes in the path of a particle before it hits a final panel. If a particle knows (the unreal dimension) two state changes are going to occur, it goes back to being unreal quantum waves. When you see fringes appear on the final panel, it is because the quantum waves when through polarizers at the slits and the additional polarizer at unreal quantum waves.

QFT assumes spacetime is involved ..it's not. It uses points (x, y, z) Cartesian coordinates. They are assuming those points are in spacetime. There isn't anything saying it has to be locations from spacetime.

Spacetime is fine for points in space when the object in question is observed/has a physical state.

Unobserved quantum fields do not need spacetime to function.

You never get anything faster than light ..when spacetime is involved.

Observation/state change, gets spacetime involved. You are not considering speeds from unobserved objects.

If what I'm describing isn't spacetime ..it would be something completely new/undiscovered. A property of nature. I have no doubt something is going on here and it's the key to the theory of everything.

Higgs is lie because particles are not assigned mass by some random particle they decided to name the higgs boson. The higgs field is just spacetime and the matter field. Particles are real when they are given a physical state.

[MigL]

Didn't have the will or the time to read all of your crap.
Skipped to the last sentence.

The Higgs particle doesn't 'assign' mass to anything.
Rather certain specific particles interact with the Higgs field ( of which the Higgs particle is a manifestation ) and thereby gain the property we call mass.
Much like the interaction of a particle with a field can give it the property we call kinetic energy or potential energy.

But you are right, space-time is a field ( look up what that means ) in GR, and 'matter' is simply manifestations of their specific QuantumElectroDynamic and QuantumChromoDynamic fields, as per quantum field theory.

Edited November 3, 2019 by MigL

[Mordred]

You might also want to study QFT as it uses the same four momentum as used in GR. Secondly under particle physics rules what is observable is considered physical. 

[Mordred]

Here this will help

QFT can be described as a coupling of SR and QM in the non relativistic regime.

 

1) Field :A field is a collection of values assigned to geometric coordinates. Those values can be of any nature and does not count as a substance or medium.

2) As we are dealing with QM we need the simple quantum harmonic oscillator

3) Particle: A field excitation

 

Simple Harmonic Oscillator

[math]\hat{H}=\hbar w(\hat{a}^\dagger\hat{a}+\frac{1}{2})[/math]

the [math]\hat{a}^\dagger[/math] is the creation operator with [math]\hat{a}[/math] being the destruction operator. [math]\hat{H}[/math] is the Hamiltonian operator. The hat accent over each symbol identifies an operator. This formula is of key note as it is applicable to particle creation and annihilation. [math]\hbar[/math] is the Planck constant (also referred to as a quanta of action) more detail later.

 

Heisenberg Uncertainty principle

[math]\Delta\hat{x}\Delta\hat{p}\ge\frac{\hbar}{2}[/math]

 

[math]\hat{x}[/math] is the position operator, [math]\hat{p}[/math] is the momentum operator. Their is also uncertainty between energy and time given by

 

[math]\Delta E\Delta t\ge\frac{\hbar}{2}[/math] please note in the non relativistic regime time is a parameter not an operator.

 

Physical observable's are operators. in order to be a physical observable you require a minima of a quanta of action defined by

 

[math] E=\hbar w[/math]

 

Another key detail from QM is the commutation relations

 

[math][\hat{x}\hat{p}]=\hat{x}\hat{p}-\hat{p}\hat{x}=i\hbar[/math]

 

Now in QM we are taught that the symbols [math]\varphi,\psi[/math] are wave-functions however in QFT we use these symbols to denote fields. Fields can create and destroy particles. As such we effectively upgrade these fields to the status of operators. Which must satisfy the commutation relations

 

[math][\hat{x}\hat{p}]\rightarrow[\hat{\psi}(x,t),\hat{\pi}(y,t)]=i\hbar\delta(x-y)[/math]

[math]\hat{\pi}(y,t)[/math] is another type of field that plays the role of momentum

 

where x and y are two points in space. The above introduces the notion of causality. If two fields are spatially separated they cannot affect one another.

 

Now with fields promoted to operators one wiill wonder what happen to the normal operators of QM. In QM position [math]\hat{x}[/math] is an operator with time as a parameter. However in QFT we demote position to a parameter. Momentum remains an operator.

 

In QFT we often use lessons from classical mechanics to deal with fields in particular the Langrangian

 

[math]L=T-V[/math]

 

The Langrangian is important as it leaves the symmetries such as rotation invariant (same for all observers). The classical path taken by a particle is one that minimizes the action

 

[math]S=\int Ldt[/math]

 

the range of a force is dictated by the mass of the guage boson (force mediator)

[math]\Delta E=mc^2[/math] along with the uncertainty principle to determine how long the particle can exist

[math]\Delta t=\frac{\hbar}{\Delta E}=\frac{\hbar}{m_oc^2}[/math] please note we are using the rest mass (invariant mass) with c being the speed limit

 

[math] velocity=\frac{distance}{time}\Rightarrow\Delta{x}=c\Delta t=\frac{c\hbar}{mc^2}=\frac{\hbar}{mc^2}[/math]

 

from this relation one can see that if the invariant mass (rest mass) m=0 the range of the particle is infinite. Prime example gauge photons for the electromagnetic force.

 

Lets return to [math]L=T-V[/math] where T is the kinetic energy of the particle moving though a potential V using just one dimension x. In the Euler-Langrange we get the following

 

[math]\frac{d}{dt}\frac{\partial L}{\partial\dot{x}}-\frac{\partial L}{\partial x}=0[/math] the dot is differentiating time.

 

Consider a particle of mass m with kinetic energy [math]T=\frac{1}{2}m\dot{x}^2[/math] traveling in one dimension x through potential [math]V(x)[/math]

 

Step 1) Begin by writing down the Langrangian

 

[math]L=\frac{1}{2}m\dot{x}^2-V{x}[/math]

 

next is a derivative of L with respect to [math]\dot{x}[/math] we treat this as an independent variable for example [math]\frac{\partial}{\partial\dot{x}}(\dot{x})^2=2\dot{x}[/math] and [math]\frac{\partial}{\partial\dot{x}}V{x}=0[/math] applying this we get

 

step 2)

[math]\frac{\partial L}{\partial\dot{x}}=\frac{\partial}{\partial\dot{x}}[\frac{1}{2}m\dot{x}^2]=m\dot{x}[/math]

 

which is just mass times velocity. (momentum term)

 

step 3) derive the time derivative of this momentum term.

 

[math]\frac{d}{dt}\frac{\partial L}{\partial\dot{x}}=\frac{d}{dt}m\dot{x}=\dot{m}\dot{x}+m\ddot{x}=m\ddot{x}[/math] we have mass times acceleration

 

Step 4) Now differentiate L with respect to x

 

[math]\frac{\partial L}{\partial x}[\frac{1}{2}m\dot{x}^2]-V(x)=-\frac{\partial V}{\partial x}[/math]

 

Step 5) write the equation to describe the dynamical behavior of our system.

 

[math]\frac{d}{dt}(\frac{\partial L}{\partial\dot{x}}-\frac{\partial L}{\partial x}=0[/math][math]\Rightarrow\frac{d}{dt}[/math][math](\frac{\partial L}{\partial\dot{x}})[/math][math]=\frac{\partial L}{\partial x}\Rightarrow m\ddot{x}=-\frac{\partial V}{\partial x}[/math]

 

recall from classical physics [math]F=-\nabla V[/math] in 1 dimension this becomes [math]F=-\frac{\partial V}{\partial x}[/math] therefore [math]\frac{\partial L}{\partial x}=-\frac{\partial V}{\partial x}=F[/math] we have [math]m\ddot{x}-\frac{\partial V}{\partial x}=F[/math]

As you can see from the above QFT deals with observable's.  So does QM for that matter despite your claims.

 Now maybe you will better understand that for the delay choice quantum erasure the probability functions with regards to entanglement do in fact describe observable action. QM may use position and momentum but all operators describe physical observable's.  Here is a paper with those mathematics 

https://www.google.com/url?sa=t&source=web&rct=j&url=https://arxiv.org/pdf/quant-ph/9903047&ved=2ahUKEwj31NOSrs3lAhUBHzQIHYoxBAcQFjABegQIBxAB&usg=AOvVaw3VgxY0z5huFoSMko4AYC9e

Edited November 3, 2019 by Mordred