7 Reasons To Abandon Quantum Mechanics-And Embrace This New Theory

[andrewgray]

So, OceanBreeze.  I am curious.  How would YOU explain the perpendicular photoelectric ejections along the polarization of the incoming light wave?  So for example, how do YOU explain that if you shine a horizontally polarized laser onto a metal, then no electrons will come out?  How do you explain that?  Wouldn't "horizontal light particles"  (oh my goodness!) still be absorbed by the electrons and get knocked out?   How are you going to bring in the wave characteristic (polarization) of the light wave if you think "particles"?  Do the "horizontal light particles" not get absorbed by the electrons for some reason?

 

I'm curious.  I am digging in here.  Since you have insulted my theory, I suppose you have a better answer.

 

Andrew Ancel Gray

 

[devin553344]

So, OceanBreeze.  I am curious.  How would YOU explain the perpendicular photoelectric ejections along the polarization of the incoming light wave?  So for example, how do YOU explain that if you shine a horizontally polarized laser onto a metal, then no electrons will come out?  How do you explain that?  Wouldn't "horizontal light particles"  (oh my goodness!) still be absorbed by the electrons and get knocked out?   How are you going to bring in the wave characteristic (polarization) of the light wave if you think "particles"?  Do the "horizontal light particles" not get absorbed by the electrons for some reason?

 

I'm curious.  I am digging in here.  Since you have insulted my theory, I suppose you have a better answer.

 

Andrew Ancel Gray

What makes you think the photons aren't reflecting off the surface and pulling electrons with them? I'm not versed on the photoelectric effect. But that would be my first guess knowing the small amount about it that I do.

 

And if that's the case then my ideas would explain it better with a strong force of attraction equal to Planck's constant.

 

PS I have yet to find angle of ejection in the articles that I read. Where are you getting this information from?

Edited June 25, 2020 by devin553344

[OceanBreeze]

 

 

I'm curious.  I am digging in here.  Since you have insulted my theory, I suppose you have a better answer.

 

Andrew Ancel Gray

 

First off, there is some misunderstanding here, so let me clarify that I didn’t “insult your theory” as I haven’t watched any of your videos and I’m not really sure what your theory is.

 

My post, that you took offense to, was directed to the poster who was going by the handle “Dalton Chance” aka Polymath or Superpolymath or at least two dozen other handles. He is a banned troll and I deleted his post in your thread that’s all. I apologize for any confusion this caused.

 

 

So, OceanBreeze.  I am curious.  How would YOU explain the perpendicular photoelectric ejections along the polarization of the incoming light wave?  So for example, how do YOU explain that if you shine a horizontally polarized laser onto a metal, then no electrons will come out?  How do you explain that?  Wouldn't "horizontal light particles"  (oh my goodness!) still be absorbed by the electrons and get knocked out?   How are you going to bring in the wave characteristic (polarization) of the light wave if you think "particles"?  Do the "horizontal light particles" not get absorbed by the electrons for some reason?

 

 

Well, now that you asked, I would explain it by saying it is not established by any science I know of.

 

You are basically denying the conservation of angular momentum; not something I recommend.

 

As far as I know, experiments of the photoelectric effect with metals (you did say metals, remember) have never shown any strong anistrophy of the outgoing photoelectrons with regard to the polarization of the incident light. The resulting direction of the outgoing electrons is  statistical and can depend on whether or not the electrons in the target material are moving randomly in all directions or they have a bias, as in a crystal structure, but you specified a metal.  

 

If you have evidence to the contrary, I would be interested in seeing it, but I am not going to watch a slew of silly videos! If you can present your evidence, with links, so that I can read up on it, that would be appreciated.

 

By the way, I did have a laugh at your boasting of having your work published in “Physics Essays”

[OceanBreeze]

You're rude and insulting and this is just another example of that.

 When you have no other reason to enter a thread except to troll, then that is a clear rules violation. Consider this a warning if you want to continue posting here.

[andrewgray]

Ok, OceanBreeze, sorry for the misunderstanding. And thanks for removing trolls from this thread!  I had no idea.  Now back to the understanding!

 

Yes, photoelectrons do show a big preference for ejection along the polarization of the incident light wave.

Here is the most stark evidence for that:

 

https://journals.aps.org/pr/abstract/10.1103/PhysRev.37.1233

The Angular Distribution of Photoelectrons Ejected by Polarized Ultraviolet Light in Potassium Vapor

 

Milton A. Chaffee Phys. Rev. 37, 1233 – Published 15 May 1931

Quote

 

...the most probable direction of ejection is that of the electric vector and that the angular distribution varies as the square of the cosine of the angle between the electric vector and the direction in question.

 

 

Another Example:

 

https://escholarship.org/uc/item/01t182mz

Evidence of vectorial photoelectric effect on copper

2005   Pedersoli, E.
 

Quote

 

...enhancement in p polarization which can not be explained in terms of optical absorption,a phenomenon known as vectorial photoelectric effect.  

 

...enhancement is found for light with electric field perpendicular to the sample’s surface, showing a vectorial photoelectric effect.

So you see, OceanBreeze, in the first example cos²(90^o) = 0,  so no electrons come out at 90^o to the polarization.  Most come out along the polarization, and it drops off as cos².

 

In the second example, there is a clear preference for vertically polarized ejections, as can be seen in the plot!

 

Now, you should see  Episode 2 The PhotoElectric Effect  to find out why this is so!  QM is incapable of an explanation! :

 

Quote

 

E. Pedersoli:

 

...dependence on angle of incidence and light polarization is a long standing problem [4–8] that largely remains to be understood."

 

Intermittent Electrons explains both!  

 

Andrew Ancel Gray

Edited September 29, 2020 by andrewgray

[andrewgray]

Devin,

 

 

the photons...  ...reflecting off the surface and pulling electrons with them

 

Now come on Devin.  "Light Particles pulling electrons out?"  By what mechanism do these fictitious light particles "pull on" electrons (oh my goodness!)???  I sure hope you start to see that "light particles" are never needed for any explanation in physics.

 

Andrew Ancel Gray

[devin553344]

Ok, OceanBreeze, sorry for the misunderstanding. And thanks for removing trolls from this thread!  I had no idea.  Now back to the understanding!

 

Yes, photoelectrons do show a big preference for ejection along the polarization of the incident light wave.

Here is the most stark evidence for that:

 

https://journals.aps.org/pr/abstract/10.1103/PhysRev.37.1233

The Angular Distribution of Photoelectrons Ejected by Polarized Ultraviolet Light in Potassium Vapor

 

Milton A. Chaffee Phys. Rev. 37, 1233 – Published 15 May 1931

 

 

Another Example:

 

https://escholarship.org/uc/item/01t182mz

Evidence of vectorial photoelectric effect on copper

2005   Pedersoli, E.

 

 

So you see, OceanBreeze, in the first example cos²(90^o) = 0,  so no electrons come out at 90^o to the polarization.  Most come out along the polarization, and it drops off as cos².

 

In the second example, there is a clear preference for vertically polarized ejections, as can be seen in the plot!

 

Now, you should see  Episode 2 The PhotoElectric Effect  to find out why this is so!  (QM is incapable of an explanation!)

 

Andrew Ancel Gray

And how do the papers explain the results? But their answer is "An explanation in terms of non local conductivity tensor is proposed."

 

P.S. OK I had a chance to review the paper at escholarship and it looks reflective with polarization taken into account. Although the author indicates absorption as the mechanism. Noting that the photoemission angle is connected to the incidence angle (a type of reflection). The mechanism is most likely atomic in this case for reflection, and the perpedicular to incidence plane would allow greater reflected energy increasing the chance of an emission. It really isn't that difficult to explain.

Edited June 26, 2020 by devin553344

[devin553344]

Devin,

 

 

Now come on Devin.  "Light Particles pulling electrons out?"  By what mechanism do these fictitious light particles "pull on" electrons (oh my goodness!)???  I sure hope you start to see that "light particles" are never needed for any explanation in physics.

 

Andrew Ancel Gray

Have you looked at my theory? Apparently not, it's in the lineup: http://www.scienceforums.com/topic/36816-unification-of-the-forces/, but basically I might explain it with a strong force of attraction that matches E=hv. Noting that the wavelength for photoelectrics is around 1E-7 which could reach into several atoms before reflecting.

 

This does not take into account polarization as my calculations are raw for the total energy of Planck's constant from a Plane solution. In order to describe polarization one would have to expand on that.

 

Thanks for your time :)

Edited June 26, 2020 by devin553344

[OceanBreeze]

Ok, OceanBreeze, sorry for the misunderstanding. And thanks for removing trolls from this thread!  I had no idea.  Now back to the understanding!

 

Yes, photoelectrons do show a big preference for ejection along the polarization of the incident light wave.

Here is the most stark evidence for that:

 

https://journals.aps.org/pr/abstract/10.1103/PhysRev.37.1233

The Angular Distribution of Photoelectrons Ejected by Polarized Ultraviolet Light in Potassium Vapor

 

Milton A. Chaffee Phys. Rev. 37, 1233 – Published 15 May 1931

 

 

Another Example:

 

https://escholarship.org/uc/item/01t182mz

Evidence of vectorial photoelectric effect on copper

2005   Pedersoli, E.

 

 

 

So you see, OceanBreeze, in the first example cos²(90^o) = 0,  so no electrons come out at 90^o to the polarization.  Most come out along the polarization, and it drops off as cos².

 

In the second example, there is a clear preference for vertically polarized ejections, as can be seen in the plot!

 

Now, you should see  Episode 2 The PhotoElectric Effect  to find out why this is so!  QM is incapable of an explanation! :

 

 

Intermittent Electrons explains both!  

 

Andrew Ancel Gray

 

 

Glad the misunderstanding is cleared up Andrew.  I will try to keep the trolls out by removing the off topic posts as well as my replies to them. My replies just make the problem worse.

 

Well, OK, back to the topic.

 

From the first paper you cited:

 

 “This result is in accord with predictions of the wave mechanics for a spherically symmetrical atom and incidentally therefore constitutes additional evidence that molecules do not play an appreciable part in the observed photo-ionization of potassium vapor”

 

Whoa! This is photo-ionization in a gas! Nothing to do with the photoelectric effect with metals. This is exactly why I stressed, in my earlier post, that your claim is “if you shine a horizontally polarized laser onto a metal, then no electrons will come out”. Now you are citing a paper on the photo-ionization of a gas to support that claim? That doesn’t work for me.

 

The second paper:

 

“Quantum Efficiency (QE) measurements of single photon photoemission from a Cu(111) single crystal and a Cu polycrystal photocathodes, irradiated by 150 fs-6.28 eV laser pulses, are reported over a broad range of incidence angle, both in s and p polarizations. The maximum QE (approx. = 4x10-4) for polycrystalline Cu is obtained in p polarization at an angle of incidence theta = 65 deg. We observe a QE enhancement in p polarization which can not be explained in terms of optical absorption, a phenomenon known as vectorial photoelectric effect”

 

Whoops, this is about photon photoemission from a crystal. In my post I already addressed this  quote: “the result is also is statistical and can depend on whether or not the electrons in the target material are moving randomly in all directions or they have a bias, as in a crystal structure, but you specified a metal”

 

It seems that neither of the papers support your claim that “if you shine a horizontally polarized laser onto a metal, then no electrons will come out”

 

I hope you can find some better sources than these to make your case.

[andrewgray]

OceanBreeze,

 

Thanks for the interesting discussion!

 

Now, I doubt crystal orientation has much to do with electron ejection along the light wave polarization:

 

Quote

 

OceanBreeze:

The resulting direction of the outgoing electrons is statistical and can depend on whether or not the electrons in the target material are moving randomly in all directions or they have a bias, as in a crystal structure,

 

If the crystal structure gave a bias in direction, then the poly-crystalline copper, which would have crystals aligned in all directions, would not have such a bias.  But we see the same "along-the-polarization" direction of ejection from poly-crystalline copper:

So again, I doubt that it is the crystal structure giving the bias towards the polarization. So the bottom line is this:  "light comes into stuff, and ejects the electrons along its polarization E field".  No matter what you call it, the electrons prefer the ejection along the polarization of the light wave  (here in this plot at 60^o, by 8 to 1).

 

Now since you do not want to watch my Episode 2 video explaining why this is, let me summarize it for you.

 

1.  The intermittent electrons in the copper are pulsating their electrical influence ON and OFF.

2.  If these electrons are pulsating in such a way to be in resonance with just the peaks of a vertical  light wave (ON with peaks, OFF with valleys)  then they will be radically ejected upwards along the polarization E field of the light wave.

3.  If these electrons are pulsating in such a way to be in resonance with just the peaks of a horizontal light wave (ON with peaks, OFF with valleys)  then they will be pushed sideways along the surface of the copper (not ejected much).

4.  Now as the electrons (from 2.) get accelerated upwards, they increase their intermittent pulsation rate in accordance with De Broglie.  As the pulsation rate increases, eventually these electrons will come into pulsation resonance with both the peaks and valleys of the incident light wave.  In other words, the acceleration is over and the electrons gain just a finite amount of energy that we see in the photoelectric effect.

 

That's the explanation for the bias towards polarization ejection and energy limitation of the photoelectric effect. 

 

Now the only thing so far from the QM world that I have seen to explain this is:

 

Quote

 

This result is in accord with predictions of the wave mechanics for a spherically symmetrical atom...

 

An explanation in terms of non local conductivity tensor is proposed...

 

Huh?

 

So let me again put you on the spot!  Let us hear the explanation for the bias towards polarization ejection from poly-crystalline copper (and from the potassium vapor).

 

Andrew Ancel Gray

Edited September 29, 2020 by andrewgray

[andrewgray]

Well,

 

I guess it is time to move on from the Episode1/Episode2 videos to Episode3_TheAtom.   Let's discuss the Intermittent Electron Atom in Episode 3, please.

 

What do you think about the "New-Wisdom-Non-Radiating-Planetary-Orbits-In-The-Episode3-Atom"?

 

Episode 3: The Atom! (Part 1)

https://drive.google...iew?usp=sharing

 

Episode 3: The Atom! (Part 2)

https://drive.google...iew?usp=sharing

 

 

Andrew Ancel Gray

[devin553344]

Well,

 

I guess it is time to move on from the Episode1/Episode2 videos to Episode3_TheAtom.   Let's discuss the Intermittent Electron Atom in Episode 3, please.

 

What do you think about the "New-Wisdom-Non-Radiating-Planetary-Orbits-In-The-Episode3-Atom"?

 

Episode 3: The Atom! (Part 1)

https://drive.google...iew?usp=sharing

 

Episode 3: The Atom! (Part 2)

https://drive.google...iew?usp=sharing

 

 

Andrew Ancel Gray

I just calculated gravitation from a vibrating particle, the formula for the oscillation is: ue^2f^2r^2 where u is the permeability of free space, e is the elementary charge, f is the particle frequency and r is the particle wavelength. Now if you can use that for intermittence then maybe your idea works in my opinion. This vibration represents a temperature.

Edited September 7, 2020 by devin553344

[andrewgray]

Devin,

 

I am not sure what your "gravitation" is that you are calculating here.  But I am glad you mentioned it!

 

In order to understand gravitation, the key is to understanding the neutron. The key is the neutron's mass.  Hopefully we can have a friendly discussion about the neutron mass here.

 

In the 1930's James Chadwick and the Joliot-Curies were arguing over the mass of the neutron.  Chadwick calculated and insisted that the neutron mass was less than that of the hydrogen atom, and the Joliot-Curies calculated and insisted that the neutron was greater than the mass of the hydrogen atom.

 

Then along came Maurice Goldhaber, Chadwick's new assistant, who came up with the idea of using gamma radiation to split a deuteron!  Here is the equation that they used:

Here they mistakenly assumed that the gamma electromagnetic waves were "gamma particles", and mistakenly put E_γ=hv into this equations so that the neutron mass gets calculated incorrectly!  Let us emphasize that there are no gamma particles by writing this equation:

There are no "gamma particles"!  There are no EM particles of any kind.  Especially no gamma particles.  How do we know there are no EM particles? Let us review.  There are no EM particles because we know:

 

1.  The X-ray frequency limit is a Nyquist Frequency Limit (and has nothing to do with fictitious "X-ray particles".)

2.  The photoelectric effect is a transverse acceleration resonance with pulsating electrons and has nothing to do with fictitious "light particles".

3.  The Compton change in wavelength is due to a Doppler shift from an acceleration resonance with a receding electron orbit and has nothing to do with EM particles.

4.  The hydrogen spectra is the disturbance of resonant electron orbits and has nothing to do with energy level changes and light particles.

 

There really is No Reason to Use EM particles in any Explanation in Physics!!! 

 

So we clearly see the mistakes made by using them!  So if the above neutron mass equation does not use a fictitious gamma particle mass, then the gamma EM radiation can be any amount of energy...   And since the neutron ALWAYS decays into an electron and proton, we can go back to James Chadwick's original calculations of the neutron mass (not using fiction!) and we, like Chadwick, get that the neutron's mass is less than the hydrogen atom!  (yes, I know that this means that quarks are also fictitious.  Yes they are).  

 

Phew!  That was a lot.  So let's take a break here and get back to where we were originally going.

 

Quote

 

Andrew Gray:

"Since there are no fictitious "gamma particles", we go back to Chadwick's original calculation without them and conclude, along with Chadwick, that the neutron mass is less than the hydrogen atom.  This means that the neutron is indeed a composite particle made up of a proton and electron, and then this explains gravity!"

 

Before we go on... I am sure you guys have some friendly push back on this.  SO LET's HEAR IT!

Edited September 29, 2020 by andrewgray

[devin553344]

Devin,

 

I am not sure what your "gravitation" is that you are calculating here.  But I am glad you mentioned it!

 

In order to understand gravitation, the key is to understanding the neutron. The key is the neutron's mass.  Hopefully we can have a friendly discussion about the neutron mass here.

 

In the 1930's James Chadwick and the Joliot-Curies were arguing over the mass of the neutron.  Chadwick calculated and insisted that the neutron mass was less than that of the hydrogen atom, and the Joliot-Curies calculated and insisted that the neutron was greater than the mass of the hydrogen atom.

 

Then along came Maurice Goldhaber, Chadwick's new assistant, who came up with the idea of using gamma radiation to split a deuteron!  Here is the equation that they used:

Here they mistakenly assumed that the gamma electromagnetic waves were "gamma particles", and mistakenly put E_γ=hv into this equations so that the neutron mass gets calculated incorrectly!  Let us emphasize that there are no gamma particles by writing this equation:

 

There are no "gamma particles"!  There are no EM particles of any kind.  Especially no gamma particles.  How do we know there are no EM particles? Let us review.  There are no EM particles because we know:

 

1.  The X-ray frequency limit is a Nyquist Frequency Limit (and has nothing to do with fictitious "X-ray particles".)

2.  The photoelectric effect is a transverse acceleration resonance with pulsating electrons and has nothing to do with fictitious "light particles".

3.  The Compton change in wavelength is due to a Doppler shift from an acceleration resonance with a receding electron orbit and has nothing to do with EM particles.

4.  The hydrogen spectra is the disturbance of resonant electron orbits and has nothing to do with energy level changes and light particles.

 

There really is No Reason to Use EM particles in any Explanation in Physics!!! 

 

So we clearly see the mistakes made by using them!  So if the above neutron mass equation does not use a fictitious gamma particle mass, then the gamma EM radiation can be any amount of energy...   And since the neutron ALWAYS decays into an electron and proton, we can go back to James Chadwick's original calculations of the neutron mass (not using fiction!) and we, like Chadwick, get that the neutron's mass is less than the hydrogen atom!  (yes, I know that this means that quarks are also fictitious.  Yes they are).  

 

Phew!  That was a lot.  So let's take a break here and get back to where we were originally going.

 

 

Before we go on... I am sure you guys have some friendly push back on this.  SO LET's HEAR IT!

I'm really not desiring to get into a lengthy discussion as I haven't the time for it. I just wanted to push an idea your way that I have made a mathematical connection between electric charge and temperature pressure. Basically mass is a super vacuum and gravity well of sorts, and charge is a super pressure and has a temperature equivalent. This means charge appears to be vibrating with temperature to push other objects away for like charges.

 

The gravitation I have proposed is that of a Shockley diode equation. Such that there is a diode-like connection between the matter valley and the charge hill for particles. The magnetic moment of the particle can be considered to be an amperage and so the gravitation is a reduced "reversed biased" amperage. Using this idea I have calculated the gravitation of the electron and proton. I don't need to get into neutrons for my idea to be valid.

 

For more information you can look at my pdf file in the op here (http://www.scienceforums.com/topic/37097-unification-of-the-forces-a-new-aproach/) and see if it relates to your work. Cheers :)

[andrewgray]

On 6/26/2020 at 11:41 AM, OceanBreeze said:

Devin,  You lose me here:

 

devin553344, on 12 Sept 2020 - 4:41 PM, said:

It keeps the wave-particle system together...

 

Like I explained earlier in this thread:

 

 

Quote

 

1.  Light is a wave and only a wave.

2.  Electrons are particles and only particles.

 

So "wave-particle duality" is gone.  It really is silly once you know what is going on!

So Devin "No!". I am not with you. Devin, don't go there!

Now can we get back to gravity?

 

 

 

[andrewgray]

Quote

Apparently not together. Your views are too unorthodox for me and go against the grain.

Yes, Devin, this New Wisdom Unifies Physics theory "goes against the grain".  There will be no more...

 

Light Particles

Matter Waves

Virtual Particles

Wormholes

Multiverses

NonLocality

Entanglement.....

etc...

 

Is it so bad to be against this grain???    The grain of current modern physics is very coarse and needs a good sanding!   Our new theory is what actually goes with the grain and will become the orthodox.  Perhaps Devin, if we have to leave you behind, you will at least listen to what we are trying to tell you, and consider it!

 

Now on with gravity!

 

Andrew Ancel Gray

[andrewgray]

OK.  Gravity.  Like we were saying, the key to understanding gravity is the neutron's mass.  Since we have shown that there are no gamma particles, this equation:

cannot be used, and we must revert back to Chadwick's original calculations for the neutron mass, in which he found that the neutron WAS ACTUALLY LIGHTER than a hydrogen atom!   This means that the neutron actually IS a composite particle, made up of a bound electron and proton.

 

Finally, this explains why the inertial mass and the gravitational mass are the same!  How? 

 

Well first we must digress momentarily... Consider two hydrogen atoms sitting 1 meter apart out in interstellar space, H_A and H_B. There are 4 electrical forces at play here between the two atoms.  Two repulsive electrical forces of the "like charges" repelling each other:  ProtonA↔ProtonB and ElectronA↔ElectronB.  And two attractive forces of "unlike charges" attracting each other:  ProtonA→←ElectronB and ElectronA→←ProtonB.  We will denote these 4 forces as P_AP_B, E_AE_B, P_AE_B and E_AP_B.   Now conventionally,

 

P_AP_B+ E_AE_B+ P_AE_B+ E_AP_B = 0.

 

But is this sum really zero?  Now the magnitude of each of these four electrical forces at 1 meter separation is given by

But suppose that the sum is REALLY SMALL, but not quite zero.  The electron and proton are not antimatter pairs and are very different things.  So suppose the sum is REALLY REALLY SMALL but

 

P_AP_B+ E_AE_B+ P_AE_B+ E_AP_B = -1.87 x 10^-64  Newtons       

 

 

An unimaginably small quantity!  This minuscule sum is experimentally not different from zero!

But if this were true, then you would have gravity!  How so?

 

Well, now let us consider a Hydrogen Atom and a Deuterium Atom out in interstellar space separated by 1 meter, H_A and D_B. Since the deuterium's neutron is a composite of an electron and proton, all we have in these two atoms are electrons and protons!  So we know how to compute the force between the Hydrogen and the Deuterium since we know the above sum!   There are 8 forces at play in this case between the two atoms!  We can write them as

 

(P_AP_B1+ E_AE_B1+ P_AE_B1+ E_AP_B1) + (P_AP_B2+ E_AE_B2+ P_AE_B2+ E_AP_B2)

 

where E_B1 and P_B1 denote the forces involving the "regular" electron and proton in the Deuterium, and E_B2 and P_B2 denote the forces involving the electron and proton in the Deuterium's neutron!   But we know how to sum these forces!

(P_AP_B1+ E_AE_B1+ P_AE_B1+ E_AP_B1) + (P_AP_B2+ E_AE_B2+ P_AE_B2+ E_AP_B2)

= 2 x (-1.87 x 10^-64) N

That is, the gravitational force between Hydrogen→←Deuterium is twice that of the gravitational force between Hydrogen→←Hydrogen.  We know this because we now know how to handle the Deuterium's neutron!  And... notice that the inertial mass, which is proportional to how many electron-proton pairs an atom has, is the same as the gravitational mass, because the gravitational mass also depends on how many electron proton pairs an atom has as well! 

 

That means that gravity is really a "residual electrical force", and not some independent gravitational force and it means that gravity is not due to "curved spacetime"!  So gravity is really a vector force, and that means that we will have to deal with Einstein, his equivalence principle, and his General Relativity.

 

Any pushback or comments before I go on? 

 

 

 

Edited September 29, 2020 by andrewgray