Gravity could still be a pushing force

[ydoaPs]

I don't know anything about vectors sorry

I'm sorry, I assumed that since you're throwing around phrases like "Schrödinger Hypothesis" you knew what they meant. In QM, we represent the state of a quantum system via what we call a state vector. When we act on a system, represented by using a linear operator on the state vector, we get a new observable property (represented by a scalar called the 'eigenvalue').

 

[latex]A|\Psi>={\lambda}|{\Psi}>[/latex] where A is the linear operator, lambda is the eigenvalue for the state vector Psi. To find the probability that lambda is observed, we take [latex]|<\Psi|\Psi>|^2[/latex] where <Psi| and |Psi> are complex conjugates of each other. THAT is how probability is incorporated into QM.

 

This, however, has absolutely nothing to do with your 'equation'. In fact, as I said before, multiplying a probability by a scalar with a unit is meaningless. It literally doesn't mean anything to say something is 18i kg/m² times more probable than something else.

[Popcorn Sutton]

But if you choose a point in space, any point that had no observable unit, and you put a probability if a unit being there, say 10% chance that there is an atom of hydrogen at that point. G will be influenced because the hydrogen is both there and not there at the same time, so you can observe G wrt earth as approximately .00000000001 at a probability of .1 that it exists at that point, so the gravity from that point has an area if effect on our planet (too small to notice), but then we observe that there is, in fact, a hydrogen atom at that point, then the probability becomes 1 and G gets amplified after observation (approx .000000001 wrt earth). Still barely noticeable, probably unnoticeable, but amplified by observation, but the area of effect would be the same.

 

Plus, your straw manned me because I never said that the mass of the unit would be the probability, just that the probability affects G. I could be wrong, but it coincides with QM.

 

G = P(u|o)*len(u)|len(poi)

 

Simple enough

 

Now for the aoe (area of effect)

This one is tricky because the two edges of the length of the unit get infinitesimally close to the center but never absolutely converge, so how do I represent that in math?

 

Aoe= len(space to u)|len(poi)*(len(u)*.9(repeating))

 

So this should give you a result like this. Given the distance between us and the asteroid, the gravitational influence it has can be pinpointed at (location) at 1/1000 the size of the planet. The gravitational influence at time t is equal to .001 the gravity of earth, and the rotation of the asteroid could cause dust clouds, potentially a tornado at time t that will move in this direction until the object is altered or earth moves out of the spiralling gravitational tension. (hypothetical scenario of use)

 

Through complexity comes clarity.

[Bignose]

Through complexity comes clarity.

Nope. Not clarity at all. And still no comparison between mainstream, your idea, and measurement...

[ydoaPs]

But if you choose a point in space, any point that had no observable unit, and you put a probability if a unit being there, say 10% chance that there is an atom of hydrogen at that point. G will be influenced because the hydrogen is both there and not there at the same time, so you can observe G wrt earth as approximately .00000000001 at a probability of .1 that it exists at that point, so the gravity from that point has an area if effect on our planet (too small to notice), but then we observe that there is, in fact, a hydrogen atom at that point, then the probability becomes 1 and G gets amplified after observation (approx .000000001 wrt earth). Still barely noticeable, probably unnoticeable, but amplified by observation, but the area of effect would be the same.

That's not how you would use probability there. In fact, I wouldn't use probability there. I'd use a range of values for position and generate a range of values for the force. Once you get the range of values, however, you can do things like calculate the probability of the force being a certain value.

 

With a proper equation, like [latex]|F|=G\frac{{m_1}{m_2}}{r^2}[/latex], we get definite values. So, let's say we have four possible distance values: r₁, r₂, r₃, r₄. Given constant masses, they give us four constant forces: F₁, F₂, F₃, F₄. So, now we have something to work with. To find the probability of a given F, we need to find P(F_n&r_n). Basic probability theory tells us that P(A&B)=P(A|B)xP(B). Since F_n is a deductive consequence of r_n, we know that P(F_n|r_m)=1. This means that P(F_n)=P(r_n). So, the probability of any given force value is equal to the probability of the corresponding distance between the interacting objects. This, of course, means you need to know the probability distribution of the position.

Plus, your straw manned me because I never said that the mass of the unit would be the probability, just that the probability affects G. I could be wrong, but it coincides with QM.

I didn't say it did. What I did say, though, is that multiplying a probability by a number with a unit on it is gibberish. It literally means nothing. And there's nothing in QM that I know of which co-incides with the way you're doing things.

G = P(u|o)*len(u)|len(poi)

Now, this is better, since the scalar factor is unitless. However, we still have the problem of "What, exactly, is G the probability of?". Your resultant probability is vague to the point of not meaning anything at all. I'm also not sure what the lengths here are. I just went back, and still have no idea what you mean by 'poi'.

Simple enough

 

Now for the aoe (area of effect)

This one is tricky because the two edges of the length of the unit get infinitesimally close to the center but never absolutely converge, so how do I represent that in math?

 

Aoe= len(space to u)|len(poi)*(len(u)*.9(repeating))

Leaving aside the arbitrary use of a value equivalent to one as a multiplier, this doesn't work. Again, do the dimensional analysis. A unit divided by itself is unitless. A unitless value multiplied by a value with a unit means the resulting value has the unit of the united value. So, your area, then, is a length.

[Bignose]

With a proper equation, like [latex]|F|=-G\frac{{m_1}{m_2}}{r^2}[/latex], we get definite values.

Yep. This is why the current model is very, very, very good. PS has a very high hurdle to jump to demonstrate his model makes better predictions.

 

Not an impossible hurdle to jump, mind you. But, the current model is very, very, very good.

[ydoaPs]

Oops, shouldn't have had the negative in there, since I put the absolute value bars. My bad.

[swansont]

!

Moderator Note

Moderator consensus is that this has stayed on the shelf well past its expiration date.

 

It's not a good example of making your case and accepting scientific feedback (though at least we have avoided the drama of personal attacks and appeals to conspiracy)