Antigravity existtence demonstration?

[frankuitaalst]

Guys, if you continue doing this, it'll continue to be A MESS!
THis thing is complicated by itself, there's no need to make it even more complex by changing rerference frame depending on what you are looking at!

Let's just get a cartesian frame (is this the english name?!?), let's put a mass A in the origin and a mass B on the x+ axis ("+" is toward right); let's attribute "+" and "-" signs to the 2 masses in all 4 combinations.
Let's also suppose the mass in the origin can't move.

Ok, now, what happens to mass B in the 4 cases?
1) A=+, B=+ (known case)
2) A=+, B=-
3) A=-, B=+
4) A=-, B=-

Just "by guess" I suppose in the last case we'll have a behaviour similar to case A: B moves toward A.
In cases 2 and 3, B "escapes".
But it's just intuition: what do formulas say instead?

Jumpjack ... It isn't difficult ...
If you write down the equation F = GmM/rē , you get a positive force for positive masses , yes ?
This means that the masses are attracted to each other . So you get : O----> <-----O
While the acceleration is a= F/m and the masses are positive the acceleration has the same direction as the force . So the masses move towards each other . If a mass is negative the accleration has the opposite direction of the force .
If you write down the same equations for the other cases I'm sure you'll come to the answer of your question..

[Tobin Dax]

Guys, if you continue doing this, it'll continue to be A MESS!
THis thing is complicated by itself, there's no need to make it even more complex by changing rerference frame depending on what you are looking at!

Calm down. I did nothing wrong. If you want to know what formulas say, read mine again. I think I explain it clearly.

I did what (it seems) that you are trying to do: I put one mass at the origin of a coordinate system, and describe the relative motion of the other mass. I also looked at the system in an arbitrary reference frame at the end of my post. My explanation holds in that case.

However, Mugaliens makes a very good point, which I agree with, in his reply to me.

[Jeff Root]

I have thought about this question for many years, and have an answer
which is quite different from the one described above. However, it is an
against-the-mainstream speculation, so rather than explain it, I'll just ask
these loaded questions:

If F=ma implies that a negative mass accelerates in the direction opposite
the applied gravitational force, in which direction does it accelerate if the
force is electrostatic rather than gravitational? Why?

-- Jeff, in Minneapolis

[parejkoj]

Jeff: it doesn't matter what the force is. For a negative (inertial) mass, the acceleration is always opposite in sign to the applied force.

Push a negative (inertial) mass away, and it moves toward you, just as Chris said.

As for why, look at the equation you just wrote (in full vector form, of course).

[publius]

Yes, negative mass(-energy) and the Equivalence Principle(EP)will blow your mind, and things would be exactly like Chris said.

The way I like to phrase is that positive mass attracts everything, including negative mass, and negative mass repels everything, including positive mass.

If inertial mass remained positive (and this would mean the EP is out and we've got a non-metric behavior), the gravity would behave like EM but with the signs flipped: like signs attract and opposites repel. GR would be out for such a theory, as the EP is the very foundation.

You can think of positive mass as making a "downhill" gravity well. Geodesics roll toward it. Negative mass makes an uphill gravity well -- geodesics roll away (and stationary clock rates *increase* above the far away observer-- put a -M in Schwarzschild).

What I didn't know was that the relativistic version of the positive and negative mass accelerating has the accelerating in perfect Rindler fashion. That's just cool, and it make perfect sense. Constant acceleration in Newton translates to constant *proper* acceleration.

Now, if you want to have even more fun with this, consider the external sum of the g-fields of the positive and negative mass. Are there any locations where a test particle would be "dragged" along in unison?

If so, you've got yourself a neat little "gravity drive" that can accelerate something without it feeling a force.

Consider a hollow sphere made out of equal halves, hemispheres of positive and negative mass. What would the g field at the center of the sphere be? Now with a positive mass sphere (and even a negative mass one), g at the center is 0. But would this hold for with two equal and opposite hemispheres?

What would the relativistic version of that be?

-Richard

[What Max]

if u push a box with negative mass, it moves toward you.