Inverse Area Intensity on a Point Particle

So the derivation and intuition behibd it feels good and makes sense. However, lets say we had an infinitely small particle sharing the same space as a particle that is emitting a field. What do? We know a point particle covers 0 area, no matter how close it is to the source. So should 1/(r^2) not apply to point particles? Or do perhaps a gravitational field produces a probability field for gravitons to appear, in which gravitons are assumed to be larger than an infinitesmally small point and they could perhaps nestle in planck wells or pocket? Or perhaps, the gravity source will change the properties of space surrounding it, so any other particles within that modified space are accelerated..... which wouldn't make sense to me. Should I treat a gravitational field like light, and assume it is both a particle and a wave?

Other urls found in this thread:

en.wikipedia.org/wiki/Classical_electron_radius
einstein-online.info/spotlights/gravity_of_gravity/index.html@searchterm=None.html
en.wikipedia.org/wiki/Vacuum_polarization
sureshemre.wordpress.com/2015/12/19/difference-between-a-point-particle-and-an-extended-particle/
twitter.com/AnonBabble

en.wikipedia.org/wiki/Classical_electron_radius

point charges have infinite energy

Explain more please?

A sphere of finite dimension ACTS as if its mass was all at the center, but that only applies when you're outside the surface of the sphere. If you tunnel down, the part of the sphere at a greater radius than you're at no longer affects you. Newton proved that.

So you just ignore the "overlying" mass and consider yourself standing on the surface of a smaller planet. When you reach the center point, you're weightless again. (Probably crushed, but that's another issue.)

You could only experience finite gravity from a point particle if the particle still had finite mass; that means infinite density, which is what black holes are supposed to be at their centers.

As noted, any form of inverse-square force, electrical or gravitational, means infinite force and energy if exerted at zero distance. That's why physicists like the notion of string theory. There aren't any point particles and the problem goes away.

Potential energy (I'll use gravitation here) is -Mm/R.
If M and m are infinitely separated, the energy is zero. As you fall towards a planet, you gain kinetic energy while your potential energy becomes increasingly negative. The sum of the two remains constant.
You can see that if R went to zero, PE would become minus infinity and KE plus infinity.

In the theory of em, you use the Dirac delta function to address this.

In practice, it doesn't matter.

Sorry I just don't think we are on the same page...
What I'm trying to say is that a point particle takes up zero volume and hence zero area. So if you have another gravitational source that propagates it's "gravitational sphere" outwards in all directions, the surface area of it obviously becomes greater, and hence the intensity becomes less. Now lets say we placed a point particle, A, that could be accelerated by gravity, near gravity source B. Depending on how close A is to B, B will have a greater/smaller "gravitational sphere", which kisses A. But since A is a point particle, it covers zero area; So no matter how close A is to B, A will still cover no surface area of the gravity sphere that propagates out from B. Hence, I WOULD think that point particles are not affected by gravity... Which begins to mess with my brain!

>Now lets say we placed a point particle, ...

This is your problem, point particles don't exist in real life like this, and you can't make one.

Ignoring the fact that eventually assembling matter in a dense enough volume would make a black hole, according to classical gravity or electrodynamics it would take infinite energy to compress any amount of matter or charge to an infinitely dense point.

Point particles still have fields even though they have undefinable area.
Mathematically, this is resolved by the delta Dirac function. Also, you can use greens theorem and stokes theorem to show that even though it's a point particle, you can still measure the field Flux and curl if it's dynamic by using an arbitrary surface.

There is no problem with point particles, in fact it is still a matter of high doubt whether elementary particles are point or not whats the matter?

Einsteinian gravity is space-time curvature. EVERYTHING must follow the curve, whether it has mass or dimensions or not.
Back to the equivalence principle.
An apple and a dimensionless point are floating within an box far from any planet. Rockets drive the box upwards. Apple and point thud into the floor. Right?
Alternatively, we put the box on the surface of a planet. The apple falls, If the point didn't, we've a way of distinguishing between inertia and gravitation without looking outside the box.

Gravity results from mass (or energy or pressure) and a zero-dimensional object can have none of those attributes. Even a black hole, though it may CONTAIN a dimensionless point with mass, still has a size, that of the event horizon.

Your problem is you're assuming an "impossible" object. There's an old puzzler: what happens when an irresistible object (a cannonball which will break anything it hits) meets an immovable object (a wall which cannot be breached by any projectile)?
The objects, individually, might exist (in theory) but BOTH of them cannot. Logical paradox. What if I told you there was a man who was 5 feet tall and also 30 feet tall? How can this be? It can't. I'm lying to you. There is no such man and cannot be.

So you don't know whether point particles can emitt gravity or not? It wasn't really the highlight of my problem I am trying to propose, but still intrigues me. Nonetheless, Einstein's view of gravity bending peacetime changes my perspective of things, but then also brings up another problem.

At first I wasn't thinking about space curvature. Instead, I was thinking about a source of gravity (Point or not, doesn't matter) that fires out gravitons. The Inverse-square law perfectly represents the intensity of gravitons per square meter. However, when a point test-particle was floating around, i would assume that there is an infinitely-small chance that a graviton would occupy the same location as our test particle, no matter how close the test particle is to the gravity source.

Now, under Einstein's view, the point particle shall accelerate along the curvature of space; all is good now. However, another question arises: How can we prove that the curvature of space is warped via inverse-square from the gravity source? I hope I explained that correctly and we are all on the same page... If not, I'll try to explain differently.

Depending whether or not the gravitons are points or not*
(And also assuming that gravitons don't generate a field and have a solid cut-off radius).

Let me clarify:

Classical point particles don't and cant exist. Quantum mechanics makes the situation a bit more complicated.

Gravity is not exactly inverse-square. If it was, Mercury's motion could be described exactly by Newton's laws. The orbit wouldn't have that slight precession which drove 19th century astronomers mad (or sent them hunting for Vulcan, an imaginary "inferior" planet.)
Einstein's equations are non-linear. That's why they're so difficult to solve. Any form of mass or energy exerts gravity. A gravitational field represents energy. Therefore, a gravity field itself creates MORE gravity. And so on. (The sequence converges. Fortunately. Or we'd all be immediately yanked into singularities.)

To return to your original question, I showed that a zero-dimensional point (which I'll call a ZP to save typing) must "fall" like everything else. Equivalence principle.
Now, you're still asking if a ZP has a gravity field; if it attracts other objects.
Let's assume, for a moment, that it doesn't. Place a ZP and an ordinary mass (OM) one meter apart in some location far from all other masses. The ZP will fall towards the OM. But the OM is not attracted to the ZP. The mass-center of the system has moved! And that is impossible. Violates momentum conservation. So our assumption (ZPs do not exert a gravity field) must be wrong.
If a stiff wire connects the ZP and the OM, then the whole conglomerate will slide away, accelerating forever. Now we've violated energy conservation!

Convinced???
ZPs MUST have gravity fields. At least they would if they could exist. But they can't.

I'm sure I would be convinced if this made sense to me and/or if I knew whether or not you were on the same page. Nonetheless, I see this:
> If a stiff wire connects the ZP and the OM, then the whole conglomerate will slide away, accelerating forever

Sounds good, that sounds about right. It also leads me to conclude that in order for something to produce gravity, it should also be able to produce repulsion.

What's unclear?

I actually stole the ZP and OM from arguments about "negative mass". NOT antimatter.

Would an antigravity-apple fall "up" if released? No! If it goes up in a gravity field, then it has to go "up" in a closed box being accelerated between stars by a rocket under the floor. That's silly. Equivalence principle again.

An AGA must fall down in a gravity field.
So how does it behave?
All objects (AG or regular) must fall down under normal gravity.
All objects (AG or regular) must fall up in the field surrounding an AGA.

So a regular apple and an AGA will accelerate indefinitely to one side. In theory, this does not violate energy or momentum conservation. The apple gains positive KE and momentum. The AGA gains negative KE and momentum. (If the two don't have equal masses, then you need the stiff connecting wire again and the situation is more complicated. There are elaborate analyses o this. Look up Dr. Robert Forward, for example )

So ONLY a negative mass object can have a repulsive field. No such objects have ever been observed and there's no theoretical reason to believe they exist. Even if they did, they wouldn't coalesce into planets and stars since every single particle tries to flee from all forms of matter.

I mean repulsive as in electrostatics.

What is the new equation for gravity? Did Einstein derive the new equation similar to inverse-square?

Or not even electrostatics, neutron degeneracy pressures too.

You'll have to look up General Relativity.
When the field is weak, it reduces to the inverse-square law.
Only in the case of Mercury is it even detectible -- a tribute to the 19th century mathematicians who accounted for all the perturbations introduced by the other planets. And they did it all with pencil and paper. (Makes my head ache.)

In VERY strong fields, the deviations become obvious. But we're talking neutron stars and black holes by that point.

einstein-online.info/spotlights/gravity_of_gravity/index.html@searchterm=None.html


As the above link notes, ONLY gravity deviates from the inverse-square law.
However, when you get down to the level of individual subatomic particles, quantum effects kick in.
When you have a capacitor, charges separate in the dielectric between the plates in an effort to partially neutralize the "strain". Similarly, the strong field around a proton polarizes the virtual particles which pop out of the surrounding vacuum. These partially neutralize the charge of the proton when measured from a distance.
en.wikipedia.org/wiki/Vacuum_polarization
The article, unfortunately, isn't very clear on this.

sureshemre.wordpress.com/2015/12/19/difference-between-a-point-particle-and-an-extended-particle/
is somewhat better (though less technical) and explicitly discusses "point particles".

It would be adequate for you to look up the schwarzchild metric and its corresponding equations of motion, as the schwarzchild metric describes the curvature of space around a black hole, which is basically a gravitational point mass. But the problem is that the similarities with an inverse square law would be difficult to see without experience in general relativity, for several reasons. First, there is no force being applied, the equations of motion are found by tracing the natural path a particle would take in the curved spacetime using the geodedic equation and the metric. Secondly, because any object simply follows a geodesic with no forces applied, the concept of normal energy no longer works like it does in classical physics, and instead you comb through particular solutions and look for terms that have a similar form to things like rotational kinetic energy, potential energy, and radial kinetic energy. However, you have to be careful conceptually because spacetime is so warped at low r values that many of these energy-like terms no longer act the same as the energies you are probably familiar with from classical physics.

I would suggest checking out Moore T A's "A General Relativity Workbook". It is very clear and there are plenty of conceptual discussions, examples, and exercises.