# The absence of Power of Gravity at fundamental level

Discussion in 'Physics & Math' started by Iskcon, Feb 12, 2019.

1. ### IskconRegistered Member

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As per the present science, every object is compressed by the gravity while internal forces create the counter balance. Such internal forces are present due to constituent particles or internal exothermic reactions like fusion in any active star. If there is any fundamental particle, that is having no internal sub particles or no internal reactions, then gravity must compress it. to what? This suggests that a fundamental particle cannot exist in stand alone mode or gravity is absence for a single fundamental particle till it interacts with others.

3. ### NotEinsteinValued Senior Member

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The radius of fundamental particles is taken to be the minimum it can be (zero or planck length); in other words, they cannot be compressed: they are already point particles.

Why would compressing a fundamental particle change it to something else?

Which is obviously not the case in the Standard Model of particles.

Why would gravity be absent? Couldn't it be that gravity is restricted to interaction between two different fundamental particles?

Also, you've missed a speculative possibility. Look at the Bohr atomic model. The electron can only orbit at certain radii. Perhaps the radius of a fundamental particle is similarly only allowed to take certain values. In such a case, gravity would not be able to compress it beyond its smallest value either.

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5. ### exchemistValued Senior Member

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The notion of "particles" is an approximation that breaks down at atomic scale, given the wavelike nature of matter. So you have a problem defining any sort of dimension of a "particle". If you can't do that, you have no way to calculate a "separation" between opposite "sides" of a particle and hence calculate a gravitational force between them.

To quote from the Wiki article on Point Particles, in which I have highlighted in bold the passage that seems most relevant:-

"In quantum mechanics, there is a distinction between an elementary particle (also called "point particle") and a composite particle. An elementary particle, such as an electron, quark, or photon, is a particle with no internal structure. Whereas a composite particle, such as a proton or neutron, has an internal structure (see figure). However, neither elementary nor composite particles are spatially localized, because of the Heisenberg uncertainty principle. The particle wavepacket always occupies a nonzero volume. For example, see atomic orbital: The electron is an elementary particle, but its quantum states form three-dimensional patterns.

Nevertheless, there is good reason that an elementary particle is often called a point particle. Even if an elementary particle has a delocalized wavepacket, the wavepacket can be represented as a quantum superposition of quantum states wherein the particle is exactly localized. Moreover, the interactions of the particle can be represented as a superposition of interactions of individual states which are localized. This is not true for a composite particle, which can never be represented as a superposition of exactly-localized quantum states. It is in this sense that physicists can discuss the intrinsic "size" of a particle: The size of its internal structure, not the size of its wavepacket. The "size" of an elementary particle, in this sense, is exactly zero.

For example, for the electron, experimental evidence shows that the size of an electron is less than 10−18 m.[6] This is consistent with the expected value of exactly zero. (This should not be confused with the classical electron radius, which, despite the name, is unrelated to the actual size of an electron.)"

It seems to me that attempting to infer internal effects of gravitation within an individual subatomic quantum entity is not feasible.

7. ### IskconRegistered Member

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Pl provide a link which puts the radius of a fundamental particle at Planck length.

Why a fundamental particle (if of non zero radius) be not compressed by gravity, as there is no counter force present to resist.

Can you elaborate pl?

Why gravity is not able to compress it beyond its smallest value?

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correct

9. ### IskconRegistered Member

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Surely, you are not implying that "internal effects of gravitation within an individual subatomic (you mean sub-nucleonic!) entity" are absent. Our present theory of gravity does not claim so. I have not understood the use of word "feasible" here by you. Is it the limitation of theory or is it flaw in the theory?

10. ### exchemistValued Senior Member

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I am not implying something I did not say. But all theories have limitations. What I am suggesting is that QM and gravitation, between them, do not enable you to deduce anything about the hypothetical effect of gravity on a fundamental wave-particle entity. Gravity is a large scale phenomenon. We have no idea if gravity remains distinct from other interactions at the sub-atomic scale. And then there is the QM problem of wave-particles being in a smeared out distribution, subject to the uncertainty principle.

But just for fun, I thought I'd apply Newton's law of gravitation to the dimensions and mass of a proton and see what order of force it predicts. The answer is around 10⁻²⁴ N. I might be out by a factor of ten or so as I did not do a full integration. But still, seeing as Avogadro's Number is 6 x 10⁻²³, it suggest that scaled up it is about a newton's-worth of compressive force on a mole of protons. So not huge, and not something you would expect to have dramatic consequences, in terms of compressing quarks or something.

Last edited: Feb 13, 2019
11. ### NotEinsteinValued Senior Member

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I never claimed that it was? But here you go for an explanation why I mentioned it: https://www.quora.com/What-is-the-smallest-measurable-distance-technically

Why are you dodging my question?

Why would gravity be entirely absent if there is only one particle? Think about virtual photons: they don't (meaningfully?) exist when there is only one particle, but it doesn't mean QED or electrical charges don't exist in that case.

Because it literally isn't possible for it to be compressed. For example, let's say the radius can only take positive integer values: 1, 2, 3, etc. What radii smaller than 1 are possible? Answer: none. There is no positive integer smaller than 1, so the radius cannot possibly be smaller than 1.

12. ### NotEinsteinValued Senior Member

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(I didn't know you practiced cosmology! Or: check your orders of magnitude; you'll be negatively surprised...

)

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13. ### IskconRegistered Member

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You claimed
You backed out

14. ### IskconRegistered Member

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Can you back it up?

15. ### IskconRegistered Member

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Any inward compression, irrespective of its magnitude, will have some effect, if there is no outward counter. The compression will continue and equilibrium will not be achieved as long as there is no resistance.

You are right to the extent that for sub protonic particles, the compression (gravitation) potential energy is very very small but it is non zero.

16. ### NotEinsteinValued Senior Member

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No, I in fact did not back out, but I see where your confusion is coming from. The word "or" in English indicates that one of the things that's written to the left and right of it is true, or both. In other words, I never claimed that the radius is Planck length, but that the radius can be one of the following: zero, Planck length. Also, the parenthesis in this case indicate that the more important statement is that the radius is the minimum value it can take; the two mentioned values are mere examples. So even is it isn't the Planck length or zero, the significant part of my claim would still stand.

But more importantly, I did provide a link that explains why the radius of a fundamental particle can be taken to be the Planck length: because it meaningless to argue that it's less, since that difference cannot be measured. Why have you ignored that? You do know it's intellectually dishonest to ignore me proving my claims, and then you claiming I didn't back them up?

I have already done that, in post #8, in the very paragraph you decided to only partially quote. Let me repeat the part you ignored:
"For example, let's say the radius can only take positive integer values: 1, 2, 3, etc. What radii smaller than 1 are possible? Answer: none. There is no positive integer smaller than 1, so the radius cannot possibly be smaller than 1. "

17. ### exchemistValued Senior Member

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Whoops thanks for pointing out the typo - serves me right for using copy/paste when composing.

18. ### NotEinsteinValued Senior Member

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(No problem; I figured that was what happened. Although, you could have gone the "defend your mistake" route, by claiming that "^-1" came from the units, seeing as the units are "mol^-1".

)

19. ### exchemistValued Senior Member

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You are making a series of unjustified assumptions.

First you are not entitled to conclude that, if there were a compressive force of the order of 10⁻²⁴ N on a proton, the force would not be resisted. Quarks have electric charge, remember.

Second, the exercise was just a bit of fun, as I made clear, because, again as I made clear, gravitation is a large scale phenomenon and we have no idea whether it works at the subatomic scale.

As a general point, if you decide to stretch the applicability of a model into conditions remote from those it was designed to model, you must recognise you are merely speculating and that no firm conclusions can be drawn.

In a way I am agreeing with one idea in your original post, to the extent that we do not know whether or how gravity operates at these scales. But your other idea in that post, that it somehow means particles cannot exist on their own, is unjustified.

20. ### exchemistValued Senior Member

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I like to think I have enough self-confidence not to defend mistakes.

But it's a pain using superscripts on my computer. I have to treat them as "special characters" and copy and paste them in, character by character. Hence when I've done it once I tend to copy and paste it for use elsewhere and then edit it to fit - which I failed to do properly on this occasion.

The reason I did the estimation was just to satisfy myself that Newton's law didn't predict a tremendously high force. I had wondered if iskcon might have an agenda based on having done this and found a huge force. But it doesn't work out that way.

I have a recollection of a rather circular discussion with someone about the electron radius, in which it was the classical( i.e. fictitious but occasionally useful) radius that was being talked about, as if it were real. A bit like trying to work with the Rutherford-Bohr model of the atom, pushing a simple model too far leads to unreliable conclusions. Seems to be what iskcon is doing here.

21. ### NotEinsteinValued Senior Member

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I either don't both layouting them properly, or I switch to TeX. Which I suppose isn't less of a pain.

I'm getting the feeling there aren't many calculations behind Iskcon's reasoning: (s)he would have posted those by now, if (s)he had them.

Oh by the gods, yes, I remember that. It was interesting seeing somebody being proved wrong by even the Wikipedia articles on the subject, and then trying to continue to defend their claims.

22. ### IskconRegistered Member

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Wow!
You are fooling around.

The purpose of my ignoring was that you showed lack of understanding of subject.
First of all Bohr model has an applicability issue.
Secondly the electron can be further pushed to nucleus (Read White Dwarf and beyond).
Thirdly subject to applicability the radius is dependent on Z

You are taking an analogy that since Bohr radius can take certain values then the minimum radius of a fundamental particle should also take certain fix values like r1, r2, r3, without even establishing any connect. Moreover there is an angular momentum aspect coupled with equilibrium of electrostatic forces with centripetal forces in case of orbiting electron, so these radii are well understood but you are just blindly throwing some random numbers.

And yes, desist teaching English on this sub forum. If you feel you have good knowledge of English or any other language, please go to appropriate forum.

23. ### IskconRegistered Member

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No, I am making no assumption.
Please tell how this compression will be resisted.
In Newtonian gravity charge is not considered.
In Relativity charge contributes to gravity.
So charge on a fundamental particle is not shown to resist the gravitational compression.

This is not the case. Gravitation is an universal phenomenon, applicable to both large as well as sub atomic level; yes it is true that present theory may not be able to explain the gravitation at sub nucleonic level.

My point was a bit different.
The present theory of gravity kind of suggests collapse of a particle if there is no resistance. Either there must be some resistance or gravity is absent in case of stand alone particle or all such particles must annihilate if they fail to form a complex particle.