The absence of Power of Gravity at fundamental level

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

  1. NotEinstein Valued Senior Member

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    Erm, you are the one that doesn't seem to understand what the word "or" means...

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    How very intellectually honest and mature of you!

    And I think you got the two of us mixed up. You have demonstrated in this very thread a lack of understanding of the radius of fundamental particles, Planck length, the concept of a minimum value, and the role of gravity in the quantum domain.

    What applicability issue? Also, that is irrelevant, because I'm not talking about the Bohr model as it applies in this specific case; I'm using it as an example, an analogue.

    Source please. And I'm talking specifically about a source that shows that the electron can be in a stable situation closer to the nucleus, because that's what we are talking about.

    Please provide a source or calculation which gives the value of Z needed for the electron orbital with any arbitrary radius.

    Aha, so you do know I'm just using it as an example! Now I'm really confused as to what your "applicability issue" is about?

    I have never claimed there is a "connect"?

    In English, the word "analogue" means that I'm going to use a similar situation to describe the currently-under-discussion one. This similar situation will have some feature(s) that are very much like the one under discussion, but this does not imply there is a connection or direct relationship between the two.

    Yes, and these do not explain the electron orbitals at all. In fact, they predict that all atoms are extremely unstable, and that the electron will spiral into the nucleus. Yet, that doesn't happen.

    This is very basic knowledge if you look up how atoms "work". Why are you showing a lack of understanding of the subject?

    Yes, they indeed are.

    In English, the term "for example" means (in this specific case) that I'm going to use arbitrary values for illustrative purposes; the actual values themselves are just picked at random. They don't mean anything. It's the demonstration of their usage that's important, not their values.

    I will "desist teaching English" once you do not need to have basic English words explained to you.

    And I have seen you have once again chosen to ignore most of my post, including critical parts where I point out your mistakes and errors. Indeed, how very intellectually honest and mature of you!
     
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  3. NotEinstein Valued Senior Member

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    But Iskcon did unintentionally raise a valid point: perhaps it is time this thread is moved to a more appropriate sub-forum. He's made it quite clear he isn't willing to discuss lacks of understanding in good faith (anymore?).
     
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  5. Iskcon Registered Member

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    I must put you on ignore, you are quite a drain.
     
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  7. NotEinstein Valued Senior Member

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    Well, if learning basic English is such a drain for you, then why are you on an English forum in the first place?
     
  8. exchemist Valued Senior Member

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    I'm afraid you are making unjustified assumptions.

    My example was a proton, a composite particle for which compression has a meaning, in that there is some sort of of interquark distance that could hypothetically be shortened. I see from the Wiki article on the proton one can indeed speak of a "pressure" due to the quarks. However quarks do resist being pushed together. I see this is described as a feature of the Strong Force rather than simple Coulomb repulsion, but either way there most certainly is a resisting force that will oppose compression.

    As for electrons and quarks, ie. particles that are not composite, the idea of "compression" has no meaning. They are modelled as pointlike, or as close to pointlike (cf. Planck Length) as makes discussion of dimensions - and hence the notion of "compression" - irrelevant.

    When you say:_
    " 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",

    I am unable to follow your logic. Just because gravity "does not consider charge" does not mean that charge ceases to apply, obviously. When you say charge is not "shown" to resist this presumed gravitational compressive force of yours, what you must mean is that Newtonian gravitation does not show this. Well clearly not, but that does not mean that neither it nor the Strong Force cease to apply. Both Coulomb repulsion and the Strong Force were discovered a century or more after Newton.

    And finally you make an unsupported assertion, namely that "Gravitation is an universal phenomenon, applicable to both large as well as sub atomic level". Can you provide evidence of this? Bear in mind we already know from GR that Newtonian gravitation gives the wrong answer under some circumstances, so the inverse square law is not a universally correct expression of the interaction. There is also a hypothesis that the phenomenon known as "dark matter" may be due to gravitation following a different distance law at large separations (MOND etc). At the sub-atomic level, it seems to me that we lack any evidence at all on how gravity may behave.

    It seems to me you are trying to extend a three-century-old model of gravitation into a realm for which we have no evidence it is relevant, on the basis of a billiard-ball model of subatomic particles which was abandoned a century ago.
     
    Last edited: Feb 15, 2019
  9. exchemist Valued Senior Member

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    You do need to realise that Indian English can sometimes be a bit hard to follow for those unfamiliar with it. (I have an advantage, having worked in the Middle East for a some years.

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  10. James R Just this guy, you know? Staff Member

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    Moderator note: Iskcon turned out to be the sock puppet of a banned user, so this identity has also been banned.
     
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  11. el es Registered Senior Member

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    GR is a field theory.

    Spacetime is the field.

    A stand alone particle with mass-energy content has an effect on spacetime, a gravitational effect.

    An interaction with another particle is not required.

    Is a minimum amount of mass required to affect the "fabric" of spacetime? I don't know.
     
  12. exchemist Valued Senior Member

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    So at least three of us were right in our suspicions, then.

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  13. exchemist Valued Senior Member

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    Interesting way of phrasing it.
     
  14. TheFrogger Banned Valued Senior Member

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    I'm uncertain of a lack of gravity at a fundamental level, but I have a query about sound and gravity. When noise is made in the universe, the volume decreases and the noise disappears if the friction is not maintained. When I increase the volume on my television, the level is regulated: it remains at that volume. Just as a dropped object falls, volume decreases without maintainence. Why does this not happen to my television or stereo? Also, if one could pause time when a sound is made, would that sound continue, or fade (As in the real world)?
     
  15. TheFrogger Banned Valued Senior Member

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    Scrap that last question, no-one can answer that. My other questions remain valid though!

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  16. QuarkHead Remedial Math Student Valued Senior Member

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    Interesting but wrong.

    Spacetime is not a field, it is a 4-manifold.

    In applications ( like physics) a field is the assignment to every point on an arbitrary manifold a value - scalar, vector or tensor.

    Do not confuse the manifold (essentially a set of points) with the values assigned to these ponts.
     
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  17. TheFrogger Banned Valued Senior Member

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    That reads like digital, rather than analogue. If every point on a speaker is valued via x and y, and a value stored AT that location, a sound can be reproduced, digitally!
     
  18. exchemist Valued Senior Member

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    Yes indeed, good clarification.

    Nice to see someone who knows what he's talking about still sometimes reads these posts.
     
  19. QuarkHead Remedial Math Student Valued Senior Member

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    Well somebody has to keep you guys honest.
     
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  20. el es Registered Senior Member

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    GR is a classical geometric FIELD theory. The gravitational FIELD is determined by solving the Einstein FIELD equations.

    There is no proper distinction between gravity and spacetime in general relativity.

    So...gravitational field = spacetime field

    Spacetime field may be non-standard usage, but technically not incorrect.
     
  21. QuarkHead Remedial Math Student Valued Senior Member

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    Yes, this quite true.

    But there is, as I explained in an earlier post.

    I think I see where your confusion lies.

    I repeat myself; spacetime is a 4-manifold. Manifolds need not have a metric, in which case we can think of them as a sort of amorphous set of points.

    But a manifold, in order to be of interest to a physicist, is better with a metric - specifically a metric field. Loosely this makes it possible to make measurements at every point on our spacetime manifold.

    It is a regrettable fact that many pop-sci treatments say that the metric field IS the gravitational field.

    The truth is that, given a gravitational source, the metric field may not be constant throughout our manifold.

    It simply remains to define another field, called the curvature, that is a second order derivative of the metric field (with respect to spacetime coordinates), and to remember that, from elementary calculus, these second derivatives are zero when the metric field is constant. That is, in the absence of a gravitational source and hence a constant metric field, then spacetime is "flat"

    Otherwise it is not. That is your gravitational field. It is quite distinct from spacetime, Which I repeat is not NOT a field
     
  22. el es Registered Senior Member

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    Pop-science is all I can handle. Onward and upward.
     
  23. exchemist Valued Senior Member

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    Let me try to paraphrase to see if I have understood.

    Gravitation, in the GR model, is the curvature (2nd derivative wrt the 3 space and 1 time coord) of the metric field. But spacetime itself is not the metric field.

    Is that it?
     

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