QM + GR = black holes cannot exist

Discussion in 'Physics & Math' started by RJBeery, Sep 24, 2014.

Thread Status:
Not open for further replies.
  1. brucep Valued Senior Member

    Messages:
    4,098
    The last stable orbit in the Schwarzschild geometry is at r=6M. Between the photon sphere at r=3M and r=6M the possibility of knife edge orbits exist. These orbits require energy to be summed to the orbit to remain stable.
    Get a clue. You can't identify anything. Pay attention. The rate has nothing to do with the mass of the object in free fall. That's what the guy in the link (you think doesn't apply) said. Galileo figured that out. His experiment is based on weak field Earth parameters so the measurements reflect that. Get it? You think the parameter for M is the spacetime curvature associated with the Earth for objects like the Sun and black holes? Let's do it for an object whose mass is < the mass of the earth. The moon.

    dr_shell/dt_shell = (2M/r)^1/2

    M_earth = .00444meter

    r_earth = 6.371E6meter

    = (.00888m/6.371E6)^1/2

    = .00003733384159 (c=1)

    The rate objects fall in the near earth spacetime is dependent on the mass of the earth and the distance r. The mass of the falling object has no bearing on this. So your analysis claiming Galileo was wrong is crank nonsense. You need to learn the physics if you want folks to quit calling you names. Idiot wind.
     
  2. Google AdSense Guest Advertisement



    to hide all adverts.
  3. RJBeery Natural Philosopher Valued Senior Member

    Messages:
    4,222
    So, you're saying that all objects would take 1.4 seconds to fall 10 meters to the Earth unless they are "very very big"? That seems scientific to you, does it?
     
  4. Google AdSense Guest Advertisement



    to hide all adverts.
  5. Layman Totally Internally Reflected Valued Senior Member

    Messages:
    1,001
    If something was in a black hole it would already have had to have accelerated faster than the speed of light to even get to the event horizon. The escape velocity at the event horizon is the speed of light, but closer to the center of the black hole the escape velocity is greater than the speed of light. Then light inside of the event horizon wouldn't even make it to the event horizon, because the escape velocity of where it was inside of the black hole would have already been greater than the speed of light.
     
  6. Google AdSense Guest Advertisement



    to hide all adverts.
  7. brucep Valued Senior Member

    Messages:
    4,098
    The last stable orbit in the Schwarzschild geometry is at r=6M. Between the photon sphere at r=3M and r=6M the possibility of
    Quit trolling.
     
  8. Stryder Keeper of "good" ideas. Valued Senior Member

    Messages:
    13,105
    RJBeery,
    I'd reason (feel free to find evidence to the contrary) that Gravity is a unification of field fluctuations surrounding all matter. (It can be disputed whether matter is the cause of the gravitation field, or whether that a distortion in space-time creates a pocket where gravity exists and matter accumulates.) In either sense there is going to be a limitation on the speed of gravity based upon the unified fluctuation. I couldn't tell you what the value is specifically however it's what defines the base Unified Rate of Attraction that both Galileo and Newton were familiar with.

    Increasing or decreasing the volume of mass doesn't increase the Rate of Attraction it just factors in on the distance at which the Rate of Attraction effects other bodies. The further the distance the greater the time period that a small body is effected by Attraction, which is what effect's the speed of it's relative "Fall". However that fall could only ever accelerate as fast as that fluctuation propagates.

    This is why some of the members here have likely come across a little rude to you in what you've been postulating because you are trying to argue against something that isn't just a belief, a hypothesis or even a theory, it's something that has been test (and continually tested) to be true.
     
  9. Stryder Keeper of "good" ideas. Valued Senior Member

    Messages:
    13,105
    To All,
    It's come to the attention of the moderation that some posters have been overly hostile in regards to how they conduct themselves here. We ask if you could please refrain from name calling and seemingly hostile interactions so that the report button can be given a rest, as it threatened to quit this morning and we don't have a replacement one.
     
    OnlyMe likes this.
  10. rpenner Fully Wired Valued Senior Member

    Messages:
    4,833
    ???? \(\sqrt{2} \, c \; \color{red}{\gt} \; c\)
     
  11. RJBeery Natural Philosopher Valued Senior Member

    Messages:
    4,222
    First of all, being wrong does not merit ad hominem attacks on any science forum. Secondly, I'm simply not wrong. Double shame on the closed minds here AND the moderator who attempts to justify their behavior. Please click here to see my analysis and give your feedback if desired.
     
  12. RJBeery Natural Philosopher Valued Senior Member

    Messages:
    4,222
    Gah! You'd have to accelerate to sqrt(2)c if you wanted to break free of a circular orbit at the photon sphere (not possible, obviously).

    The equation for escape velocity is
    sqrt(2GM/r)

    I've made a calculator at Wolfram Alpha for a 100 Solar Mass black hole HERE. Change the "1.0" to the multiplier of the R_s desired.

    The photon sphere is at 1.5 R_s, and the result is 2.448E8 m/s which is clearly < c.
     
  13. RJBeery Natural Philosopher Valued Senior Member

    Messages:
    4,222
    rpenner, since you've posted here you must know of the Galileo issue. Would you mind taking 10 seconds to explicitly let others know that the mathematical proof is valid? Your response in the other thread clearly meant that you passively agreed with it, but brucep is not getting the message and I'm tiring of trying to penetrate his skull.
     
  14. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

    Messages:
    23,198
    I have a few questions (sort of on topic but not much related to BH.):
    First is: Do photons have a very tiny gravitational field?
    They have no rest mass so I guess the answer is no* (and at end of post support that with simple observation). Here is the problem that makes me have some doubt:

    An electron and positron far from all other mater and 2 cm from each other, I think, could be orbiting about their mass center. A meter from it I think there should be a weak and slightly modulated gravitational field. (Modulated as in 1/4 the orbit period the field at 1M from the center of mass changes in magnitude only (not direction) from:
    M/(1.01)^2 + M/(0.99)^2 To 2M/(1.0...0X)^2
    (Yes I'm too lazy to find X and how many zeros between X and the 1.0 as I'm just justifying "modulated")

    Second question: Does this modulation at point 2 meters form the mass center have speed of light delay? - I. e. does not have its max strength (about 25% as strong) exactly when field at 1M from mass center does have its max? - I think "yes.

    Third question: These two charges are accelerating so are radiating. Hence spiral closer together will they not? Again, I think "yes."

    Fourth question: Eventually they will transform into two ~ 511Mev gamma rays, that travel away from the old mass center point (perhaps the still mass center point if answer to question 1 was yes) will they not?

    Fifth question: Does the slightly modulated gravitational field speeding off into space continue but now as a growing "no field hole" centered on the old mass center point? If answer to Q1 was yes, I think there is no hole as that remains forever the mass center of the two gamma rays, but if answer to Q1 is no then I guess there is an expanding field free hole.

    Now for my observation supporting the No answer to Q1:
    On a clear night I look at some stars - first with one eye then with the other. It is fact that both eye see the same star field. But consider the small (incredibly small) angle the path of the photons reaching my right eye differs from the path of the photons coming to my left eye. If these photons did have mutual gravitational attraction for say 100,000 light year long paths, then the "photo light field" should be "quills of light" with no light voids between. I. e. the field of stars seen by my right eye should not be the same as that seen by my left eye. Likewise, If I did not like the star field one eye was seeing, (other closed) just move head a little to see a different star field.

    Comments (or at least what answers do you think correct to my 5 questions?)

    * Also the photon's M = E/(c^2) mass is "frame dependent" so that would seem to be a problem with their "bending of space" view of gravity, I think. I do lean to the POV that all energy which is the same in all frames does make gravity. I.e. the gravity from a hot brick deceases slightly as it cools - Its temperature (random KE) is same in all frames. Any arguments on this POV? Note that jar of half ice & half water has T = 0C in all frames.
     
    Last edited by a moderator: Oct 2, 2014
  15. OnlyMe Valued Senior Member

    Messages:
    3,914
    Billy, you present a series of questions that are very difficult to answer from a single theoretical basis. It seems your questions assume an inherent definition of gravity consistent with GR, and yet most of your questions/thought experiment are formed within the context of an interaction between fundamental particles, which is the domain of QM and an area where the charge relationship dominates the interactions. Gravity has not yet been successfully incorporated or explained within the context of QM. Still...

    To the issue of whether a photon has a gravitational field, I would say no.., at least not directly and/or independently. But even this depends on the theoretical context you approach the issue from. If you approach that question from some interpretations of GR all energy contributes to a gravitational field. Some would interpret this to mean that even the photon has (or contributes to) a gravitational field. If on the other hand you begin from some of the theoretical work of Haisch, Rueda & Puthoff dealing with inertia and gravitation as emergent from an interaction between vacuum energy (specifically an EM spectrum of the ZPF) and charged particles (which could be extrapolated to matter), photons would have no independent gravitational field, while they would play a role in how inertia and gravitation emerges from quantum phenomena. (Note that Haisch et al are only dealing with the issue as boundary conditions between charged fundamental particles and an isotropic EM ZPF background. Extending their theoretical work to a classical scale is entirely speculative on my part... But it would suggest that the photons associated with the EM ZPF spectrum do not independently gravitate.)

    I have a lot of trouble addressing the questions that follow, as individual questions, because they seem to me to be questions which would be dominated by QM rather than GR and gravity (Even though my comment above does just that.., crosses that divide). As example, when you begin with an orbiting electron/positron pair, even if those particles generate any independent gravitational field, they are point particles and their charge interaction would be far greater than any possible gravitational field.

    While gravitation seems to have a far greater reach than, charge related interactions, it is also generally a far weaker force. It is only the extreme gravitational forces found at the centers of stars and black holes where gravity overcomes those charge related forces associated with atomic stability. This would not be the case when dealing with individual fundamental particles.

    It does not seem to me that gravity can logically be thought of as playing a role in the hypotheticals you present.

    Your last paragraph brought to mind a quote, which I am sad to say, I cannot find the reference for right now. It went something like this,


    Keeping all of the above in mind, it seems that your thought experiment is attempting to ask what happens to the gravitational field if the source, the electron/positron pair in your example, annihilate one another radiating away as two gamma rays. To that my answer would be that the gravitational field dissipates at the speed of light outward from the center of mass of the initial electron/positron pair. The mass has been converted completely to a non-mass form of energy, the gamma rays. From the Haisch et al perspective, the gamma photons would have no independent gravity, so the initial field would just dissipate at the speed of light.

    If the gamma photons, do have independent gravitational fields, the change in the combined field would still occur at the speed of light, photons propagate at the speed of light, but the change in the gravitational field would be the same as if you were dealing with two stars, initially orbiting one another.., and then flying off away from their initial center of mass. The only difference would be how fast the change happens.

    If this last accurately describes your overall question, it really did take me all of the above rambling to get there.
     
    Last edited: Oct 2, 2014
  16. tashja Registered Senior Member

    Messages:
    715
     
  17. RJBeery Natural Philosopher Valued Senior Member

    Messages:
    4,222
    Hi Tashja, thanks again for your independent research. Unfortunately, your wording was incorrect and led the Professors astray. I have explicitly restricted the question to masses being dropped separately, at different times. With the wording you provided, I would agree with "poster #2" as well.

    I would also point out that they use phrases like "much smaller than that of the Earth" and "for small objects it is negligible", but brucep's assertion was that
    brucep is simply wrong on all accounts and his meltdown is occurring because of his inability to accept that fact.
     
  18. brucep Valued Senior Member

    Messages:
    4,098
     
  19. tashja Registered Senior Member

    Messages:
    715
    Hi RJ, but what is the difference if the two masses are dropped separately at different times? They make it quite clear that as long as the masses are smaller than the Earth, Galileo's right. Don't you agree?
     
  20. RJBeery Natural Philosopher Valued Senior Member

    Messages:
    4,222
    No!! Tashja, you are clever enough to understand my explanation. Please read it in the other thread. If you drop the items at the same time then they essentially become a "single item". If you drop them separately and time the descent you will get different results with different masses because the Earth has been pulled up towards the falling objects at correspondingly different rates.

    Send a link of the other thread to your mentors, if you wish. They will surely agree with my analysis.
     
  21. brucep Valued Senior Member

    Messages:
    4,098
    Hi Billy T

    The energy of the photon is a component of the tensor descri
    Hi Billy T
    The photons energy is a component of the stress-energy tensor.
    http://en.m.wikipedia.org/wiki/Stress–energy_tensor

    As the photon follows it's natural path (null geodesic) through the gravitational field (spacetime curvature) it's energy contributs to the local spacetime curvature over it's path. It's a very small contribution and it's fleeting since the photons energy remains with the photon as it continues along the natural path of light.
     
  22. OnlyMe Valued Senior Member

    Messages:
    3,914
    RJ, that is taking only the effect of the different masses of the falling objects, on the earth's mass and inertial resistance, into account. You are correct in stating that the larger mass will "pull" the earth toward it more than the smaller mass would, but what is being neglected is that the earth "pulls" the smaller mass toward it more than it does the larger mass. You have to account for the inertial resistance of all objects.

    The net result is that if you begin with the same separtion distance and assume both objects are at rest relative to oneanother, it does not matter what the mass of each object is, the total mass and distance between the masses, is all that matters. They, a small and large object dropped toward a larger object, fall toward the larger object at the same rate. This will hold true as long as you are comparring how two objects with masses smaller than the third object, interact with the larger mass gravitationally.
     
    Last edited: Oct 2, 2014
    Dr_Toad likes this.
  23. brucep Valued Senior Member

    Messages:
    4,098
    If I was wrong you'd be writing

    dr/dt = (2M/r)^1/2 (Newton, Einstein, and everybody else in physics)

    As

    dr/dt = (2M+m)^1/2

    LOL

    Regardless your juvenile analysis it's

    dr/dt = (2M/r)^1/2

    LOL
     
Thread Status:
Not open for further replies.

Share This Page