Asymptotic Freedom in QCD and...

Discussion in 'Pseudoscience' started by RajeshTrivedi, Sep 14, 2017.

  1. NotEinstein Valued Senior Member

    And I can't figure out where that factor of 3 difference is coming from.

    From my post #132: "Note that I never claimed yours was wrong; all I said was that our formulae were different, and that because I have a derivation showing and you haven't pointed out any mistakes in it, I suspect your derivation is different (for example, using different assumptions)."
    Please stop misrepresenting what I said.

    So according to you, the factor 3 difference is coming from the fact that I used a constant density, and you did too? That's incoherent. Please show your calculations, so I can see what you actually did!

    From my post #135: "It's not the factor 3 itself that worries me: it's that there is any difference."

    Please point out where I have committed the straw-man fallacy.

    That's not a proper proof, it's just the starting point of one. But I guess I'll do your homework for you.

    \(dm=4\pi r^2 d(r) dr\) is the mass of an infinitesimal spherical shell of thickness \(dr\) at radius \(r\). To get the inner mass at a radius R, we thus have to integrate this from r = 0 to r = R. Thus we get: \(M=\int_0^R {4\pi r^2 d(r) dr}\)
    Since I don't know the functional form of \(d(r)\) the integral cannot be evaluated. But let's assume a constant/uniform density, so that \(d(r) = d\). Then the integral becomes: \(M=d \int_0^R {4\pi r^2 dr}=d [\frac{4}{3}\pi r^3]_0^R=d (\frac{4\pi R^3}{3} - \frac{4\pi 0^3}{3})=\frac{4\pi d R^3}{3}\), which is identical to my result in post #100. From that point onwards, I can just copy-paste my post #100. In other words, I get the result I got in post #100, and not yours.

    Please post your calculations, because you clearly are doing something different than what I am doing.
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  3. RajeshTrivedi Valued Senior Member

    The color highlighted part is the mistake you are making. Please note that in my#129, I showed you that even if you take non-uniform density, as you are insisting, then my claim would be untrue only for very high and unrealistic density profiles, and hence uniform or realistic non-uniform density profiles will tally with my conclusion.

    Proof: Since I am not ok with /tex, please parse it accordingly;
    [Somewhat elaborated for clarity]

    For the entire inner fraction also to be beneath or just at their event horizon,

    r = 2 Gm/c^2, Where m is the mass of the inner sphere of radius r.

    rearranging this will become

    m = rc^2/2G

    (since m is variable with r)
    dm/dr = c^2/2G

    that is
    4*pi*r^2 * d(r) = c^2/2G

    thus d(r) = c^2/(8*pi*G*r^2).

    There is no factor 3., for the inner fraction to be beneath its event horizon this will have the sign of > in place of =.

    This will give you d(r) > 5.3 * 10^25/r^2.

    Now please substantiate which celestial object will have this kind of density profile when at EH?
    and also please let me know that why any realistic non-uniform density function as long as < 5.3*10^25/r^2 will also conclude as per my claim.
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  5. NotEinstein Valued Senior Member

    If that's a mistake, you've made it in your texts and all throughout this thread as well.

    (I'll comment on this later, first let's see your calculations.)

    This is the Schwarzschild radius. Let's call it \(r_s\) to distinguish it from the coordinate \(r\). And \(m\) is the enclosed mass at coordinate \(r\), so let's call that: \(m(r)\).

    \(m(r)=\frac{r_s c^2}{2G}\)

    \(\frac{dm(r)}{d r_s}=\frac{c^2}{2G}\)

    This derivative shows how the mass of an object exactly at the Schwarzschild condition changes as the Schwarzschild radius changes.

    Here you plug in the shell's mass formula. However, that formula uses \(r\), but here we have \(r_s\); that's why I made that distinction clear in my derivation. Here you thus set \(r = r_s\). Your equation is therefore only valid at the very radius where the enclosed mass exactly reaches the Schwarzschild condition.

    Also, by setting to derivates equal to each other like that, there might be a missing constant. Though it is an addition constant, not a multiplication factor.

    This thus is the mass-density in the shell at the radius where the Schwarzschild condition is hit exactly. However, the missing constant makes me weary, as it might be non-zero...

    Well, you've computed something totally different, and might be missing a constant (you've at the very least not shown why it would be zero). See, this is why I asked you to show your calculations: the factor 3 difference is irrelevant, because we are comparing apples to oranges!

    At the Schwarzschild radius, because you've set \(r = r_s\).

    You have not given a density profile; you've given the density at the point where the Schwarzschild condition is hit. That's (typically) only one point, not an entire density profile.

    (I'll defer formulating an answer to this question until we've cleared up what you have actually calculated in the first place.)
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  7. RajeshTrivedi Valued Senior Member

    You are playing or you have lost track of your argument.

    Ok, once more;

    What are you countering : My argument and
    My argument is that when an object is just at its EH, the inner fractions will not be beneath their respective EHs.

    1. In my post#107, I proved that my conclusion is good for uniform density.
    2. In my post#129, I proved that my conclusion is good for non-uniform realistic density profiles.

    For my conclusion to be false, for non uniform density profiles,

    rs > r (for all the shells of various r from 0 - outer EH.)

    Where 'r' is the radius of some arbitrary inner fractional sphere of mass m (when the bigger object is at EH) and rs is the schwarzschild radius of this inner shell of radius r.

    this will give:

    2Gm/c^2 > r
    m > rc^2/2G
    dm/dr > c^2/2G
    4*pi*r^2*d(r) > c^2/2G
    d(r) > c^2/(8*pi*G*r^2)
    d(r) > (5.3 * 10^25)/r^2

    (is it realistic density profile?? If yes, pl show, if no then my conclusion prevails and retract your objection.)

    What's your problem with this derivation?
  8. NotEinstein Valued Senior Member

    If you think I'm playing, feel free to invite the moderators to take action.

    I do not disagree with this, at least in specific cases.

    You do remember that I proved it in post #100 already?

    Post #129 doesn't contain a full proof, as the derivation is missing.

    Please read my post #143 for my objections to this derivation.
  9. RajeshTrivedi Valued Senior Member

    I have read your #143, I have further eased out the derivation for you in #144, I have not set r = rs, it is just that the condition is rs > r (beneath EH) or rs = r (just at EH).

    1. Please go through post#144 again and again and again, till you find the problem in the derivation in #144.

    2. Or alternative if you are a genuine poster, not here to pester others, please provide a density profile for rs > r for all the r when the outermost object is at EH.

    3. Or accept that the derivation # 144 is correct and you withdraw your objection.

    I think I am being very fair to you by seeking #1 or #2 or#3
    So, no more word games please.
  10. NotEinstein Valued Senior Member

    Yes you have. You've taken the Schwarzschild radius, and used it for the shell. That means the shell you are calculating the density for has to be at exactly the Schwarzschild radius. In other words: \(r = r_s\).

    No matter how many times I do that, my objections won't change. You are setting \(r = r_s\).

    If you genuinely believe I'm here to pester others, please take this up with the moderation.

    That's easy: a black hole singularity.

    Your derivation is that of the density at the point where the Schwarzschild condition is hit. It's just not a derivation of the entire density profile.

    And I think I'm being quite fair, giving you the opportunity time after time after time to explain what you are doing, and giving you a chance after chance after chance to respond to my criticism. Similar to how I asked you again and again and again for your calculation earlier, because I suspected something went wrong somewhere. (Which it did; you calculated the density at the point where the Schwarzschild condition is hit, not the density profile.)

    If you think I'm playing word games, please involve the moderation.
  11. NotEinstein Valued Senior Member

    Let me make this more explicit. Take the Schwarzschild equation:

    and rewrite it like you did:
    \(\frac{dM}{d r_s}=\frac{c^2}{2G}\)

    Then, let's do the same with the shell equation:
    \(\frac{dm}{dr}=4\pi r^2 \rho(r)\)
    (now that I think about it, I think there should be a \(\frac{d\rho(r)}{dr}\)-term in there too...)

    Your next step is: "4*pi*r^2 * d(r) = c^2/2G". In other words: \(\frac{dM}{d r_s}=\frac{dm}{dr}\)
    How can this be true, if not for the assumption that \(r=r_s\)?
  12. RajeshTrivedi Valued Senior Member

    Thats meaningless, I did not use drs. (rs = Schwarzschild radius). You are using it incorrectly again and again.
    \(\{d r_s}\)

    If you think or confused that there should be \(\frac{d\rho(r)}{dr}\)-term in there, then please refresh your Physics/maths basics.

    Thats not assumption, it is foolish to insist after such prolonged word game that it is the assumption. Thats the condition to arrive at the density profile. You are not even keeping track of the main argument. The main argument is that when the object is "just at its EH" then all other fractional shells will be out of their respective schwarzschild radius.
    Try putting this condition mathematically, and see what you get.

    End of Discussion for now. Go back to #144, come to the final correct conclusion, take help of others if required, we will see after that.
  13. NotEinstein Valued Senior Member

    Your derivation starts with the equation for the Schwarzschild radius. To avoid confusing it with the radial coordinate \(r\) I've called it \(r_s\).

    If your derivation doesn't start with the Schwarzschild radius, please explain where: "r = 2 Gm/c^2, Where m is the mass of the inner sphere of radius r." comes from in post #142.

    You are taking derivates to \(r\), and \(\rho\) has an \(r\)-dependence. The "just stick a d in front of it"-method of making derivatives is very easily misapplied, and I think that's what's happening here.

    And just so you know, you made a very basic mistake here:
    The last step is wrong. The derivative does not need to have the same inequality. Perhaps you should refresh your Physics/maths basics as well.

    Perhaps you should refresh your Physics/maths basics. You say: \(r_s\leq r\). How does this lead to: \(\frac{dm}{dr_s}\leq\frac{dm}{dr}\)? It doesn't in general; you need to still prove this to be true (at least under the right circumstances) for your derivation to make any sense.

    If that's the sole condition you used, your answers is plain wrong.

    Please explain to me how that excuses the basic Physics/maths mistakes you're making.

    I know that I do not get \(\frac{dm}{dr_s}\leq\frac{dm}{dr}\).

    I just went back, and found another mistake; see above. Please address that as well.
  14. RajeshTrivedi Valued Senior Member

    Now, you are objecting that

    dm/dr > c^2/2G is wrong if m > rc^2/2G.

    That is a linear relationship, in first quadrant of the form
    y > kx (where x,y,and k all are positive real numbers).
    There is nothing wrong in this conclusion.

    [Alternatively please work with equality r = rs (for every r) to get the density profile and apply the proper condition at the end.]
  15. NotEinstein Valued Senior Member

    Let's take two functions:
    and restrict ourselves to the domain where \(x>0\).

    For every \(x\), \(g(x)>f(x)\). Just plot the graphs if you don't believe me.

    The derivatives of these functions are:
    You can clearly see that for every \(x > 0\), \(g'(x)<f'(x)\). The sign flipped! But is this generally true? Nope, because if we take two different functions:
    Thus always: \(g(x)>f(x)\)


    So for \(x<0\): \(g'(x)<f'(x)\), for \(x=0\): \(g'(x)=f'(x)\), and for \(x>0\): \(g'(x)>f'(x)\). It's all over the place!

    This demonstrates that you must provide justification why this step in your derivation is allowed:
    m > rc^2/2G
    dm/dr > c^2/2G
    because it clearly is not true in general.

    And just to remind you of the other major issue I brought up:
    dm = 4*pi*r^2 d(r) dr

    I invite you to prove that this derivative is correct, by integrating it. I'll even provide the first steps:
    \(\frac{dm}{dr}=4\pi r^2 d(r)\)
    \(m=\int_0^R 4\pi r^2 d(r) dr\) (or just the indefinite integral, that's fine too)
    Please pay specific attention to the integration of \(d(r)\), as that's the point under discussion. In other words, show me the equation that, if I take a derivative of it with respect to \(r\), I get your derivative. (Hint: don't forget

    As I said in post #143: "Your equation is therefore only valid at the very radius where the enclosed mass exactly reaches the Schwarzschild condition." And you can't change it to an equality, do the substitution and calculations (including taking derivatives!), and then put the inequality back somewhere. Come on, this is high school mathematics, if even that!
    Last edited: Nov 27, 2017
  16. RajeshTrivedi Valued Senior Member

    Linear relationship in first quadrant.
    Radius = Positive
    Mass = Positive
    c^2/2G = Positive

    it is like y > kx
    dy/dx > k, holds here;
    you do your exercise with this function, why do with some unrelated quadratic or inverse function?
    Don't obfuscate; You are being a drain.
  17. NotEinstein Valued Senior Member

    The mass can remain constant while \(r\) increases (for example, outside the object, but also if there's a spherical shell of vacuum inside the object), so dm/dr can be zero, and thus your "dy/dx > k" argument is proven wrong. Care to try again?
  18. RajeshTrivedi Valued Senior Member

    But I am not talking about " outside the object" and neither about this unrealistic "spherical shell of vacuum inside the object".
    I am talking about a very specific case, when the object is "just at its EH".

    Be a sport, brother, say your objection on this point is resolved. Nothing wrong in that. It is meaningless to counter a simple derivation of linear first quadrant inequality with inverse function, quadratic function, mass out side the object and finally creating vacuum inside the object.
  19. NotEinstein Valued Senior Member

    True, and that will be reflected in your d(r). But note that this formula doesn't contain any d(r).

    Which doesn't mean the object can't be hollow.

    You can't resolve a point my misunderstanding it. The calculations leading up to your conclusion are incorrect; there are several mistakes/oversight that need to be addressed. That you seem unwilling/unable to address them doesn't make your conclusion correct.

    Actually, it is. Your derivation is obviously incorrect, as the "hollow object" example demonstrates. I've also pointed out where the match of that part of the derivation is dodgy.

    Perhaps you should refresh your Physics/maths basics.
  20. paddoboy Valued Senior Member

    As long as you fail to recognise that once the Schwarzchild radius is reached and an EH formed, further collapse is compulsory and all paths lead to the singularity. There is no other outcome according to GR.
  21. RajeshTrivedi Valued Senior Member

    Don't be dishonest.

    Your immediate objection/claim is that if y > kx then dy/dx > k is incorrect.
    This has been resolved. I have clarified that for first quadrant operation (where x,y and k are +ive), this holds.
    Do you admit that your this objection is removed?
    No, drama and no word games, simple yes or no, pl.

    Once you acknowledge above, then we will move over to next objection. One by one.
  22. RajeshTrivedi Valued Senior Member


    I understand this much GR.
    Pl understand I am not still disputing it.
    What I am simply saying is that when the core is collapsing to form a singularity (not yet formed), then during this transit motion a photon produced at or near the center of the core can move away from the center. I am not saying yet that this will come out of original EH. So no contradiction with GR as yet. Only when you understand this aspect then we can move over to next. And mind you GR does not talk about photon production in the core of a collapsing object.

    You just cannot create any realistic density profile which will ensure that when an object is "just at its EH" all other inner spherical shells will also be beneath their respective EHs, so photon can move away from center. For how long and what is the use, is the next part.
  23. paddoboy Valued Senior Member

    Of course you are disputing it! You are suggesting as you have many times that BHs do not form.
    You seem obsessed with this singularity rajesh. What singularity are you suggesting? The one where GR is non applicable and quantum domain takes over, or the one that entails infinite quantities such as density and spacetime curvature?"The latter by the way is generally not accepted by most cosmologists. Irrespective, other then the GR edict that once any EH is formed,(Shcwarzchild radius) nothing will get back out, we really have no other knowledge about what goes on inside, except that the compulsory collapse as dictated should occur at least up to the region where quantum domain takes over. So talking about photon production at the core is not known one way or the other.
    I certainly do and I suggest most cosmologists do also. The EH of any specific mass takes place as a whole, so to speak and yes, all spherical shells beneath that EH are beneath that EH. And by the way, speaking of BH density is rather silly in that any BH is no more then critically curved spacetime, with the mass squeezed to at least within or at the quantum domain.

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