Formal Debate: Gravitational Shift and the Least Action Principle

Discussion in 'Pseudoscience' started by Trapped, Oct 14, 2013.

  1. Trapped Banned Banned

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    I don't get the impression I am speaking to an idiot, so I doubt you have the right person. You might be getting confused with some other resident idiots here, perhaps?
     
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  3. origin Heading towards oblivion Valued Senior Member

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    Nope.

    If you are not going to formilize a debate you should move to a different section.
     
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  5. Trapped Banned Banned

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    I think you should act on what you preach first.

    And to conduct a proper conversation, you first need to spell correctly. I doubt your rendition of the word ''formilize'' was a matter of typo.

    Nevertheless, you have demonstrated your loyalty for Brucep, but I will point out again, his errors have been exposed. Your continuation of hate against me, looks immature, perhaps ill-directed.
     
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  7. Pete It's not rocket surgery Registered Senior Member

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    A better purpose might be to discover the truth. It's not scientific to go into an exploration of an idea with the dogmatic belief that you already know the outcome, right? If you discover that you were wrong, and bruce was right, that would be a happy ending for you, would it not?

    Also, this should be a proposal thread, not a debate thread. See [thread=74142]How the Formal Debates forum works[/thread]
    Once you and brucep agree on the terms of debate, then a debate thread can be opened and no one else can post in it. A discussion thread will also be opened in which other posters can discuss the debate.

    Can you change this to a "Proposal" thread, Trapped? If not, I'll ask a mod with rights to this forum to fix it for you.
     
  8. Trapped Banned Banned

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    I don't believe I can. Could a mod sort it?
     
  9. brucep Valued Senior Member

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    Let me be the first to 'kiss up' to you today. Anybody who can't recognize your talents and your willingness to share them with others is a scientifically illiterate dolt. You and Mr_Homm. I know you'll like that comparison.
     
  10. brucep Valued Senior Member

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    Maybe I should take him off my ignore list? What was that thing we'd recite as children? Oh yeah: Sticks and stones ...... but 'cranks' can never hurt me..... Especially when there's folks like you and origin around to smack them around intellectually.
     
  11. brucep Valued Senior Member

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    I'm not interested in a formal debate with you. Whatever that says 'to you' it's irrelevant to me.

    Sorry to the folks who felt they needed to respond since I didn't. Also thanks.
     
  12. brucep Valued Senior Member

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    Your denominator reduces to 1. It doesn't predict anything. If you wanted to make an equation with a denominator that doesn't reduce to 1 then you could have wrote down the 'following'. The bath of radiation observed from the shell will increase as you approach r=2M. Because of the relativistic effect we describe using gamma. g doesn't change since you could only measure g locally. You can predict what it is but that's one reason we have relativistic physics. g is an invariant.

    (1-2M/r + [the other crap])^1/2
     
  13. Trapped Banned Banned

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    Again, how do you reach this conclusion exactly? The equation for the shift is

    \(1+z = \sqrt{\frac{g_{tt}(r)}{g_{tt}(s)}}\)

    This is essentially what the author wrote in the equations. It doesn't reduce to 1 at all. You keep showing yourself up again Bruce.


    If the redshift is zero, meaning \(z=0\) then what we have is 1. We don't have it reducing to 1, when there is a non-zero z.

    I have explained this, a number of times now. Your trolling is relentless.
     
  14. Trapped Banned Banned

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    You also claim it doesn't do anything. That is simply not true. The work which inspired the author to rewrite the metric and then present it in this form came from studying equations by Motz. He then found out you reach the same form of the equation for the bolometric luminosity, which turned out to be derived from solid math, albeit simple.

    The equation is fully relativistic and predicts simultaneous measure for large redshift using the equivalence principle. It has massive applications you are just ignorant of.
     
  15. Trapped Banned Banned

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  16. Trapped Banned Banned

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    He put it nicely in a post to me:

    ''It occurred to me that the paper by Motz was kind of revolutionary in it's approach. Often things like this just get overlooked. Looking back on old papers can sometimes give you a gee-whiz kind of moment.

    The equations had a new form in my mind: Rewrite the metric in terms of energy, (which can be relativistic or not) and then express it as a ratio of metrics in the form which is similar (in fact they are essentially identical) that measures the redshift.

    His equation was considered to account for the luminosity of certain stars, we have taken this further to account for the redshift of distant moving charges; generally-speaking, it has a good application when describing quasars since all detected have very high red-shift values. What my equation does, is that it measures this shift in the denominator. It will spit out real values as well when you calculate the variables.

    It is fully relativistic, it also has a possible application to black holes, since virtual particles are boosted by the gravitational field. When this happens, these virtual particles will give off Synchroton radiation which has an application to detecting these black holes.''
     
  17. James R Just this guy, you know? Staff Member

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    Moderator note: Thread moved out of Formal Debates subforum. Members are advised to read the rules of the Formal Debates subforum before posting there.
     
  18. Tach Banned Banned

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    You are right, brucep is incorrect, the gravitational redshift z can be larger than 1:

    \(z=\frac{1}{\sqrt{1-r_s/r}}-1\)
    This happens for \(r_s<r<\frac{4r_s}{3}\)
     
    Last edited: Nov 10, 2013
  19. pmb Banned Banned

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    First - Bruce never made that statement. He said a^2_g / divided by two frame dependent remote measurements could result in over unity.

    Second – You’re terminology is inconsistent and as such is not used as it is normally defined.

    Gravitational redshift - A term used in astrophysics and is the process which causes light from a source in a gravitational field to be observed with a reduced frequency when the light moves into a weaker region of the gravitational field.

    See http://en.wikipedia.org/wiki/Gravitational_redshift

    Cosmological Redshift – This is what you’re talking about. It’s what you wrote the expression for. It too refers to the reduction in the frequency of light but as it moves across cosmological distances and is caused by the expansion of the universe.

    See http://en.wikipedia.org/wiki/Redshift

    The expression you gave is for cosmological redshift and not for gravitational redshift cosmological redshift. The numerical value of gravitational redshift is the value z not 1 + z.

    If I’m accurate on this then I believe a great deal of Bruce’s knowledge of general relativity came from his studies of the subject. The textbook I believe he learned from is Exploring Black Holes by Taylor and Wheeler. It’s a really nice text. It was proof read by a very handsome man from Boston’s North Shore. Lol!

    I think you're way off here. I think you misunderstood what Bruce was saying. When he get's here he'll explain what he meant.
     
  20. Trapped Banned Banned

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    *tach

    Yes, I know I was right.
     
  21. Trapped Banned Banned

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    Let's be clear on one thing, the author of the work is using 1 + z.

    Bruce said that the gravitational shift cannot be over unity. He did say this and he is wrong. The redshift is only unity when there is a zero z.

    The redshift in question is for a stationary spacetime (except we are using the equivalence principle, obviously a stationary spacetime is accelerated relative to us in a gravitational field) - we don't even need to bring in any other concepts. The author is simply rewriting the Schwarzschild metric and it can be used for distant cosmological objects in an accelerated frame of reference.
     
  22. Trapped Banned Banned

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    This becomes a gravitational acceleration when we considered \(a_g\) to begin with. The distant source and the local receiver does measure a gravitational shift in a distant accelerating charged object described by a Schwarzschild metric. The paper the author was inspired by the work came from this paper

    http://gravityresearchfoundation.org/pdf/awarded/1968/motz.pdf

    A metric becomes related to the gravitational redshift when you notice that any Schwarzschild metric describes the gravitational field strength and more so when we involved the gravitational acceleration of the charge due to the gravitational field.
     
  23. Trapped Banned Banned

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    Rpenner also agreed with Brucep but I am sure for different reasons. He or she stated that the author cut and paste this ratio

    \(\frac{1}{\frac{\sqrt{1 - 2\frac{Gm}{\Delta E} \frac{M}{r} + \frac{GQ^2}{c^4 R^2}}}{\sqrt{1 - 2\frac{Gm}{\Delta E} \frac{M}{r} + \frac{GQ^2}{c^4 R^2}}}}\)

    And stated that it comes to unity.

    I explained I wasn't sure why they were saying this but to also note some subscripts have been omitted, for simplicity. I have the authors permission to post these equations. What we have above is just

    \(1 + z = \sqrt{\frac{g_{tt}(r)}{g_{tt}(s)}}\)

    Knowing this, I can't fathom why anyone would say gravitational redshift cannot be over unity... gravitational redshift HAS to be over unity if z is the shift and if the shift is zero then all you are left with is unity.

    *edit* and yes, this redshift is written for a schwarzschild metric and is totally allowed - only different is that he has kept the same Schwarzschild rules and wrote it for the charged metric case of the same arguments.
     

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