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  • #16
    Originally posted by pancreasman View Post
    Sorry, I'm just mildly cranky from talking to Mickiel all morning. Any science topic is interesting in and of itself. It's just this one seems a little abstruse and specific.
    Pro-tip: Unsubscribe from threads which make you cranky. Not reading threads which make you cranky will at the very least curb the rate at which your crankiness grows.








    Sincerely,
    Chrawnus, Ph.D. in Applied Crankiness

    Comment


    • #17
      Originally posted by Chrawnus View Post
      Pro-tip: Unsubscribe from threads which make you cranky. Not reading threads which make you cranky will at the very least curb the rate at which your crankiness grows.








      Sincerely,
      Chrawnus, Ph.D. in Applied Crankiness
      This is true, but it also means I end with about 3 threads a day to read. There is also some perverse correlation between being amused and being annoyed.

      Comment


      • #18
        Originally posted by sylas View Post
        Of course. The gravitational redshift of this case can be written ν'/ν = sqrt(1-R/r), where R is the Schwartzchild radius and r is the radius at which the photon was emitted with frequency ν
        Actually the equation for relativistic shift is sqrt((1-v/c)/(1+v/c))


        This invites all kinds of horrible confusion. The LHS of that is the time dilation for a moving observer, using v as a VELOCITY. This is bound to get mixed up with ν that we have been using as a frequency. You are going to get awfully confused if you don't stick to a specific problem and consistent naming of terms. Nothing previously involved the time dilation of a moving observer; this is unrelated to your original post. If you are going to talk about moving observers AS WELL as observers in gravitational wells, then use f for frequency, so you don't confuse v as velocity with ν as frequency.
        Yes, the Greek letter nu ν can be mix up with our letter vee v. So f for frequency. And lambda λ can be still used for wavelength.

        Where fλ = c

        37818, I am not really joining in to "talk about" this. There's nothing difficult or contentious here, and I can explain some points if you are confused or would like to know more about anything. Ask a question if you like, and if you think it would help. But the formula v = sqrt(2GM/r) isn't what you appear to think. It's a VELOCITY calculation; and specifically it happens also to be the escape velocity for a particle in a gravitational field. It has nothing at all to do with photon frequencies.

        Cheers -- sylas
        I'm of the opinion that event horizons are not closed. Light can always escape from gravitational wells. To be able to see this is true. Requires an open mind to truth.

        Escape velocity is a matter of kinetic energy not velocity. Since the limiting velocity is c.

        We get v = sqrt(2GM/r) and the Schwartzchild radius as 2GM/c^2 if only the first term of Ke as mv^2/2.

        So if instead of mv^2/2 = GMm/r we use mc^2(1/sqrt(1 - v^2/c^2) - 1) = GMm/r

        Then sqrt(1 - v^2/c^2) = (1 /(1 + GM/(rc^2))

        Where v = sqrt(2GM/r + (GM/(rc))^2)/(1 + GM/(rc^2))
        Last edited by 37818; 01-24-2015, 09:27 PM.
        . . . the gospel of Christ: for it is the power of God unto salvation to every one that believeth; . . . -- Romans 1:16 KJV

        . . . that Christ died for our sins according to the scriptures; And that he was buried, and that he rose again the third day according to the scriptures: . . . -- 1 Corinthians 15:3-4 KJV

        Whosoever believeth that Jesus is the Christ is born of God: . . . -- 1 John 5:1 KJV

        Comment


        • #19
          Originally posted by 37818 View Post
          I'm of the opinion that event horizons are not closed. Light can always escape from gravitational wells. To be able to see this is true. Requires and open mind to truth.

          OK. I had a feeling this was on the cards from the previous posts; goodbye, good luck.

          If you want to take this up some more you'd need to work on learning a bit more about kinetic energy. Escape velocity is the velocity at which a particle has zero energy (the kinetic energy matches the negative potential energy). The problem with velocity is that you need to specify the co-ordinate system in which it is to be defined. We could work on your formulae and clean then up to be more useful and more correct; but that would mean letting go of opinions. To be honest, it's pretty clear that you don't yet know enough about the basics to do that correctly; that is, you are disagreeing with physics that you have not even learned enough about to describe. Disagreeing with something you cannot express is vacuous.

          Working with a gravitation field is going to need general relativity. You can't just plug in the SR formulae without knowing what you are doing. In your first sentence just above, you've quoted a non-gravitational form of the redshift, arising from motion, which simply a complete non-sequitur to everything that's gone before. You are badly mixed up, and making a horrible mess of things. I gave you the correct formula for gravitational redshift, which was the topic of your OP.

          If you stick with study and learning, and hold your opinions a lot more loosely, then you'll be able to learn a lot of really cool and interesting things; it's hard work but well worth it. I'm not, however, interested in debating with "opinions".

          Adios -- sylas
          Last edited by sylas; 01-24-2015, 09:41 PM.

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          • #20
            And the point of this thread is ...?

            K54

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            • #21
              Originally posted by klaus54 View Post
              It's a great calculation if you're falling into a Black Hole.

              Relevance to age of Earth/History?

              K54
              I just want to discuss what I stated. Motive: I happen to not believe in black holes in space. The problem, the algebra solution is wrong if there are black holes. Why? The only answer I can find out, is "it just is." Do what you want. I think this is going to prove to be futile.
              . . . the gospel of Christ: for it is the power of God unto salvation to every one that believeth; . . . -- Romans 1:16 KJV

              . . . that Christ died for our sins according to the scriptures; And that he was buried, and that he rose again the third day according to the scriptures: . . . -- 1 Corinthians 15:3-4 KJV

              Whosoever believeth that Jesus is the Christ is born of God: . . . -- 1 John 5:1 KJV

              Comment


              • #22
                Originally posted by sylas View Post
                OK. I had a feeling this was on the cards from the previous posts; goodbye, good luck.

                If you want to take this up some more you'd need to work on learning a bit more about kinetic energy. Escape velocity is the velocity at which a particle has zero energy (the kinetic energy matches the negative potential energy). The problem with velocity is that you need to specify the co-ordinate system in which it is to be defined. We could work on your formulae and clean then up to be more useful and more correct; but that would mean letting go of opinions. To be honest, it's pretty clear that you don't yet know enough about the basics to do that correctly; that is, you are disagreeing with physics that you have not even learned enough about to describe. Disagreeing with something you cannot express is vacuous.

                Working with a gravitation field is going to need general relativity. You can't just plug in the SR formulae without knowing what you are doing. In your first sentence just above, you've quoted a non-gravitational form of the redshift, arising from motion, which simply a complete non-sequitur to everything that's gone before. You are badly mixed up, and making a horrible mess of things. I gave you the correct formula for gravitational redshift, which was the topic of your OP.

                If you stick with study and learning, and hold your opinions a lot more loosely, then you'll be able to learn a lot of really cool and interesting things; it's hard work but well worth it. I'm not, however, interested in debating with "opinions".

                Adios -- sylas
                You are welcome. And thank you for what you attempted to explain.


                Ke = mv^2/2

                Ke = mc^2(1/sqrt(1 - v^2/c^2) - 1)

                Ke = mv^2/(1 + sqrt(1 - v^2/c^2) - v^2/c^2) = mc^2(1/sqrt(1 - v^2/c^2) - 1)

                I'm trying to find out why mc^2(1/sqrt(1 - v^2/c^2) - 1) = GMm/r, is not valid.

                And sqrt(1 - v^2/c^2) = sqrt(1 - 2GM/(rc^2), is valid.

                This whole thing is futile.
                . . . the gospel of Christ: for it is the power of God unto salvation to every one that believeth; . . . -- Romans 1:16 KJV

                . . . that Christ died for our sins according to the scriptures; And that he was buried, and that he rose again the third day according to the scriptures: . . . -- 1 Corinthians 15:3-4 KJV

                Whosoever believeth that Jesus is the Christ is born of God: . . . -- 1 John 5:1 KJV

                Comment


                • #23
                  See? I was right after all! Reminds of the conversations I had with Magellan where he didn't believe in instantaneous velocity.

                  Comment


                  • #24
                    Originally posted by 37818 View Post
                    Actually the equation for relativistic shift is sqrt((1-v/c)/(1+v/c)).
                    For relative motion yes, but in the General Theory of Relativity there's also one you can derive for gravitational fields.

                    I'm of the opinion that event horizons are not closed. Light can always escape from gravitational wells. To be able to see this is true. Requires an open mind to truth.
                    Alright, but you won't be able to establish this using Newtonian mechanics. You'd need to establish this from the General Theory of Relativity. It wouldn't even work to say that you dismiss, GR (I'm only guessing at what you're implying), because GR is clearly a better mechanics than Newtonian mechanics only works now as a low-velocity approximation (over short spaces and time intervals and low gravitational gradients). So when you're dealing with a theoretical entity like a blackhole which is an extreme you won't be able to rely on Newtonian mechanics.

                    We get v = sqrt(2GM/r) and the Schwartzchild radius as 2GM/c^2 if only the first term of Ke as mv^2/2.
                    Yup, correct, you get the Schwartzchild radius from Newtonian mechanics if you plug the speed of light into the formula for escape velocity v = sqrt(2GM/r). Its sort of a neat trick, which can be ultimately explained by two errors cancelling each other out in the right way to give this result. If you've stumbled on this trick on your own, that's kinda neat.

                    There are problems with doing this conceptually however. First of all, again, the photon simple doesn't have any mass. It just doesn't. If you try to treat the photon as a particle with mass, you're going to get wrong results when you try to calculate how much a photon is deflected by passing a gravitational field.

                    Also you can't explain using Newtonian mechanics why a photon is always, and only, moving at the speed of light in all frames of reference. In the theory relativity this can be explained by the photon not having mass, since you can derive that all massless particles must move at the speed of light and no other velocity.

                    So if instead of mv^2/2 = GMm/r we use mc^2(1/sqrt(1 - v^2/c^2) - 1) = GMm/r

                    Then sqrt(1 - v^2/c^2) = (1 /(1 + GM/(rc^2))

                    Where v = sqrt(2GM/r + (GM/(rc))^2)/(1 + GM/(rc^2))
                    So let me understand your logic. You're dubious about blackholes for some reason. And you want to disprove their existence, by proving that photons can always leave a gravitional well. You know enough of the theory of relativity to know that mass changes due to a particle's velocity relative to another frame of reference, so you figure that the equation for kinetic energy must be replaced by the one from the the theory of relativity correct? And its simple a matter now of finding when the kinetic energy would be equal to the potential energy, showing that it'll always be capable of overcoming it?

                    bc293276-5e2d-441c-95ec-458f9af73a92_zpsdd983842.png

                    Then its correct that if you use replace equation for newtonian kinetic energy, with relativistic kinetic energy,

                    CodeCogsEqn1_zps686b761c.png

                    You can derive a velocity (your result can be simplified),

                    CodeCogsEqn2_zps6414b91b.png

                    The problem is 37818, what on earth is this velocity? Its not the velocity of the photon, because photons only ever move at the speed of light. See in your first equation where you derived the gravitional red-shift (not the doppler redshift), using Newtonian mechanics with a fictious mass for the photon, you folded away the erroneous result of the velocity changing. This can happen sometimes, but you have to understand that its an error.

                    Now you're deriving an escape velocity, again using the fictious mass, and if you ask whether this proves anything. It doesn't. You're mixing and matching different kinds of physics here, and playing loose and ugly with velocities and frequencies, plugging them into the equations and hoping something falls out.

                    In short, since a photon is massless and since a photon always moves at the speed of light, your basic assumptions here are wrong. That's the short of it. Sometimes you can still crank out a true result using wrong assumptions, that happens sometimes in physics, but this result constitutes no proof.
                    Last edited by Leonhard; 01-25-2015, 05:54 AM.

                    Comment


                    • #25
                      Correction, I made a mistake.

                      Comment


                      • #26
                        Originally posted by 37818 View Post
                        I just want to discuss what I stated. Motive: I happen to not believe in black holes in space. The problem, the algebra solution is wrong if there are black holes. Why? The only answer I can find out, is "it just is." Do what you want. I think this is going to prove to be futile.
                        Why do you not believe in black holes? The science of black holes are based on direct observations in our universe, and predicted by standard sound physics and math models before they were discovered by direct physical observations of the effects of black holes.
                        I believe the math behind the prediction and discovery of Black Holes goes far beyond basic algebra.
                        Glendower: I can call spirits from the vasty deep.
                        Hotspur: Why, so can I, or so can any man;
                        But will they come when you do call for them? Shakespeare’s Henry IV, Part 1, Act III:

                        go with the flow the river knows . . .

                        Frank

                        I do not know, therefore everything is in pencil.

                        Comment


                        • #27
                          Originally posted by Leonhard View Post
                          Correction, I made a mistake.

                          If that equation is functionally correct, escape velocity is always less than c. Event horizons would never be closed where light cannot escape.
                          . . . the gospel of Christ: for it is the power of God unto salvation to every one that believeth; . . . -- Romans 1:16 KJV

                          . . . that Christ died for our sins according to the scriptures; And that he was buried, and that he rose again the third day according to the scriptures: . . . -- 1 Corinthians 15:3-4 KJV

                          Whosoever believeth that Jesus is the Christ is born of God: . . . -- 1 John 5:1 KJV

                          Comment


                          • #28
                            Originally posted by pancreasman View Post
                            Let's do a new thread on 2 +2 = 5. No it doesn't.
                            Originally posted by sylas View Post
                            2 + 2 = 5, for sufficiently large values of 2.
                            2 + 2 = 5 exactly in Q/Z.

                            Comment


                            • #29
                              Originally posted by 37818 View Post
                              I just want to discuss what I stated. Motive: I happen to not believe in black holes in space. The problem, the algebra solution is wrong if there are black holes. Why? The only answer I can find out, is "it just is." Do what you want. I think this is going to prove to be futile.
                              You got that right.

                              K54

                              Comment


                              • #30
                                The most vicious math thread I ever saw was on the old II board.

                                Topic: Is .9999.... = 1?

                                Call me jaundiced, but I just can't see this one measuring up.

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