Declan Posted July 8, 2016 Author Share Posted July 8, 2016 To Swansont: But that is what the centripetal acceleration equation and my paper shows. As the flow is constant with distance, it's affect is an extra velocity vector towards the center for every such vector calculation along r from the center outward. This constant extra vector is equivalent to a constant extra centripetal acceleration towards the center. The graph in my paper shows the required orbital speed at each distance r from the center when an extra constant acceleration is applied to objects in orbit around a Galaxy. To ajb: The the gravitational field is just the curvature of space time, what has spin got to do with it? In your opinion in GR is the coordinate system a fixed grid or does the actual coordinate system deform with the curvature of space? Link to comment Share on other sites More sharing options...

ajb Posted July 8, 2016 Share Posted July 8, 2016 The the gravitational field is just the curvature of space time, what has spin got to do with it? The gravitational field is not the curvature - the field is usually understood as the metric. Spin has a lot to do with it, all fields have a property we call spin, which is really to do with the nature of there quanta. Even if gravity doe snot have quanta, the basic dgeree of freedom here is the metric and this is a rank 2 tensor. You cannot endoce all the information of the metric into a 'smaller' tensor - if we could we would be doing it already. In your opinion in GR is the coordinate system a fixed grid or does the actual coordinate system deform with the curvature of space? You are asking about coordinate systems on smooth manifolds, quite independent of physics and general relativity. A coordinate system is a diffeomorphism from an (open) neighborhood of a point to an open subset of R^n (n is the dimension of the manifold). A collection of such things that covers all the manifold is an atlas. So, forgive me, but I don't really uunderstand your question. Coordinates are not fixed, we always have a lot of choice. It is always possible to find local coordinates about a point known as Riemann normal coordinates such that the Levi-Civita conenction vanishes and the metric is just the flat metric at that point. The Riemann curvature tensor simplifies in these coordinates. Link to comment Share on other sites More sharing options...

Declan Posted July 8, 2016 Author Share Posted July 8, 2016 The rank or degree of the tensor is 2 which means it is a two dimensional matrix (4x4). There are 3 spatial dimensions and one time dimension - agreed, space time requires all these dimensions to describe the field (metric). But just because is is rank 2 does not mean it is spin 2. The matrix has nothing to do with spin, unless the word spin has been used to mean rank or degree. There may be other applications where rank 2 tensor a are describing spin, but it seems irrelevant or misleading to say that for the space time metric. So you are saying that any coordinate system can be used at any point in space to suit whatever you are trying to do? What is the curvature measured against if not a fixed grid coordinate system? What is it curving relative to? Link to comment Share on other sites More sharing options...

swansont Posted July 8, 2016 Share Posted July 8, 2016 To Swansont: But that is what the centripetal acceleration equation and my paper shows. As the flow is constant with distance, it's affect is an extra velocity vector towards the center for every such vector calculation along r from the center outward. This constant extra vector is equivalent to a constant extra centripetal acceleration towards the center. The graph in my paper shows the required orbital speed at each distance r from the center when an extra constant acceleration is applied to objects in orbit around a Galaxy. No, that's not true. The centripetal acceleration equation shows that for a specific ∆v and r you get a specific acceleration. You have shown for one case that adding in a v', you get a different acceleration. What you have not shown is that when you add v' for another r, you get the correct acceleration. You can't extrapolate from a single example. You will get an acceleration, but you haven't shown it to be the correct one. e.g. for some r_{0 }you have a ∆v which is the result of a combination of the gravitational acceleration and this flow giving v' a = v_{t}^2/r v_{t} is the tangential (i.e. rotation) speed, which is basically constant, but shouldn't be if all we had was gravity and observed mass e.g. http://www.astronomy.ohio-state.edu/~thompson/1144/RotCurve2.gif Now let's double the radius. The acceleration from gravity drops by a factor of 4, since gravity is a 1/r^2 force, mitigated by some additional observable mass. The overall acceleration has to drop in half, since we have the same tangential speed. v' is the same. The problem is that contribution from gravity doesn't seem to match up with what we need. As we keep increasing the radius, v' is a larger and larger fixed fraction of ∆v, which has to keep getting smaller. That's the problem you haven't properly addressed. (Also the fact that the curve is Keplerian out to 5 kpc or so is probably an issue) Using that curve, at 10 kpc: The overall centripetal acceleration is 5065 (km/s)^2/kpc, while the gravitational acceleration is 4000 (km/s)^2/kpc. The v' in some ∆t is 1065 in these units at 20 kpc, the overall acceleration is 2530 (km/s)^2/kpc, while the gravitational contribution is 780 (km/s)^2/kpc. Now the v' "flow" is 1750 in these units. You might note that these two numbers are not equal. Link to comment Share on other sites More sharing options...

ajb Posted July 8, 2016 Share Posted July 8, 2016 The rank or degree of the tensor is 2 which means it is a two dimensional matrix (4x4). It depends on how you like to think of tensors and where you put the indices. So in physics we usually think about the components of a tensor - for a rank 2 tensor we need two indices. So the metric, which we can think of in different ways - the most clear way is as a symmetric pairing between vector fields [math] g : Vect(M) \times Vect(M) \rightarrow C^{\infty}(M) [/math] [math] (X,Y) \mapsto g(X,Y) = X^{a}(x)Y^{b}(x)g_{ba}(x)[/math] Where I have used local coordinates on the right hand side - I hope you can follow this. As the g has two indices is it rank 2. Or I can say that the metric is rank two as it 'eats' two vector fields. This is very improtant when it comes to looking at how the components of the metric change under changes of coordinates. There are 3 spatial dimensions and one time dimension - agreed, space time requires all these dimensions to describe the field (metric). Okay, space-time is 3+1 dimensional But just because is is rank 2 does not mean it is spin 2. The matrix has nothing to do with spin, unless the word spin has been used to mean rank or degree. So, if we accept that the gravitational field is given by the metric, then locally we have a spin 2 representation of the Poincare group with zero mass. Weinberg showed this carefully in Phys.Rev. 138 (1965), B988-B1002. So you are saying that any coordinate system can be used at any point in space to suit whatever you are trying to do? Almost, as long as we do have a coordinate system - that is a local diffeomrophism from our 4-d manifold to R^4. There maybe choices of coordinates that make certian problems more clear. What is the curvature measured against if not a fixed grid coordinate system? What is it curving relative to? The curvature is a measure how how different locally the manifold is to standard Euclidean space (or Minkowski space-time in the context of gravity). Link to comment Share on other sites More sharing options...

Declan Posted July 8, 2016 Author Share Posted July 8, 2016 To ajb: Precisely: compared to Euclidean space, which is effectively a fixed grid coordinate system. A spin-2 representation, maybe, but does it really mean spin-2 in physical terms? Link to comment Share on other sites More sharing options...

ajb Posted July 8, 2016 Share Posted July 8, 2016 Precisely: compared to Euclidean space, which is effectively a fixed grid coordinate system. If you pick standard Cartesian coordinates - which we often do - then you can think like this. A spin-2 representation, maybe, but does it really mean spin-2 in physical terms? Via general theorems of QFT this means that a gravition, if realised in nature, will be massless and have helicity 2. Link to comment Share on other sites More sharing options...

Declan Posted July 8, 2016 Author Share Posted July 8, 2016 To Swansont: I think you are misunderstanding me: the velocity vector of the inflow doesn't go into the velocity variable of the centripetal acceleration equation, this is the orbital velocity of the star. The inflow velocity represents an apparent acceleration. This is because the inflow is consistently pulling the star towards the centre and changing the star's orbital velocity vector by bending it towards the centre. The rate of inflow is constant with distance from the centre because the field density decreases with inverse square law as we move further from the center, but the volume of the spherical she'll increases by inverse square law as we move further out. The two changes cancel out such that the flow rate is constant with distance. Constant flow rate = constant inward velocity of space = constant apparent centripetal acceleration with distance. To ajb: Ok, if you are thinking in terms of particles - but I am not, and I don't think it is relevant to in the case of gravitation. Ok, so if the curvature is measured relative to standard Cartesian coordinates (a fixed grid) then how is this different to having a fixed grid filled with a field with variable density and the metric describes the curvature of this field? See it is the same as GR. Link to comment Share on other sites More sharing options...

ajb Posted July 8, 2016 Share Posted July 8, 2016 (edited) Ok, so if the curvature is measured relative to standard Cartesian coordinates (a fixed grid)... It is not really measured against coordinates. We know by set-up that the Riemann curvature tensor for Euclidean space (irrespective of the coorrdinates used) is globally zero. If the Riemann curvature tensor is locally not zero on a manifold then we know that that manifold cannot be Euclidean space. The actual local values of the Riemann curvature do depend on the coordinate system employed - other than when it is zero. If all the components of a tensor are zero in one coordinate system then they will also be zero in another coordinate system. then how is this different to having a fixed grid filled with a field with variable density and the metric describes the curvature of this field? What is a fixed grid? Is it just a coordinate system? How does the metric describe the curvature of the field? (Which is a strange term in this context - we usually speak of connections as having curvature, and when we mean curvature of a manifold we really mean the curvature of some given connection. In the context of GR, the connection is the Levi-Civeta connection. Unless this field is a local connection one-form?) Edited July 8, 2016 by ajb Link to comment Share on other sites More sharing options...

swansont Posted July 8, 2016 Share Posted July 8, 2016 To Swansont: I think you are misunderstanding me: the velocity vector of the inflow doesn't go into the velocity variable of the centripetal acceleration equation, this is the orbital velocity of the star. The inflow velocity represents an apparent acceleration. This is because the inflow is consistently pulling the star towards the centre and changing the star's orbital velocity vector by bending it towards the centre. How is that different from what I've done? I looked at the difference in radial velocity from a purely gravitational system and from what is observed. The value needed to equalize the effects is not constant. I used the tangential velocity to calculate the centripetal acceleration, because that's how you calculate it. I took those values from the rotational curve plot, without modification. The rate of inflow is constant with distance from the centre because the field density decreases with inverse square law as we move further from the center, but the volume of the spherical she'll increases by inverse square law as we move further out. The two changes cancel out such that the flow rate is constant with distance. Claimed but not demonstrated. Constant flow rate = constant inward velocity of space = constant apparent centripetal acceleration with distance. Centripetal acceleration varies with distance for a constant tangential speed, which is what we have. a = v^2/r A constant radial velocity can't give a constant centripetal acceleration. This is my point, and the demonstration you keep avoiding. Centripetal acceleration has to drop as r increases. Your analysis worked with one value of r, because we have a free variable, but it won't work when r takes on another value. Two curves can intersect at a point, but that doesn't make them identical curves. Link to comment Share on other sites More sharing options...

Mordred Posted July 8, 2016 Share Posted July 8, 2016 (edited) The rate of inflow is constant with distance from the centre because the field density decreases with inverse square law as we move further from the center, but the volume of the spherical she'll increases by inverse square law as we move further out. The two changes cancel out such that the flow rate is constant with distance. . [latex]v=\frac {a^2}{r}[/latex] this equation cannot nor does not show a constant velocity as you increase r. Not unless your doing your math wrong. Edited July 8, 2016 by Mordred Link to comment Share on other sites More sharing options...

Declan Posted July 9, 2016 Author Share Posted July 9, 2016 To Swansont/Mordred: Ok, I see what you are saying now - I missed the link to the curve you are talking about in your earlier post. The curve I was using was the M33 spiral galaxy. It shows the curve flattening out with distance but still climbing as r increases. This curve matches the constant acceleration I am referring to. What type of Galaxy is the curve you posted? There may be more than one effect contributing to the final curve: there could be *some* dark matter, or the flow rate could be dimishing with distance due to space expanding too. It would also depend on the distribution of black holes in the Galaxy. Not every situation in every Galaxy will be the same. I am not in a position to be able to account for the details of each Galaxy. The point is that the effect I am suggesting can account for the bulk of the extra effect required, rather than having to invoke 80% of the mass of the Universe being invisible Dark Matter. Link to comment Share on other sites More sharing options...

Mordred Posted July 9, 2016 Share Posted July 9, 2016 (edited) that isn't what I'm saying at all. Without reading ANY of your papers. Nor any other post. If someone were to look at your formula. How can that person possibly use that formula to describe a system to the dynamics your describe? [latex]v=\frac {a^2}{r}[/latex] by the way why didn't you leave it in units of force or gravitational potential? the only detail shown is two scalar values. one for radius. the other is the acceleration. the dynamics you describe is increased acceleration closer to the BH. That means velocity will be greater closer to the BH. Not CONSTANT. Lastly based on baryonic matter distribution of ALL SPIRAL galaxies. The greatest mass concentration is the galactic bulge. The mass distribution of baryonic matter distribution tapers off as r increases. this includes M33. This is one reason why this equation doesn't work. [latex]v=\sqrt{\frac{GM}{r}}[/latex]. it works great for solar systems and even BHs until you need time dilation. Even without time dilation it gives the Keplar curve. However you want a Keplar curve for a solar system around a BH. OR star. But not for galaxies. Edited July 9, 2016 by Mordred Link to comment Share on other sites More sharing options...

Declan Posted July 9, 2016 Author Share Posted July 9, 2016 I think you meant to type a=v^2/r rather than v=a^2/r. Right? Close to the center shell theorem is required which complicates things a bit. I am mainly referring to the more distant region of r where the main discrepancy exists and the calculation is simpler. There are two effects: (1) the normal gravitational acceleration (keplerian and shell theorem) and (2) the extra acceleration due to inflow which is constant with r as I previously explained. Close to the center the effect (1) will be greatest and the extra acceleration of (2) will only be ~1x10^-11 msec^2 so won't affect it much. I just noticed on the wiki page about Galaxy rotation that there is typically an asymmetry in the orbital speeds on each side of a Galaxy - which would mean that a Dark Matter halo would have to be lopsided in most galaxies to explain the observations! Why would this be? Alternatively if you add the Galaxy's velocity through space to the inflow vector, then this is exactly the sort of result one would expect. On one side of the Galaxy the velocity would add, and other side it would subtract. Link to comment Share on other sites More sharing options...

Mordred Posted July 9, 2016 Share Posted July 9, 2016 I think you meant to type a=v^2/r rather than v=a^2/r. Right? Close to the center shell theorem is required which complicates things a bit. I am mainly referring to the more distant region of r where the main discrepancy exists and the calculation is simpler. There are two effects: (1) the normal gravitational acceleration (keplerian and shell theorem) and (2) the extra acceleration due to inflow which is constant with r as I previously explained. Close to the center the effect (1) will be greatest and the extra acceleration of (2) will only be ~1x10^-11 msec^2 so won't affect it much. I just noticed on the wiki page about Galaxy rotation that there is typically an asymmetry in the orbital speeds on each side of a Galaxy - which would mean that a Dark Matter halo would have to be lopsided in most galaxies to explain the observations! Why would this be? Alternatively if you add the Galaxy's velocity through space to the inflow vector, then this is exactly the sort of result one would expect. On one side of the Galaxy the velocity would add, and other side it would subtract. I quoted your formula in the last post. you tell me Link to comment Share on other sites More sharing options...

Declan Posted July 9, 2016 Author Share Posted July 9, 2016 (edited) Oh really - sorry, my mistake then... Which post (number) did I give that equation? Edited July 9, 2016 by Declan Link to comment Share on other sites More sharing options...

Mordred Posted July 9, 2016 Share Posted July 9, 2016 (edited) ok if a=v^2/r in order to keep v constant over r is to gradually increase mass as you increase radius. Based on point mass via shell theorem. Baryonic matter distrubution alone doesn't support a constant velocity curve. thats why the Keplarian curve is referred to as the missing mass problem Edited July 9, 2016 by Mordred Link to comment Share on other sites More sharing options...

Declan Posted July 9, 2016 Author Share Posted July 9, 2016 Yes - understood. If normal gravity is the only effect, then there is the missing mass problem. So we can invoke an extra 80% of the Universe to explain the problem, or we can make a small change to the Schwartzchild metric and explain (most) of the problem. Link to comment Share on other sites More sharing options...

Mordred Posted July 9, 2016 Share Posted July 9, 2016 (edited) Yes - understood. If normal gravity is the only effect, then there is the missing mass problem. So we can invoke an extra 80% of the Universe to explain the problem, or we can make a small change to the Schwartzchild metric and explain (most) of the problem. then do so. So far you haven't. Your end equation in this paper certainly doesn't show a solution. I have an idea that explains Galaxy Rotation rates without requiring Dark Matter. The URL of my short paper explaining the idea is here: http://vixra.org/pdf/1606.0218v2.pdf your paper on SR only shows a vacuum solution of gravitational potential. It doesn't show any added dynamic. mathematically. Not verbally. Though mathematical wise you SR paper is decent. Your verbal descriptions in it is another matter lol (though we already covered some of that aspect this thread) lets put it in simple English. In order to maintain a constant velocity term. your sum of forces or gravitational accekeration must stay consistent as a function of radius from a point source. The BH itself. Now we both know this isnt the case in the Schwartchild metric. Which definetely has an increase in gravitational acceleration as you aproach it. However it is the case using the NFW profile with a halo dark matter distribution. Edited July 9, 2016 by Mordred Link to comment Share on other sites More sharing options...

Declan Posted July 9, 2016 Author Share Posted July 9, 2016 (edited) Which is why I added a constant 'f' term to the Schwartzchild metric to account for the inflow.The inflow is effectively contracting space (divergence) and its effect on an orbiting star is equivalent to a constant acceleration towards the center of the Galaxy. I have another paper on SR that includes a notion of space flow: http://gpcpublishing.com/index.php?journal=gjp&page=article&op=view&path%5B%5D=503 Edited July 9, 2016 by Declan Link to comment Share on other sites More sharing options...

Mordred Posted July 9, 2016 Share Posted July 9, 2016 (edited) did you derive that addition? If so please point out that equation. Ive looked for your added field in your papers. I couldn't find it. that last paper is even worse grr... time dilation is neglibible in galaxy rotation curves. Even if it wasn't we can already account for it. your added f brings us back to the problem of adding energy to the system. Which changes the metric tensor via the stress tensor. Edited July 9, 2016 by Mordred Link to comment Share on other sites More sharing options...

Declan Posted July 9, 2016 Author Share Posted July 9, 2016 Do you mean the vector addition? If so I posted an example of the vector addition with and without the extra inflow vector (I think it was post #101). I wasn't talking about time dilation, I was referring to the velocity vector of the field associated with moving mass in that case. Where is the added energy? I am just talking about the space field moving. Objects moving through the space field with a higher velocity would mean more energy, but that is what we are talking about: stars that are moving faster through space than expected. So the extra energy is already there in the galaxies we observe. Link to comment Share on other sites More sharing options...

Mordred Posted July 9, 2016 Share Posted July 9, 2016 (edited) you don't move a static field that already accounts for how particles move without adding energy. You aleady know the vacuum solution already correlates particle movement. You also know it already accounts for Time dilation and redshift. You know those same metrics do not solve the missing mass problem Which is why I added a constant 'f' term to the Schwartzchild metric to account for the inflow. The Schwartzchild metric already accounts for infalling matter. (matter field) despite being a vacuum solution. adding more inflow just makes the Kepler problem even worse. That will be like adding greater mass at the BH. You need your additional mass in the OPPOSITE direction. Thats why your descriptives and equations still have a Kepler decline. You added to the WRONG vector. Edited July 9, 2016 by Mordred Link to comment Share on other sites More sharing options...

Declan Posted July 9, 2016 Author Share Posted July 9, 2016 If you have a laser (i.e. energy in a standing wave between two mirrors) then you suddenly change one mirror to a black body absorber (i.e. black hole) then the existing energy will flow in one direction (inwards) but not be reflected back in the other direction (outwards). The energy of the waves flowing inwards has not changed, but the outward waves have been removed (eaten by the black hole). The effect of the space flowing inwards is a greater effect at, say 20,000 light years from the galaxy center than the Keplerian gravitational acceleration due to mass/energy consumed by the black hole & the black hole getting bigger. Anyhow, the mass/energy eaten by the black hole would already be contributing to the Keplerian gravitational acceleration experienced at 20,000 light years distance even before it is consumed. There would be a change in energy of any matter accelerated relative to the space it is in, but for an already formed galaxy this is already accounted for in the orbital speeds of the stars in stable orbits. When a black hole first forms and space starts to flow the situation is dynamic and there would be some changes to energy distribution. I am concerning myself with the static situation of an already formed Galaxy. Link to comment Share on other sites More sharing options...

Mordred Posted July 9, 2016 Share Posted July 9, 2016 (edited) What does mirrors or observation influence have to do with rotation curves? We already account for observational redshift influence. Either way your missing mass is in the wrong direction. Why would a BH be any different than any other gravitational body.? How does a volume flow? Come on man we already showed the error in thinking spacetime flow. Your model shows none of these dynamics other than your misinterpreted data. How does a blackhole eat a volume? except for increasing its event horizon? You really need to go back and study. The terminology usage you use is horrendous. Edited July 9, 2016 by Mordred Link to comment Share on other sites More sharing options...

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