Mordred Posted July 7, 2016 Share Posted July 7, 2016 Cross edit see above What about my other questions? I should rather note a geodesic is a type of worldline. The metric tensor contains the worldline metric Link to comment Share on other sites More sharing options...

Declan Posted July 7, 2016 Author Share Posted July 7, 2016 So the geodesic equation is essentially based on the gravitational acceleration (gradient in the gravitational potential) and determines the metric (i.e. The length, and time taken for light to travel the curved line between two points). It will be exactly the same if calculated for a space filled with a field (medium for light/matter) that is proportional to -phi (i.e. A positive gravitational potential field) that determines the speed of light and rate of time (higher density field = slower speed of light, and slower rate of time). Therefore the GR equations will work in the same way for the energy field I proposed in my Energy Field Theory paper, with the added benefit that space is now viewed as filled with an energy field, and so this field can be consumed by black holes and explain the additional acceleration required to account for Galaxy rotation rates. Link to comment Share on other sites More sharing options...

Mordred Posted July 7, 2016 Share Posted July 7, 2016 (edited) You can seperate a medium mass flow by the term "Interstellar medium" Which is all your dust, particles etc. Now the solution you just studied won't work too well for BH dynamics. Now you need to goto the Scwartzchild metric. You'll also learn the stress tensor on that metric. Edited July 7, 2016 by Mordred Link to comment Share on other sites More sharing options...

Declan Posted July 7, 2016 Author Share Posted July 7, 2016 Yes I have had a look at the metric already and made a suggested modification for field inflow in post #38. It might be better to define the modification to the metric elements for the time (t) and distance ® dimensions as: Originally: 1-2Gm/rc^2 Suggested: 1-(2Gm/rc^2 + f) where f represents a steady inflow independent of distance r. Of course as the black hole grows larger (through consumption of matter) the rate of inflow 'f' would slowly increase over time too. Link to comment Share on other sites More sharing options...

Mordred Posted July 7, 2016 Share Posted July 7, 2016 (edited) Field as in particle field. The metrics has that already. Which brings up an earlier question I asked quite some time ago. If the the particle field or interstellar medium is already following those geodesics. What distinquishes your field from that dynamic. Without adding any energy/mass density? I would suggest studying a radiation solution and vacuum solution to GR. You need to learn the pressure relations. Aka equations of state. Pressure may increase gravity but exerts pressure opposing gravity in localized amistrophy regions. Edited July 7, 2016 by Mordred Link to comment Share on other sites More sharing options...

ajb Posted July 7, 2016 Share Posted July 7, 2016 It will be exactly the same if calculated for a space filled with a field (medium for light/matter) that is proportional to -phi (i.e. A positive gravitational potential field) that determines the speed of light and rate of time (higher density field = slower speed of light, and slower rate of time). You say that, but you need to actually show that. If this 'energy field' is acting as a source of gravity - i.e., you include its energy-momentum tensor in the Einstein Field Equations, then the geometry will be different as compared to it not being there. You no longer have a vacuum solution. Or, you may decide that your 'energy field' is a gravitational field and not include the energy-momentum in the field equations. If so what role does it actually play? Do you couple it to test particles? If so there will be another term in the geodesic equations related to this extra force. 1 Link to comment Share on other sites More sharing options...

Mordred Posted July 7, 2016 Share Posted July 7, 2016 You say that, but you need to actually show that. If this 'energy field' is acting as a source of gravity - i.e., you include its energy-momentum tensor in the Einstein Field Equations, then the geometry will be different as compared to it not being there. You no longer have a vacuum solution. Or, you may decide that your 'energy field' is a gravitational field and not include the energy-momentum in the field equations. If so what role does it actually play? Do you couple it to test particles? If so there will be another term in the geodesic equations related to this extra force. Absolutely correct. 1 Link to comment Share on other sites More sharing options...

Declan Posted July 7, 2016 Author Share Posted July 7, 2016 The field I am talking about is identical to the Gravitational field and so will be exactly the same as the space described in GR - by definition. The energy in the Gravitational field is not part of the Stress Momentum Tensor. This quote from the wiki page on "Mass in General Relativity": ' "gravitational field energy" is not a part of the energy–momentum tensor; instead, what might be identified as the contribution of the gravitational field to a total energy is part of the Einstein tensor on the other side of Einstein's equation (and, as such, a consequence of these equations' non-linearity).' source : https://en.wikipedia.org/wiki/Mass_in_general_relativity Conceptually, each quantum particle in the Universe is a 3D standing wave that can be thought of as comprising spherical IN and OUT waves that combine to form a standing wave (that satisfies the Classical and Schrodinger wave equations - and so therefore is a stable wave structure). All of these wave functions add together in space (superposition principle) - resulting in a large sum of essentially random wave energy in empty space. In a normal volume of space as much wave energy enters the region as exits it (zero divergence). But In the vicinity of a black hole (the event horizon), waves can flow inwards but not outwards - thus creating a deficit in the energy balance of the space outside of the event horizon. (i.e. move wave energy exits the region than enters it, on the side towards the black hole). Thus the energy field of space begins to flow towards the black hole. Link to comment Share on other sites More sharing options...

ajb Posted July 7, 2016 Share Posted July 7, 2016 The field I am talking about is identical to the Gravitational field and so will be exactly the same as the space described in GR - by definition. Meaning that you can go from your 'energy field' to the metric (or the Levi-Civita connection or one of the other ways of encoding this information)? Please show this. The energy in the Gravitational field is not part of the Stress Momentum Tensor. Okay. You seem to say a lot without really being about to show anything... you really need to think about this. Link to comment Share on other sites More sharing options...

Strange Posted July 7, 2016 Share Posted July 7, 2016 The field I am talking about is identical to the Gravitational field and so will be exactly the same as the space described in GR - by definition. If that were true, then you would get exactly the same results as GR, in all circumstances. Thus the energy field of space begins to flow towards the black hole. So not the same as GR then. 2 Link to comment Share on other sites More sharing options...

Declan Posted July 7, 2016 Author Share Posted July 7, 2016 Well being able to see GR as resulting from a space-filling field is really the only change and is conceptual rather than a mathematical change - however it does allow one to see the situation differently and understand why the Galaxy rotation rates are as they are. This is different to GR as it cannot explain the Galaxy rotation rates. The difference only affects black holes due to gravitational collapse and resulting inflow of the field - for this change I have provided an explanation via the centripetal acceleration equation. It doesn't really require any new or complex maths, although there must be some as-yet unknown equation that relates black hole size to rate of inflow. Link to comment Share on other sites More sharing options...

ajb Posted July 7, 2016 Share Posted July 7, 2016 Well being able to see GR as resulting from a space-filling field is really the only change and is conceptual rather than a mathematical change - however it does allow one to see the situation differently and understand why the Galaxy rotation rates are as they are. This is different to GR as it cannot explain the Galaxy rotation rates. So it is not the same as GR, or is it? I don't understand the key difference between 'conceptual' and 'mathematical' change. The concepts of phsyics are stated mathematically - there is no great divide here. The difference only affects black holes due to gravitational collapse and resulting inflow of the field - for this change I have provided an explanation via the centripetal acceleration equation. Then it is not the same as GR... right? (Not that you have really presented any details of a theory so that people can check) It doesn't really require any new or complex maths, although there must be some as-yet unknown equation that relates black hole size to rate of inflow. This suggests you really have nothing... Anyway you really must sort out the mathematical framework and clearly show that your theory is equivalent to GR, or show that it is not but by only 'small corrections'. So far you have said plenty without backing it up. Link to comment Share on other sites More sharing options...

swansont Posted July 7, 2016 Share Posted July 7, 2016 The difference only affects black holes due to gravitational collapse and resulting inflow of the field - for this change I have provided an explanation via the centripetal acceleration equation. It doesn't really require any new or complex maths, although there must be some as-yet unknown equation that relates black hole size to rate of inflow. Did I miss the post where you showed the math behind this? That a constant radial velocity will look like a centripetal acceleration? Link to comment Share on other sites More sharing options...

ajb Posted July 7, 2016 Share Posted July 7, 2016 Did I miss the post where you showed the math behind this? I can assure you that you have not missed anything. It is an interesting idea trying to put all the gravitational information into a single scalar field (I think that is what we have here - but nothing seems to be explained properly). However, it seems doomed to fail as we cannot capture curvature of a 4-dimensional manifold properly with a single scalar. There are several curvature scalars that one can build, but these do not capture everything - they are a 'coarse' measures, although they maybe useful. From a physics perspective, it is also going to fail as we know that single scalar - so spin-0 - cannot capture gravity. Gravity is spin-2. Link to comment Share on other sites More sharing options...

Mordred Posted July 7, 2016 Share Posted July 7, 2016 (edited) Did I miss the post where you showed the math behind this? That a constant radial velocity will look like a centripetal acceleration? Thats another aspect that makes little sense in his model. [latex]f=ma[/latex] [latex]a_c=\frac{v^2}{R}[/latex] [latex]F=\frac{GMm}{R^2}[/latex] combining the above [latex]F=\frac{GMm}{R^2}=\frac{mv^2}{R}=ma[/latex] Gives [latex]v=\sqrt{\frac{GM}{R}}[/latex] The last formula is what gives the Keplar curve. With baryonic mass distribution in our system the curve is an inverted slope. As most of the mass is the sun. So velocity decreases the further from Sun you get. Now applying that same formula to galaxy curves. Most of the mass is in the galactic bulge (strictly baryonic mass,no dark matter). So once again you get the Keplarian curve. So lets isolate the mass component [latex]m=\frac{v^2R}{G}[/latex] This last formula tells us you require a uniform mass distribution to maintain a uniform curve. However visible baryonic matter mass distribution isn't uniform. So you end up with a Keplarian curve. The velocity formula already incorperates centripetal acceleration. Edited July 7, 2016 by Mordred Link to comment Share on other sites More sharing options...

Mordred Posted July 7, 2016 Share Posted July 7, 2016 (edited) This is the formula you give. [latex]v=\frac{a^2}{r}[/latex] [latex]v=\frac{GM}{r}[/latex] Your interpretation on galaxy curves doesn't match the model with dark matter. Nor does it match any verbal description you've given. Edited July 7, 2016 by Mordred Link to comment Share on other sites More sharing options...

Declan Posted July 7, 2016 Author Share Posted July 7, 2016 To Swansont/Mordred: I gave an example of the vector addition that results in the orbital speed in post #101. There is a velocity vector towards the galaxy center due to normal gravitational acceleration (i.e. Gm/r) for a unit time. There is also an additional velocity vector towards the galaxy center due to inflowing space. The two vectors add, thus requiring the orbital velocity to increase in order to maintain the same orbit that would exist if only normal gravitational acceleration was acting. To ajb: An example of a conceptual change is the Earth going around the Sun rather than the Sun going around the earth. The actual numbers in both models are the same, but the earth going around the sun results in a better model. This might not be the best example, but the conceptual change I am suggesting for GR is just in the understanding of what space curvature means: i.e. curvature due to the space-time field changing in density, rather than deforming in actual geometry. The maths capturing the curvature is identical - its just how we think about it that has changed. As the space/energy field can be flowing it would be a vector field rather than scalar. This helps capture other curvature issues such as frame dragging (i.e. Kerr metric etc). Can you explain more about the spin-2 aspect? The real problem with GR may be that the energy of the gravitational field is always assumed to be connected to matter (i.e. it is part of the Einstein Tensor, rather than part of the Stress-Momentum Tensor). Perhaps There should be a version of GR that treats all energy in the system equally. Link to comment Share on other sites More sharing options...

Bignose Posted July 8, 2016 Share Posted July 8, 2016 (edited) An example of a conceptual change is the Earth going around the Sun rather than the Sun going around the earth. The actual numbers in both models are the same, but the earth going around the sun results in a better model. This might not be the best example, This is actually a terrible example. Because in order to describe the motion of the sun and the planets asuming the earth was the center, the best model used the concepts of epicycles. That is, the planets moved in small circles as they moved around the earth. This was decent enough for records keeping for the ancient Greeks. But as the tools and the records keeping got better and better, in order to work, they were having to introduce epicycles inside of epicycles and so on... sometimes up to 3 or 4 layers deep to match observations. What you're really missing here is that when the model was introduced of earth and the rest of the planets going around the sun, suddenly the accuracy of all those measurements agreed very well with the new model. So, you're actually quite, quite wrong in this example. The measurements were the same, but the models were quite different. The real problem with your saying that you're using the exact same math as GR is that you're immediately ceding away any reason to even consider your model. At least from a scientific point of view. You seem to forget that science is almost wholly about making as accurate prediction as possible. If your math is the same as GR (a big if, since you haven't demonstrated it at all), your predictions are identical too. All you're arguing about is philosophy or interpretation. Your model has to be different in some way so that it can be tested to really see if it is better or not. So, to bring this all back, to actually show us your model is better, you should use the observations we have. And show us a plot of the current best data, the best current model or models, and your model. Show us how closely your predictions fall on the measurements. If you're closer than the current best models, then you'll have some scientific interest. Edited July 8, 2016 by Bignose Link to comment Share on other sites More sharing options...

Declan Posted July 8, 2016 Author Share Posted July 8, 2016 Fair point - it was not a good example. The curvature of space is the same in my model. The only un-modeled effect is the inflow affecting black holes. If I make my small change to the Schwartzchild Metric to account for the extra steady inward flow, then use the new metric in GR it should have the desired effect. The amount of the new term would need to be adjusted to a particular rate of inflow for a given galaxy to match the observations. Link to comment Share on other sites More sharing options...

Bignose Posted July 8, 2016 Share Posted July 8, 2016 The amount of the new term would need to be adjusted to a particular rate of inflow for a given galaxy to match the observations. So you get to tune your model for each galaxy? It will have awfully limited predictive power for an unknown situation then, won't it? The real power of models is that they make predictions about situations that they haven't seen, yet. Link to comment Share on other sites More sharing options...

Declan Posted July 8, 2016 Author Share Posted July 8, 2016 Well it might take analysis of numerous galaxies to work out the equation relating black hole size to rate of inflow. Once this equation is known it would have predictive power to all situations in all galaxies. I would need a supercomputer to model a Galaxy. My main point is that it provides a mechanism to explain the rotation rates without requiring Dark Matter and with only minimal change from existing theory. Link to comment Share on other sites More sharing options...

ajb Posted July 8, 2016 Share Posted July 8, 2016 An example of a conceptual change is the Earth going around the Sun rather than the Sun going around the earth. The actual numbers in both models are the same, but the earth going around the sun results in a better model. This might not be the best example, but the conceptual change... Okay, that is not a bad example. However, this is really no more than a change in the coordinate syatems uses. You are quite free to put any point in the Solar System as origin and then use Newton's theory of gravity or general relativity. Once we are past two bodies the problems grow, but the point is that we have this choice and the physics does not change - though calculations may be much harder for some choices. ...I am suggesting for GR is just in the understanding of what space curvature means: i.e. curvature due to the space-time field changing in density, rather than deforming in actual geometry. So we have a flat space-time and then this extra field on top that phenomenologically gives (almost?) the same results as GR? Then you have to start with a flat background and cannot discuss black holes. You need to show that you theory produces a situation that looks the same as a black hole. The maths capturing the curvature is identical - its just how we think about it that has changed. You say this, but you need to show this. However, it looks to me that you really have changed the mathematics. As the space/energy field can be flowing it would be a vector field rather than scalar. You have not defined flow very carefully yet - you should do that. My idea of what you could mean by a flow does not need the field to be a vector field. This helps capture other curvature issues such as frame dragging (i.e. Kerr metric etc). Well, show this. Can you explain more about the spin-2 aspect? Loosley, as general relativity understands - in the standard formulation - the gravitaional field as the metric which is a rank 2 tensor, the quanta associated with gravity is spin-2. We also know that spin-1, so a vector field cannot capture gravity either. So you cannot replace GR with a flat space-time and a vector field. What you can do, and people have thought about this - and this loosley seems a bit like what you are doing - is include a spin-0 and spin-1 field in a gravity theory. But as you have not really written a theory out clearly we don't really know what you are thinking of. The real problem with GR may be that the energy of the gravitational field is always assumed to be connected to matter (i.e. it is part of the Einstein Tensor, rather than part of the Stress-Momentum Tensor). This has been a question for a long time. However, all attempts to build a meaningful energy-momentum tensor for the gravitational field fail. We just don't seem to be able to find a local notion of the energy of the gravitational field in a general and reasonable way. There are some partial solutions, these are not tensors and so how meaningful they are is not at all clear. For example there is the Landau–Lifshitz pseudotensor. Perhaps There should be a version of GR that treats all energy in the system equally. People for sure have thought about how to work with the energy of the gravitational field locally. The problem is that energy is something to do with the physics being invariant with changes in time. Thus, the notion of energy is not at all clear in general relativity, not even globally. You should look up ADM mass and Bondi mass. Link to comment Share on other sites More sharing options...

Declan Posted July 8, 2016 Author Share Posted July 8, 2016 In the field I am describing a black hole will form just as it does in GR. It is simply due the the gradient in the potential increasing past a point of no return for any matter/light/energy (where the event horizon is). Why does the gravitational field have to have a quantum particle associated with it? This is just the view of the Standard Model that says every force has a particle that carries the force. This view is part of the problem - I don't think there are gravitons. Every particle is a structure within the space field. My phone battery is nearly depleted (out of the house at the moment) so I have to stop typing now - more later.... Link to comment Share on other sites More sharing options...

swansont Posted July 8, 2016 Share Posted July 8, 2016 To Swansont/Mordred: I gave an example of the vector addition that results in the orbital speed in post #101. There is a velocity vector towards the galaxy center due to normal gravitational acceleration (i.e. Gm/r) for a unit time. There is also an additional velocity vector towards the galaxy center due to inflowing space. The two vectors add, thus requiring the orbital velocity to increase in order to maintain the same orbit that would exist if only normal gravitational acceleration was acting. Which would be the first step in the derivation. Now show that this works for an arbitrary r and that it matches the rotation curves. I think you can make it do one but not the other, but both have to work. Link to comment Share on other sites More sharing options...

ajb Posted July 8, 2016 Share Posted July 8, 2016 In the field I am describing a black hole will form just as it does in GR. Again, you say a lot, but can you actually show this? We are still very confused about this - is your 'theory' phenomenologically identical to GR or not? Some times you say yes and other times no. Why does the gravitational field have to have a quantum particle associated with it? It may be the case that a quantum theory of gravity is not a theory of gravitons at all. This is the famous symptotic safety property of gravity in terms of renormalisation group flow. (This is common knowledge for those interested in extensions and quantisation of GR) Basically we know that the methods of perturbation theory do not work for general relativity (and similar actions), but it still maybe possible to have a quantum theory that is well defined, just not using the methods that worked for say QED. There is evidence that GR maybe like this. This is just the view of the Standard Model that says every force has a particle that carries the force. The standard model says nothing about gravity. It is the general framework of quantum field theory (on flat space-time) that gives this association. But for gravity, as I hinted to above is more subtle. This view is part of the problem - I don't think there are gravitons. You are not alone here - but other have some reasoning behind this. [ Link to comment Share on other sites More sharing options...

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