michel123456 Posted April 5, 2015 Posted April 5, 2015 In each case, the other is approaching. We each determine the other's speed to be negative, if you are defining "approaching" as negative. otherwise, all we have to do is reverse the coordinate system (or the observers) and get the opposite answer. But even when we reverse the observers, the other is still approaching.
swansont Posted April 5, 2015 Posted April 5, 2015 But even when we reverse the observers, the other is still approaching. Yes, that was my point. But I was addressing the possibility that this is tied to a coordinate system (the "otherwise"), where one velocity is positive and one is negative, e.g. a fixed Cartesian system. Since the choice of + vs - there is arbitrary, it can't be relied on for this argument.
beejewel Posted April 6, 2015 Author Posted April 6, 2015 (edited) We're talking about orbital speeds. How are you defining a positive velocity? In each case, the other is approaching. We each determine the other's speed to be negative, if you are defining "approaching" as negative. otherwise, all we have to do is reverse the coordinate system (or the observers) and get the opposite answer. Except that Kepler's equations aren't wrong. They work as expected for the solar system. GPT differs from current and past understanding of the world, in that it sets an upper and lower limit to the observers world domain, and defines the observers GP within that domain, further we understand from GPT that this world domain is quite asymmetric. due to this asymmetry of the energy, speed can not strictly scalar, as the body in motion must either have a lower or higher energy potential than the observer. Because of the asymmetry in the energy field around a massive body the sum of velocities for forward moving bodies differs from the sum of velocities for backwards moving bodies. Three forward moving bodies velocities at r1, r2, r3 sum as 3+2+1 = 6, while the same bodies concidered to be moving backwards sum as (-3) + (-5) + (-6) = -13 But even when we reverse the observers, the other is still approaching. As mentioned above the world according to GPT is not symmetric, which is also consistent with our understanding that time only frows in one direction. Yes, that was my point. But I was addressing the possibility that this is tied to a coordinate system (the "otherwise"), where one velocity is positive and one is negative, e.g. a fixed Cartesian system. Since the choice of + vs - there is arbitrary, it can't be relied on for this argument. We are not dealing with a cartesian plane, the space-time around a massive body is non euclidean. Steven PS: adding this little graphic at the end, which makes it more obvious how we see the spiral arms of galaxies in terms of real velocity rather than apparent velocity. Edited April 6, 2015 by beejewel
ajb Posted April 6, 2015 Posted April 6, 2015 due to this asymmetry of the energy, speed can not strictly scalar, as the body in motion must either have a lower or higher energy potential than the observer. I am happy with things that look like scalars, but are not quite. So, what kind of object is 'speed'? How does it transform under changes of coordinates?
beejewel Posted April 6, 2015 Author Posted April 6, 2015 I am happy with things that look like scalars, but are not quite. So, what kind of object is 'speed'? How does it transform under changes of coordinates? Good question, GPT defines the four-velocity or absolute velocity of a body as a function of it's potential, so technically the proton is the only particle who's four-velocity is excactly 'c', here at ground potential our four-velocity is c(930/938) or 99.14 % the speed of light. The difference works out to be around 2500 km/s, so I guess we can call this for speed. Ultimately you, me and the proton are all heading to the same place, the black hole in the centre of the galaxy, but the proton being a light body with a high surface potential must take a longer path down, hence relative speed. The transformation of coordinates are fascinatingly perfect, because as we change our potential, everything in our world changes accordingly. All the electrons in your world instantly change mass, objects that were travelling in straight lines curve and everything just works without a glitch, and things seem great, that's until we start looking at the really tiny things, on a quantum scale, and realize that looking at our experiment changes the outcome. The normal Lorenz transformation should correctly transform the coordinates when using this form of gamma [latex] \gamma = \frac{1}{\sqrt{1- \frac{\phi^2}{\Phi^2}}}[/latex] Steven
ajb Posted April 6, 2015 Posted April 6, 2015 (edited) ...so technically the proton is the only particle who's four-velocity is excactly 'c', You mean then magnitude of the four-velocity. Is this what you are using as speed? The transformation of coordinates are fascinatingly perfect... The normal Lorenz transformation should correctly transform the coordinates when using this form of gamma [latex] \gamma = \frac{1}{\sqrt{1- \frac{\phi^2}{\Phi^2}}}[/latex] So if I employ some admissible (as defined by your theory) local coordinates [math]x^{\mu}[/math] then how are they related to [math]x^{\mu'}[/math]? What happens to 'speed' and 'potential' under general coordinate changes? Edited April 6, 2015 by ajb
swansont Posted April 6, 2015 Posted April 6, 2015 GPT differs from current and past understanding of the world, in that it sets an upper and lower limit to the observers world domain, and defines the observers GP within that domain, further we understand from GPT that this world domain is quite asymmetric. due to this asymmetry of the energy, speed can not strictly scalar, as the body in motion must either have a lower or higher energy potential than the observer. How do you decide who is at rest, and who is moving?
michel123456 Posted April 6, 2015 Posted April 6, 2015 PS: adding this little graphic at the end, which makes it more obvious how we see the spiral arms of galaxies in terms of real velocity rather than apparent velocity. revolution.png (...) Ultimately you, me and the proton are all heading to the same place, the black hole in the centre of the galaxy, but the proton being a light body with a high surface potential must take a longer path down, hence relative speed. (...) Steven But, correct me if I don't understand correctly, in your diagram the hypothetized "real motion" goes outwards and you are stating at the same time that we are heading to the black hole in the centre of the galaxy, which contradicts the graph.
swansont Posted April 6, 2015 Posted April 6, 2015 Because of the asymmetry in the energy field around a massive body the sum of velocities for forward moving bodies differs from the sum of velocities for backwards moving bodies. Three forward moving bodies velocities at r1, r2, r3 sum as 3+2+1 = 6, while the same bodies concidered to be moving backwards sum as (-3) + (-5) + (-6) = -13 So if I pick one direction for my z axis in a cylindrical coordinate system, and then invert the system (z —> -z and phi —> -phi), the speeds will spontaneously change?
Strange Posted April 6, 2015 Posted April 6, 2015 Good question, GPT defines the four-velocity or absolute velocity of a body as a function of it's potential, so technically the proton is the only particle who's four-velocity is excactly 'c', here at ground potential our four-velocity is c(930/938) or 99.14 % the speed of light. The difference works out to be around 2500 km/s, so I guess we can call this for speed. Ultimately you, me and the proton are all heading to the same place, the black hole in the centre of the galaxy, but the proton being a light body with a high surface potential must take a longer path down, hence relative speed. At 2,500 km/s it would take about 3 million years to reach the galactic centre. As the Earth is about 4 billion years old, how come we are still here?
Mordred Posted April 6, 2015 Posted April 6, 2015 I'm confused, from what I've seen thus far is all you've done is replace v and c with your ground potential symbols into standard formulas. Then you claim This describes characteristics that is different than what the original formulas describe. How does that even work? If your describing properties that differ from what the original describe. Simply replacing the symbols cannot describe different relations or coordinate changes. Then you claim your formulas remove the need for dark matter...... Yet the original formulas that you simply changed the symbols in, require dark matter on the global scale. Sorry but you can't simply replace v and c, then describe dynamics that differ from the relationships in the original formulas. This doesn't work. Especially since in those formulas c is invariant. Yet neither of your GPT variables are. 1
Sensei Posted April 6, 2015 Posted April 6, 2015 Good question, GPT defines the four-velocity or absolute velocity of a body as a function of it's potential, so technically the proton is the only particle who's four-velocity is excactly 'c', here at ground potential our four-velocity is c(930/938) or 99.14 % the speed of light. The difference works out to be around 2500 km/s, so I guess we can call this for speed. Proton doesn't have 938 MeV, but 938.272046 MeV (+- couple eV) 930/938=0.99147121535181236673773987206823 but 930/938.272046 = 0.99118374459170448311533731870341 Honestly, I doubt that you even know how to calculate proton mass-energy 938272046 eV.. If I am mistaken, show me it.
michel123456 Posted April 6, 2015 Posted April 6, 2015 On the overall my feeling is that there is something right in Beejewel thoughts. -1
swansont Posted April 6, 2015 Posted April 6, 2015 On the overall my feeling is that there is something right in Beejewel thoughts. This being science, what's important is being able to objectively demonstrate it. 2
beejewel Posted April 7, 2015 Author Posted April 7, 2015 You mean then magnitude of the four-velocity. Is this what you are using as speed? So if I employ some admissible (as defined by your theory) local coordinates [math]x^{\mu}[/math] then how are they related to [math]x^{\mu'}[/math]? What happens to 'speed' and 'potential' under general coordinate changes? Not sure how I can explain it any clearer, the four velocity is the body-potential divided by the proton-potential multiplied by speed of light, while the relative velocity is simply the difference in potential between two bodies divided by c multiplied by the speed of light. From this function alone we can only determine the sign of the velocity not the direction of the vector. The sign indicates weather an object is approaching from the past or receeding into the past. How do you decide who is at rest, and who is moving? Technically the observer is always the one at rest, at least this makes the transformations simpler, weather sitting at the station or sitting on the train, you are at rest, the rest, let the rest of the world do the moving. But, correct me if I don't understand correctly, in your diagram the hypothetized "real motion" goes outwards and you are stating at the same time that we are heading to the black hole in the centre of the galaxy, which contradicts the graph. It's not really a contradiction, think of the world as a great big funnel leading to the black hole in the cetre of our galaxy, we are all going in there eventually, but the bodies enjoying higher potential i.e higher up the funnel, are moving downwards at a slower rate than you, meaning space between you and the bodies of higher potential is stretching, so they are receeding from you. Another way to put it is you will be going down the gurgler before them So if I pick one direction for my z axis in a cylindrical coordinate system, and then invert the system (z —> -z and phi —> -phi), the speeds will spontaneously change? Sorry, I can't visualize your proposed cylindrical system.. At 2,500 km/s it would take about 3 million years to reach the galactic centre. As the Earth is about 4 billion years old, how come we are still here? 2500 km/s is the relative velocity between ground potential and a proton, so protons don't (can't) stand still, they have to move all the time (electrons too for the same reason), this constant movement is what I understand as the coloumb force. The effect is very similar to a gas under pressure, in a pressurised gas the molecules also give the appearance of repelling each other, but in reality is is just thermal velocity of the particles that manifests itself as repulsion. The absolute velocity is 99% of c and this motion is fortunately perpendiculat to time, so instead of heading straight down the drain, we enjoy spiralling around for a few billion years more (I hope). I'm confused, from what I've seen thus far is all you've done is replace v and c with your ground potential symbols into standard formulas. Then you claim This describes characteristics that is different than what the original formulas describe. How does that even work? If your describing properties that differ from what the original describe. Simply replacing the symbols cannot describe different relations or coordinate changes. Then you claim your formulas remove the need for dark matter...... Yet the original formulas that you simply changed the symbols in, require dark matter on the global scale. Sorry but you can't simply replace v and c, then describe dynamics that differ from the relationships in the original formulas. This doesn't work. Especially since in those formulas c is invariant. Yet neither of your GPT variables are. phi and Phi are pro numerals, member Sensei below has kindly defined the value of Phi to 6 decimal places for you... Proton doesn't have 938 MeV, but 938.272046 MeV (+- couple eV) 930/938=0.99147121535181236673773987206823 but 930/938.272046 = 0.99118374459170448311533731870341 Honestly, I doubt that you even know how to calculate proton mass-energy 938272046 eV.. If I am mistaken, show me it. Thanks for defining the value of Phi. here is ground potential defined to three decimal places. http://www.wolframalpha.com/input/?i=0.511%3D%28%28938.272-x%29%2F2%29*√%281-%28x%5E2%2F938.272%5E2%29%29 Not sure why I have to justify the proton mass experiment in this thread, it is well known. On the overall my feeling is that there is something right in Beejewel thoughts. hey, I like this guy This being science, what's important is being able to objectively demonstrate it. Swanson, you are my favorite antagonist, you give me sleepless nights, but ultimately help me improve the theory Steven
Robittybob1 Posted April 7, 2015 Posted April 7, 2015 Thanks for defining the value of Phi. here is ground potential defined to three decimal places. http://www.wolframalpha.com/input/?i=0.511%3D%28%28938.272-x%29%2F2%29*√%281-%28x%5E2%2F938.272%5E2%29%29 What were the three decimal places? Show me Ground potential defined to 3 decimal places please?
Mordred Posted April 7, 2015 Posted April 7, 2015 (edited) Not sure how I can explain it any clearer, the four velocity is the body-potential divided by the proton-potential multiplied by speed of light, while the relative velocity is simply the difference in potential between two bodies divided by c multiplied by the speed of light. From this function alone we can only determine the sign of the velocity not the direction of the vector. The sign indicates weather an object is approaching from the past or receeding into the past. Technically the observer is always the one at rest, at least this makes the transformations simpler, weather sitting at the station or sitting on the train, you are at rest, the rest, let the rest of the world do the moving. It's not really a contradiction, think of the world as a great big funnel leading to the black hole in the cetre of our galaxy, we are all going in there eventually, but the bodies enjoying higher potential i.e higher up the funnel, are moving downwards at a slower rate than you, meaning space between you and the bodies of higher potential is stretching, so they are receeding from you. Another way to put it is you will be going down the gurgler before them Sorry, I can't visualize your proposed cylindrical system.. 2500 km/s is the relative velocity between ground potential and a proton, so protons don't (can't) stand still, they have to move all the time (electrons too for the same reason), this constant movement is what I understand as the coloumb force. The effect is very similar to a gas under pressure, in a pressurised gas the molecules also give the appearance of repelling each other, but in reality is is just thermal velocity of the particles that manifests itself as repulsion. The absolute velocity is 99% of c and this motion is fortunately perpendiculat to time, so instead of heading straight down the drain, we enjoy spiralling around for a few billion years more (I hope). phi and Phi are pro numerals, member Sensei below has kindly defined the value of Phi to 6 decimal places for you... Thanks for defining the value of Phi. here is ground potential defined to three decimal places. Steven Doesn't matter what the value is. It is a variant quantity, that you replaced into an invariant quantity. Then claim it explains everything, without further mathematical detail. Other than simply replacing v and c with phi and Phi. You even claimed this into the lorentz factor formula. You have not mathematically shown any of the graphs you posted, as far as I see none of the formulas you posted result in a sinusoidal wave form. Do you even know how that formula is derived? Do you know the tensor matrix relations that are involved with the Lorentz factor? None of your descriptives matches those relations. Yet you simply plug phi and Phi into those equations. This far none of your descriptions mathematically show how to derive your variables. maybe you can demonstrate it in phase space coordinates? Handy to show particle velocities in coordinate space (Hint this is why x and x' was asked for) phase space 6 coordinate. [latex]dV=dx_1,dx_2,dx_3,dv_1,dv_2,dv_3[/latex] you already mentioned your theory doesn't work in Cartesian coordinates. And yet the substitution you did was on a formula that DOES use Cartesian coordinates. "Consider two observers O and O′, each using their own Cartesian coordinate system to measure space and time intervals. O uses (t, x, y, z) and O′ uses (t′, x′, y′, z′). " http://en.m.wikipedia.org/wiki/Lorentz_transformation In other words you don't know what is invloved or understand what the current models are describing. Your just replacing variables then adding a description, without a property mathematical analysis. Take a good look at the vector components on that link above then tell me your descriptions match. [latex] \gamma = \frac{1}{\sqrt{1- \frac{\phi^2}{\Phi^2}}}[/latex] when [latex]\Phi [/latex] is not invariant. by your definition Good question, GPT defines the four-velocity or absolute velocity of a body as a function of it's potential, so technically the proton is the only particle who's four-velocity is excactly 'c', here at ground potential our four-velocity is c(930/938) or 99.14 % the speed of light. The difference works out to be around 2500 km/s, so I guess we can call this for speed. Ultimately you, me and the proton are all heading to the same place, the black hole in the centre of the galaxy, but the proton being a light body with a high surface potential must take a longer path down, hence relative speed. The transformation of coordinates are fascinatingly perfect, because as we change our potential, everything in our world changes accordingly. All the electrons in your world instantly change mass, objects that were travelling in straight lines curve and everything just works without a glitch, and things seem great, that's until we start looking at the really tiny things, on a quantum scale, and realize that looking at our experiment changes the outcome. The normal Lorenz transformation should correctly transform the coordinates when using this form of gamma [latex] \gamma = \frac{1}{\sqrt{1- \frac{\phi^2}{\Phi^2}}}[/latex] Steven This above is complete garbage. Edited April 7, 2015 by Mordred
beejewel Posted April 7, 2015 Author Posted April 7, 2015 What were the three decimal places? Show me Ground potential defined to 3 decimal places please? oops, apologies, here it is to the nearest volt + or - about 20 volts. http://www.wolframalpha.com/input/?i=510998%3D%28%28938272046-x%29%2F2%29*√%281-%28x%5E2%2F938272046%5E2%29%29 The margin of error translates to a height of around 7-10 meters, as electrical potential increases by around 3 volts per meter height above ground. Steven
Robittybob1 Posted April 7, 2015 Posted April 7, 2015 oops, apologies, here it is to the nearest volt + or - about 20 volts. http://www.wolframalpha.com/input/?i=510998%3D%28%28938272046-x%29%2F2%29*√%281-%28x%5E2%2F938272046%5E2%29%29 The margin of error translates to a height of around 7-10 meters, as electrical potential increases by around 3 volts per meter height above ground. Steven You lost me there sorry. All I see is what looks like an equation that is being rearranged. 510998=((938272046-x)/2)*√(1-(x^2/938272046^2)) So there is still an X in it, so how can you say you've solved it?
beejewel Posted April 7, 2015 Author Posted April 7, 2015 (edited) You lost me there sorry. All I see is what looks like an equation that is being rearranged. 510998=((938272046-x)/2)*√(1-(x^2/938272046^2)) So there is still an X in it, so how can you say you've solved it? Sorry, you should be seing this, maybe you need to give Wolfram Alpha a bit more time.. There are two solutions for x, one is your ground potential and the other solution is the antimatter solution for the electron. The argument for this solution is that the mass per nucleon i.e. surface potential of Ni-62 is 930.417 MV, which is the element with the highest binding energy, and I claim this is because it is already at ground potential, so it can't decay any further, at least not for the moment. http://en.wikipedia.org/wiki/Nickel-62 Steven Edited April 7, 2015 by beejewel
Robittybob1 Posted April 7, 2015 Posted April 7, 2015 (edited) Sorry, you should be seing this, maybe you need to give Wolfram Alpha a bit more time.. Screen Shot 2015-04-07 at 2.13.45 pm.png There are two solutions for x, one is your ground potential and the other solution is the antimatter solution for the electron. The argument for this solution is that the mass per nucleon i.e. surface potential of Ni-62 is 930.417 MV, which is the element with the highest binding energy, and I claim this is because it is already at ground potential, so it can't decay any further, at least not for the moment. http://en.wikipedia.org/wiki/Nickel-62 Steven It probably has to do with being unregistered at the Wolfram site. Are those x values read where the graph crosses the x axis? Edited April 7, 2015 by Robittybob1
beejewel Posted April 7, 2015 Author Posted April 7, 2015 It probably has to do with being unregistered at the Wolfram site. Are those x values read where the graph crosses the x axis? Yes, you get two solutions for x where the graph crosses the line one solution is the exact negative value of the proton, which I assume is the invariant electron mass in the antimatter domain, and the other is 930,377,217 Volts which I understand to be ground potential. I have been asked by other members how I derived this equation, and all I can say is that it was an intuitive educated guess, I knew what I was looking for, I had already worked out the hypothesis of how the proton and electron should to be related, so I tried a whole bunch of different formulas before hitting on this one and not only did it look right, it worked. Inspiration, perspiration and some derivation Steven PS: I am not ignoring Mordred's request above, just need to work on it a bit before I reply.
ajb Posted April 7, 2015 Posted April 7, 2015 Not sure how I can explain it any clearer, the four velocity is the body-potential divided by the proton-potential multiplied by speed of light, while the relative velocity is simply the difference in potential between two bodies divided by c multiplied by the speed of light. From this function alone we can only determine the sign of the velocity not the direction of the vector. The sign indicates weather an object is approaching from the past or receeding into the past. This looks more like a speed, i.e. the units are okay by there is no vector bit here. It looks a bit like the magnitude of a velocity, and if we are using a pseudo-Riemannian metric (which we probabily are) then this could be -,0,+ upon normalisation. How do you relate this to the more standard notion of a velocity. I take some curve and define its velocity vector etc. The best way to show me how this changes is to write down a coordinate change explicitly. Technically the observer is always the one at rest, at least this makes the transformations simpler, weather sitting at the station or sitting on the train, you are at rest, the rest, let the rest of the world do the moving. Now what about two observers, which is a common situation. Which one is at rest? Your description here seems to be that "an inertial observe is always at rest with respect to himself". Okay, but this does not define some absolute notion of 'at rest'.
swansont Posted April 7, 2015 Posted April 7, 2015 Technically the observer is always the one at rest, at least this makes the transformations simpler, weather sitting at the station or sitting on the train, you are at rest, the rest, let the rest of the world do the moving. So each one sees the other at a higher potential? Sorry, I can't visualize your proposed cylindrical system.. http://mathworld.wolfram.com/CylindricalCoordinates.html It uses theta instead of phi for the azimuthal coordinate
beejewel Posted April 7, 2015 Author Posted April 7, 2015 I feel we are reaching the limits of what my current theory and I can do in this area, so I request an adjournment to think, learn and extend it a little further. A few questions remain unanswered for now, but don't worry I'll be back. Steven 2
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