Unified Field Posted December 30, 2017 Posted December 30, 2017 (edited) According to science, there's absolutely no correlaton between the size of an object and the gravitational field, which it produces. Science tells, that changing the size of Sun wouldn't have no effect on the planets in Solar System. I don't agree with this. I think, that density of an object has significant importance for the properties of g. field, which it produces. What makes me thinks this way? Well, I've noticed, that celestial objects, which have low density, like a nebula, don't attract other objects with the same force, as planets or stars. If density of objects has no influence on gravity, then try this: - take a beach ball and a golf ball with similar masses - throw them both into a pond - think... I say, that big objects attract matter at larger distance, than small objects, but the force of attraction is weaker. Mass concentrated in small area generates a g. field with higher magnitude, but distributed over smaller space. Mass scattered over large area generates bigger, but weaker g. field. It should be rather obvious, that objects with higher concentration of mass, create different curvatures of "space-time", than objects with low density of matter. Those two images show, what according to me, is the difference of "space-time curvatures", for 2 objects with the same mass, but different sizes: And this is, what makes the difference, between those images and the mainstream science: According to science, force of attraction is always directed towards the center of mass. According to me, force is directed towards the object, which is the source of g. field. Such difference seems to be insignificant, but it changes a lot. Imagine, that Earth is hollow - what would be the direction of attraction for objects, which are placed inside the planet? Science say, that towards the center. I say, that towards the inner surface... I will now use the rubber sheet model of "space-time", to compare the difference between a force, which is concentrated in a point and a force, which is distributed over some area: In the rubber sheet model, mass of objects works, as a force directed downwards - thanks to the gravity of Earth. Force and mass can be used in this case as the same variable value. This allows me to compare the distance, at which objects with different sizes attract other bodies, when a force is applied: Of course, I'm aware, that this experiment is far from perfect and can't be used, to collect any valuable readings. However, it should be rather clear, that force, which is distributed over larger area, attracts objects at bigger distances, than force, which is concentrated in a single point. If you are interested in some actual readings and real values, you should recreate this experiment in a laboratory - because sadly I don't have necessary equipement. If you think, that I alternated the results in any way, I suggest you to make this experiment by yourself (it shouldn't be too complicated). If your results will contradict somehow my claims, feel free to present your evidence in this thread. In my next movie, I will compare the force of attraction of an heavy and small object with an lighter, but bigger object - it's already recorded, but first I want to hear, what can you say about this part... Edited December 30, 2017 by Unified Field -1
Mordred Posted December 30, 2017 Posted December 30, 2017 (edited) If you compress the sun below its Scwartzchild radius. The amount of mass remains unchanged so the amount of gravity remains unchanged. These videos of yours is meaningless as evidence. You have been told before your rubber sheet analogy is just that. An analogy to get a complex concept across to those with little to poor math skills. They are not tests of GR. Edited December 30, 2017 by Mordred
Unified Field Posted December 30, 2017 Author Posted December 30, 2017 3 minutes ago, Mordred said: If you compress the sun below its Scwartzchild radius. The amount of mass remains unchanged so the amount of gravity remains unchanged. These videos of yours is meaningless as evidence. You have been told before your rubber sheet analogy is just that. An analogy to get a complex concept across to those with little to poor math skills. "The height as a function of position of the elastic sheet is a solution to Poisson's equation, which is the same equation that gives you the (classical) gravitational potential. So the demonstration is an analogue computation that solves the analogous problem in gravity" Exactly the same calculations are being used for the rubber sheet and the space-time curvature. Curvature of rubber sheet is fully consistent with the expected curvature of space-time.
Mordred Posted December 30, 2017 Posted December 30, 2017 (edited) Fine post your calculations then. However keep the radius of the observer constant. I have a specific reason to mention that. You do realize a rubber sheet is two dimensional while spacetime under Poisson is 4d? Edited December 30, 2017 by Mordred
Unified Field Posted December 30, 2017 Author Posted December 30, 2017 Rubber sheet shows incorrect outcome, when it comes to simulate orbits of bodies, which are moving around the center of gravity, but can be used without any problems, to visualise the shape of space-time curvatures.
Unified Field Posted December 30, 2017 Author Posted December 30, 2017 8 minutes ago, Mordred said: Fine post your calculations then. However keep the radius of the observer constant. I have a specific reason to mention that. First I need to figure out something to measure the inward force - I don't own a laboratory. Then I can calculate, how this force differs, depending on the size of central object... 7 minutes ago, Mordred said: no it can't its 2d not 4d. But still the same equations are being used in both cases... Quote You do realize a rubber sheet is two dimensional while spacetime under Poisson is 4d? Yes. But it doesn't mean, that 4D won't behave in similar way... There are many reasons, why I can assume it...
Mordred Posted December 30, 2017 Posted December 30, 2017 (edited) Who ever said you need a lab ? Start with a central potential force. [math]F=\frac{GMm}{r^2}[/math] Work from there to Poisson in 3d then figure out how time dilation and length contraction occurs due to field interactions. Don't forget to apply Shell theorem Edited December 30, 2017 by Mordred
Unified Field Posted December 30, 2017 Author Posted December 30, 2017 Just now, Mordred said: Who ever said you need a lab ? Start with a central potential force. GMmr2 Work from there to Poisson in 3d then figure out how time dilation and length contraction occurs due to field interactions. Thanks!
Mordred Posted December 30, 2017 Posted December 30, 2017 (edited) Thanks may be premature. Central potential and Newton limit In the presence of matter or when matter is not too distant physical distances between two points change. For example an approximately static distribution of matter in region D. Can be replaced by the equivalent mass [latex]M=\int_Dd^3x\rho(\overrightarrow{x})[/latex] concentrated at a point [latex]\overrightarrow{x}_0=M^{-1}\int_Dd^3x\overrightarrow{x}\rho(\overrightarrow{x})[/latex] Which we can choose to be at the origin [latex]\overrightarrow{x}=\overrightarrow{0}[/latex] Sources outside region D the following Newton potential at [latex]\overrightarrow{x}[/latex] [latex]\phi_N(\overrightarrow{x})=-G_N\frac{M}{r}[/latex] Where [latex] G_n=6.673*10^{-11}m^3/KG s^2[/latex] and [latex]r\equiv||\overrightarrow{x}||[/latex] According to Einsteins theory the physical distance of objects in the gravitational field of this mass distribution is described by the line element. [latex]ds^2=c^2(1+\frac{2\phi_N}{c^2})-\frac{dr^2}{1+2\phi_N/c^2}-r^2d\Omega^2[/latex] Where [latex]d\Omega^2=d\theta^2+sin^2(\theta)d\varphi^2[/latex] denotes the volume element of a 2d sphere [latex]\theta\in(0,\pi)[/latex] and [latex]\varphi\in(0,\pi)[/latex] are the two angles fully covering the sphere. The general relativistic form is. [latex]ds^2=g_{\mu\nu}(x)dx^\mu x^\nu[/latex] By comparing the last two equations we can find the static mass distribution in spherical coordinates. [latex](r,\theta\varphi)[/latex] [latex]G_{\mu\nu}=\begin{pmatrix}1+2\phi_N/c^2&0&0&0\\0&-(1+2\phi_N/c^2)^{-1}&0&0\\0&0&-r^2&0\\0&0&0&-r^2sin^2(\theta)\end{pmatrix}[/latex] Now that we have defined our static multi particle field. Our next step is to define the geodesic to include the principle of equivalence. Followed by General Covariance. Ok so now the Principle of Equivalence. You can google that term for more detail but in the same format as above [latex]m_i=m_g...m_i\frac{d^2\overrightarrow{x}}{dt^2}=m_g\overrightarrow{g}[/latex] [latex]\overrightarrow{g}-\bigtriangledown\phi_N[/latex] Denotes the gravitational field above. Now General Covariance. Which use the ds^2 line elements above and the Einstein tensor it follows that the line element above is invariant under general coordinate transformation(diffeomorphism) [latex]x\mu\rightarrow\tilde{x}^\mu(x)[/latex] Provided ds^2 is invariant [latex]ds^2=d\tilde{s}^2[/latex] an infinitesimal coordinate transformation [latex]d\tilde{x}^\mu=\frac{\partial\tilde{x}^\mu}{\partial x^\alpha}dx^\alpha[/latex] With the line element invariance [latex]\tilde{g}_{\mu\nu}(\tilde{x})=\frac{\partial\tilde{x}^\mu \partial\tilde{x}^\nu}{\partial x^\alpha\partial x^\beta} g_{\alpha\beta}x[/latex] The inverse of the metric tensor transforms as[latex]\tilde{g}^{\mu\nu}(\tilde{x})=\frac{\partial\tilde{x}^\mu \partial\tilde{x}^\nu}{\partial x^\alpha\partial x^\beta} g^{\alpha\beta}x[/latex] In GR one introduces the notion of covariant vectors [latex]A_\mu[/latex] and contravariant [latex]A^\mu[/latex] which is related as [latex]A_\mu=G_{\mu\nu} A^\nu[/latex] conversely the inverse is [latex]A^\mu=G^{\mu\nu} A_\nu[/latex] the metric tensor can be defined as [latex]g^{\mu\rho}g_{\rho\nu}=\delta^\mu_\mu[/latex] where [latex]\delta^\mu_nu[/latex]=diag(1,1,1,1) which denotes the Kronecker delta. Finally we can start to look at geodesics. Let us consider a free falling observer. O who erects a special coordinate system such that particles move along trajectories [latex]\xi^\mu=\xi^\mu (t)=(\xi^0,x^i)[/latex] Specified by a non accelerated motion. Described as [latex]\frac{d^2\xi^\mu}{ds^2}[/latex] Where the line element ds=cdt such that [latex]ds^2=c^2dt^2=\eta_{\mu\nu}d\xi^\mu d\xi^\nu[/latex] Now assume that the motion of O changes in such a way that it can be described by a coordinate transformation. [latex]d\xi^\mu=\frac{\partial\xi^\mu}{\partial x^\alpha}dx^\alpha, x^\mu=(ct,x^0)[/latex] This and the previous non accelerated equation imply that the observer O, will percieve an accelerated motion of particles governed by the Geodesic equation. [latex]\frac{d^2x^\mu}{ds^2}+\Gamma^\mu_{\alpha\beta}(x)\frac{dx^\alpha}{ds}\frac{dx^\beta}{ds}=0[/latex] Where the new line element is given by [latex]ds^2=g_{\mu\nu}(x)dx^\mu dx^\nu[/latex] and [latex] g_{\mu\nu}=\frac{\partial\xi^\alpha}{\partial\xi x^\mu}\frac{\partial\xi^\beta}{\partial x^\nu}\eta_{\alpha\beta}[/latex] and [latex]\Gamma^\mu_{\alpha\beta}=\frac{\partial x^\mu}{\partial\eta^\nu}\frac{\partial^2\xi^\nu}{\partial x^\alpha\partial x^\beta}[/latex] Denote the metric tensor and the affine Levi-Civita connection respectively I could do the above in an easier format but it would end up lacking important details. Did you want a strictly Calculus version under 3d? Edited December 30, 2017 by Mordred
Unified Field Posted December 30, 2017 Author Posted December 30, 2017 (edited) Ok... I wanted to start with measuring the force of attraction of a body in 4 scenarios: 1. r and m 2. r and 2*m 3. 2*r and m 4. 2*r and 2*m Where r is for radius of a central body and m is for it's mass... Then I wanted to calculate the results, using standard equations and compare them with measured results. I don't need the exact values, only the ratio, how much stronger should be the force, when mass of central body is 2 times so high. I think, that the best would be to measure all 4 scenarios twice - for close and distant locations of attracted body. And finally create a formula, which would express the correlation between size and attraction... Edited December 30, 2017 by Unified Field
Mordred Posted December 30, 2017 Posted December 30, 2017 (edited) Here lets do an easier methodology [math]\vec{F}=\frac{GMm}{r^2}\hat{r}[/math] in differential form which has two parts [math]m_1\frac{d^2x}{dt^2}=\vec{F}=m_g\vec{G}(x)[/math] and [math]m_g=m+i[/math] the gravitational field is then [math]\vec{g}(x)=-\frac{\sum Gm_i}{|\vec{x}-\vec{x_i}|^2}\hat{x}[/math] which gives [math]\phi(x)=-\frac{GM}{r^2}=-\frac{GM}{|\vec{x}-\vec{x_i}|}[/math] so the integral is [math]\phi(x)=-G\int \frac{\rho(\vec{\acute{x}})d^3\acute{x}}{|\vec{x}-\vec{x_i}|}[/math] the differential form becomes [math]\nabla^2\phi=4G\rho[/math] [math]\vec{\nabla}=\frac{\partial}{\partial x}+\frac{\partial}{\partial y}+\frac{\partial}{\partial z}[/math] [math]{\nabla}^2=\frac{\partial}{\partial x^2}+\frac{\partial}{\partial y^2}+\frac{\partial}{\partial z^2}[/math] 21 minutes ago, Unified Field said: Ok... I wanted to start with measuring the force of attraction of a body in 4 scenarios: 1. r and m 2. r and 2*m 3. 2*r and m 4. 2*r and 2*m Where r is for radius of a central body and m is for it's mass... Then I wanted to calculate the results, using standard equations and compare them with measured results. I don't need the exact values, only the ratio, how much stronger should be the force, when mass of central body is 2 times so high. I think, that the best would be to measure all 4 scenarios twice - for close and distant locations of attracted body. And finally create a formula, which would express the correlation between size and attraction... well the formulas are commonly known you should be able to use the above for that. As far as real tests use a Newton scale with a known weights at different heights. Newton scales are dirt cheap, This is actually a high school lab project. Edited December 30, 2017 by Mordred
Unified Field Posted December 30, 2017 Author Posted December 30, 2017 (edited) Quote well the formulas are commonly known you should be able to use the above for that. As far as real tests use a Newton scale with a known weights at different heights. Newton scales are dirt cheap, This is actually a high school lab project. Exactly. I don't need sophisticated equipement and 4D geometry equations. I would "borrow" a Newton scale from school, but I'm not a student anymore - so I will have to buy one. I think, that everything can be done, using calculations on vectors... Something like this... Edited December 30, 2017 by Unified Field
Mordred Posted December 30, 2017 Posted December 30, 2017 (edited) Everything in Physics involves scalars, vectors and spinors. This includes tensors which is a functional of a vector. You don't need time dilation to have intrinsic curvature aka Newton limit above. Its real simple take two objects and drop them where there is no central potential. The path of both remain parallel. Of course on Earth its central potential so the path will converge as it approaches COM. Aka Kronecker delta and Levi_Cevita above. Intrinsic curvature is that easy to test. Time dilation itself not so easily tested at home but you don't need time dilation to test mass to radius relations. Newtons laws works quite well for the majority of our solar system with the sole exception of Mercury orbit. An easy way to understand Lorentz transforms is Lets find a simplified expression for Lorentz shall we lets start with its coordinates. [math]x^\mu=(x^0,x^1,x^2,x^3)=(ct,x,y,z)[/math] with the subscript [math]\mu=0,1,2,3[/math] so [math]x^\mu[/math] is now my space-time coordinates. Wow ain't that just ducky one variable for all four.... So consider a Lorentz frame S where two events use [math]x^\mu[/math] and [math]\acute{x}^\mu[/math] Both observers will only agree on the Worldline invariant interval. sign convention for Lorentz (-+++). [math]-\Delta S^2=-(\Delta x^0)^2)+(\Delta x^1)^2)+(\Delta x^2)^2)+(\Delta x^3)^2)[/math] [math]-\delta S^2=-(\Delta x^0)^2)+(\Delta x^1)^2)+(\Delta x^2)^2)+(\Delta x^3)^2)=-\delta S^2=-(\Delta \acute{x}^0)^2)+(\Delta \acute{x}^1)^2)+(\Delta \acute{x}^2)^2)+(\Delta \acute{x}^3)^2)[/math] in short [math]\Delta S^2=\acute{S}^2[/math] the minus sign on the left denotes time-like separated [math]\Delta S^2>0[/math] However you want the infinitesimal coordinate variations for curve fitting under GR, QM uses quantized units but that's easy enough to add in later on. An invariant interval [math]ds^2[/math] recall I mentioned line element? so [math]ds^2=-(dx^0)^2+(dx^1)^2+(dx^2)^2+(dx^3)^2[/math] so [math]ds^2=\acute{ds}^2[/math] so lets do the subscripts vs superscripts. use x for the zeroth (a coordinate is scalar) [math]dx_\mu=dx^\mu[/math] easy enough to apply that to each coordinate. I shouldn't have to show you that. This gives us the change in the zeroth component (zeroth is a scalar value) the sign of [math]dx_\mu=(dx_0,dx_1,dx_2,dx_3)=(-dx^0,dx^1,dx^2,dx^3)[/math] so [math]ds^2=dx_0dx^0+dx_1dx^1,dx_2dx^2,dx_3dx^3[/math] we have now removed the need of the minus sign by using convariant and contravariant indices. So now we can simplify the last as [math]ds^2=\sum_{\mu=0}^{3}dx_\mu dx^\mu[/math] so in Minkowskii metric by Einstein summation above [math]\eta_{\mu\nu}[/math] [math]-ds^2=\eta_{\mu\nu}dx^\mu dx^\nu\mu[/math] notice the change in order of the basis. Now so far we are symmetric what about antisymmetric? we need to decompose a two index object [math]M_{\mu\nu}[/math] into its symmetric and antisymmetric parts. [math]M_{\mu\nu}=\frac{1}{2}(M_{\mu\nu})+M_{\nu\mu})+\frac{1}{2}(M_{\mu\nu})+M_{\nu\mu})[/math] the antisymmetric part on the right changes sign under exchange of indices so we denote this by [math]\xi_{\mu\nu}=-\xi_{\mu\nu}[/math] [math]\xi_{\mu\nu}dx^\mu dx^\nu=(-\eta_{\mu\nu}dx^\mu dx^\nu=-\xi-{\mu\nu}dx^\nu dx^\mu=-\eta_{\mu\nu}dx^\mu dx^\nu[/math] so then [math]-ds^2=\eta_{00}xx^0dx^0+\eta{01}dx^0dx^1+\eta_{11}dx^1dx^1+...[/math] Edited December 30, 2017 by Mordred
Unified Field Posted December 30, 2017 Author Posted December 30, 2017 (edited) Quote his is actually a high school lab project. And this is, what makes me wonder - how it's possible, that no one didn't think about it. I mean, it's something rather obvious, that a force behaves differently, when it's concentrated in a single point and when it's distributed over some area. If a black hole can be compared to a hole in the rubber sheet, then it's logical, that it's MUCH easier to puncture the material (or fabric of the time-space) with a nail, than with a hammer. You don't need to be a genius, to figure it out... It's the OBJECT, what creates a g. field, not it's center of mass. Inside the object, there's no attraction towards the center and matter is "floating" freely. However a force will appear, if another body will be placed inside the central object - this force can be directed outwards or towards the center of mass, depending on the DENSITY of a body. It's the density, which determines, if object will fall, or will fly up. The most dense objects are always placed in the center of a body and the concentration of matter gets smaller towards the outside. Buoyancy and gravity are created by the same force, which depends on the density, not mass... Edited December 30, 2017 by Unified Field
Mordred Posted December 30, 2017 Posted December 30, 2017 (edited) UMM this has been figured out since the 16th century ? Why would you think physicists never figured out this? How do you think we do orbits? Forget your rubber sheet ANALOGY yeesh its not realistic by any means...... Have you never learned Newtons laws of gravitational bodies and shell theorem??? Obviously not by the descriptions you gave in the last post. lets start with step 1. mass is resistance to inertia change. Not the weight of an object. ie [math]F=ma[/math] Newtons 3 laws of motion sound familiar? This doesn't change in GR. Edited December 30, 2017 by Mordred
Unified Field Posted December 30, 2017 Author Posted December 30, 2017 11 minutes ago, Mordred said: UMM this has been figured out since the 16th century ? Why would you think physicists never figured out this? How do you think we do orbits? Forget your rubber sheet ANALOGY yeesh its not realistic by any means...... Have you never learned Newtons laws of gravitational bodies and shell theorem??? Obviously not by the descriptions you gave in the last post. lets start with step 1. mass is resistance to inertia change. Not the weight of an object. ie F=ma Newtons 3 laws of motion sound familiar? This doesn't change in GR. Yes. Weight is a force, produced by gravity of Earth. But doesn't density correlate with momentum? https://www.quora.com/How-are-momentum-and-density-related It's harder to stop a bullet, than a beach ball at similar mass...
Mordred Posted December 30, 2017 Posted December 30, 2017 Only via field treatment through the coupling constants involved in this case the electromagnetic force. [math]\rho [/math] is the particle number density. The coulomb force G relates to this under newton. The central potential of a force is a direct application of the shell theorem under Newton. https://en.wikipedia.org/wiki/Shell_theorem
uncool Posted December 30, 2017 Posted December 30, 2017 Mordred - I'm guessing that you just recently learned a lot of relativity, and really liked it. Which is good, but it's not the level appropriate for this conversation. This should be possible to discuss just using Newton's laws and Newtonian gravity - and probably very little in the way of integration.
beecee Posted December 30, 2017 Posted December 30, 2017 3 hours ago, Unified Field said: I don't agree with this. GR has passed all tests thrown its way and continues to pass. It gives the same results as Newtonian within Newtonian limits, with far better accuracy. If you disagree my bet is that you are certainly wrong and probably have an agenda of sorts. But in reality science forums are all beset by "would be's if they could be's" that claim they have over thrown Einstein and his theory/s. You are just another number in that list. 7 minutes ago, Unified Field said: Yes. Weight is a force, produced by gravity of Earth. But doesn't density correlate with momentum? https://www.quora.com/How-are-momentum-and-density-related It's harder to stop a bullet, than a beach ball at similar mass... And a marshmellow hitting you at 99.9% C would pass right through you.
Mordred Posted December 30, 2017 Posted December 30, 2017 1 minute ago, uncool said: Mordred - I'm guessing that you just recently learned a lot of relativity, and really liked it. Which is good, but it's not the level appropriate for this conversation. This should be possible to discuss just using Newton's laws and Newtonian gravity - and probably very little in the way of integration. The OP specified GR but its obvious the Op doesn't understand Newton so I already stepped it down to that level, Also no I understood GR for over 10 years thanks on that. 1
uncool Posted December 30, 2017 Posted December 30, 2017 (edited) Unified Field: how much of standard physics do you accept? Do you accept Newton's laws? Newtonian gravity (as a nonrelativistic approximation)? Edited December 30, 2017 by uncool 1
uncool Posted December 30, 2017 Posted December 30, 2017 Just now, Mordred said: The OP specified GR but its obvious the Op doesn't understand Newton so I already stepped it down to that level, Also no I understood GR for over 10 years thanks on that. My apologies - I mistook your enthusiasm for that of the newly-"initiated". Good on you for keeping up that enthusiasm! I would say that even in the OP, the thread should clearly have been at the level of Newton's laws, and simple integration. I think that there already as a problem just with intrinsic vs extrinsic properties, as I plan to explain.
Unified Field Posted December 30, 2017 Author Posted December 30, 2017 Quote mass is resistance to inertia change. Not the weight of an object. ie F=ma Newtons 3 laws of motion sound familiar? Ok, but it is also a source of force directed towards it. Outside the g. field of Earth this force can be measured. If every particle of matter is a source of this force, then it's obvious, that the larger is the number of particles in a specified area, the stronger is the force...
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