Mordred

mass below and mass above. Is the force of G less or greater than at sea level ? I will leave you with that thought experiment (google shell theorem to answer correctly) https://en.m.wikipedia.org/wiki/Shell_theorem Has the important equations

see here for example "A tensor is a particular type of function." http://www.physics.miami.edu/~nearing/mathmethods/tensors.pdf second example Roughly speaking, the metric tensor is a function http://mathworld.wolfram.com/MetricTensor.html every tensor will have associated functions A mixed can have several "In mathematics, the modern component-free approach to the theory of a tensor views a tensor as an abstract object, expressing some definite type of multi-linear concept " any multilinear concept has a function. https://en.wikipedia.org/wiki/Tensor_(intrinsic_definition) Though I should be more specific a rank 1 tensor is a set of vector functionals

You need to look at the sum of forces. If your at the precise center of the Earth the force of attraction is zero as your surrounded by equal matter on all facings. Its the blooming distribution.

Ignoring what I stated isn't going to change anything. You can change density without changing the force of attraction. Its as simple as that...

No your still missing the point its a matter of using the proper terminology. Density isn't appropriate because you can change density without changing the force of attraction. Mass as resistance to inertia change is the appropriate term via f-ma ( you cannot ignore this by finding appropriate situation. A definition must follow all potential situations). Simply ignoring the fact that temperature can increase and decrease the density of a field of plasma without causing a stronger force of attraction isn't appropriate. Newtons basic laws of inertia. this is why both uncool and I mentioned the Shell theorem as its directly applicable.

No your missing the point. Your equating this to density but its the distribution itself that is important. You can have an incredibly dense field that is uniform that has no inherent direction of force. Take a plasma gas and just increase the temperature. You decrease the density. Now cool it down the density increases Doesn't mean you develop a greater attraction to a COM.

Not exactly, its the distribution of those particles that matter. This is part of the Shell theorem which is going to be tricky to explain at the moment till we get you straightened on basic kinematics. Lets put it this way if I have a uniform distribution of particles then there is no net sum of force in any direction. So if I have a space time field of uniform distribution there is no inherent motion in any direction. We can arbitrarily choose any COM in this type of field. Same would apply to an electromagnetic field. It is the potential difference (voltage) that leads to current flow. In you OP you have a fixed number of particles that comprise the Sun for example. Simply changing the density via volume change does not change the number of particles which relate to G (in the electromagnetic field Coulombs law). However you can change the particle number density in a finite volume to increase density but this is equivalent to increasing the mass term. two different scenarios of altering density. the first does not increase particle number while the latter does. We can alter density one other way and that is via temperature change. Now if you apply the ideal gas laws you can see that density isn't nearly as appropriate as mass via resistance to inertia change, which is the physics definition. https://www.britannica.com/science/Coulombs-law

No worries, the added details on intrinsic to what I have will help

Good question there Uncool.

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.

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

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.

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]

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] 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.

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?

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

no it can't its 2d not 4d.

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?

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.

Looks good, if you learn to understand that vid, your doing good.

Universe via whiteholes has been proposed before. However they don't work for well to observational evidence. The difficulties due to the above is far more complex than you realize with CMB evidence via thermal hydrodynamics. LCDM models are far more accurate.

Well you need to At least present some effort other than mere disbelief. So obviously you don't have anything remotely examinable.

So what is your gradiose solution to solve every problem you described in your posts? Or the grandiose claims? You don't have one on any of them yet. Simple babble of disbelief is pointless. The rubber sheet is nothing but a laymens genetalization for those with no math skill analogy. It isn't anything like that in reality.

No actually I am not. MHD has its relativistic equations as well. I understand how they work as well as GR. So go ahead show me tell me how you generated those videos without applying any equations?

You haven't got anthing yet. A magneto hydrodynamic field has spin 1 statistics which has none of needed characteristics the observational tests for GR fall under. Which is spin zero. Start with the electromagnetic field. If you like I can readily show this under Maxwell. There is very distinctive differences between magnetohydronics and how it would show up under measurement compared to the tests under GR. Start with basic kinematics if you must under calculus.